V5B7 — ADVANCED TOPICS IN ANALYSIS: SOBOLEV SPACES

OLLI SAARI

Contents

1. Short motivation 21.1. Duals of normed spaces 21.2. Convergence of smooth functions 21.3. Distributions 31.4. Weak approximation by smooth functions 31.5. Derivative 32. Sobolev spaces 32.1. Definition and approximation 42.2. Basic properties and examples 52.3. Poincare’s inequality 53. Maximal functions 64. Self-improving of generalized Poincare inequalities 74.1. Weak Lp spaces 74.2. Rearrangement 84.3. Controlling oscillation 84.4. Truncation argument 115. Hardy–Sobolev spaces 135.1. Motivation 135.2. Hardy spaces 135.3. Hardy–Sobolev spaces 135.4. Characterization with Calderon maximal function 155.5. Pointwise characterizations 166. Embeddings for Calderon–Hardy spaces 166.1. Higher order spaces 166.2. Fractional integral 176.3. Inequality for Calderon functions 177. Littlewood–Paley theory 187.1. Schwartz class 187.2. Fourier transform 187.3. Littlewood–Paley decomposition 197.4. Calderon–Zygmund theory 197.5. Khintchine’s inequality 198. Function spaces 208.1. Riesz potentials 218.2. Potential spaces 218.3. Besov and Triebel–Lizorkin scales 228.4. Finite differences 229. Real interpolation 259.1. Preliminaries 259.2. K functional 259.3. J functional 269.4. Equivalence of the methods 269.5. Basic properties and duality 279.6. Explicit computations 2810. Complex interpolation 30

Date: February 4, 2020.

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10.1. Basic properties 3110.2. Scalar and vector valued Lp spaces 3110.3. The method of retracts 3210.4. Sneiberg’s stability theorem 3210.5. Surjectivity 3310.6. Injectivity 3310.7. Consistency 3411. Direct method in the calculus of variations 3411.1. Problems with boundary values 3511.2. Application I: Holder continuity for n-Laplacian 3611.3. Application II: Very weak solutions to linear equations 3811.4. Fractional Laplacians 3811.5. Caffarelli–Silvestre extension 3912. Traces and extensions 4012.1. Trace theorems 4012.2. Some explanations 4212.3. Other boundary values 4312.4. Comment on domains 4313. Capacity 4413.1. Basics 4413.2. Hausdorff measure 4713.3. Sets of capacity zero 4713.4. Exceptional sets for Sobolev spaces 4913.5. Precise representatives 5014. Modulus of a curve family 5214.1. Curves and condensers 5414.2. Sobolev spaces using curves 5714.3. Remark on metric spaces 5715. Riesz capacity and Frostman’s lemma 58References 60

1. Short motivation

This section motivates the study of weak differentiability from point of view of abstract distributiontheory. Not everything is proved, not everything is precise, and (almost) nothing is needed in the mainpart of the course. More details on distributions can be found in Rudin’s book [21].

1.1. Duals of normed spaces. Let X be a normed vector space over reals. A functional on X is amapping f ∶ X → R. Algebraic dual of X is the collection of linear functionals. The topological dual orbriefly dual is the collection of continuous linear functionals. We denote

X ′ = f ∶X → R ∶ f linear and continuous (bounded).

In addition to its norm (strong) topology, X has a weak topology, which is the topology induced bythe family X ′. Next consider the dual X ′. It is endowed by a norm topology (operator norm as X → R),but since every x ∈X defines a map X ′ → R, we can regard X as a subset i(X) ⊂X ′′ where

i(x)(f) = f(x).

The coarsest or weakest topology that makes all i(x) continuous is called the weak* topology of X ′. Inaddition, one could also consider the weak topology of X ′ induced by the family X ′′.

1.2. Convergence of smooth functions. Every topology defines a notion of convergence. We studyfunction spaces that contain C∞(Ω) as a subset (usually even a dense one). The strongest topology onemight want to use is the following. We define the “convergence in D” as follows: We say ϕi → ϕ in D if

there is K ⊂ Rn compact with suppϕi ⊂K. For all α ∈ Nn, Dαϕi →Dαϕ uniformly in K.

2

This convergence is related to the topology induced by the family of norms (N positive integer and Kcompact set from an exhaustion of Rn)

ρNK(f) ∶= max∣Dαf(x)∣ ∶ x ∈K, ∣α∣ ≤ N.It keeps C∞(Rn) closed. The space D itself we let be C∞

c (Ω). That is not closed because of therequirement on supports.

1.3. Distributions. At the other end of the world lies the space of distributions D′, the topological dualof D. Convergence in the sense of distributions is exactly the weak* convergence of D′, that is fi → f if

⟨fi, ϕ⟩→ ⟨f,ϕ⟩as real numbers for all ϕ ∈ D. The Japanese bracket denotes the duality pairing. Examples of distributionsinclude

All test functions in D. All locally integrable functions. The dirac mass ϕ↦ ϕ(0) and its derivatives that are not even measures anymore.

In the case of locally integrable f , we can define

⟨f,ϕ⟩ = ∫ fϕ ≤ ∥f∥L1∥ϕ∥∞.

for a test function ϕ,

1.4. Weak approximation by smooth functions. Let η = 1B(0,1)ce1

∣x∣2−1 be normalized to have inte-

gral one. Then it’s smooth, supported in B(0,1), even, and its dilates ηδ = δ−nη(δ−1⋅) are an approxi-mation of identity (see Section 1.2.4 in Grafakos [8]), in particular, for f ∈ Lp(Rn) with p ∈ [1,∞)

limε→0

∥f ∗ ηε − f∥Lp(Ω)

and the convergence also takes place uniformly in compact sets for continuous functions. Hence closureof D under Lp norm is Lp.

For ψ ∈ D define

τhψ(x) = ψ(x − h), h ∈ Rd

ψ(x) = ψ(−x)f ∗ ψ = ⟨f, τ ψ⟩

Proposition 1.1. Let f ∈ D′(Rn). Then ηδ ∗ f → f as distribution when ε → 0. Moreover, each ηδ ∗ fis smooth.

Sketch of the proof. Since the difference quotient is a linear operator and f is continuous, we see thatf ∗ ηδ(x) is smooth. To prove the convergence, we compute for an arbitrary test function ψ

⟨f,ψ⟩ = limδ

⟨f,ψ ∗ ηδ⟩ = ⟨f,∫ ψ(y)ηδ(⋅ − y)dy⟩

= limδ∫ ψ(y)⟨f, τyηδ⟩dy = lim

δ⟨f ∗ ηδ, ψ⟩

where we resorted to existence result of vector valued integral, see Rudin [21] Theorem 6.32 for details.

This tells that closure of D under the convergence of distributions is D′

.

1.5. Derivative. Let Dα be the differential operator corresponding to partial derivatives α. For anydistribution, we can define the distributional (weak) derivative through approximation (also without,but this is to serve as intuition). Let fi be any sequence of smooth functions converging to f in the senseof distributions. Let

⟨Dαf,ϕ⟩ ∶= limi⟨Dαfi, ϕ⟩ = lim

i(−1)∣α∣⟨fi,Dαϕ⟩ = (−1)∣α∣⟨f,Dαϕ⟩.

The last expression is clearly independent of the sequence, and it is the usual defintion of distributionalderivative. In general it is just a distribution and as such rather useless. However, there are still manynotions of convergense we may us in order to close C∞

c .

2. Sobolev spaces

Most of this section can be found in Evans and Gariepy [6].3

2.1. Definition and approximation. For this section, let Ω be an open set in Rn. A locally integrablefunction f defines a distribution, and hence it makes sense to speak about its distributional derivative.

Definition 2.1. Let f be a distribution. We write g = ∂kf and call it the distributional derivative if

⟨f, ∂kϕ⟩ = −⟨g,ϕ⟩.

In case all first order distributional derivatives are locally integrable functions, that is

f, ∂kf ∈ L1loc(Rn), ∀k ∈ 1, . . . , n

we call f a Sobolev function and the definition gets the expression

∫ f∂kϕ = −∫ gϕ.

For f Sobolev and p ≥ 1, we define the norms

∣f ∣W s,p = ∥∣∇f ∣∥Lp = (∫ ∣f ∣p)1/p

∥f∥W s,p = ∥f∥Lp + ∣f ∣W s,p .

and the Sobolev space as the set of Sobolev functions with finite Sobolev norm

W 1,p(Rn) = f ∶ ∥f∥W 1,p(Ω) <∞and structure as normed space.

Theorem 2.2. Sobolev space W 1,p(Ω) is a Banach space.

Proof. We show the completeness. Let fi be Cauchy in W 1,p. Then it’s Cauchy in Lp as is ∂kfi. Bycompleteness of Lp, there are limit functions f and g such that fi → f and ∂kfi → g in Lp. It remains toshow that g is the distributional derivative of f . Take ϕ ∈ C∞

c (Ω). Then

∫ f∂kϕ = limi∫ fi∂kϕ = − lim

i∫ (∂kfi)ϕ = −∫ gϕ

where we used convergence in Lp.

Theorem 2.3 (Meyers-Serrin). C∞(Ω) is dense in W 1,p(Ω) whenver p ∈ [1,∞).

Lemma 2.4. Let f ∈W 1,p(Ω), ε > 0 and fε = ηε ∗ f be the standard mollification. Then

∥fε − f∥W 1,p(Ω′) → 0

if p ∈ [1,∞) and d(Ω′,Ω) > 0.

Proof. Take ε < 10−10d(Ω′,Ω). Since suppηε ⊂ B(0, ε), we have suppηε(⋅ − x) ⊂ Ω so that fε is welldefined. Then for any partial derivative ∂

∂fε(x) = ∫ ∂(ηε)(x − y)f(y)dy = −∫ ∂(ηε(x − y))f(y)dy = ∫ ηε(x − y)∂f(y)dy = ηε ∗ ∂f(x)

and the right hand side converges to ∂f in Lp.

Proof of the Meyers-Serrin theorem. Fix δ > 0. Let Ωi = Ω2−i−1 ∖Ω2−i+1 so that Ωi form a finitely over-lapping cover of Ω. Let ψi be the associated smooth partition of unity. For each i, choose εi so that

∥(ψif)εi − ψif∥W 1,p(Ω) < 2−iδ,

here the subscript εi denotes convolution by the mollifier ηεi .For l > 0 and x ∈ Ωl, we note that

f(x) =∞∑i=1

(ψiu)(x) =l+10

∑i=1

(ψiu)(x), v(x) =∞∑i=1

(ψif)εi(x) =l+10

∑i=1

(ψif)εi(x).

Hence

∥u − v∥W 1,p(Ωl) ≤l+10

∑i=1

∥ψiu − (ψif)εi∥W 1,p(Ωl) ≤l+10

∑i=1

∥ψiu − (ψif)εi∥W 1,p(Ω) ≤∞∑i=1

2−iδ = δ.

Sending l → ∞, we obtain ∥u − v∥W 1,p(Ω) ≤ δ by monotone convergence. The function v is obviously

smooth by convolution and finite overlap of Ωi. 4

2.2. Basic properties and examples.

Proposition 2.5. The following hold:

i. Let p ≥ 1. If f ∈W 1,p and g ∈W 1,p′ with p′ = p/(p − 1), then ∇(fg) = f∇g + g∇f ∈ L1.

ii. If F ∈ C1, F′ ∈ L∞ and F (0) = 0, then ∇F (f) = F ′(f)∇f ∈ Lp.

iii. Let f+ = 1f>0f and f = f+ − f−. Then ∇f+ = 1f>0∇f almost everywhere.

Proof. First and second items follow by approximating with smooth functions to which rules of calculusapply. The third follows from the second by choosing Fε(r) = (r2 + ε2)1/2 − ε and sending ε → 0. SeeEvans and Gariepy section 4.2.2 for more details.

Example 2.6. i. Any smooth and compactly supported is trivially Sobolev.ii. f ∶ R→ R defined via f(x) = ∣x∣ is not smooth, but it is Sobolev.

iii. Let n ≥ 2 and take smooth ϕ with 1B(0,1) ≤ ϕ ≤ 1B(0,2). Then

f(x) = ϕ(x)∣x∣1/2 ∈W 1,1(R2)

though it is unbounded.iv. Let f be as previously. Let qi be an enumeration of rationals in B(0,1). Then

g(x) =∞∑i=1

2−if(x − qi) ∈W 1,1(R2)

though it is unbounded in any open subset of B(0,1), discontinuous at every point in B(0,1)and definitely not differentiable there either.

v. If n = 1, then W 1,1(R) is subset of absolutely continuous functions.

2.3. Poincare’s inequality. For bounded measurable E and a locally integrable f denote

fE = ⨏Ef = 1

∣E∣ ∫E f.

Theorem 2.7 (Poincare’s inequality). Let f ∈ W 1,1loc (Rn). Then for all balls B = B(x, r) with center

x ∈ Rn and radius r > 0 we have

⨏B∣f − fB ∣ ≤ Cr⨏

2B∣∇f ∣

where 2B = B(x,2r) is the concentric dilate.

Proof. By approximation, we may assume f is smooth. Then

⨏B∣f − fB ∣ ≤ ⨏

B⨏B∣f(x) − f(y)∣dxdy ≤ ⨏

B⨏B∫

1

0∣x − y∣∣∇f(x + t(x − y))∣dtdxdy

≤ Cr⨏B∫

1

0⨏

2B∣∇f(x)∣dxdtdy = Cr⨏

2B∣∇f ∣

with constant only depending on the dimension of the ambient space.

Definition 2.8. For α ∈ (0,1], define the Holder seminorm and norm by

∣f ∣Cα(Ω) = supx,y∈Ωx≠y

∣f(x) − f(y)∣∣x − y∣α

∥f∥Cα(Ω) = ∥f∥L∞(Ω) + ∣f ∣Cα(Ω)

Theorem 2.9 (Morrey’s embedding). Suppose f ∈ W 1,p(Rn) and p > n. Then f can be redefined in aset of measure zero to a bounded and Holder continuous function (still denoted by f) satisfying

∥f∥Cα(Rn) ≤ C∥f∥W 1,p(Rn)

for α = 1 − n/p.5

Proof. Suppose z is a Lebesgue point and take R > 0. Then by telescoping, Poincare and Holder

∣fB(z,R) − f(z)∣ = ∣∞∑i=1

fB(z,21−iR) − fB(z,2−iR)∣ ≤ C∑i>1⨏B(z,21−i)

∣f − fB(z,21−iR)∣

≤ C∑i>1

(21−iR)⨏B(z,21−i)

∣∇f ∣ ≤ C∑i>1

(21−iR)1−n/p (∫B(z,21−iR)

∣∇f ∣p)1/p

≤ CR1−n/p (∫B(z,R)

∣∇f ∣p)1/p

Then choose x, y Lebesgue points and R = ∣x − y∣. It follows by triangle inequality, Poincare and theprevious property

∣f(x) − f(y)∣ ≤ ∣f(x) − fB(x,R)∣ + ∣fB(x,R) − fB(y,R)∣ + ∣fB(y,R) − f(y)∣ ≤ CR1−n/p∥∇f∥Lp(Rn).Taking supremum over x, y, we obtain the bound for ∣f ∣Cα .

To prove boundedness, take an arbitrary Lebesgue point z. By the previous reasoning and Holder’sinequality

∣f(z)∣ ≤ ∣f(z) − fB(z,1)∣ + ∣fB(z,1)∣ ≤ C∥∇f∥Lp(Rn) + ∥f∥Lp(Rn).

3. Maximal functions

Next we aim at showing that Poincare inequality captures the essential of what it is to be a Sobolevfunction. To put this idea into the right framework, we will need several maximal functions. The firstone is classical.

Definition 3.1 (Hardy–Littlewood maximal function). For f ∈ L1loc, define

Mf(x) = supB∋x

⨏B∣f ∣

where the supremum is over the Euclidean balls.

The maximal function satisfies Mf ≥ f almost everywhere (Lebesgue differentiation theorem), it’slower semicontinuous (exercise), Mf = 0 if only if f = 0, and Mf ≳ ∣x∣−n whenever f ≠ 0 in a set ofpositive measure. Hence Mf ∉ L1 for any non-trivial f .

Theorem 3.2 (Hardy–Littlewood–Wiener). For f ∈ L1loc we have ∥Mf∥Lp ∼ ∥f∥Lp whenever p > 1 and

if p = 1, it holds

∣Mf > λ∣ ≤ Cλ∫ ∣f ∣.

Proof can be found for instance in Grafakos [8] as Theorem 2.1.6.

Definition 3.3 (Calderon maximal function). For p ∈ (0,∞), f ∈ Lploc(Rn), and α ∈ [0,1], we define

Nαp f(x) = sup

B∋xinfc∈R

rα (⨏B∣f − c∣p)

1/p.

Warning: this is not a standard name for this maximal function and is sometimes used to refer toanother quantity (that is of comparable size though). This operator and its relatives are aextensivelystudied in [5].

Proposition 3.4. If p > 1 and f ∈W 1,p(Rn), then N11 f ∈ Lp(Rn).

Proof. By Poincare’s inequality 2.7 and Hardy–Littlewood maximal theorem 3.2

∥N11 f∥Lp(Rn) ≲ ∥M ∣∇f ∣∥Lp(Rn) ≲ ∥∇f∥Lp(Rn).

In order to reverse the implication, we have to use a result of Haj lasz [10] (extending earlier work ofCalderon):

Theorem 3.5. Let f ∈ L1loc(Rn) and suppose there exists g ∈ L1

loc(Rn) such that

⨏B∣f − fB ∣ ≤ r(B)⨏

2Bg, for all balls B.

Then f ∈W 1,1loc (Rn) and ∣∇f ∣ ≤ Cg for some constant C only depending on the dimension n.

6

Proof. Let ϕ be the usual smooth function ϕ ≥ 0, suppϕ ⊂ B(0,1) and ∫ ϕ = 1 so that ϕε(x) = ε−nϕ(xε−1)with ε > 0 is an approximation of identity. Now ϕε(x) ≠ 0 only if ∣x∣/ε ≤ 1 so that suppϕε ⊂ B(0, ε). Also,by compact support and fundamental theorem on calculus

∫ Dϕ = 0, where D = ∂k for any k.

Then

∣D(f ∗ ϕε)(x)∣ = ∣f ∗Dϕε(x)∣ = ε−1∣f ∗ (Dϕ)ε(x)∣ = ε−1 ∣∫ f(y)(Dϕ)ε(x − y)dy∣

= ε−1 ∣∫B(x,ε)

(f(y) − fB(x,ε))(Dϕ)ε(x − y)dy∣ ≤ Cε−1−n ∫B(x,ε)

∣f(y) − fB(x,ε)∣dy

≤ C ⨏B(x,2ε)

g

where the last inequality followed by the assumption.Then take any test function ψ ∈ C∞

c (Rn) with suppψ =K. Let Kε = x ∶ d(x,K) < ε. Then

∣⟨D(f ∗ ϕε, ψ⟩∣ ≤ C ∫K

∣ψ(y)∣⨏B(y,2ε)

g(x)dxdy = Cε−nC ∫K2ε

∫K

∣ψ(y)∣1K(y)1B(y,2ε)(x)g(x)dydx

≤ ∥ψ∥L∞(Rn) ∫K2ε

g

Taking the limit ε→ 0, the we see that the distribution on the left extends to a boudned linear functionalon any Cc(Ω) for Ω an open subset of Rn. By the Riesz representation theorem (Section 1.8 in Evansand Gariepy [6]) it is given as integration against a signed Radon measure µ.

To see that µ is absolutely continuous with respect to the Lebsgue measure, suppose for contradictionthere existed K compact with µ(K) > 0 = ∣K ∣. Let χi(x) = (1 − id(x,K))+ so that these functions areLipschitz and χi → 1K . Then

0 < ∣µ(K)∣ ≤ limi

∣∫ χi dµ∣ ≤ C limi

∥χi∥L∞(Rn) ∫suppχi

g = 0

so that µ is absolutely continuous with respect to Lebesgue measure. By Radon-Nikodym theoremdµ = h(x)dx for a measurable h. By Lebesgue differentiation theorem we conclude h ≤ Cg.

Corollary 3.6. If p ≥ 1 and N11 f ∈ Lp(Rn), then ∇f ∈ Lp(Rn).

Proof. For any ball B, we have

⨏B∣f − fB ∣ ≤ 2r inf

z∈BN1

1 f(z) ≤ r⨏BN1

1 f.

Now N11 f ∈ Lp(Rn) ⊂ L1

loc(Rn). By the previous theorem f is weakly differentiable and ∣∇f ∣ ≤ CN11 f ∈

Lp(Rn)

4. Self-improving of generalized Poincare inequalities

4.1. Weak Lp spaces. For f measurable and t > 0, let

df(t) = ∣x ∈ Rn ∶ ∣f(x)∣ > t∣.

We define for p > 0

∥f∥Lp,∞(Rn) supt>0

t[df(t)]1/p.

Standard examples are Lp functions and power functions. If f ∈ Lp(Rn), then

df(t) = ∫∣f ∣>t

dx = ∫ ( ∣f ∣t)p

Ô⇒ ∥f∥Lp,∞(Rn) ≤ ∥f∥Lp(Rn).

On the other hand ∣x∣−n/p ∉ Lp(Rn) but

∣x∣−n/p > t = B(0, t−p/n)

whose measure is t−p. Consequently ∣x∣−n/p ∈ Lp,∞(Rn).7

4.2. Rearrangement. Infimums and supremums are to be understood as essential ones. Still f beingmeasurable and t > 0, let

f∗(t) = infλ ∶ df(λ) < t.Here is a list of some useful properties, assuming f ≥ 0 (in Grafakos [8] Section 1.4.1 you will find morebut mind the difference in the definition):

f∗ is non-increasing. Indeed, λ ∶ df(λ) < t1 ⊂ λ ∶ df(λ) < t2 whenever t2 ≥ t1 so thatf∗(t2) ≤ f∗(t1).

df(f∗(t)) ≤ t. There is an decreasing sequence λi → f∗(t) so that df(λi) < t. But

df(f∗(t)) = ∣∞⋃i=1

f > λi∣ = limidf(λi) ≤ t.

∣f ≥ f∗(t)∣ ≥ t. If the set has infinite measure, there is nothing to prove so assume it’s finite.Then

∣f ≥ f∗(t)∣ = ∣∞⋂i=1

f > f∗(t) − 2−i∣ = limk

∣k

⋂i=1

f > f∗(t) − 2−i∣ = limk

∣f > f∗(t) − 2−k∣ ≥ t

since each member of the sequence is below infimum of what would give measure above t. f∗(0) ∶= supt>0 f

∗(t) = ∥f∥L∞(Rn). Firstly

f∗(0) = supt>0

infλ > 0 ∶ df(λ) < t ≤ infλ > 0 ∶ df(λ) = 0 = ∥f∥L∞(Rn)

just by inclusion. The other direction follows by contradiction. Suppose ∥f∥L∞(Rn) > f∗(0).Assume f∗(0) <∞ since otherwise there is nothing to prove. There is a sequence of ti → 0 andf∗(ti)→ f∗(0). If f∗(0) < ∥f∥L∞(Rn), then

0 < df(f∗(0)) = limidf(f∗(ti)) ≤ lim

iti = 0

which is a contradiction. The last line follows along the same lines as the previous items. infz∈E f(z) ≤ (1Ef∗)(∣E∣). For any 0 < λ < infz∈E f(z) we have d1Ef(λ) = ∣E∣. Hence d1Ef(λ) <

∣E∣ only if λ ≥ infz∈E f(z) and consequently f∗(∣E∣) ≥ infz∈E f(z). (f + g)∗(t) ≤ f∗(θt)+ g∗((1− θ)t) for all θ ∈ (0,1). Let A = λ ∶ df(λ) < θt and B = λ ∶ dg(λ) <

(1 − θ)t. If then a ∈ A and b ∈ B, then

df+g(a + b) ≤ df(a) + dg(b) < tand

(f + g)∗(t) ≤ inf(A +B) = inf A + infB = f∗(θt) + g∗((1 − θ)t). df(t) < s if and only if f∗(s) < t so df(t) ≥ s if and only if f∗(s) > t. It holds

∥f∥Lp,∞(Rn) = supt>0

f∗(t)t1/p

as can be seen from the previous items.

4.3. Controlling oscillation.

Definition 4.1. Let Q be the family of all cubes with sides parallel to coordinate axes and p > 0. Leta ∶ Q → [∞) be in Dr (or satisfy the condition Dr) if for all cubes Q and all collections Qi of itspairwise disjoint subcubes

∑i

a(Qi)r ∣Qi∣ ≤ Caa(Q)r ∣Q∣

where Ca is a constant only depending on the function a but not on Q or the family Qi.

Examples include

The usual right hand side of Poincare inequality for f ∈W 1,p(Rn) with p > 1:

a(Q) = `(Q) (⨏2Q

∣∇f ∣p)1/p

is in Dr with r = pn/(n − p) The fractional averages with p > 0 and αp ∈ [0, n) and µ a locally finite positive Borel measure

a(Q) = `(Q)α (µ(Q)∣Q∣ )

1/p

are in Dr with r = np/(n − αp).8

The argument for the latter is as follows: Fix Q and Qi as in the definition

∑i

a(Qi)r ∣Qi∣ =∑i

∣Qi∣α/n−1/p+1µ(Qi)r/p ≤ ∣Q∣α/n−1/p+1∑i

µ(Qi)r/p

≤ ∣Q∣α/n−1/p+1 (∑i

µ(Qi))r/p

≤ ∣Q∣α/n−1/p+1µ(Q)r/p = a(Q)r ∣Q∣.

We used the exponent in the Lebesgue measure of the cube being positive and r/p > 1.There is a slight complication when using

a(Q) = `(Q) (⨏2Q

∣∇f ∣p)1/p

because of the dilation on the right hand side and the above argument must be refined in order toverify the Dr condition for such an function a. We will omit that and move on to the main theoremabout generalized Poincare inequalities. The theorem as well as the proof are here in the case of Rn andLebesgue measure, but the original theorem in [18] is more general. It applies in spaces of homogeneoustype and for weighted measures.

Theorem 4.2 (Lerner and Perez [18]). Let r > 0 and suppose a satisfies Dr. Suppose

infc∈R

[1Q(f − c)]∗(λ∣Q∣) ≤ Cλa(Q), λ ∈ (0,1)

holds for all cubes Q. Then

infc

∣Q∣−1/r∥1Q(f − c)∥Lr,∞(Rn) ≤ Ca(Q)for all Q and a constant Q only depending on r, Ca and n.

The strength of the assumption is best understood through the following

Proposition 4.3. Let f be measurable, Q a cube and λ ∈ (0,1). Then

(f1Q)∗(λ∣Q∣) ≤ λ−1/δ (⨏Q∣f ∣δ)

1/δ

for any δ > 0.

Proof. Take s so that s < (f1Q)∗(λ∣Q∣). Then df1Q(s) ≥ λ∣Q∣ and for any δ > 0

λ∣Q∣ ≤ ∣x ∈ Q ∶ ∣f ∣ > s∣ < ∫ ( ∣f ∣s

)δ

so that

s < ( 1

λ⨏Q∣f ∣δ)

1/δ.

Taking limit s→ (f1Q)∗(λ∣Q∣) concludes the proof.

Proof of the theorem. Fix Q. We first show for ε > 0 small enough that if t ∈ (0, ε∣Q∣/2), then

(4.1) (1Qf)∗(t) ≤ ca(Q)( ∣Q∣t

)1/r

+ (f1Q)∗(2t).

The left side is the level above which there is t mass, the right hand side is the smaller level with mass2t above plus an oscillation term paid for the difference.

Set-up and CZ cover. Define E = x ∈ Q ∶ ∣f(x)∣ ≥ (f1Q)∗(t). We first claim that t ≤ ∣E∣ ≤ 2t. Thelower bound is the third bullet in the property list. For the upper bound, note that we may assumef∗(2t) < f∗(t) since otherwise (4.1) is trivially true. Now

∣E∣ ≤ df(f∗(2t)) ≤ 2t

by the second bullet of the property list.Next we let Qi be the Calderon-Zygmund cubes of 1E , that is, the maximal relatively dyadic

subcubes of Q so that ∣E ∩Qi∣ > ε∣Qi∣. We have the properties

ε∣Qi∣ < ∣E ∩Qi∣ ≤ 2nε∣Qi∣. E ⊂ ⋃iQi ⊂ Q Qi ∩Qj = ∅ unless i = j.

9

This is possible, since ∣E ∩Q∣ = ∣E∣ ≤ 2t ≤ ε∣Q∣ by our restriction on values of t to be studied. Indeed, westart dividing Q into dyadic subcubes. If a cube gives high density of E, we choose it to be some Qi andleave it aside from further subdivisions. The large cube Q gives density of E small enough so that theprocess is well defined.

Control over Q by CZ cubes. First, by the definition of E, the fifth bullet in the property list of rear-rangements and the fact that f∗ is non-increasing, we see

(f1Q)∗(t) ≤ infx∈E

∣f(x)∣ = infi

infx∈E∩Qi

∣f(x)∣ ≤ infi(f1E∩Qi)∗(∣E ∩Qi∣) ≤ inf

i(f1E∩Qi)∗(ε∣Qi∣)(4.2)

where the last inequalty used the lower bound from CZ decomposition. Think ε as being small so thatthe value above is almost the sup. We next control it by almost the inf plus an oscillation.

Precisely, for any c ∈ R(f1Qi)∗(ε∣Qi∣) = ((f − c)1Qi)∗(ε∣Qi∣) + ∣c∣

and

∣c∣ ≤ infx∈Qi

(∣f(x) − c∣ + ∣f(x)∣) ≤ [(∣f − c∣ + ∣f ∣)1Q]∗(∣Qi∣) ≤ [∣f − c∣1Q]∗(ε∣Qi∣) + [f1Qi]∗((1 − ε)∣Qi∣).

The first line is clear from definition and the second line used some of the properties of rearrangements.Now altogether

(f1Qi)∗(ε∣Qi∣) ≤ 2 infc[∣f − c∣1Qi]∗(ε∣Qi∣) + [f1Qi]∗((1 − ε)∣Qi∣)

and plugging this in (4.2), we arrive at

(f1Q)∗(t) ≤ infi

(2 infc[∣f − c∣1Q]∗(ε∣Qi∣) + [f1Qi]∗((1 − ε)∣Qi∣))

Reduction to small oscillation cubes. Divide the CZ cubes in two families. We say i ∈ I if

a(Qi) ≤M1/ra(Q)( ∣Q∣t

)1/r

where M is a large constant to be chosen. Since we are estimating an infimum over i, it will suffice tofind even one index i such that i ∈ I but eventually we will need that there are many. This is proved asfollows. For i ∉ I, we have the reversed inequality so

∑i∉I

∣Qi∣ ≤∑i∉I

∣Qi∣a(Qi)r (Ma(Q)r ∣Q∣t

)−1

≤ tCaM

where the last inequality used the Dr condition. On the other hand, the CZ condition on the cubes give

∑i

∣Qi∣ ≥∑i

2−nε−1∣Qi ∩E∣ = 2−nε−1∣E∣ ≥ 2−nε−1t.

Putting the inequalities together, we see

(4.3) ∑i∈I

∣Qi∣ =∑i

∣Qi∣ −∑i∉I

∣Qi∣ ≥ t(1

2nε− CaM

) > 2t

by choice of M large enough only depending on Ca and ε and n. Hence there must be at least one indexin I. We conclude it is legal to restrict attention to cubes in I having low oscillation in comparison tothe large cube Q, and(4.4)

(f1Q)∗(t) ≤ infi

(2 infc[∣f − c∣1Q]∗(ε∣Qi∣) + [f1Qi]∗((1 − ε)∣Qi∣)) ≤ Ca(Q)( ∣Q∣

t)

1/r

+infi∈I

[f1Qi]∗((1−ε)∣Qi∣)

so that the first term is of the desired form.10

Iteration. It remains to estimate T = infi∈I[f1Qi]∗((1− ε)∣Qi∣), which is of the same form as the startingpoint in (4.2) but now restricted to a smaller cube. Set

Ei = x ∈ Qi ∶ ∣f(x)∣ ≥ (f1Qi)∗((1 − ε)∣Qi∣)Then using again the third bullet in the list about rearrangements as well as the lower bound for themeasure of Qi cubes in the family I (equality (4.3)), we conclude

∣⋃i

Ei∣ =∑i

∣Ei∣ ≥∑i

(1 − ε)∣Qi∣ > 2t

provided M is large enough. Now

T ≤ infi

infx∈Ei

∣f(x)∣ = infx∈⋃iEi

∣f(x)∣ ≤ (f1Q)∗(∣⋃i

Ei∣) ≤ (f1Q)∗(2t).

Plugging this in (4.4) we have proved (4.1).Next choose a positive integer k such that ε2−k−1∣Q∣ ≤ t < ε2−k ∣Q∣ and iterate (4.1):

(f1Q)∗(t) ≤ Cε,aa(Q)( ∣Q∣t

)1/r k

∑j=0

2−j/r + (f1Q)∗(2k+1t) ≤ Cε,aa(Q)( ∣Q∣t

)1/r

+ (f1Q)∗(ε∣Q∣).

Then note that the function f = f − c with c ∈ R satisfies the hypothesis of the theorem whenever f does.Hence we conclude

infc((f − c)1Q)∗(t) ≤ Cε,aa(Q)( ∣Q∣

t)

1/r

+ infc((f − c)1Q)∗(ε∣Q∣) ≤ Cε,aa(Q)( ∣Q∣

t)

1/r

.

which concludes the proof for t < ε∣Q∣/2. For t ≥ ε∣Q∣/2 the claim is trivial.

4.4. Truncation argument. The previous theorem is an example of an argument yielding an estimatewith weak Lr norm on the left hand side. In certain situations of practical interest, such estimates can be

improved to strong Lr bounds. In particular, if a(Q) = `(Q) (⨏Q ∣∇f ∣p)1/p

, we can apply what is called

Maz’ya’s truncation argument. Presentation follows mostly Haj lasz [12].Take t1 < t2 and a measurable function u ≥ 0. Let

ut2t1(x) =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

t1, u(x) ≥ t2u(x), t1 < u(x) < t2t2, t1 ≥ u(x)

and

ut2t1(x) = ut2t1− t1.

Proposition 4.4. Let Ak = x ∈ Ω ∶ ∣u(x)∣ > 2k for some measurable Ω ⊂ Rn. Fix q > p and assume fora pair of measurable functions (u, g) and all k ∈ Z

∥1Ωu2k+1

2k ∥Lq,∞(Rn) ≤ C∥g∥Lp(Ak∖A+1).

Then

∥u∥Lq(Ω) ≤ C∥g∥Lp(Ω)

where the constants C need not be same but they only depend on q, n and p.

Proof.

∫Ω∣u∣q ≤ ∑

k∈Z2qk ∣x ∈ Ω ∶ 2k−1 < ∣u(x)∣ ≤ 2k∣ ≤ ∑

k∈Z2qk ∣x ∈ Ω ∶ 2k−2 < ∣u(x)∣ − 2k−2∣

≤ ∑k∈Z

2qk ∣x ∈ Ω ∶ 2k−2 < u2k−1

2k−2∣ ≤ C∑k∈Z

(∫Ak∖Ak+1

∣g∣p)q/p

≤ C (∑k∈Z∫Ak∖Ak+1

∣g∣p)q/p

≤ ∥g∥q/pLp(Ω).

Proposition 4.5. If u ∈W 1,ploc (Rn), p ≥ 1 and 0 ≤ a < b ≤∞, then uba ∈W 1,p

loc and

(4.5) inf ∣Q∣−1/p∗∥1Q(uba − c)∥Lp∗,∞(Rn) ≤ C`(Q)( 1

∣Q∣ ∫2Q∩a<∣u∣≤b∣∇u∣p)

1/p

where p∗ = np/(n − p).11

Proof. We only prove the case p > 1. Recall the truncation property ∇u± = 1u±≠0∇u from Proposition2.5. We restrict the attention to positive u and write M = max(u, a) = (u − a)+ + a and m = min(u, b) =−(u − b)− + b so that uba =M +m − u ∈W 1,p

loc . Hence by Poincare’s inequality 2.7 and Holder’s inequality

⨏Q∣uba − (uba)Q∣ ≤ C`(Q)⨏

2Q∣∇uba∣ ≤ `(Q) (⨏

Q(M ∣12Q∇uba∣)p)

1/p=∶ a(Q).

The right hand side clearly satisfies Dp∗ so by Theorem 4.2 and the maximal function theorem

inf ∣Q∣−1/p∗∥1Q(uba − c)∥Lp∗,∞(Rn) ≲ C`(Q)( 1

∣Q∣ ∫2Q∩a<∣u∣≤b∣∇u∣p)

1/p

.

Theorem 4.6 (Sobolev-Poincare). Let u ∈W 1,ploc (Rn) with p ≥ 1. Then

(⨏Q∣u − uQ∣p

∗

)1/p∗

≤ C`(Q) (⨏2Q

∣∇u∣p)1/p

for 1/n = 1/p − 1/p∗ and all cubes Q.

Proof. We only present proof for the case p > 1. We can restrict the attention to positive part of u andhence asssume u ≥ 0. Take λ such that

∣u ≥ λ ∩Q∣ ≥ ∣Q∣2

and ∣u ≤ λ ∩Q∣ ≥ ∣Q∣2.

Let v = (v − λ)+ so that ∣u − λ∣ = v + (u − λ)−. We aim at estimating ⨏Q ∣v∣p∗ .

Suppose v is such that ∣v = 0 ∩Q∣ ≥ ∣Q∣/2. Then for any t > 0 and c ∈ R it follows

∣v > t∣ ≤ ∣∣v − c∣ > t∣ + ∣v ≠ 0, ∣c∣ > t/2∣ = ∣∣v − c∣ > t∣ + ∣v ≠ 0∣1∣c∣>t/2

≤ ∣∣v − c∣ > t∣ + ∣v = 0∣1∣c∣>t/2 = 2∣∣v − c∣ > t∣

This holds in particular for all truncations vba with 0 < a < b. Then by equation (4.5)

∣vba > t∣tp∗

≤ 2 infctp

∗

∣∣vba − c∣ > t/2∣ ≲ C`(Q)p∗

( 1

∣Q∣ ∫2Q∩a<∣u∣≤b∣∇u∣p)

p∗/p

whence by Proposition 4.4

⨏Qvp

∗

≲ `(Q)p∗

( 1

∣Q∣ ∫2Q∣∇u∣p)

p∗/p

.

Since the same argument applies to (u − λ)−, we have

⨏Q∣u − λ∣p

∗

≲ `(Q)p∗

( 1

∣Q∣ ∫2Q∣∇u∣p)

p∗/p

from which the claim follows since

⨏Q∣u − uQ∣p

∗

≲ ⨏Q∣u − λ∣p

∗

+ ∣λ − uQ∣p∗

.

Corollary 4.7. Suppose u ∈W 1,p(Rn) and p ≥ 1. Then

∥u∥Lp∗(Rn) ≲ ∥u∥W 1,p(Rn).

Proof. The previous inequality is equivalent to

(∫Q∣u − uQ∣p

∗

)1/p∗

≤ C (∫2Q

∣∇u∣p)1/p

where we can send `(Q)→∞ to obtain the claim. Note that ∣uQ∣ ≤ (∣u∣p)1/pQ → 0 as `(Q)→∞.

The value p = 1 can be included in the Sobolev–Poincare inequality but this requires either morecareful proof of the Poincare inequality with no dilation on the right hand side or a more general versionof the Lerner–Perez theorem, none of which were done. Note also that for mere application to Sobolev–Poincare inequality the tools in this section are unnecessarily complicated. However, they will find usealso elsewhere.

12

5. Hardy–Sobolev spaces

5.1. Motivation. We have been trying to understand smoothness as something that allows to controlmean oscillations in small scales or approximate by constant with good error bound. When p ≥ 1, thisproperty of Sobolev functions was made precise by Poincare’s inequality. When p < 1, the Lp norm ofthe derivative is not enough to control mean oscillation. Indeed, let φ ∈ C1(R) satisfy

φ(x) =⎧⎪⎪⎨⎪⎪⎩

0, x ≤ −1

1, x ≥ 1

Take the dilate ψε = ε−1φ(⋅ε−1). Then for arbitrarily small ε > 0

infc

(∫1

−1∣φε − c∣)

1/p≳p 1

but

∫1

−1∣(φε)′(x)∣p = ε1−p ∫

1

−1∣φ′(x)∣dx ∼ ε1−p → 0

as ε → 0. Hence a Poincare inequality with a derivative on the right hand side is not possible unlessp ≥ 1.

On the other hand, we saw in connection with the Calderon maximal function that a maximal functioncontrol is always true. It will turn out, that N1

pf being in Lp is equivalent with distribution (or actuallyfunction) f ∈ Lp with p > n/(n + 1) having derivatives in Hardy space.

5.2. Hardy spaces. Let

CN = ψ ∈ C∞ ∶ suppψ ⊂ B(0,1), supx

∣∂αψ(x)∣ ≤ 1, for all ∣α∣ ≤ N, ∫ ψ = 1.

For ϕ ∈ CN with N ≥ 2, let a > 0 and define the maximal functions acting on distributions f

Mϕf(x) ∶= supt>0

∣ϕt ∗ f(x)∣

Maϕf(x) ∶= sup

∣x−y∣<at∣ϕt ∗ f(y)∣,

MNf(x) ∶= supM1ϕf ∶ ϕ ∈ CN.

Theorem 5.1. Fix p > 0 and a > 0. Let ϕ,ψ ∈ CN with N large enough depending only on a and p.Then

∥Mϕf∥Lp(Rn) ∼a ∥Maψf∥Lp(Rn) ∼a,N ∥MNf∥Lp(Rn).

For proof, see Grafakos [9] Theorem 6.4.4.

Definition 5.2. Let η = c1B(0,1)e−1/(1−∣x∣2). For f a distribution, define

∥f∥Hp(Rn) ∶= ∥Mηf∥Lp(Rn)and Hp = f ∶ ∥f∥Hp(Rn) <∞.

The few properties and observations:

Hp is a quasi-normed space. Continuous functions are dense in Hp. For ϕε a smooth approximation of identity, the limit limε→0 ϕε ∗ f(x) exists almost everyhwere

independently of the exact choice of ϕ. For p > 1, Hp = Lp. For p ≤ 1, this is not the case. A compactly supported and bounded f fails to be in Hp unless ∫ f = 0. In the case of mean zerof ∈Hp with p > n/(n+ 1). Membership in lower index Hardy spaces comes with more vanishingmoments.

5.3. Hardy–Sobolev spaces. What follows is mostly a simple case of more general treatment in Miy-achi [20]. He deals with domains more general that the full space and classes of functions differentiablemore than once.

Definition 5.3. A distribution f is said to be in the homogeneous Hardy–Sobolev space H1,p(Rn) if∂kf ∈Hp(Rn) for all k = 1, . . . , n. We define a seminorm as

∥f∥H1,p(Rn) =n

∑k=1

∥∂kf∥Hp(Rn).

13

The seminorm does not see difference by constant. There are ways to define a proper norm, but thechoice of the non-homogeneous part dependes in the application one has in mind. For our purposes, themost natural definition will be Lp∩H1,p. Once could also think about Hp∩H1,p, but this choice imposesa zero-cancellation condition on the function itself. In the study of differentiability, such a conditionwould just be an unnecessary complication.

Lemma 5.4. Let Q be a cube and ψ smooth with suppψ ⊂ Q and ∫ ψ = 0. Then there are vk smoothand supported in Q such that ψ = `(Q)∑nk=1 ∂kψ. Moreover

If ∥∂αψ∥∞ ≤ `(Q)−n−∣α∣, then also ∥∂αvk∥∞ ≤ C`(Q)−n−∣α∣

for all ∣α∣ ≤m and some m.

Proof. We normalize Q = [−1,1]n. The general claim follows by change of variables. Let g ∈ C∞,supported in [−1,1] and with ∫ g = 1.

For f with ∫ f = 0 and support in [−1,1], let T 11 f = ∫

t−∞ f(s)ds. Hence we have the claim for

n = 1 and can assume it for dimension n − 1 as we prove it for higher dimensions by induction. Suppose we have defined operators Tn−1

k for all k = 1, . . . , n − 1. For f ∈ C∞c ([−1,1]n) having

mean zero. Set

f(x′) = ∫Rf(x′, t)dt x′ ∈ Rn−1,

Tnk f(x′, t) = g(t) ⋅ Tn−1k f(x′), for k = 1, . . . , n − 1

Tnn f(x′, t) = ∫t

−∞(f(x′, s) − g(s)f(x′))ds.

We take vk = Tnk ψ. Given the construction, checking the properties is easy.Suppose we already have the claim for

dimension n − 1. Then

n

∑k=1

∂Tnk ψ(x′, xn) = ψ(x′, xn) − g(xn)ψ(x′) + g(xn)n−1

∑k=1

∂Tnk ψ = ψ(x′, xn)

by induction. Derivative bounds are deduced similarly.

Proposition 5.5. Let f ∈ H1,p. Then

For any sequence εi → 0 and smooth, compactly supported ϕ ≥ 0 with ∫ ϕ = 1, the limit limi ϕϕε ∗ f(x) exists almost everywhere and is independent of the precise choice of ϕ.

For ϕ as above and 0 < ε < δ, it holds

∣f ∗ (ϕε − ϕδ)(x)∣ ≲ δM∂kf(x)

for every x.

Proof. We start with the second item. Fix k to be an integer such that 2−kδ ≤ ε < 2−k+1δ and let ϕi = ϕ2−iδ

for all i ∈ 0, . . . , k − 1. Write

ϕε − ϕδ = ϕε − ϕk +k

∑i=1

(ϕi − ϕi−1) =k

∑i=0

Φi

where Φi = ϕi − ϕi−1 for i < k and Φk = ϕε − ϕk. Now each Φi satisfies the assumptions of the previouslemma so there are functions vj,i with

Φi = 2−iδn

∑j=1

∂jvj,i.

Hence

∣(ϕε − ϕδ) ∗ f(x)∣ ≤ δk

∑i=0

2−in

∑j=1

∣∂jvj,i ∗ f(x)∣ ≲ δ∑k

MN∂kf(x).

This concludes the proof of the second item. It also shows that the sequence ϕεi(x) is Cauchy wheneverMN∂kf are finite. Hence there is a limit. Similarly, it can be used to show that the limit does notdepend on the approximate identity in question.

14

Lemma 5.6. Let f ∈ H1,p(Rn) ∩ Lp(Rn) with p > 0. Then for Q a cube, c(Q) its center, ϕ ∈ CN andalmost every x ∈ Q

∣f(x) − ϕ`(Q) ∗ f(c(Q))∣ ≲ `(Q)n

∑k=1

MN∂kf(x).

Proof. We may assume ∣Q∣ = 1. Choose a point x such that the approximations ϕε∗f(x)→ f(x) convergeas ε → 0. Take a sequence of nested cubes Qi ⊂ Qi−1 such that Q0 = Q, `(Qi) = 2−i and ∣x − ci∣ whereci = c(Qi) is always minimized so that from some index i0 on ci = x.

∣f(x) − c∣ = ∣∞∑i=0

(f ∗ ϕ2−i−1(ci−1) − f ∗ ϕ2−i(ci))∣ = ∣∞∑i=0

f ∗Φi(ci)∣

whereΦi(y) = ϕ2−i−1(y + ci−1 − ci) − ϕ2−i(y).

By the previous lemma, there are vk such that Φ = 2−i∑nk=1 ∂kvk with vk satisfying good bounds.Consequently

∣∞∑i=0

f ∗Φi(ci)∣ ≤∞∑i=0

2−in

∑k=1

∣(∂kvk ∗ f)(ci)∣ ≲n

∑k=1

MN∂kf.

5.4. Characterization with Calderon maximal function.

Proposition 5.7. Let f ∈ H1,p(Rn) ∩Lp(Rn). Then ∥N1pf∥Lp(Rn) ≲ ∥f∥H1,p(Rn)

Proof. Take q < p so that p > nq/(n − q). By the previous lemma,

infc

1

∣Q∣1/n (⨏Q∣f − c∣q)

1/q≲

n

∑k=1

(⨏QMN∂kf

q)1/q

.

After multipication by ∣Q∣1/n, the right hand side satisfies the summability condition Dq∗ with q∗ =nq/(n − q). Hence we can apply Theorem 4.2 to conclude

infc

∥1Q(f − c)∥Lq∗,∞(Rn)

∣Q∣1/q∗ ≲n

∑k=1

(⨏QMN∂kf

q)1/q

.

Denote f = 1Q(f − c). For

∫Q∣f ∣p = p∫

∞

0df(t)t

p−1 dt = p∣Q∣T p + ∥f∥q∗

Lq∗,∞(Rn) ∫∞

Ttp−q

∗−1 dt ≲ ∣Q∣T p + ∥f∥q∗

Lq∗,∞(Rn)Tp−q∗ ,

one may optimize the right hand side to see

infc

1

∣Q∣1/n (⨏Q∣f − c∣p)

1/p≲ inf

c

∥1Q(f − c)∥Lq∗,∞(Rn)

∣Q∣1/q∗

so that

N1pf ≲

n

∑k=1

MqMN∂kf.

Here Mqf = (M ∣f ∣q)1/q and now the claim follows form the maximal function theorem as q < p.

As useful piece of the previous proof is the following

Corollary 5.8 (of the proof.). Let Nαp f ∈ Lp(Rn) and q < pn/(n − αp)

(5.1) Nαq f(x) ≲MpN

αp f(x)

for all x.

Proposition 5.9. Suppose p > n/(n + 1) and N1pf ∈ Lp(Rn). Then ∂kf ∈Hp(Rn) for all k = 1, . . . n.

Proof. First we use (5.1) to see N11 f ≲MqN

1q ≤MqN

1p for any q ∈ (n/n + 1, p). By the assumption, the

right hand side is in Lp(Rn) and hence almost everywhere finite. The same goes for the left hand sidean f ∈ L1

loc(Rn) as a consequence. Take ϕδ to be the usual approximation of identity. Then

∣ϕδ ∗ ∂kf(x)∣ = infc∈R

δ−n−1 ∣∫ (f(y) − c)(∂kϕ)(y − xδ

)dy∣ ≲ infcδ−1 ⨏

B(x,δ)∣f − c∣ ≤ N1

1 f(x).

NowMϕ∂kf ≲ N1

1 f ≲MqN1p ∈ Lp(Rn).

15

At this point we give a name to the function space defined through Nαp .

Definition 5.10. Let α > 0 and let Pα be the space of all polynomials of degree at most k < α. Define

Nαp f(x) ∶= sup

B∋xinfP ∈Pα

r−α (⨏B∣f − P ∣)

1/p.

We set

Cαp = f ∈ Lp ∶ Nαp f ∈ Lp.

This is sometimes called Calderon–Hardy space

We will be mostly interested in the case α ≤ 1 so that the polynomials are of order zero. The previouspropositions can be summarized now as

Corollary 5.11.

Lp(Rn) ∩ H1,p(Rn) = C1p

whenever p > n/(n + 1).

5.5. Pointwise characterizations. Next we write a pointwise formula in the spirit of the Morreyembedding theorem to characterize the spaces Cαp . In the case of usual Sobolev spaces W 1,p(Rn) withp > 1, the formula below was used given in Haj lasz in [10] and later used as a definition for Sobolevspaces in more general metric measure spaces. The meaning of the formula for p = 1 remained open fora while until it was connected to Hardy–Sobolev space by Koskela and Saksman in [16]

Theorem 5.12. Let α ∈ (0,1] and p > n/(n + α). Suppose f ∈ Lp(Rn). Then Nαp f ∈ Lp(Rn) for all k if

and only if there is g ∈ Lp(Rn) such that for almost all x, y ∈ Rn it holds

∣f(x) − f(y)∣ ≤ C ∣x − y∣α(g(x) + g(y)).

Proof. Assume first Nαp f ∈ Lp. By (5.1) Nα

1 f ∈ Lp. For any Lebesgue point z (almost every point bylocal integrability) and any R > 0, we write

∣f(x) − fB(x,R)∣ ≤∞∑k=1⨏B(x,2−kR

∣f − fB(x,2−kR)∣ ≲ Rα∞∑k=1

2−kαNα1 f(x).

Then for R = ∣x − y∣ and almost every x and y

∣f(x) − f(y)∣ ≤ ∣f(x) − fB(x,R)∣ + ∣fB(y,R) − fB(x,R)∣ + ∣fB(y,R) − f(y)∣ ≲ ∣x − y∣α(g(x) + g(y))where g = Nα

1 f .To prove the converse, take q < p and note that for any cube Q with center c

∣Q∣−α/n (⨏Q∣f − f(c)∣q) ≲ (⨏ ∣g − g(c)∣q)

1/q≲Mqg(c).

If qn/(n − qα) > p as it can be chosen to be, we conclude

Nαp f ≲MqMqg ∈ Lp(Rn).

6. Embeddings for Calderon–Hardy spaces

6.1. Higher order spaces. This is just a brief comment on values of the smoothness parameter α > 1we have been neglecting. As indicated in definition 5.10, there is an extension of Nα

p operator for thesevalues using polynomial approximation in stead of mean oscillation. It is routine to verify that Sobolevspaces W k,p can be defined here, their members satisfy a Poincare inequality of type

infP ∈Pk−1

⨏B∣f − P ∣ ≲ rk ∑

∣β∣≤k⨏

2B∣∂βf ∣,

where Pk−1 are the polynomials of degree at most k − 1, and there is an analogue of Haj lasz’ lemma3.5. The coincidence with the corresponding Calderon–Hardy spaces Ckp is also obvious as p > 1 andeventually doable as p > n/(n−kp) though not as a verbatim repetition as the the case p > 1. From pointof view of Cαp , we do not study these higher order spaces in detail, but if need be, the book [5] as well asthe paper [20] contain a lot of additional information.

16

6.2. Fractional integral.

Definition 6.1. Let s > 0. Define ks(x) = ∣x∣−n+s. For f ∈ Cc a priori, we define

Isf = ks ∗ f.ks is called Riesz kernel and Is Riesz potential or fractional integral.

If s < n and f ∈ Lp with p > 1, we can also define Isf directly.

Proposition 6.2. Let p > 1 and sp ∈ (0, n), p∗ = np/(n − sp). Then for all f ∈ Lp

∥Isf∥p∗ ≲ ∥f∥Lp .

Proof. Let R > 0. Then Isf(x) = Is(1B(x,R)f +E. Let Bi = B(x,2−i+1R). Then

∣Is(1B(x,R)f ∣ ≤∞∑k=1∫Bk∖Bk+1

∣f(y)∣∣x − y∣ dy ∼

∞∑k=1

(2kR)n−s ∫Bk∖Bk+1

∣f(y)∣dy ≲ R−s∞∑k=1

2−ksMf(x) = R−sMf(x)

and

∣E∣ ≤ ∥f∥Lp (∫∣x−y∣>R

∣x − y∣p(s−n)p−1 )

1− 1p

≲ ∥f∥LpRs−n/p.

Choosing R such that R−sMf(x) ∼ ∥f∥LpRs−n/p and applying the Hardy–Littlewood maximal functiontheorem we conclude the proof.

Definition 6.3. Let s ∈ [0, n). For f ∈ L1loc, we set

Msf(x) = supB∋x

rs ⨏B∣f ∣.

The maximal function shares the Lp mapping properties of the fractional integral.

Proposition 6.4. Msf ≲ Is∣f ∣Proof. First not that the supremum B ∋ x can be replaced by supremum over r > 0 for B = B(x, r) inthe definition of the fractional maximal function at cost of a dimensional constant. Take B(x, r) and letBk = B(x,2−k+1r). Then

rs ⨏B∣f ∣ = Crs−n

∞∑k=1∫Bk∖Bk+1

∣f(y)∣dy ≲∞∑k=1∫Bk∖Bk+1

∣f(y)∣∣x − y∣n−s dy ≤ Is∣f ∣

whence the claim follows by taking a supremum over radii.

Corollary 6.5. Ms ∶ Lp∗ → Lp whenever sp ∈ (0, n) and p∗ = np/(n − ps).

Proof. Follows from the same bound for the fractional integral.

6.3. Inequality for Calderon functions.

Theorem 6.6. Let p > 0, α ≥ β > 0 and q > p such that

α − βn

= 1

p− 1

q.

Then∥Nβ

1 f∥Lq ≲ ∥Nα1 f∥Lp

Proof. Let B = B(x, r). Then for any δ > 0

infc∈R

r−β ⨏B∣f − c∣ ≤ rα−β inf

y∈BNα

1 f(y) ≤ rα−β (Nα1 f

δ)1/δ ≤ (Mγ(Nα1 f

δ))1/δ

where γ = δ(α − β). Now choose δ < p so that Mγ ∶ Lp/δ → Lq/δ as γ/n = δ(α − β)/n = δ(1/p − 1/q). Then

∥Nβ1 f∥Lq ≤ ∥Mγ(Nα

1 fδ)∥1/δ

Lq/δ≲ ∥Nα

1 f∥Lp .

We can list some interesting special cases easiest using the homogeneous spaces defined using only theseminorms taking into account the derivatives of the function.

Corollary 6.7. The following continuous embeddings hold:

For k > l ≥ 0 integers and q > p > 1 with (k − l)/n = 1/p − 1/q we have W k,p W l,q.17

For k/n > 1/p, denote l = maxj ∈ Z ∶ j < k/n − 1/p and σ = k/n − 1/p − l. Then W k,p Cl,σ (ltimes continuously differentiable with σ-Holder continuous lth derivatives).

For 1/n = 1/p − 1/q with p > n/(n + 1) we have H1,p Lq.

7. Littlewood–Paley theory

7.1. Schwartz class. Here is a list of most important properties to be recalled. A good source foreverything in this section is Grafakos [8], chapters 2 and 5 in particular. For the purposes of this section,we start assuming our functions are complex valued in contrast to the real valued considerations in theprevious sections.

For α,β ∈ Nn and f ∈ C∞, we let

ρα,β(f) = supx∈Rn

∣xα∂βf(x)∣

where xα =∏i xαii . If ρα,β(f) <∞ for all α and β, then we write f ∈ S and call it a Schwartz function. A

sequence fj → f in the sense of Schwartz class if ρα,β(f − fj)→ 0 for all α and β. Clearly S is completeunder this notion of convergence. Also, in topology consistent with this convergence, multiplication bypolynomials as well as differentiation are continuous operations. Finally C∞

c ⊂ S and in particular it isdense in all Lp spaces. When dealing with linear or sublinear operators, it suffices to work with Schwartzfunctions. If not explicitly otherwise stated, all functions are supposed to be Schwartz.

Together with the Schwartz class comes the class of tempered distributions that has the same role inthis theory as the distribtuions have for compactly supported functions. The crucial difference is thattempered distribution cannot grow too fast. For instance, et defines a distrbution on real line but not atempered one.

7.2. Fourier transform. For f ∈ L1, we write

Ff(ξ) = f(ξ) = ∫ e−2πix⋅ξ dx

for ξ ∈ Rn. Basic properties that are repeatedly used are

for f, g ∈ S, F(f ∗ g) = f g.

for α ∈ Nn, F(∂αf)(ξ) = (2πiξ)αf(ξ) and F((−2πix)αf) = (∂αf)(ξ). Define for p ≥ 1 and δ > 0 the dilation (Dilpδ f)(x) = δ−nf(xδ−1). Then F(Dilpδ f)(ξ) = Dil

1/p′δ−1

fwhere p′ = p/(p − 1).

For translations and modulations we have that for a ∈ Rn it holds e−2πia⋅ξ f(ξ) = F(f(⋅ − a))(ξ). ∫ fg = ∫ fg There is inverse F−1f(ξ) = f(ξ) = f(−ξ). Fourier transform is a homeomorphism of the S.

∫ fg = ∫ f g and ∥f∥L2 = ∥f∥L2 so that it is isometry in L2 (a priori this is norm inequality forSchwartz functions which allows to extend to L2).

It holds ∥f∥L∞ ≤ ∥f∥L1 and by interpolation ∥f∥Lp′ ≤ ∥f∥Lp whenever p ∈ [1,2]. No nontrivial compactly supported function has its Fourier transform compactly supported.

Proposition 7.1 (Bernstein). Let R > 0 and suppose f is supported in B(0,R). Then for all q > p ≥ 1and a dimensional constant C

∥f∥Lq ≤ CRn(1/p−1/q)∥f∥Lp .For any multi-index α ∈ Nn

∥∂αf∥Lp ≤ CR∣α∣∥f∥Lp .

Proof. Let ϕ ∈ C∞ be such that 1B(0,1) ≤ ϕ ≤ 1B(0,2). Let η = Dil∞2R ϕ. Then by Young’s convolutioninequality

∥f∥Lq = ∥η ∗ f∥Lq ≤ ∥η∥Lr∥f∥Lpwhere 1 + 1/q = 1/r + 1/p. Now

∥η∥Lr = ∥Dil1(2R)−1 ϕ∥Lr = Rn(1−1/r)Cϕ

which proves the first claim as 1 − 1/r = 1/p − 1/q. The second claim is done similarly and the proof isomitted.

18

7.3. Littlewood–Paley decomposition. Let ϕ be smooth with 1B(0,1) ≤ ϕ ≤ 1B(0,2) and set ϕ0(x) =ϕ(2x) − ϕ(x) and for j ∈ Z set ϕj = Dil∞2j ϕ0. Then

suppϕj ⊂ 2j−1 ≤ ∣x∣ ≤ 2j+1 and ∑j

ϕj(x) = 1 whenever x ≠ 0.

We also denote Kj = ϕj and Pjf = Kj ∗ f is the Littlewood–Paley operator projecting f to frequenciesaroung 2j . Note that each Kj is an L1 normalized kernel living at scale 2−j and having mean value zero(even arbitrarily many vanishing moments). The formula

∑j

Pjf = f

is very similar to the telescoping sum of mean values argument we have been using many times before.

7.4. Calderon–Zygmund theory. There are certain instances of Calderon–Zygmund theory that areneeded. Here are the main theorems. The proofs are not repeated here though they were discussed inlectures. The books [8] and [9] contain all. Another good source is the lecture notes for “HarmonicAnalysis” here https://sites.google.com/site/ioannisparissis/teaching.

Theorem 7.2. Let T ∶ L2 → L2 be linear and bounded. Suppose

∫ Tf ⋅ g =∬ f(x)K(x, y)g(y)dxdy, whenever supp f ∩ supp g = ∅

where K(x, y) ∈ L1loc(Rn ×Rn ∖ (x,x)) and there are constants and α ∈ (0,1] with

∣K(x, y)∣ ≲ ∣x − y∣−n, for x ≠ y

∣K(x, y) −K(x′, y)∣ ≲ ∣x − x′∣α∣x − y∣n+α , for∣x − x′∣ ≤ 1

2∣x − y∣

and the same properties holds for the adjoint T ∗ (that is, T is Calderon–Zygmund operator). ThenT ∶Hp →Hp for all p > 0 and T ∶ L1 → L1,∞

We will not prove this but just quote it. The region p > 1 and the weak endpoint are explained inTheorem 8.2.1 in [9]. The Hp bounds are, for instance, Theorem 6.4.14 in the same book. There are twospecial cases we will need frequently.

Corollary 7.3 (Mikhlin-Hormander theorem). Let m ∈ L∞ ∩ Cn/2+1 be such that ∣∂αm(ξ)∣ ≲ ∣ξ∣−∣α∣ for

all multi-indices ∣α∣ ≤ n/2 + 1. Then f ↦ F−1(mf) is bounded Lp → Lp whenever p > 1.

Corollary 7.4. Let a ∈ `∞. Then f ↦ ∑j ajPj is bounded Lp → Lp for p > 1.

7.5. Khintchine’s inequality. We want to understand a characterization of the Lp norm through thefollowing. There is a continuous embedding of the sequence spaces `p ⊂ `q whenever q > p. Introducingrandom signs results in un-doing the size difference between `1 and `2 sums. The `2 way of saying thisis Khintchine’s inequality.

Proposition 7.5 (Khintchine). Let (ai)Ni=1 be complex numbers and ωi independent random variablestaking values −1,1 with equal probability. Let p > 0. Then

(E∣∑i

ωiai∣p)1/p ∼p (∑i

∣ai∣2)1/2

where the hidden constant is in particular independent of N .

Proof. For simplicity, assume ai are real. We start by estimating the expectation. Let f = ∑i ωiai. Forany t > 0, by independence and zero expectation of the random variables

Eetf = E∏i

etωiai =∏i

Eetωiai =∏i

1

2(etai + e−tai) ≤∏

i

e12 t

2a2i = e 12 t

2∑i a2i .

Also, P(f > λ) = P(−f > λ) so by Tschebyschev and the preceding remark

P(∣f ∣ > λ) = 2P(f > λ) = P(tf − tλ > 0) = P(etf−tλ > 1) ≤ 2e−tλEetf ≤ 2e−tλe12 t

2∑i a2i = e 1

2λ2∑i a

2i

when we choose t = λ(∑i a2i )−1. Then by Cavalieri principle

E∣f ∣p = p∫∞

0λp−1P(∣f ∣ > λ)dλ ≤ 2p∫

∞

0λp−1e−λ

2/(∑i a2i ) dλ = Cp (∑

i

a2i)p/2

19

which concludes the estimate.For the other direction, we note that for p = 2 the comparability holds as an equality. Also, by Holder’s

inequality for probablitity spaces the case p > 2 is clear. For p ∈ (0,2) take s > 2 and θ ∈ (0,1) such that1/2 = θ/s + (1 − θ)/p. Then (the following are with respect to the probablity measure

∥f∥L2 ≤ ∥f∥1−θLp ∥f∥θLs ≤ Cθs ∥f∥1−θ

Lp ∥f∥θL2

where the last step used the first part of the theorem. Now hiding the L2 norm on the left concludes theproof.

Theorem 7.6 (Littlewood–Paley theorem). Let Pj be the Littlewood–Paley pieces. If f ∈ Lp with p > 1,then

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp≲ ∥f∥Lp .

If f is a tempered distribution and the left hand side is finite, then there is a polynomial P such that

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp≳ ∥f − P ∥Lp .

Proof. By Fatou, Khintchine, and Corollary 7.4,

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp≤ lim inf

N→∞

XXXXXXXXXXXXX

⎛⎝ ∑∣j∣≤N

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp≲ lim inf

N→∞(E

XXXXXXXXXXXX∑

∣j∣≤NωiPjf

XXXXXXXXXXXX

p

Lp

)1/p ≲ lim infN→∞

(E ∥f∥pLp)1/p

= ∥f∥Lp .

Conversely, we know that Pj = (Pj−1 + Pj + P j+1)Pj so set Fj = Pj−1 + Pj + P j+1. For f ∈ S ′ we know

that supp f(1 −∑j ϕ) = 0. Such and expression must be a polynomial P (this would require a proof,not too complicated, but we skip it, see [8] Section 2.4.1). Let ⟨⋅⟩ be the duality pairing of Schwartzclass and let g be a Schwartz function. The Littlewood–Paley decomposition converges as tempereddistribution so the following can be justified

⟨f − P, g⟩ = ⟨∑j

FjPjf, g⟩ = limN

⟨ ∑∣j∣≤N

Pjf,Fjg⟩ ≤ ∫⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2⎛⎝∑j∈Z

∣Fjg∣2⎞⎠

1/2

≤XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Fjg∣2⎞⎠

1/2XXXXXXXXXXXXXLp′≤XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp∥g∥Lp′

hence f − P extends to a bounded linear functional on Lp′

which proves the claim.

A fact: One can use the Littlewood-Paley square function when p ∈ (0,1]. Then

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp∼ inf

P∥f − P ∥Hp .

We won’t prove this, but the argument contains few additions to what we have seen. However, the theoryof singular integrals on Hardy spaces should be used instead of that for Lp spaces.

8. Function spaces

The Littlewood–Paley theorem and Hormander–Mikhlin theorem give a systematic way to generatefunction spaces with varying smoothness requirements. We mostly list different definitions and thenstudy in detail what turns out to be most commonly used fractional Sobolev space or Sobolev–Slobodeckijspace.

20

8.1. Riesz potentials. Recall the fractional integrals defined through Isf = ks ∗f where ks(x) = ∣x∣s−n.

Suppose s ∈ (0, n/2) so that ks is locally in L1 and decays at infinity faster than ∣x∣−n/2. Then ks ∈ L1+L2

and we can tell its Fourier transform is in L∞+L2. Then recall (or check) that Fourier transform of radialdistribution is radial and Fourier transform of distribution homogeneous of degree d is homogeneous ofdegree n − d. These symmetries nail down the form C ∣x∣s for the Fourier transform of ks. By Fourierinversion we can extend this to s ∈ (n/2, n).

More generally, we can define the Is as multiplier with symbol ∣ξ∣−s when it acts on Schwartz functions

f such that ∣ξ∣−sf(ξ) is locally integrable or with a lenghtier computation we can identify it with aconvolution by C ∣x∣s−n that is well defined for all Schwartz functions.

F(Isf) = c∣ξ∣−sf(ξ) as was discussed. This is the Riesz potential. See above for issues with thesingularity at origin.

F((−∆)s/2f) = c∣ξ∣sf(ξ) as can be seen to be consistent with the case s an integer. Fractionalpowers of Laplacian can be inverted up to polynomials by Riesz potential.

F(Rjf) = cξj ∣ξ∣−1f(ξ). This operator is called the Riesz transform and it is Calderon–Zygmundoperator in the sense of the Theorem 7.2.

F(Jsf) ∶= (1+ ∣ξ∣2)−s/2f(ξ) is called the Bessel potential. It has the effect of increasing smooth-ness by s derivatives but still being bounded Lp → Lp unlike Is.

8.2. Potential spaces. We start with the homogeneous spaces. Denote by W k,p the space of tempereddistributions modulo polynomials so that the kth order derivatives are in Lp.

Proposition 8.1. Let k > 0 be integer and p > 1. Then W k,p = Ikf ∶ f ∈ Lp.

Proof. Take first f ∈ Lp. Consider the tempered distribution Ikf . Let α be a multi-index with ∣α∣ = k.Then ∂αIkf = Rα1

1 ⋯Rαnn f so that

∥∂αIkf∥Lp = ∥Rα1

1 ⋯Rαnn f∥Lp ≲ ∥f∥Lp .Conversely, take f ∈ W k,p. Then there is a polynomial P of degree at most k − 1 so that f = f − P hasFourier transform zero at zero. Then

(−∆f)k/2f = (−∆f)k/2f = c(−∆f)k/2(n

∑j=1

R2j)kf ∈ Lp

where we used the identity for Riesz transforms clear from the Fourier symbol as well as boundedness ofthe Riesz transforms.

More generally, we can give a definition of potential spaces of non-integer order of smoothness.

Definition 8.2. We let Hs,p be the collection of tempered distributions that are of the form Isf forf ∈ Lp. This space is normed through ∥f∥Hs,p = ∥(−∆)s/2f∥Lp for p > 0. This is the homogeneouspotential space

We define the inhomogeneous potential space as image of Js when p > 1: f ∈H1,p if f = Jsg for someg ∈ Lp. We norm this through ∥f∥H1,p = ∥(1 + (−∆)1/2)s/2f∥Lp

The inhomogeneous space can be proved to be Hs,p ∩ Lp. We deliberately omit the definition ofinhomogeneous potential space when p ≤ 1 as there are at least two reasonable candidates for definitionand not so clear reason to favor one of them.

Theorem 8.3. Let p > 1 and s ∈ R and f ∈ Hs,p. Then

∥f∥Hs,p ∼XXXXXXXXXXXXX

⎛⎝∑j∈Z

(2js∣Pjf ∣)2⎞⎠

1/2XXXXXXXXXXXXXLp.

Let f ∈Hs,p, then

∥f∥Hs,p ∼ ∥f∥Lp +XXXXXXXXXXXXX

⎛⎝∑j≥0

(2js∣Pjf ∣)2⎞⎠

1/2XXXXXXXXXXXXXLp.

Proof. This is very similar to the characterization of Lp. The upper bound for the square function isobtained through replacing the Littlewood–Paley piece ϕ0(ξ) by ϕ0(ξ)∣ξ∣−s. The lower bound followsthe lines of the lower bound for Lp.

Theorem 8.4. Let s − n/p = σ − n/q with 0 < s − σ < n and p, q ∈ (0,∞). Then Hs,p ⊂ Hσ,q

21

Proof. This follows from Hp →Hq bounds for the fractional integrals. We only deal with the case p > 1for which we have proved the relevant bounds for fractional integral

∥(−∆)σ/2∥Lq = ∥(−∆)(σ−s)/2(−∆)s/2f∥Lq = ∥Is−σ(−∆)s/2f∥Lq ≤ ∥(−∆)s/2f∥Lp = ∥f∥Hs,p .

8.3. Besov and Triebel–Lizorkin scales. The characterization of Lp was an Lp norm of `2 norm ofthe Littlewood–Paley pieces. Replacing the Khintchine–given 2 by something else, we get some otherspaces, which go by the name of Triebel–Lizorkin spaces:

∥F ∥F sp,q =XXXXXXXXXXXXX

⎛⎝∑j∈Z

(2js∣Pjf ∣)q⎞⎠

1/qXXXXXXXXXXXXXLp, s ∈ R, p, q ∈ (0,∞).

The most important case is q = 2 which gave Hardy spaces and Lp spaces F 0p,2 and homogeneous potential

spaces F sp,2. One could also define the non-homogeneous version by cutting the sum and adding the Lp

norm on the right hand side. However, we will not use that in future.Whenever p = q one may change the order of sum and integral. This brings us to Besov scale which

can be defined in its own right for more general values. Here we prefer the non-homogeneous version

∥f∥Bsp,q = ∥f∥Lp +⎛⎝∑j≥1

(2js∥Pjf∥Lp)q⎞⎠

1/q

, s ∈ R, p, q ∈ (0,∞).

The most importand value of q is equal to p.Embeddings for Besov spaces are easily proved.

Proposition 8.5. Let s − n/p = σ − n/q, s > σ and p, q ∈ [1,∞). Then Bsp,r ⊂ Bσq,r.

Proof. For each piece ∥Pjf∥Lq ≲ 2jn(1/p−1/q)∥Pjf∥Lp = 2j(s−σ)∥Pjf∥Lp by Bernstein’s inequality. Hence

⎛⎝∑j≥1

(2jσ∥Pjf∥Lq)r⎞⎠

1/r

≲⎛⎝∑j≥1

(2js∥Pjf∥Lp)r⎞⎠

1/r

.

For the other part, we first write ψ = 1 −∑j≥1 ϕj . Then we decompose f = f1 + f2 where f1 = ψ ∗ f . Forf1, we have by Bernstein and Young ∥f1∥Lq ≲ ∥f∥Lp . For the other term, we take M > max(r, p, q,2) anduse the Littlewood–Paley theorem as well as Holder’s inequality and Minkowski to obtain

∥f2∥Lq ∼XXXXXXXXXXXXX

⎛⎝∑j≥1

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLq≲XXXXXXXXXXXXX

⎛⎝∑j≥1

(2jε∣Pjf ∣)M⎞⎠

1/MXXXXXXXXXXXXXLq≤⎛⎝∑j≥1

(2jε∥Pjf∥Lq)M⎞⎠

1/M

≤⎛⎝∑j≥1

(2j(ε+s−σ)∥Pjf∥Lp)M⎞⎠

1/M

≤ ∥f∥Bsp,r .

8.4. Finite differences. For f a measurable function y ∈ Rn and m > 1 integer, let the differenceoperators be

∆yf(x) = ∆1yf(x) = f(x) − f(x + y), ∆m

y f(x) = ∆y(∆m−1y f)(x)

and for p > 0 and t > 0 define the modulus of smoothness

ωmp (t, f) = sup∣y∣<t

∥∆my f∥Lp .

Theorem 8.6. Assume s > 0, m,N ∈ N with m+N > s and 0 ≤ N < s. Then for p ∈ [1,∞] and q ∈ [1,∞]

∥f∥Bsp,q ∼ ∥f∥Lp +n

∑j=1

(∫∞

0(tN−sωmp (t, ∂Nj f))q

dt

t)

1/q.

Proof. We will prove the theorem only in the case s ∈ (0,1) and p, q ∈ (1,∞). The more general casefollows along the same lines but with an additional layer of technicalities. However, before starting, letus notice the derivatives in the right hand side exist under the assumption of the left hand side beingfinite. Indeed, we have the embedding Bsp,q ⊂ FNp,2 = HN,p because of the condition s > N : Let fj = Pjf

22

be the Littlewood–Paley projections. Suppose first p < 2, then by the continuous embedding `p ⊂ `2 ofsequences

∥(∞∑k=0

22Nk ∣fk ∣2)1/2

∥Lp ≤ ∥(∞∑k=0

2pNk ∣fk ∣p)1/p

∥Lp = (∞∑k=0

2pNk∥fk∥pLp)1/p

.

If p > 2, then by Minkowski’s inequality

∥(∞∑k=0

22Nk ∣fk ∣2)1/2

∥Lp = ∥∞∑k=0

22Nk ∣fk ∣2∥1/2Lp/2

≤ (∞∑k=0

22Nk∥fk∥2Lp)

1/2

.

Fix then r = max(q/p, q/2,2) > 1 so that 2r ≥ q and pr ≥ 2. For the case p > 2, we use Holder as well as`q ⊂ `2r to see

(∞∑k=0

22Nk∥fk∥2Lp)

1/2

= (∞∑k=0

22k(N−s)22sk∥fk∥2Lp)

1/2

≤ (∞∑k=0

22k(N−s)r′)1/r′

(∞∑k=0

22rsk∥fk∥2rLp)

1/(2r)

≤ Cr,N−s (∞∑k=0

2qsk∥fk∥qLp)1/(2r)

<∞.

The case p ≤ 2 is done exactly the same way but pr ≥ 2 being used instead of 2r ≥ q. Using these twoestimates we see

∥f∥FNp,2 = ∥f∥Lp + ∥(∞∑k=0

22Nk ∣fk ∣2)1/2

∥Lp <∞

which proves that f ∈ FNp,2 =HN,p so that it has weak Nth derivatives in Lp.To start the actual proof, we first not that t↦ ω(t, f) is increasing. Hence

∫∞

0(t−sωp(t, f))q

dt

t= ∑k∈Z∫

2k+1

2k(t−sωp(t, f))q

dt

t≤ ∑k∈Z

(2−skωp(2k+1, f))q ∫2k+1

2k

dt

t

≲ ∑k∈Z

(2−s(k+1)ωp(2k+1, f))q.

The same computation can be done the other way round so we find

(8.1) ∫∞

0(t−sωp(t, f))q

dt

t∼ ∑k∈Z

(2−ksωp(2k, f))q)

so that it suffices to deal with the discrete sum instead of the integral on the left.Left controlls right. Let Pkf = ϕk ∗ f be the Littlewood–Paley pieces and ψ a smooth function

with 1B(0,1) ≤ ψ ≤ 2B(0,2). To prove an Lp bound for ∆yPkf , it suffices to assume f is Schwartz as theoperators are linear, and we can take limits to upgrade the a priori bound for nice functions to a generalresult Lp functions. Now the Fourier transform can be written as

F(∆yPkf)(ξ) = ϕk(ξ)f(ξ)1 − e−2πiξ⋅y

ξ ⋅ y ξ ⋅ y

so that

∆yPkf = Ty(y ⋅ ∇Pkf), where F(Tyg) =1 − e−2πiξ⋅y

ξ ⋅ y g(ξ)

for any Schwartz function g. Suppose that ∣y∣ = 1. Using a Hormander–Mikhlin type argument, one canverify my defines an Lp bounded operator for all p. By dilation symmetry of the Fourier transform, wealso have for general y = ∣y∣e with ∣e∣ = 1

g ∗ my(x) = ∫ g(x − z) 1

∣y∣n me(z

∣y∣ )dz = ∫ g(x − ∣y∣z)me(z)dz

= ∫ (Dil∞∣y∣−1 g)(∣y∣−1x − z)me(z)dz = ((Dil∞∣y∣−1 g) ∗ me)(x∣y∣−1)

so that∥g ∗ my∥Lp = ∣y∣n/p∥Dil∞∣y∣−1 g ∗ me∥Lp∥ ≲e ∣y∣n/p∥Dil∞∣y∣−1 g∥Lp = ∥g∥Lp .

Hence∥∆yPkf∥Lp = ∥Ty(y ⋅ ∇Pkf)∥Lp ≲ min(1, ∣y∣2k)∥Pkf∥Lp

where the estimate with one follows from plain Minkowski ∥∆yPkf∥Lp ≤ 2∥Pkf∥Lp and the other estimateuses the previous multiplier bound as well as Bernstein’s inequality (Proposition 7.1).

23

Now return to the modulus of smoothness. For any i ∈ Z2−isωp(2i, f) ≤∑

k

2−isωp(2i, Pkf) ≲∑k

2−ismin(1,2i+k)∥Pkf∥Lp

=∑k

min(1,2i+k)2−s(i+k)2sk∥Pkf∥Lp = a ∗ b(i)

where the convolution is with respect to the counting measure on integers and

a(l) = 2−sl∥P−lf∥Lp , b(l) = 2−sl min(1,2l).Recall the Young’s inequality: ∥a ∗ b∥`r ≤ ∥a∥`q∥b∥`s provided that s, r, q ∈ [1,∞] and 1 + 1/r = 1/q + 1/s.Taking the `q norm of the inequality above and applying Young’s inequality with r = q and s = 1, weobtain

(∑k∈Z

(2−ksωp(2k, f))q))1/q

≲ ∥a ∗ b∥`q ≤ ∥a∥`q∥b∥`1 ≲ ∥a∥`q ≤ ∥f∥Bsp,q .

Right controlls left. Since all ϕj giving rise to the Littlewood–Paley decomposition are Schwartzfunctions, we have for any integer M the bound (1+ ∣x∣2)Mϕ(x) ≤ CM . Also, ϕj have Fourier transforms

supported away from the origin so their mean value is zero (recall f(0) = ∫ f(x)e−2πix⋅0 = ∫ f forintegrable functions). Consequently

∣ϕj ∗f(x)∣ = ∣∫ ϕj(x−y)f(y)dy∣ = ∣∫ ϕj(x−y)(f(y)−f(x))dy∣ ≲M∞∑k=1

2−Mk ⨏B(x,2k−j)

∣f(y)−f(x)∣dy

where the last inequality used that ϕj = Dil12j ϕ0. Now

∥ϕj ∗ f∥Lp ≲∞∑k=1

2−kM∥⨏B(0,2k−j)

∣f(z − ⋅) − f(⋅)∣dz∥Lp ≤∞∑k=1

2−kMωp(2k−j , f)

and

⎛⎝∞∑j=1

(2js∥ϕj ∗ f∥Lp)q⎞⎠

1/q

≤ ∑k≥1

⎛⎝∑j≥1

(2js2−kMωp(2k−j , f))q⎞⎠

1/q

= ∑k≥1

(∑l∈Z

(2(k−l)s2−kMωp(2l, f))q)1/q

= ∑k≥1

2k(s−M) (∑l∈Z

(2−lsωp(2l, f))q)1/q

≲ (∑l∈Z

(2−lsωp(2l, f))q)1/q

.

From the proof it is possible to read that the modulus of smoothness can actually be replaced by

the quantity ωp(t, f) = (⨏B(0,t) ∥∆yf∥pLp dy)1/p

. This is useful with the following definition. Let f be a

measurable function. For s ∈ (0,1) and 1 ≤ p <∞ define the Gagliardo seminorm

∣f ∣Bsp,p ∶= (∬∣f(x) − f(y)∣p∣x − y∣n+sp dxdy)

1/p

.

The following proposition or actually the previous theorem shows that this is indeed a seminorm forBesov space when p = q. In this special case the Besov space is usually called fractional Sobolev spaceor Sobolev–Slobodeckij space.

Proposition 8.7. Let f be a measurable function and s ∈ (0,1) and 1 ≤ p, q <∞. Then

(∬∣f(x) − f(y)∣p∣x − y∣n+sp dxdy)

1/p

∼ (∫∞

0(t−sωp(t, f))p

dt

t)

1/p.

Proof. Let Bk = B(0,2k). Then

∬∣f(x) − f(y)∣p∣x − y∣n+sp dxdy ∼ ∫ ∑

k∈Z2−k(n+ps) ∫

Bk∖Bk−1∣∆hf ∣p dhdx ≲ ∑

k∈Z(2−ksωp(2k, f))p.

To reverse the last inequality, note that

∑k∈Z

(2−ksωp(2k, f))p = ∫ ∑k∈Z

2−k(n+ps) ∑j≤k∫Bj∖Bj−1

∣∆hf ∣p dh = ∫ ∑j∈Z∑j≤k

2−k(n+ps) ∫Bj∖Bj−1

∣∆hf ∣p dh

≲ ∫ ∑j∈Z

2−j(n+ps) ∫Bj∖Bj−1

∣∆hf ∣p dh.

24

9. Real interpolation

9.1. Preliminaries. This section mostly follows the exposition and notation of [2]. In what follows,A0 and A1 are Banach spaces that are contained in tempered distribtutions (say, but this assumption

could be modified or weakened). We denote the ordered pair by A = (A0,A1) and use the notations

Σ(A) = A0 +A1 and ∆(A) = A0 ∩A1. They come with natural norms infa=a0+a1(∥a0∥A0 + ∥a1∥A1) andmax(∥a0∥A0 , ∥a1∥A1).

9.2. K functional. The first interpolation method starts with the sum space. Define for t > 0 anda ∈ Σ(A)

K(t, a,A) = infa=a0+a1

(∥a0∥A0 + t∥a1∥A1).

This is a norm on the sum space. An arbitrary portion of the arguments of K will be omitted in thenotation depending on which ones of them are in focus at given time. The following proposition is easyto verify, and the proof is omitted.

Proposition 9.1. For a ∈ A0 +A1, the mapping t ↦ K(t, a,A) is positive, increasing, concave, and itsatisfies K(t, a) ≤ max(1, t/s)K(s, a).

Define the functional

Φθ,q(f) = (∫∞

0(t−θf(t))q dt

t)

1/q

for θ ∈ [0,1] and q > 0 though the important values will be θ ∈ (0,1) and q ≥ 1. We define the norm

∥a∥Aθ,q,K ∶= Φθ,q(K(⋅, a,A)), Aθ,q,K = a ∈ Σ(A) ∶ ∥a∥Aθ,q,K <∞.

This is indeed a norm (as the K functional and Φθ,q are).

Theorem 9.2. For θ ∈ (0,1), q ≥ 1, the following hold

∆(A) ⊂ Aθ,q,K ⊂ Σ(A) with continuous inclusions.

K(s, a,A) ≤ Csθ∥a∥Aθ,q,K for all s > 0 and aΣ(A).

If ∥T ∥Ai→Bi ≤Mi for i ∈ 0,1 and T is linear, then ∥Ta∥Bθ,q,K ≤M1−θ0 Mθ

1 ∥a∥Aθ,q,K .

Proof. By the previous proposition

min(1, t/s)K(s, a) ≤K(t, a).

Considering the quantities above as functions of t and applying the functional Φθ,q on both sides, weobtain

∥a∥Aθ,q,K ≥K(s, a) (∫∞

0(t−θ min(1, t/s))q dt

t)

1/q= s−θK(s, a) (∫

∞

0(t−θ min(1, t))q dt

t)

1/q= s−θK(s, a)C

where the constant C is positive and finite. This proves the second claim. The first inclusion in the firstclaim is obvious from the definition of the K functional. The second inclusion follows from the inequalityjust proved.

In order to prove the bounds for the operator T as in the hypothesis of the theorem. Note that sinceB0 +B1 ⊃ T (A0) + T (A1) and a0 + a1 = a implies Ta0 + Ta1, we can write

K(t, Ta,B) ≤ infa=a0+a1

(∥Ta0∥B0 + t∥Ta1∥B1) ≤ infa=a0+a1

(M0∥a0∥A0 + tM1∥a1∥A1) =M0K(tM1/M0, a,A)

whence the claim follows by applying Φθ,q to both sides of the inequality.

In practice it is often easier to work with a sum instead of an integral as in the finite differencecharacterization of the Besov spaces. For a sequence (aj) indexed over integers, define

(9.1) ∥(aj)∥λθ,q =⎛⎝∑j∈Z

(2−jθ ∣aj ∣)q⎞⎠

1/q

, θ ∈ [0,1], q > 0.

Proposition 9.3. Let a ∈ Σ(A) and αj =K(2j , a,A) for j ∈ Z. Then ∥a∥Aθ,q,K ∼ ∥(α)j∥λθ,q .

Proof. This is the same computation as (8.1). 25

9.3. J functional. The second interpolation method starts with the intersection space and attempts atrepresenting general functions as sums of functions in intersection space that converge in the norm ofthe sum space. For t > 0 and a ∈ da, let J(t, a,A) = max(∥a∥A0 , t∥a∥A1). The following proposition is assimple as its K functional counterpart and the proof is omitted.

Proposition 9.4. For a ∈ ∆(A), the mapping t ↦ J(a, t) is positive, increasing and convex and thefollowing inequalities are valid for all t, s > 0

J(t, a) ≤ max(1, t/s)J(s, a) and K(t, a) ≤ min(1, t/s)J(s, a).Define a ∈ Aθ,q,J if there are uj ∈ ∆(A) for j ∈ Z such that

a =∑j

uj ,⎛⎝∑j

(2−θjJ(2j , uj))q⎞⎠

1/q

<∞

where the convergence is in the norm of Σ(A). We let

∥a∥Aθ,q,J = inf(uj)

∥J(2j , uj)∥λθ,q

wher the infimum is taken over all decompositions as above.

Proposition 9.5. For s > 0, θ ∈ [0,1] and q ≥ 1 and a ∈ ∆(A), it holds ∥a∥Aθ,q,J ≤ Cs−θJ(s, a,A).

Proof. This is actually obvious. As the J functional is increasing, it suffices to restrict the attention ofs = 2j for some integer j. Choose the sequence uk = a for j = k and uj = 0 for j ≠ k. Then

∥a∥Aθ,q,J ≤ ∥(J(2k, uk))∥λθ,q = 2−jθJ(2j , a).

9.4. Equivalence of the methods. The interpolation spaces obtained via J andK functionals coincide,but it is practical to have both definitions at ones disposal.

Theorem 9.6. Let θ ∈ (0,1) and q ≥ 1. Then Aθ,q,J = Aθ,q,K .

Proof. Take a ∈ Aθ,q,J and some (uk) ⊂ ∆(A) with ∑j uk = a, convergence in the norm of Σ(A). Thenby the basic properties of the J functional and a shift of the summation variable

K(2j , a) ≤∑k

K(2j , uk) ≤∑k

min(1,2j−k)J(2k, uk) =∑k

min(1,2k)J(2j−k, uj−k).

Multiplying both sides with 2−jθ and applying `q norm, we obtain

∥a∥Aθ,q,K ≤ ∥∑k

min(1,2k)2−θk2−(⋅−k)θJ(2⋅−k, u⋅−k)∥`q ≤ ∥(J(2j , uj))∥λθ,q∑k

min(1,2k)2−θk

where the sum in k converges to a numerical constant. Since (uj) was an arbitrary sequence, we cantake infimum and finish the proof of the first inclusion Aθ,q,J ⊂ Aθ,q,K .

To prove the converse, we need the following lemma

Lemma 9.7. Assume min(1, t−1)K(t, a) → 0 as t → 0 or t → ∞. Then there exists (uj) ⊂ ∆(A) with

a = ∑j uj converging in Σ(A) and J(2j , uj) ≲K(2j , a).

Before proving the lemma, note the following: If a ∈ Aθ,q,K , then K(s, a) ≲ sθ∥a∥Aθ,q,K so that thehypothesis of the lemma is valid. The conclusion of the lemma is enough to finish the proof of thetheorem just by taking λθ,q norms of both sides of the inequality.

Proof of the lemma. Let ε > 0. For every j ∈ Z we can find aj0 ∈ A0 and aj1 ∈ A1 so that a = aj0 + aj1 and

∥a0j∥A0 + 2j∥a1

j∥A1 ≤ (1 + ε)K(2j , a). Taking limits and possibly multiplying by 2−j we see

∥a0j∥A0 ≤K(2j , a)→ 0, as j → −∞

and∥a1j∥A1 ≤ 2−jK(2j , a)→ 0, as j →∞.

Let uj = aj0 − aj−10 = aj−1

1 − aj1. This equality is because of these pairs forming a decomposition of a. On

the other hand, from this identity it is obvious that uj ∈ ∆(A). Moreover, for any pair of large integersM and N a telescoping sum argument shows

a −M

∑j=−N

uj = a + a−N−10 − aM0 = a−N−1

0 + aM126

so that

K(1, a −M

∑j=−N

uj) ≤ ∥a−N−10 ∥A0 + ∥aM1 ∥A1 → 0

as M,N →∞. Hence ∑j uj = a with convergence in Σ(A). Finally,

J(2j , uj) ≤ max(∥aj0∥A0 + ∥aj−10 ∥A0 ,2

j(∥aj1∥A1 + ∥aj−11 ∥A1)) ≲K(2j , a)

which concludes the proof of the lemma and hence the equivalence theorem.

9.5. Basic properties and duality. Before studying the dual space of the interpolation space, wecheck some elementary properties of the interpolation spaces. The elementary proof of the followingauxiliary proposition is omitted.

Proposition 9.8. A normed space X is complete if and only if the following holds: Let xi ∈ X be asequence. Then ∑i ∥xi∥X <∞ implies ∑∣i∣≤N xi → x ∈X as N →∞.

Proposition 9.9. Let A be an interpolation couple and θ ∈ (0,1) and q ∈ (1,∞). Then

If A0 and A1 are complete so is Aθ,q.

∆(A) is dense in Aθ,q.

Proof. Take a sequence of ai ∈ Aθ,q with ∑i ∥ai∥Aθ,q <∞. As ∥ai∥Aθ,q ≳ K(1, ai,A) by Theorem 9.2, we

see that ∑i ∥ai∥Σ(A). As Σ(A) is Banach, the series converges to a ∈ Σ(A) in its norm. But as Φθ,q and

K-functional are sublinear on the function variable, we conclude that a ∈ Aθ,q and the convergence alsotakes place in the Aθ,q norm.

To see the second claim, realize Aθ,q through the J method. Hence a ∈ Aθ,q can be written as a = ∑i aiwith ai ∈ ∆(A) and convergence in Σ(A) norm. Again,

∥a −N

∑i=−M

ai∥Aθ,q,J ≤ ( ∑i<−M

(2−iθJ(2i, ai))q)1/q

+ (∑i>N

(2−iθJ(2i, ai))q)1/q

→ 0

as M,N →∞.

Proposition 9.10. Suppose that ∆(A) is dense in both A0 and A1. Denote A′ = (A′

0,A′1). Then

Σ(A)′ = ∆(A′), ∥a′∥∆(A′) = sup

a∈Σ(A)

∣⟨a′, a⟩∣∥a∥Σ(A)

and

∆(A)′ = Σ(A′), ∥a′∥Σ(A′) = sup

a∈∆(A)

∣⟨a′, a⟩∣∥a∥∆(A)

.

Proof. We only prove the first pair of equalities, the second one being similar. Take a′ ∈ Σ(A′). Bydefinition, there are a′i ∈ A′

i with i = 0,1 such that a′ = a′0 + a′1. Recall that as A0 and A1 are byassumption contained in the same linear Hausdorff space (e.g. tempered distributions), there is noambiguity in defining the sum of elements in the dual spaces and there are no issues with dualitypairings of different spaces.

Let a ∈ ∆(A). Then

∣⟨a′, a⟩∣ ≤ ∣⟨a′0, a⟩∣ + ∣⟨a′1, a⟩∣ ≤ ∥a∥A0∥a′0∥A′

0+ ∥a∥A1∥a′1∥A′

1≤ (∥a′0∥A′

0+ ∥a′1∥A′

1)max(∥a∥A1 , ∥a∥A1)

so that a′ ∈ ∆(A)′ and ∥a′∥Σ(A′) ≤ ∥a′∥∆(A)′ .

To prove the other direction of the equality, take k ∈ ∆(A)′. Let D = (a, b) ∈ A0 ⊕A1 ∶ a = b. Theform (a0, a1)↦ k(a0+a1

2) is bounded and linear functional on D. By Hahn–Banach theorem, there is an

extension k of k on A0 ⊕A1 with ∥k∥A′

0⊕A′

1≤ ∥k∥∆(A)′ , k = a′0 ⊕ a′1. Since ∆(A) is dense in A0 and also

A1, the functional a′0 + a′1 (though not the summands) is uniquely determined by its values on D. Weconclude that k has a unique extension to A′

0 +A′1 (though not necessarily A′

0 ⊕A′1) and it satisfies the

claimed norm inequality.

Recalling that K and J functionals at 1 are norms of ∆(A) and Σ(A) and using the homogeneityproperty, we get the following variant of the previous proposition.

27

Corollary 9.11. Let A be an interpolation couple such that ∆(A) is dense in both A0 and A1, and lett > 0. Then

K(t, a′,A′0,A

′1) = sup

a∈∆(A)

∣⟨a′, a⟩∣J(t−1, a,A0,A1)

, J(t, a′,A′0,A

′1) = sup

a∈Σ(A)

∣⟨a′, a⟩∣K(t−1, a,A0,A1)

.

Theorem 9.12. Let A be a couple such that ∆(A) is dense in both A0 and A1. Let q ∈ (1,∞) andθ ∈ (0,1). Then (A0,A1)′θ,q = (A′

0,A′1)θ,q′ where q′ = q/(q − 1).

Proof. Step 1: (A0,A1)′θ,q,J ⊂ (A′0,A

′1)1−θ,q′,K . First take a′ ∈ (A0,A1)′θ,q,J . Let ε > 0. By the previous

corollary, for each j ∈ Z there is bj ∈ ∆(A) such that

(9.2) K(2−j , a′,A′0,A

′1) − εmin(1,2−j) ≤ ∣⟨a′, bj⟩∣

J(2j , bj ,A0,A1).

To evaluate the λ1−θ,q′ norm, we use duality of sequences. Let α ∈ λθ,q and for any such sequence defineaα = ∑j J(2j , bj)−1αjbj . Each term in the sum is obviously in ∆(A). Also,

∥aα∥Aθ,q,J ≤⎛⎝∑j

(2−jθJ(2j , αjbj

J(2j , bj)q⎞⎠

1/q

≤⎛⎝∑j

(2−jθ ∣αj ∣)q⎞⎠

1/q

= ∥α∥λθ,q <∞

and one can see the sequence converges not only in Σ(A) but even in Aθ,q,J . Moreover, as a directconsequence

(9.3) ∣⟨a′, aα⟩∣ ≤ ∥a′∥A′

θ,q,J∥α∥λθ,q .

On the other hand, by (9.2)

(9.4) ⟨a′, aα⟩ ≥∑j

(K(2−j , a′) − εmin(1,2−j))αj .

Let βj =K(2j , a′,A′1,A

′0) = 2jK(2−j , a′,A′

0,A′1). Using (9.3) as well as (9.4), we see

∑j

βjαj2−j − ε ≤ ∥a′∥A′

θ,q,J∥α∥λθ,q .

Taking supremum over all sequences α ∈ λθ,q with norm bounded by one, we can invoke the duality if

λθ,q and λ1−θ,q′ to conclude

∥K(2j , a′,A′1,A

′0)∥λ1−θ,q′ = sup

α∑j

2−jK(2j , a′,A′1,A

′0)αj ≤ ∥a′∥A′

θ,q,J+ ε

which conclude the proof of the first inclusion as ε→ 0.Step 2: (A′

0,A′1)1−θ,q′,K = (A′

0,A′1)1−θ,q′,J ⊂ A′

θ,q,K . The equality is from the equivalence theorem ofthe J and K methods. Take a from the J space above. By the definition, there are aj in the intersection

of the duals with ∑j aj = a converging in Σ(A′) = ∆(A)′. Hence for any a ∈ ∆(A) by Proposition 9.10

∣⟨a′, a⟩∣ =∑j

∣⟨aj , a⟩∣ ≤∑j

J(2−j , aj ,A′)K(2j , a,A) ≤∑

j

2jJ(2j , aj ,A′1,A

′0)K(2j , a,A)

≤ ∥a′∥1−θ,q′,J∥a∥Aθ,q,K .Finally, A′

θ,q,K = A′θ,q,J so also their duals must coincide so the proof is complete.

Corollary 9.13. If A0 and A1 are reflexive, so are all the real interpolation spaces Aθ,q with q ∈ (1,∞)and θ ∈ (0,1).

9.6. Explicit computations. The next examples show how to compute some K functionals. This ispart of what is told in [3]

Example 9.14. Let A0 = Lp and A1 =W 1,p with p > 1. We show that

K(t, f,A) ∼ min(1, t)∥f∥Lp + ωp(t, f)

where ωp is the modulus of smoothness from section 8.4. Fix f ∈ Σ(A). To get an upper bound forthe K functional, take ϕ to be a smooth bump supported in B(0,10−10) and integral one, and denoteϕt(x) = t−nϕ(xt−1). Let g = f ∗ ϕt and b = f − g. Clearly g ∈W 1,p as it is actually smooth. Now

K(t, f,A) ≤ ∥f∥Lp trivially.28

By Minkowski and a linear change of variables,

∥b∥Lp ≤ t−n ∫B(0,t)

∣ϕ(y)∣∥f(⋅ − y) − f(⋅)∥Lp dy ≤ ωp(t, f).

By Minkowski and the integral one property of ϕ, we see ∥g∥Lp ≤ ∥f∥Lp . For any k ∈ 1, . . . , n, the usual tricks show

∥∂kg∥pLp =1

tp∫ ∣f ∗ (∂kϕ)t∣p dx =

1

tp∫ ∣(f − f(x)) ∗ (∂kϕ)t(x)∣p dx ≲ t−pωp(t, f).

Putting the estimates together, we get the desired upper bound for the K functional:

K(t, f) ≤ ∥b∥Lp + t∥g∥W 1,p ≲ min(1, t)∥f∥Lp + ωp(t, f).To verify the converse inequality, one has to work with an arbitrary decomposition f = b + g with

b ∈ Lp and g ∈W 1,p. First note that

min(1, t)∥f∥Lp ≤ min(1, t)(∥b∥Lp + ∥g∥Lp) ≤ ∥b∥Lp + t∥g∥W 1,p

so it suffices to estimate the modulus of smoothness. Here, however, we can again note that ωp(b) ≤ 2∥b∥Lpso it remains to estimate g.

As g ∈W 1,p, we can find a sequence of smooth gj → g with convergence in W 1,p. Then for h ∈ Rn

∫ ∣∆hgj ∣p = ∫ ∣gj(x)−gj(x−h)∣p dx = ∫ ∣∫1

0h⋅∇gj(x−sh)ds∣p dx ≤ (∫

1

0∥∇gj∥Lp ∣h∣ds)

p

≤ ∣h∣p∥∇gj∥pLp .

Taking limit and supremum over ∣h∣ ≤ t we obtain ωp(g) ≤ t∥g∥W 1,p . All in all,

min(1, t)∥f∥Lp + ωp(f) ≲ ∥b∥Lp + t∥g∥W 1,p .

Taking infimum over all decompositions b + g = f we obtain the lower bound for the K functional. Also,together with Theorem 8.6 this shows that

Bsp,q = (Lp,W 1,p)s,q.

Example 9.15. The second example is Lp spaces, but we need to review one more fact about rear-rangements (recall the others from 4.2). Namely, for a measurable f , it holds for all t > 0

df(t) = df∗(t).Indeed, let E = ∣f ∣ > t. Consider the open set (0, ∣E∣) and take s from there. Suppose we had

f∗(s) = infλ > 0 ∶ df(λ) < s ≤ t.Then there would exist a decreasing sequence λk → f∗(s) with df(λk) < s for all k. But the consequent

∣E∣ > s > df(λk)→ df(f∗(s)) ≥ df(t) = ∣E∣is a contradiction so we must have (0, ∣E∣) ⊂ f∗ > t. Conversely, if s > ∣E∣ = df(t), then f∗(s) ≤ t so that[0, ∣E∣]c ⊂ f∗(s) ≤ t or equivalently f∗ > t ⊂ [0, ∣E∣]. Which concludes the proof that the distributionfunction of f and its non-increasing rearrangement are the same. In particular, we have that

∫ ∣f ∣ = ∫∞

0f∗(s)ds.

Consider then the couple A0 = L1 and A1 = L∞. Take f ∈ Σ(A) and t > 0. We will see that

K(t, f,A) = tf∗∗(t) = t(1

t∫

t

0f∗(s)ds) .

Consider first an arbitrary decomposition f = f1 + f∞. Then for any θ ∈ (0,1)

tf∗∗(t) = ∫t

0f∗(s)ds ≤ ∫

t

0(f∗1 (sθ) + f∗∞(s(1 − θ)))ds = 1

θ∫

tθ

0f∗1 (s)ds + t 1

t(1 − θ) ∫t(1−θ)

0f∗∞(s)ds

≤ ∥f1∥L1θ−1 + t sups∈(0,t)

f∗∞(s) ≤ ∥f1∥L1θ−1 + ∥f∞∥L∞ .

We can take θ → 1 and take infimum over decompositions to control the right hand side by the Kfunctional.

To prove the other inequality, choose E = ∣f ∣ > f∗(t) and let f1 = f1E and f∞ = f − f1. Then it isobvious from definition as well as the fact the rearrangement is non increasing that

∥f∞∥ ≤ f∗(t) ≤ f∗∗(t).29

Also,

∫ f1 = ∫∞

0∣f1 > s∣ = ∫

∞

f∗(t)∣∣f ∣ > s∣ = ∫

∞

f∗(t)∣f∗ > s∣ = ∫

∞

f∗(t)∣1[0,t]f

∗ > s∣ds ≤ ∫t

0f∗(s)ds.

Hence the K functional is bounded by tf∗∗(t) as claimed.As a concluding remark, by choosing θ = 1/p′ and q = p, it is easy to verify

∫∞

0(t−θK(t, f,A))q dt

t= ∫

∞

0f∗∗(t)p dt ≲ ∫

∞

0f∗(t)p dt = ∫ ∣f ∣p

by the Hardy–Littlewood maximal function theorem.More generally, one may define the Lorentz spaces Lp,q as interpolation spaces of L1 and L∞ by setting

θ = 1/p′ and q = q so that the norm becomes

∫ (t−θK(t))q dtt∼ ∫ (t1/pf∗(t))q dt

t.

Example 9.16. Assume s0 < s1 and 1 < p, q <∞ and θ ∈ (0,1). Let s = (1 − θ)s0 + θs1. Then

(Hs0,p,Hs1,p)θ,q = Bsp,q.

Not too detailed proof. Let f be in the interpolation space on the left hand side. To compute the Kfunctional, take f0 and f1 from the relevant potential spaces above. Fix an integer k ≥ 0. By Hormander–Mikhlin theorem 7.3 one may verify that the operator 2skJsPk is bounded on Lp with operator normcontrolled by a dimensional constant. Recall that Pk is the Littlewood–Paley projection to frequenciesabout 2k and Js is the Bessel potential mapping Lp →Hp,s →Hp,2s. Now

∥Pkf∥Lp ≤ ∥Pkf0∥Lp + ∥Pkf1∥Lp = 2−s0k∥2s0kJs0J−s0Pkf0∥Lp + 2−s1k∥2s1kJs1J−s1Pkf1∥Lp

≲ 2−s0k∥J−s0Pkf0∥Lp + 2−s1k∥J−s1Pkf1∥Lp = 2−s0k(∥f1∥Hs0,p + 2(s0−s1)k∥f1∥Hs1,p)

so that ∥Pkf∥Lp ≲ 2−s0kK(2k(s0−s1), f). Multiplying by 2sk and taking the `q sum over k we get the firstbound for high frequencies. The piece with frequencies close to zero is even simpler and the argument isomitted here.

To estimate the interpolation norm by a Besov norm, we use the J method. It is verifieable that thesum ∑j Pjf converges in Hs1,p (the sum space) so we try to compute the norm corresponding to thisdecomposition. Now

2k(s−s0)J(2−k(s1−s0), Pk) = 2k(s−s0) max(∥J−s0Pk∥Lp ,2−k(s1−s0)∥J−s1Pk∥Lp) ≲ 2ks∥Pkf∥Lpby Young’s convolution inequality and estimation of the multiplier. Taking `q norm concludes theproof.

10. Complex interpolation

We continue working on interpolation following the notations and conventions of [2] as in the previoussection. To set up the framework, we need the notion of analytic function with values in a Banach space.

Definition 10.1. Let U ⊂ C be open and X Banach. A function f ∶ U → X is said to be analytic ifz ↦ ⟨x′, f(z)⟩ is analytic as a mapping U → C for every element of the dual space x′ ∈X ′.

Basic complex analysis carries over to Banach valued setting mostly by applying classical results onmappings ⟨x′, f(z)⟩. This is how, for instance, one proves the following:

Proposition 10.2. Let U ⊂ C be a domain with non-empty boundary, z ∈ U , and f ∶ U → X analyticwith continuous extension to the closure U . Then supz∈U ∥f(z)∥X ≤ supξ∈∂U ∥f(ξ)∥.

Proof. Just take x′ ∈ X ′ with ⟨x′, f(z)⟩ = ∥f(z)∥ and apply the ordinary maximum principle to theduality pairing.

Next let S = z ∈ C ∶ Re z ∈ (0,1) and for an interpolation couple A define the subset of functions

f ∶ S → Σ(A) analytic in the interior of S

(10.1) F(A) = f ∶ f ∈ C(S) ∩L∞(S), f(j + it) ∈ Aj for t ∈ R, j = 0,1, lim∣t∣→∞

f(j + it)→ 0.

The interpolation couple in question is occasionally omitted in the notation. By the maximum principleas given in Proposition 10.2, the class F admits a norm

∥f∥F ∶= max(supt

∥f(it)∥A0 , supt

∥f(1 + it)∥A1).30

For a parameter θ ∈ (0,1), we define the complex interpolation space with respect to the couple A aswell as its norm through

Aθ = f(θ) ∶ f ∈ F, ∥a∥θ = inf∥f∥F ∶ f(θ) = a, f ∈ F.

Note that Aθ is isomorphic and isometric to the quotient space F/f(θ) = 0.

10.1. Basic properties. We start with the class F .

Proposition 10.3. F is Banach.

Proof. We use Proposition 9.8. Suppose fn is a sequence such that its sum converges absolutely in F .For any z ∈ S, maximum principle gives

∥fn(z)∥Σ(A) ≤ ∥fn∥F .

Since Σ(A) is Banach, the partial sums of ∑n fn converge to f ∈ Σ(A). As Aj ⊂ Σ(A), the same estimateholds for z ∈ ∂S. By Weierstrass M test the convergence is uniform so the limit f = ∑n fn is analytic inS (by e.g. Cauchy formula for vector valued setting) and continuous up to the boundary. Moreover, thetopology of uniform convergence is that of F so the proof is complete.

Next we record the basic interpolation properties of the space Aθ.

Proposition 10.4. Let A be an interpolation couple and θ ∈ (0,1). Then

Aθ is Banach. ∆(A) ⊂ Aθ ⊂ Σ(A) with continuous inclusions.

If T ∶ A→ B is linear with ∥T ∥Aj→Bj =Mj for j = 0,1, then ∥T ∥Aθ→Bθ ≤M1−θ0 Mθ

1 .

Proof. The first claim follows from abstract results. The pointwise evaluation z ∶ f ↦ f(z) is bounded

and linear for every z ∈ S as a mapping F → Σ(A). Its kernel N is hence closed and Aθ is isomorphicand isometric to the quotient F/N . Hence it is complete.

The second claim: By maximum principle ∥f(θ)∥Σ(A) ≤ ∥f∥F so taking infimum over f(θ) = a ∈ Aθproves the inclusion on the right. For the inclusion on the left, take a ∈ ∆(A) and set f(z) = e(z−θ)2a sothat f(θ) = a and the relevant norm inequality becomes obvious as well.

The third claim: Take a ∈ Aθ. Let ε > 0. By definition, there exists f ∈ F(A) such that ∥f∥F(A) ≤∥a∥θ + ε. Set g(z) =Mz−1

0 M−z1 T (f(z)). Then g ∈ F(B) and

∥g∥F(B) ≤ max(suptM−1

0 ∥Tf(it)∥A0 , suptM−1

1 ∥Tf(1 + it)∥A1) ≤ ∥f∥F(A) ≤ ∥a∥θ + ε

so that, noting g(θ) =Mθ−10 Mθ

1Ta,

∥Ta∥θ ≤M1−θ0 Mθ

1 ∥g∥F ≤M1−θ0 Mθ

1 (∥a∥θ + ε)

which concludes the proof after taking infimum over the functions f admissible in the definition so thatε→ 0.

10.2. Scalar and vector valued Lp spaces. The principal advantage of the complex method of in-terpolation is its precise interplay with iterated Lp spaces. The proof that scalar valued Lp spacesinterpolate as a complex scale contains the main idea. The proof for the Banach valued case is omittedto save a detour through technicalities.

Theorem 10.5. Assume p0, p1 ≥ 1 and θ ∈ (0,1). Let 1/p = (1 − θ)/p0 + θ/p1. Then (Lp0 , Lp1)θ = Lp.

Proof. We restrict our attention to bounded and compactly supported functions and prove that for themthe norms obtained through the interpolation construction as well as Lp coincide. Let a be such afunction. Set

f(z) = eε(z2−θ2)∣a∣

pp(z)

a

∣a∣where 1

p(z) =1−zp0

+ zp1

. Then f(θ) = a and

∥f(ti)∥Lp0 = e−ε(t2+θ2)∥a∥Lp , ∥f(1 + ti)∥Lp1 = eε(1−(t

2+θ2))∥a∥Lp

so that ∥f∥F ≤ ∥a∥Lpeε and at the limit ε→ 0 we have ∥a∥θ ≤ ∥a∥Lp .31

Fort the converse direction, suppose ∥a∥θ = 1. Take b bounded with compact support and normalizedthrough ∥b∥Lp′ and f almost realizing the norm of a. Define

g(z) = eε(z2−θ2)∣b∣

p′

p′(z)b

∣b∣ , F (z) = ⟨f(z), g(z)⟩.

By the Hadamard three-lines theorem

∣⟨a, b⟩∣ = ∣F (θ)∣ ≤ e10ε → 1 = ∥a∥θ, as ε→ 0.

Taking sup over all b as given, the equality of the norms follows.Finally note that as both Aθ and Lp are complete and bounded compactly supported functions are

dense in Lp, it follows Lp ⊂ Aθ. The proof of the converse norm inequality did not require any specialassumption on a so the proof is complete.

The following theorem, which is stated without proof, follows the same ideas but is more technical.

Theorem 10.6. Let A be an interpolation couple and (U,µ) a Borel measure space. Let θ ∈ (0,1),p0, p1 ∈ (1,∞) and 1

p= 1−θ

p0+ θp1

. Then

(Lp0(U,µ;A0), Lp1(U,µ;A1))θ = Lp(U,µ;Aθ).If p1 = ∞, then the same interpolation holds but with L∞(U,µ;A1) replaced by the completion of thesimple functions under the norm of L∞(U,µ;A1).

10.3. The method of retracts. A normed space B is a retract of A if the identity B → B factors asid = p i where i ∶ B → A and p ∶ A → B are bounded linear mappings. For practical purposes thismeans that B can be embedded in A in a regular way. The following proposition follows directly frominterpolation properties (boundedness) of real and complex methods, respectively.

Proposition 10.7. Suppose that B is a retract of A with maps i and p. Then Bθ,q and Bθ with all

admissible parameters are retracts of Aθ,q and Aθ with same mappings i and p.

For the rest of the section, set for a tempered distribution f and a sequence of Lp functions α

i(f) = (Pjf)j , p(α) =∑j

Pjαj

where Pj is the Littlewood–Paley projection around frequencies 2j , Pj = ∑j+1k=j−1 Pk and the sum is

understood and defined as limit of tempered distributions and undefined otherwise.

Proposition 10.8. Bsp,q with p, q > 0 is retract of λ−s,q(Lp) and Hs,p is retract of Lp(λ−s,2).

Proof. These are seen by just writing down the definitions/characterizations in terms of the Littlewood–Paley decomposition.

Proposition 10.9. For θ ∈ (0,1) and s = (1−θ)s0 +θs1 and 1/p = (1−θ)/p0 +1/p1 and 1/q = (1−θ)/q0 +1/q1, the following hold

(Bs0p0,q0 ,Bs1p1,q1)θ = B

sp,q, (Hs0,p0 ,Hs1,p1)θ =Hs,p.

Proof. Consider the Besov case. Both the left and the right hand side are retracts of λ−s,q(Lp) byinterpolating the sequence space with Theorem 10.6 where µ = ∑j 2sjδj , U = Z and Aj = Lpj . As theimage of p is both the Besov space and the interpolation space above, the identity on these spaces mapsthem to one another. Hence they must be equal. The same goes for potential spaces.

10.4. Sneiberg’s stability theorem. The moral of Sneiberg’s theorem [22] is that invertibility of alinear operator between spaces on a complex interpolation scale extrapolates. Hence being close inparameters means being close in other ways too, something not so clear for real interpolation scales. Thepresentation follows the Appendix A in [1] and Section 8 in [15].

Theorem 10.10 (Sneiberg’s theorem). Let X and Y be interpolation couples and T ∶X → Y is boundedlinear opeator (meaning here X0 → Y0 and X1 → Y1 are bounded, we refer to maximum of these bounds

as ∥T ∥). Suppose there is θ∗ ∈ (0,1) and κ > 0 such that T ∶Xθ → Y θ is surjective and ∥Tx∥Y θ ≥ κ∥x∥Xθ .

Then there is δ > 0 such that if ∣θ − θ∗∣ ≤ δ and a constant C only depending on the quantities introducedso far such that for any ∣θ − θ∗∣ ≤ δ

the operator T ∶Xθ → Y θ is surjective, the lower bound ∥Tx∥Y θ ≥ C

−1∥x∥Xθ holds (bounded inverse), and

the inverses with T −1θ Tx = x for x ∈Xθ agree on Xθ ∩Xθ∗ .

32

10.5. Surjectivity. We start with the first item, but an auxiliary proposition hopefully familiar fromBanach space theory is needed. We won’t do the proof of the proposition but proceed to the theoremdirectly.

Proposition 10.11. Let X and Y be Banach spaces and T ∶ X → Y bounded and linear operator. Ifthere is C > 0 and c ∈ (0,1) so that for every y ∈ Y with ∥y∥Y = 1 there is x ∈ X with ∥Tx − y∥ ≤ c, thenT is surjective.

Proof of the first item. Let γ > 1. Fix y ∈ Y θ with ∥y∥Y θ = 1. By definition there is g ∈ F(Y ) with

g(θ) = y and ∥g∥F(Y ) ≤ γ. Further, g(θ∗) ∈ Y θ∗ . By invertibility assumption, T −1g(θ∗) ∈Xθ. By complex

interpolation there is f ∈ F(X) with f(θ∗) = T −1g(θ∗) and ∥f∥F(X) ≤ γ∥T −1g(θ∗)∥. Set x = f(θ). We

verify the hypothesis of the proposition for this function so that surjectivity will follow.First,

∥x∥Xθ ≤ ∥f∥F(X) ≤ γ∥T−1g(θ∗)∥ ≤ γ

κ∥g(θ∗)∥Y θ∗ ≤

γ2

κas required. In particular, this upper bound is independent of y.

Second, let

h(z) = g(z) − Tf(z)z − θ∗ , z ≠ θ∗.

As T is linear and bounded, this is easily seen to be analytic when z ≠ θ∗ and the singularity is removableso h has holomorphic extension to the whole strip S and h ∈ F(Y ). Then

∥y − Tx∥Y θ = ∥h(θ)(θ − θ∗)∥Y θ ≤ ∣θ − θ∗∣∥h(θ)∥Y θ ≤ ∣θ − θ∗∣∥h∥F(Y ).

Recall that the norm on the right hand side is computed as the supremum on the boundary of the stripso

∥h∥F(Y ) ≲θ∗ ∥g∥F(Y ) + ∥Tf∥F(Y ) ≲ γ +∥T ∥γ2

κ.

Hence it is clear that for θ close enough to θ∗ we have ∥y − Tx∥Y θ ≤ 1/2. Hence the assumptions ofProposition 10.11 are valid.

10.6. Injectivity. The proof of the stability of the lower bound for the norm of the operator requires avariant of Schwarz’ lemma from complex analysis.

Proposition 10.12. Let r ≤ min(θ∗,1 − θ∗)/2. If r ≥ ∣θ − θ∗∣, then

∥f(θ)∥Xθ ≥1

2∥f(θ∗)∥Xθ −

∣θ − θ∗∣2r

∥f∥F(X)

for all f ∈ F(X)

Proof of the proposition. Take f and θ as in the claim. Let g ∈ F(X) with g(θ) = f(θ). Let h(z) =(f(z) − g(z))/(z − θ) so that h ∈ F(X) by removable singularity theorem as before. Also ∣z − θ∣ ≥ r forany z ∈ ∂S. Now

∥h∥F(X) = ∥f − g⋅ − θ ∥F(X)

≤ 1

r∥f − g∥F(X) ≤

1

r(∥f∥F(X) + ∥g∥F(X))

and

∥f(θ∗)∥Xθ ≤ ∥g + (θ∗ − θ)h∥F(X) ≤ ∥g∥F(X) +∣θ∗ − θ∣r

(∥f∥F(X) + ∥g∥F(X)) ≤ 2∥g∥F(X) +∣θ∗ − θ∣r

∥f∥F(X)

so taking infimum over all g as above proves the claim.

Proof of the lower bound in the theorem. Fix x ∈ Xθ. Take f ∈ F(X) with f(θ) = x. By continuity

Tf(⋅) ∈ F(Y ), Tf(θ) = Tx and ∥Tf∥F(Y ) ≤ C∥f∥F(X). Applying the proposition once we obtain

∥Tx∥Y θ = ∥Tf(θ)∥Y θ ≥1

2∥Tf(θ∗)∥Y θ −

C ∣θ − θ∗∣r

∥f∥F(X).

By invertibility at θ∗ and the proposition once again

∥Tf(θ∗)∥Y θ ≥ κ∥f(θ∗)∥Xθ ≥ κ(1

2∥f(θ)∥Xθ −

∣θ − θ∗∣2r

∥f∥F(X))

whence the claim follows by taking infimum over all f as above and choosing θ close to θ∗. 33

10.7. Consistency. So far we have seen that T is surjective in surrounding space and that it has aninverse in each of them with bound for the operator norm. Next weprove the inverses agree on Y θ ∩Y θ.To this end, we will find a formula for the inverses to see that it is analytic function independent of theimagynary part and hence constant.

Proof for consistency of the inverses. Take f ∈ F(Y ). We will construct a decomposition.

Now f(θ∗) ∈ Y θ and T −1θ∗ ∈ Xθ∗ . There is g1 ∈ F(X) with Tg1(θ∗) = f(θ∗) and ∥g1∥F(X) ≤

2κ−1∥f∥F(Y ).

Set ω = z−θ∗z−4

and f1 = f−Tg1ω

so that

∣ω∣ ∼ 1 on ∂S, f = Tg1 + ωf1, f1 ∈ F(Y )as the singularity of f1 at z = θ∗ is removable. It also holds

∥f1∥F(Y ) ≤ C(∥f∥F(Y ) + ∥Tg1∥F(Y )) ≤ C∥f∥F(Y )

We have constructed a decomposition f = Tg1 + ωf1.

Iterate the above construction for the output f1 and its successors to get for every integer j a pair offunctions (fj , gj) such that fj = Tgj+1 + ωfj+1 and

∥fj+1∥F(Y ) ≤ Cj+1∥f∥F(Y ), ∥gj+1∥ ≤ Cj∥f∥F(Y ).

Writing the recursion out, we see

f = T (n

∑k=1

ωk−1gk) + ωnfn.

For δ > 0 and z ∈ B(θ∗, δ) we can write

∥ω(z)nfn(z)∥Σ(Y ) ≤ ∣θ − θ∗∣n∥fn(z)∥Σ(Y ) ≤ δn∥fn∥F(Y ) ≤ (Cδ)n∥f∥F(Y )

and similarly for all the terms in the sum ∑nk=1 ωk−1gk. Choosing δ small enough we get geometric decay

and the sequences converge as

ω(z)nfn(z)→ 0, andn

∑k=1

ωk−1gk → g

uniformly in Σ(Y ) and Σ(X), respectively. Hence the function g is analytic and Tg = f . In other wordsT −1z f(z) is analytic in a neighborhood of θ∗.

To conclude the proof, we need the fact that X0 ∩X1 is dense in Xθ. This is true for general Banachspace though we did not prove it so you may take it as an assumption that can be verified usually

very easily in applications. Given the density, take x ∈ X0 ∩ X1. Set f0 = e(z−θ)2

x so that f0 is inF(X). Hence T −1

z f0(z) is analytic in a neighborhood of θ∗ by the argument so far. Consequently also

e−(z−θ)2

T −1z f0(z) = T −1

z x is analytic. This mapping is, however, independent of the imaginary part of z

so it must be constant. By density, the inverse of T −1z is the same in all Y θ for θ ∈ B(θ∗, δ).

11. Direct method in the calculus of variations

A continuous function attains its extremal values in a compact set, a lower semicontinuous its mini-mum. Minima of strictly convex functions are always unique. The direct method of calculus of variationsis based on taking thess properties to more general function spaces. We start by recalling three propo-sitions from functional analysis.

Proposition 11.1 (Separable Banach–Alaoglu). Let X be a separable Banach space. Then the closedunit ball of the dual X ′ is is weakly* sequentially compact.

Proposition 11.2. Let p, q ∈ (1,∞) and s > 0. Then Lp, W k,p and Bsp,q are separable and reflexive. Incase p = 1, they are still separable though reflexivity fails.

Proof. We take separability of Lp for given. The space W k,p is isomorphic to a closed subset in a productof Lp spaces via u↦ (∂αu)∣α∣≤k so separability follows from that of Lp spaces. Finally, Besov spaces areseparable as real interpolation spaces. Reflexivity follows along the same lines from that of Lp wheneverp > 1.

Proposition 11.3. Let fj be a bounded sequence in reflexive X. Then fj has a subsequence thatconverges weakly.

34

Proof. Just note that X is dual of X ′′ = X so that the sequence has subsequence still denoted by fjsuch that for all elements from the predual g ∈ X ′ the convergence ⟨g, fj⟩ → f . That is, f convergesweakly.

Theorem 11.4 (Direct method). Let X be a separable and reflexive Banach space and suppose F ∶X →R+ is a functional such that

F is weakly lower semicontinuous meaning lim infi F (fi) ≥ F (fi) whenever fi f , F is coercive in the sense that there are constants c,C > 0 such that for all x ∈ X it holds

∣F (x) ≥ c∥x∥X −C. F is strictly convex.

Then F has a unique minimizer x meaning that for all y ∈XF (x) ≤ F (y).

If the convexity assumption is dropped, the minimizer still exists but it need not be unique.

Proof. Let I = infx F (x). There is a sequence xj such that F (xj) I. By coercvity (second item),∥xj∥ ≤ c−1(F (xj) + C) ≤ c−1(F (x0) + C) for all j so that the sequence is bounded in B so that bypassing to a subsequence we may assume the existence of x such that xj x. By lower semicontinuityF (x) ≤ lim infj F (xj) = I so that x is a minimizer. Suppose F is convex. If there existed two distinctminimizers u and v, then (u + v)/2 ∈X would be a competitor in the definition of minimizer and

F (u) ≤ F (u + v2

) < 1

2F (u) + 1

2F (v) ⇒ F (u) < F (v)

which is a contradiction.

11.1. Problems with boundary values. Let Ω ⊂ Rn be an open and connected set. Recall thatW 1,p

0 (Ω) is the closure of C∞c (Ω) in the norm of W 1,p. For g ∈ W 1,p(Ω), we define the class with

boundary value

W 1,pg (Ω) = f ∈W 1,p(Ω) ∶ f − g ∈W 1,p

0 (Ω).Note that W 1,p

0 is always closed subspace and hence complete and reflexive.

Theorem 11.5. Let Ω be open and bounded, g as above and p > 1. Let F ∶ W 1,pg (Ω) → R+ be F (u) =

∫Ω ∣∇u∣p. Then there exists a unique minimizer of F in its domain.

Proof. Since W 1,pg (Ω) = g +W 1,p

0 (Ω), the question is equivalent to minimizing F (⋅ + g) in W 1,p0 (Ω). The

functional is clearly convex. To prove coercivity, note that for h ∈W 1,p0 (Ω)

(∫Ωhp)

1/p≲ (∫ hp

∗

)1/p∗

≲ ∥∇h∥Lp

so that from the choice h = 1Ω(f − g) ∈W 1,p with f ∈W 1,pg (Ω) we have

∥f∥Lp ≲ F (f) + ∥g∥W 1,p(Ω).

Finally, to prove weak lower semicontinuity, take uj converging to u weakly in W 1,p0 . Now ∣∇u +

g∣p−2∇(u + g) gives a bounded linear functional so

∫ ∣∇(u + g)∣p = ∫ ∣∇(u + g)∣p−2∇(u + g) ⋅ ∇(u + g) ≤ lim infi

∫ ∣∇(u + g)∣p−2∇(u + g) ⋅ ∇(ui + g)

≤ lim infi

(∫ ∣∇(u + g)∣p)1−1/p

(∫ ∣∇(ui + g)∣p)1/p

After these observations the claim follows from the direct mehtod.

Theorem 11.6. Let g, Ω and p be as above. Then u ∈W 1,pg (Ω) is a minimizer if and only if

div(∣∇u∣p−2∇u) = 0

in the weak sense, that is, for all ϕ ∈W 1,p0 (Ω) it holds

∫ ∣∇u∣p−2∇u ⋅ ∇ϕ = 0.

35

Proof. Denote J(u, t) = ∫ ∣∇(u + tϕ)∣p. Assume first ϕ ∈ C∞c (Ω). By dominated convergence

d

dtJ(u, t) = p∫ ∣∇(u + tϕ)∣p−2∇(u + tϕ) ⋅ ∇ϕ→ ∫ ∣∇u∣p−2∇u ⋅ ∇ϕ

as t→ 0. If u is a minimizer, then the expression above must vanish. The claim for general ϕ follows byapproximation in the differential equation form.

Conversely, if u is a solution, then we use the convexity inequality

∣a∣p ≥ ∣b∣p + p∣b∣p−2b ⋅ (a − b)whith the choice a = ∇v and b = ∇(v − u) where v is a competitor so that v − u is a valid test function.Then

∫ ∣∇v∣p ≥ ∫ ∣∇u∣p + p∫ ∣∇u∣p−2∇u ⋅ ∇(v − u) = ∫ ∣∇u∣p

which proves u is a minimizer.

A few remarks:

Denote ∆pu = div(∣∇u∣p−2∇u). This operator is called the p-Laplacian and it is easy to see byHolder’s inequality that it maps W 1,p into its dual.

The pairing ⟨−∆pu,ϕ⟩ = ∂t∣u + tϕ∣pW 1,p ∣t=0 is sometimes called the Gateaux derivative of theseminorm at u to direction ϕ.

If ⟨−∆pu,ϕ⟩ ≥ 0 for all positive test functions then u is said to be a supersolution. The op-posite sign gives the definition of a subsolution. These notions are best understood as su-per/subminimizers and they are useful as they satisfy comparison principle.

11.2. Application I: Holder continuity for n-Laplacian. The Holder regularity of p-Laplacians isproved in three different ways. If p > n, it is immediate from the Morrey’s theorem. If p < n, it requiresDe Giorgi/Moser iteration. We will do the case p = n here as an easy application of techniques coveredso far. We start with Caccioppoli estimate which is the information extracted from the fact that u is asolution. It amounts to knowing that in terms of size, Poincare inequality can be “undone”.

Proposition 11.7. Let u be a weak solution to −∆pu = 0 in Ω. Then for every ϕ ∈ C∞c (Ω) with 0 ≤ ϕ ≤ 1

it holds

∫ ϕp∣∇u∣p ≤ pp ∫ ∣u∣p∣∇ϕ∣p.

In particular, for r > 0 and B2r ⊂ Ω

∫Br

∣∇u∣p ≤ ppr−p ∫B2r

∣u∣p.

Proof. Let η = ϕpu be a test function so that ∇η = ϕp∇u + puϕp−1∇ϕ. Then

∫ ϕp∣∇u∣p = −p∫ uϕp−1∣∇u∣p−2∇u ⋅ ∇ϕ ≤ p∫ ∣ϕ∇u∣p−1∣u∇ϕ∣ ≤ p(∫ ∣ϕ∣p∣∇u∣p)1/p′

(∫ ∣u∣p∣∇ϕ∣p)1/p

which proves the first claim. The second one follows by approximating a Lipschitz cutoff by smoothones.

Once the derivative is gone, we use Sobolev-Poincare to bring it back and lower the exponent on theright hand side. This is essentially a reversed Holder’s inequality for the gradient.

Proposition 11.8. Let u be as before. If B4r ⊂ Ω, then there is a constant C = C(n, p) such that

(⨏Br

∣∇u∣p)1/p

≤ C (⨏B4r

∣∇u∣p∗)1/p∗

, p∗ =np

n + p .

Proof. If u is a local solution so is U = u − uB2r . By the previous estimate

⨏Br

∣∇u∣p = ⨏Br

∣∇U ∣p ≤ ppr−p ⨏B2r

∣U ∣p = ppr−p ⨏B2r

∣u −B2r ∣p = C (⨏B4r

∣∇u∣p∗)p/p∗

.

The reversed Holder inequality always admits an epsilon improvement on the left hand side. This isthe content of the following lemma.

36

Lemma 11.9 (Gehring’s lemma). Let f ∈ L1loc be positive and suppose there are q > 1 as wel as a constant

C such that for all balls with 4B compactly contained in Ω

(⨏Bfq)

1/q≤ C ⨏

4Bf.

Then there is ε > 0 and a constant D so that

(⨏Bfq+ε)

1/(q+ε)≤ C ⨏

4Bf

holds for all balls with 4B compactly contained in Ω.

Proof. Fix m > 0 to be a large number. Let fm = min(m,f). Take ε > 0 that will be made small enoughto satisfy all the requirements to appear. Denote µ(E) = ∫E fq for any measurable set E. By the layercake formula

∫Brfpf εm = ε∫

∞

0tε−1µ(Br ∩ fm > t)dt ≤ ε∫

L

0tε−1µ(Br)dt + ε∫

∞

Ltε−1µ(Et)dt = I + II

where Et = Br ∩ fm > t and L = δ−n ⨏B2rf with δ ∈ (0, r). Now

I ≤ Lεµ(Br) ≤ C ∣Br ∣δ−nε (⨏B2r

f)p+ε

by the hypothesis of the theorem.For the other term, define F (x, s) = ⨏B(x,s) f . This is continuous function in t and for almost every

x it holds F (x, s) → f(x) as s → 0. Also, for every x ∈ Et let sx = sups ∶ F (x, s) > t. By the previousobservation, the set over which the supremum is taken is nonempty. Moreover, for s small

F (x, s) ≤ ∣B2r ∣∣Bs∣ ⨏B2r

f = (2r

s)n

⨏B2r

f = (2rδ

s)n

L ≤ (2rδ

s)n

t

so that sx ≤ 2rδ. By Vitali covering theorem we may extract a countable disjoint family Bi = B(xi, si)such that 5Bi cover Et. On the other hand, for M the non-centred Hardy–Littlewood maximal function,it holds

Bi ⊂ M(1B(1+2δ)rf) > t ⊂ M(1B(1+2δ)r∩f>t/2f) > t/2

so

µ(Et) ≤∑i

µ(5Bi) ≲∑i

∣Bi∣⨏5Bi

fp ≲∑i

∣Bi∣ (⨏5Bi

f)p

≤∑i

∣Bi∣tp ≤ tp∣M(1B(1+2δ)r∩f>t/2f) > t/2∣

≲ tp−1 ∫B(1+2δ)r∩f>t/2

f

and

II ≲ ε∫m

0tε+p−2 ∫

B(1+2δ)r∩f>t/2f = ε∫

B(1+2δ)rf ∫

min(2f,m)

0tε+p−2 ≲ ε∫

B(1+2δ)rfpfεm.

In total

∫Brfpf εm ≤ C1 (⨏

B2r

f)p+ε

+C2ε∫B(1+2δ)r

fpf εm.

Setting δi = 4−i and ri = r0 +∑ij=1 δj and iterating the inequaity above, we reach the conclusion providedthat ε is small enough. Finally, taking the limit m→∞ concludes the proof.

Once Gehring’s lemma is known, the following corollary is immediate.

Theorem 11.10. Suppose u is a local solution to the n harmonic equation. Then u is locally Holdercontinuous.

Proof. We have the reverse Holder inequality for ∣∇u∣n∗ with p = n/n∗. Applying Gehring’s lemma, wesee ∣∇u∣ ∈ Ln+ε for some ε. The claim then follows from Morrey’s theorem.

37

11.3. Application II: Very weak solutions to linear equations. Next problem is W 1,2(Rn) based.Let A be a matrix valued function with bounded and measurable entries and suppose there is c > 0 sothat for all ξ ∈ Rn and almost all x ∈ Rn it holds

A(x)ξ ⋅ ξ ≥ c∣ξ∣2.Consider the Euler–Lagrange equation

div(A∇u) = 0

for the energy functional u↦ ∫ ∇u⋅A∇u posed on a domain Ω. Instead of the natural energy solutions, wecan define very weak solutions u ∈W 1,2−ε(Ω) that still satisfy the equation in the sense of distributions.Indeed, to make sense of the formulation, only integrability of gradient is needed.

Proposition 11.11. Let A be as above and Ω a domain. There exists ε > 0 only depending on parametersof the matrix c and maxi,j ∥Aij∥L∞ so that whenever u ∈ W 1,2−ε

loc (Ω) is a local solution (test functions

with compact support), then it is in W 1,2loc (Ω).

Proof. Take ϕ ∈ C∞c (Ω). Let v = uϕ. Denote L = div(A∇⋅). Then for test function ψ

−⟨Lv,ψ⟩ = ∫ ∇ψ ⋅A(u∇ϕ + ϕ∇u) = ∫ u∇ψ ⋅A∇ϕ + ∫ ∇(ϕψ) ⋅A∇u´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

=0

−∫ ψ∇ϕ ⋅A∇u = I(ψ) − II(ψ).

By Holder’s inequality I defines a bounded linear functional acting on W 1,p whenever p′ ∈ (1, (2 − ε)∗),that is p > 2− δ for some δ. Similarly, II acts boundedly on Lp for p′ ∈ (1,2− ε), that is p > 2+ δ for somedelta. Finally, v itself defines a bounded linear functional on L2. Hence I(⋅) − II(⋅) + v(⋅) ∈ W−1,2.

Second, v + L is invertible W 1,2 → W −1,2 with bounded inverse. This is omitted, but you can try todo it by the direct method or invoke the Lax–Milgram. Finally, by Holder for any w ∈W 1,p

∣∫ ∇ψ ⋅A∇w∣ ≤ ∥w∥W 1,p∥ψ∥W 1,p′

so that v + L ∶ W 1,p → (W 1,p′)′ is bounded. The spaces W 1,p as well as their duals form complex

interpolation scales (recall that the dual of Hs,p is isomorphic to H−s,p′ , the isomorphism being theBessel potential and here they coincide with the Sobolev spaces) so Sneiberg’s theorem makes I + Linvertible here for p close enough to 2 and the inverses agree. Hence v is image of W −1,2 functional under(I +L)−1, hence in W 1,2.

11.4. Fractional Laplacians. Next we go through a few notions related to calculus of variations onspaces with fractional order of smoothness. Recall that the Besov space Bsp,p with s ∈ (0,1) fromSection 8.4 had an equivalent definition given as the collection of functions in Lp such that the Gagliardoseminorm

∣f ∣W s,p(Rn) = (∬∣f(x) − f(y)∣p∣x − y∣n+sp dxdy)

1/p

is finite. We define W s,p ∶= Bsp,p for s ∈ (0,1) but note that there is no corresponding identification fors an integer. Indeed, for integer values of s the Sobolev space W s,p (weak derivatives in Lp) coincideswith the potential space and not the Besov space so the notation is misleading. For Ω ⊂ Rn an open setand g ∈W s,p(Rn), we define the class

W s,p(Ω) = f ∈W s,p(Rn) ∶ f = g a.e. in Rn ∖Ω.Note that the boundary values are given in the exterior of the domain.

The most important functional acting on the fractional Sobolev spaces as defined above is of coursethe seminorm measuring smoothness F (u) ↦ ∣u∣p

W s,p(Rn). Depending on your own interests, you might

want to check that the functional u↦ F (u+g) with is convex, lower semicontinuous and coercive as in thecase of integer order smoothness above and apply the direct method in calculus of variations to prove theexistence of the minimizer for all boundary data as defined above. The corresponding Euler–Lagrangeequation is: For all ϕ ∈ C∞

c (Ω)

p.v.∬∣u(x) − u(y)∣p−2((u(x) − u(y))((ϕ(x) − ϕ(y))

∣x − y∣n+sp = 0.

The left hand side can be seen as a fractional p-Laplacian of u acting on a test function ϕ. In generalthe fractional p-Laplacian is an element in the dual of the fractional Sobolev space (Besov space).

When p = 2, the Besov and the Bessel potential space scales meet. Hence we can see that the fractional2-Laplacian has a representation in terms of the potentials and their inverses using the Fourier transform.

38

Suppose u,ϕ are Schwartz functions. Then the fractional 2-Laplacian of u acting on ϕ can be written as(by change of variables, Fubini and Plancherel)

∬((u(x) − u(y))((ϕ(x) − ϕ(y))

∣x − y∣n+2sdxdy = ∫

1

∣h∣n+2s ∫((u(x + h) − u(x))((ϕ(x + h) − ϕ(x))

∣h∣n+2sdxdh

= ∫1

∣h∣n+2s ∫ ∣e2πiξ⋅h − 1∣2u(ξ)ϕ(ξ)dξdh

= ∫ u(ξ)ϕ(ξ)∣ξ∣2s⎛⎝∫

∣e2πi ξ∣ξ∣⋅h − 1∣2

∣h∣n+2sdh

⎞⎠dξ.

By spherical symmetry of the innermost integral, there is no dependency on ξ so the expression on theleft most part equals the s-power of Laplacian up to a constant. This computation heavily relies onPlancherel and cannot be repeated for p ≠ 2.

11.5. Caffarelli–Silvestre extension. The final remark on fractional equations is concerned with thepossibility to study them using divergence form elliptic PDE on upper half-space. This section follows[4]. As we have only treated elliptc PDE superficially, the following should be taken mostly as a heuristicdiscussion. Suppose we are given the usual fractional problem: Let s ∈ (0,1), g ∈W s,2(Rn) and Ω ⊂ Rn.Find a function u ∈W s,2(Rn) such that

(−∆)su = 0 in Ω

u = g in Rn ∖Ω.

As discussed before, this problem is equivalent to extending g∣Ωc over Ω with minimal Hs,2 =W s,2 = Bs22

energy. We want to understand this through an extension to upper half space and as boundary valuesof a problem there.

Later we will see that passage from Rn+1 based function spaces to one based on Rn costs 1/2 units ofsmoothness. If the extension to be studied sees the added direction as a scale, there must be a weightin the scale parameter to correct the mismatch in differentiability. As the boundary values (now g) livein a space with smoothness s, the extension would naturally have smoothness s + 1/2. We want it to be

one, so one needs to add a correcting weight ya/2 to the scaling so that s+1/2+a/2 = 1. Hence an energyto be compatible with the fractional problem on the boundary could be guessed to be

u↦ ∫Rn+1+

∣∇u∣2ya dxdy.

The corresponding Euler–Lagrange equation is

0 = div(ya∇u) = ∆xu +a

yuy + uyy

where x ∈ Rn and y > 0. Taking Fourier transform in x, this becomes

−∣ξ∣2u(ξ, y) + ayuy(ξ, y) + uyy(ξ, y) = 0.

For each ξ this is an ordinary differential equation. First take the variational integral

ϕ↦ ∫∞

0(∣ϕ(y)∣2 + ∣ϕ′(y)∣2)ya dy

on a weighted Sobolev space W 1,2(yady,R+) (zero at infinity) to see that its minimizer with boundarydata ϕ(0) = 1 solves

−ϕ(y) + ayϕ′(y) + ϕ′′(y) = 0.

Take any f ∈ W s,2(Rn) with f = g outside Ω (function admissible for the fractional problem). Set

u(ξ, y) = f(ξ)ϕ(∣ξ∣y) . This solves the ODE for the Fourier transform of the extension with the correctboundary data.

Finally, we can compute

∫Rn+1+

∣∇u∣2ya dxdy = ∫Rn∫

∞

0(∣ξ∣2∣u∣2 + ∣uy ∣2)ya dydξ

= ∫Rnf(ξ)∣ξ∣2 ∫

∞

0(∣ϕ(∣ξ∣y)∣2 + ∣ϕ′(∣ξ∣y)∣2)ya dydξ

= ∫Rnf(ξ)∣ξ∣1−a (∫

∞

0(∣ϕ(y)∣2 + ∣ϕ′(y)∣2)ya dy)dξ

39

where the quantity in parentheses is the energy that ϕ minimizes and the remaining part is the L2 norm

of (−∆) 1−a2 f = (−∆)sf , the energy functional for the fractional problem on the boundary.

The function K(x, y) = F−1ξ (ϕ(∣ ⋅ ∣y))(x, y) is the Poisson kernel for the extended probem (this can be

computed explicitly: try the equation with ϕ = yb and use the properties of the Fourier transform). Thevariational integrals above can be studied interchangeably by either taking functions f ∈W s,p(Ω) withfractional seminorm or the weighted first order Sobolev seminorm of their extensions ∫ K(x−z, y)f(z)dz.We know so far how to prove the existence of the solution to the fractional problem, but the extensiongives it an interpretation as a solution to an integer order equation in the upper half space or even thecylinder above Ω.

12. Traces and extensions

Traces mean restrictions of functions to low-dimensional sets (here only linear subspaces). For a givenf ∈ Lp(Rn+1) the restriction f(x, ⋅) does not make sense as f is only defined almost everywhere andall proper sub-spaces of the Euclidean space have Lebesgue measure zero. For continuous functions,however, the restriction is perfectly well defined. Next we see how Sobolev functions, which are in a waysomething between Lp and continuous, behave with respect to restrictions.

We can define traces on various hypersurfaces, but in what follows we restrict the attention to thecase of n dimensional hyperplane Rn×0 in Rn+1 or as the boundary of the upper half space. We definethe trace operator T ∶ C∞(Rn+1)→ C∞(Rn+1) a priori as

Tϕ(x) = ϕ(x,0), ∀ϕ ∈ C∞(Rn+1).

Whenever we manage to prove norm bound X(Rn+1) → Y (Rn) for two function spaces X and Y , itfollows that there is a unique extension of the trace opeartor on X, provided that C∞ is dense there,which will be the case in all what follows.

The first question is the traces of W 1,1(Rn+1) functions. Clearly they cannot be W 1,1(Rn) smooth asfor n = 1 we can build planar Sobolev functions with singularities whereas the one dimensional Sobolevfunctions have to be absolutely continuous and hence locally bounded. Another way to see (which willactually be used later) that traces are less smooth and extensions smoother is to look at the Riesz kernel∣x∣s−n whose order varies when n varies.

12.1. Trace theorems. We start with the case of W 1,1 where the trace does not have any smoothness.The presentation of this section is based on a section about traces in [17] as well as [2].

Proposition 12.1. The trace operator T has a continuous extension W 1,1(Rn+1+ ) → L1(Rn) which is

onto.

Proof. Let u ∈ Rn+1. Take uj ∈ C∞c (Rn+1

+ ) such that uj → u in W 1,1(Rn+1+ ) norm. Then

∣Tuj(x)∣ = ∣∫0

−∞∂yuj(x, y)dy∣ ≤ ∥∂yuj(x, ⋅)∥L1(Rn)

so that T is bounded and has extension.To prove T is onto, take u ∈ L1(Rn). Find uj → u smooth and compactly supported. By passing to

subsequence, assume that for all j ≥ k ∥uj − u∥L1(Rn) ≤ 2−k−1∥u∥L1(Rn). Then also

(12.1) ∥uk+1 − uk∥L1(Rn) ≤ 2−k−1∥u∥L1(Rn), ∥uk∥L1(Rn) ≤ ∥uk − u∥L1(Rn) + ∥u∥L1(Rn) ≤ 2∥u∥L1(Rn).

We use this sequence as an extension (compare to approximations of identity and the upper half space).Let tk 0 be a sequence such that

tk − tk+1 ≤ 2−k∥u∥L1(Rn)

∥∇uk∥L1(Rn) + ∥∇uk+1∥L1(Rn) + 1, t0 ≤

1

4.

Define Ik = (tk+1, tk] and

g(x, y) = ∑k≥0

1Ik(y)(tk − y)uk+1(x) + (y − tk+1)uk(x)

∣Ik ∣.

40

This will be the extension of u we are looking for. First, g is in L1(Rn+1) because the second inequalityin (12.1) gives

∫Rn+1+

∣g∣ ≤ ∑k≥0∫Ik∫Rn

∣(tk − y)uk+1(x) + (y − tk+1)uk(x)∣∣Ik ∣

≤ ∑k≥0

∣Ik ∣(∥uk∥L1(Rn) + ∥uk+1∥L1(Rn))

≤ 4t0∥u∥L1(Rn) ≤ ∥u∥L1(Rn).

Second, the vertical derivatives are in L1 as the choice of tk gives

∫Rn+1+

∣∇g∣ ≤ ∑k≥0

∣Ik ∣(∥∇uk∥L1(Rn) + ∥∇uk+1∥L1(Rn)) ≤ ∑k≥0

2−k∥u∥L1(Rn) ≤ 2∥u∥L1(Rn).

Finally, the first inequality in (12.1) gives

∫Rn+1+

∣∂yg∣ ≤ ∑k≥0∫Rn

∣uk+1 − uk ∣ ≤ ∥u∥L1(Rn).

To see these are indeed weak derivatives, it suffices to test with test function compactly supported inRn+1+ so also this point is clear.It remains to show Tg = u. First, taking gj → g smooth, we see that

∥Tg − g(⋅, y)∥L1(Rn) ≤ ∫y

0∥g∥W 1,1(Rn+1

+) → 0 as y → 0.

Second, by construction

∣u(x) − g(x, y)∣ = ∣(1 − θ)(uk+1 − u) + θ(uk − u)∣ ≤ ∣uk+1 − u∣ + ∣uk − u∣→ 0

in L1(Rn) as y → 0 and k →∞. Here θ = θ(y) ∈ [0,1]. Finally, using these two observations we see

∥Tg − u∥L1(Rn) ≤ ∥Tg − g(⋅, y)∥L1(Rn) + ∥u − g(⋅, y)∥L1(Rn) → 0

so that Tg = u almost everywhere.

Another extreme is the case of Lipschitz functions, where the trace is as smooth as the originalfunction.

Proposition 12.2. The trace operator has bounded extension W 1,∞(Rn+1) →W 1,∞(Rn) and it is sur-jective.

Proof. Both claims are clear.

Proposition 12.3. The trace operator extends to a bounded and surjective operator W 1,p(Rn+1+ ) →

B1−1/pp,p (Rn).

One might be tempted to attempt interpolating the endpoints using the real method to prove theexistence of bounded extension of the trace, but such an attempt is does not seem to work in any simpleway. Instead, we have to estimate the trace explicitly.

Proof. Suppose f is smooth function defined in the upper half space. Denote ∆hf(x, t) = f(x + h, t) −f(x, t) and ∆tf(x, τ) = f(x, t + τ) − f(x, t). Take x,h ∈ Rn and write

∣∆hf(x,0)∣ = ∣ 1

∣h∣ ∫∣h∣

0∣∆hf(x,0)∣dt∣ ≤

1

∣h∣ ∫∣h∣

0(∣∆hf(x, t)∣ + ∣∆hf(x, t) −∆hf(x,0)∣)dt

≤ 1

∣h∣ ∫∣h∣

0(∣∆hf(x, t)∣ + ∣∆tf(x + h, t)∣ + ∣∆tf(x, t)∣)dt

≤ 1

∣h∣ ∫∣h∣

0∫

1

0(∣h ⋅ ∇f(x + hs, t)∣ + ∣t∂tf(x + h, st)∣ + ∣t∂tf(x, st)∣)dsdt.

Fix ε > 0 small. Taking Lp norm in x variable and using Minkowski’s integral inequality and Holder, weget

∥∣∆hf(x,0)∣ ≤ ∫∣h∣

0∥∇f(⋅, t)∥Lp dt ≤ C ∣h∣

1−εp′

p′ (∫∣h∣

0∥∇f(⋅, t)∥p

Lp(Rn)tεp dt)

1/p

.

41

Plugging this estimate in the definition of the Besov norm we conclude

∣f(x,0)∣pB

1−1/pp,p (Rn)

∼∬∣f(x,0) − f(y,0)∣p

∣x − y∣n+p/p′ dxdy =∬∣∆hf(x,0)∣p

∣h∣n+p/p′ dxdh

≲ ∫∞

0rn−1 ⋅ 1

rn+p/p′⋅ r

p(1−εp′)

p′ ∫r

0∥∇f(⋅, t)∥p

Lp(Rn)tεp dtdr

= ∫∞

0∫

∞

t

1

r1+εp′ ∥∇f(⋅, t)∥pLp(Rn)t

εp drdt ≲ ∫∞

0∥∇f(⋅, t)∥p

Lp(Rn) dt.

Estimation of the Lp norm is similar and omitted here. This concludes the proof that the trace operatorhas a bounded extension to W 1,p with values in the Besov space.

To prove that the trace is onto, take any f ∈ B1−1/pp,p (Rn). Let ϕt be the usual approximation of

identity, that is a dilate of a positive smooth bump supported in the unit ball with integral one. Letg(x, t) = ϕt ∗ f(x). Now all components of ∇ϕt = t−1(∇ϕ)t are bounded by Ct−1, supported in B(0, t)and have zero mean value. Similarly,

∂tϕt(x) = t−n−1(−nϕ(t−1x) − t−1x ⋅ (∇ϕ)(t−1x)) = −t−1 div(xϕt(x))

has all the same properties. Hence

∣∇g(x, t)∣ ≤ 1

t⨏B(0,t)

∣f(x + h) − f(x)∣dh = 1

t⨏B(0,t)

∣∆hf(x)∣dx

and

∬Rn+1+

∣∇g∣p dxdt ≤ ∫∞

0t1−pωp(t, f)p

dt

t∼ ∣f ∣

B1−1/pp,p

where we used Theorem 8.6.

Proposition 12.4. The trace has bounded and surjective extension Hs,2(Rn+1) → Hs−1/2,2(Rn) when-ever s > 1/2.

Proof. Suppose f is Schwartz. By Fourier inversion

∣Fxf(ξ,0)∣ = ∣∫Rnf(ξ, τ)dτ ∣ = ∣∫

Rn(1 + ∣ξ∣2 + τ2)s/2f(ξ, τ) ⋅ (1 + ∣ξ∣2 + τ2)−s/2 dτ ∣

≤ (∫ (1 + ∣ξ∣2 + τ2)s∣f(ξ, τ)∣2 dτ)1/2

(∫Rn

(1 + ∣ξ∣2 + τ2)−s dτ)1/2

= C(1 + ∣ξ∣2) 12 (s−

12 ) (∫ (1 + ∣ξ∣2 + τ2)s∣f(ξ, τ)∣2 dτ)

1/2

which proves the first claim.To prove the surjectivity, we take f ∈Hs−1/2,2(Rn) and set

g(ξ, τ) = Cf(ξ)(1 + ∣ξ∣2 + τ2)−s(1 + ∣ξ∣2)−1/2+s

so that with a correct choice of the constant C in the formula above

∫Rg(ξ, τ)dτ = f(ξ).

Then

∫Rn+1

(1 + ∣ξ∣2 + τ2)s∣g(ξ, τ)∣2 dτdξ = ∫Rn+1

∣f(ξ)∣2(1 + ∣ξ∣2)−1+s(1 + ∣ξ∣2 + τ2)−s dτdξ

= C ∫Rn

∣f(ξ)∣2(1 + ∣ξ∣2)s−1/2 dξ

12.2. Some explanations. We have seen that function spaces of same type with smoothness indices s1

s2 and integrability indices p1 and p2 usually embedd provided s1−n/p1 = s2−n/p2. When proving a prioribound for the trace operator, we started usually by estimating f(x,0) by an integral expression in thelast variable, integral over the derivative or inverse Fourier transform. These integrals could have beeninterpreted as instance of embedding X ⊂ C where the space of continuous functions C has integrabilityindex infinity and smoothness zero. The examples were W 1,1(R) ⊂ C0(R) and FL1(R) ⊂ C0(R).

42

More generally, one can use the Besov scale to run the same argument. Now for p > 1 we have

B1/pp,1 (R) ⊂ B0

∞,1(R) ⊂ C(R). The last inclusion is a direct consequence of the fact that all the Littlewood–Paley pieces are uniformly continuous and the Weierstrass M-test. Hence for Schwartz functions f onemay write at any x ∈ Rn

∣f(x,0)∣ ≲ ∥f(x, ⋅)∥B0∞,1(R) ≲ ∥f(x, ⋅)∥

B1/pp,1 (R).

Recalling the the characterization of Besov spaces through finite differences (Theorem 8.6) and theaveraged modulus of smoothness discussed after the proof of that theorem, it is clear that taking Lp

norm in x from the both sides gives

∥f(x,0)∥Lp(Rn) ≲ ∥B

1/pp,1 (Rn+1)

.

Looking then at W k,p → Bk−1/pp,p bounds and interpolating, one obtains Bsp,q → B

s−1/pp,q bounds for s > 1,

and these can further interpolated with the B1/pp,1 (Rn+1) → Lp(Rn) bound to get a full range of Besov

traces.

12.3. Other boundary values. When minimizing variational integrals, we used a different concept ofboundary values. The following simple proposition tells that this was justified as the two notions ofboundary values actually coincide.

Proposition 12.5. Let T be the trace operator of the boundary of the upper half space Rn+1+ . Then

W 1,p0 (Rn+1

+ ) =W 1,p(Rn+1+ ) ∩ f ∶ Tf = 0 for p ∈ [1,∞].

Proof. By continuity of the trace, the first inclusion is clear as every compactly supported function haszero trace. For the second inclusion, we can approximate any f ∈ W 1,p(Rn+1

+ ) by compactly supportedSobolev functions so that there is no loss in generality when we assume f is supported in the unit ballaround the origin. We extend v = f(x, y)1y>0. Take smooth ui converging to u. Then for any test

function ϕ ∈ Cc(Rn+1)

∬Rn+1

v(x, y)∂yϕ(x, y)dxdy =∬Rn+1+

u(x, y)∂yϕ(x, y)dxdy = limi∬

Rn+1+

ui(x, y)∂yϕ(x, y)dxdy

= limi

(∫Rnui(x,0)ϕ(x,0)dx −∬

Rn+1+

∂yui(x, y)ϕ(x, y)dxdy)

= −∬Rn+1+

∂yu(x, y)ϕ(x, y)dxdy

where the last step used that u has zero trace. We conclude that v is in W 1,p(Rn+1). Notice that this isexactly the extension to the full space that we were using in connection with the calculus of variationsin domains. There we had the property we are proving now that made it easy. Now we had to work toextend.

Let ε > 0. Find t > 0 so that vt(x, y) = v(x, y − t) satisfies

∥vt − v∥W 1,pRn+1 <ε

2.

This is possible by continuity of translations on Lp for p <∞ (to prove the theorem for p =∞ is straigh-forward by Lipschitz continuty so we need not pay attention to that case). Take then approximation ofidentity ϕη where ϕ is supported in B(0,1) so that ϕη ∗ vt has its support contained in the upper halfspace. By choosing η small enough, we see that

∥v − ϕη ∗ vt∥W 1,p(Rn+1) ≤ ∥v − vt∥W 1,p(Rn+1) + ∥vt − ϕη ∗ vt∥W 1,p(Rn+1) < ε.As ε > 0 can be made arbitrarily small, this completes the proof.

12.4. Comment on domains. The passage to boundary traces on domains more complicated than theupper half space is omitted not done in detail here though it is admittedly an important topic. However,here is the basic idea: A domain Ω is said to be Lipschitz if its boundary can be covered with finitelymany balls so that after possible rotation and translation B ∩Ω = (x, y) ∈ Rn+1 ∶ y > bB(x) where bBis a Lipschitz function. Given a function f in the domain, the coordinate change f(x, y − b(x)) reducesB ∩Ω to the upper half space. The composition with a bi-Lipschitz map is an isomorphism of W 1,p ondomains Ω and b(Ω) so the result can be pulled through the Lipschitz map from the upper half space(this requires area formula for Lipschitz maps and the pointwise characterization from section 5.5 butnothing more complicated).

43

13. Capacity

13.1. Basics. As the trace theorems hint, sets of Lebesgue measure zero are not completely negligiblefrom the point of view of Sobolev functions. The correct tool for characterizing the negligible sets isan outer measure called capacity. The terminology comes from electrostatics. Suppose Ω is open andconnected and K ⊂ Ω is compact. The pair (K,Ω) is called a condenser. Capacity of the condenser isthe least energy needed to induce unit potential difference between the inner part K and the outer partΩ of the condenser. Most of the contents of this section follow [14].

For what follows, we let p > 1 be a fixed number. We define the set of admissible potentials W (K,Ω) =u ∈ C∞

c (Ω) ∶ u ≥ 1K. We can then define

capp(K,Ω) = infu∈W (K,Ω)

∫Ω∣∇u∣p for K compact.

capp(U,Ω) = supK⊂U

K compact

capp(K,Ω) for U open.

capp(E,Ω) = infU⊃EU open

capp(U,Ω) for E general, not compact.

The additional requirement E not compact can eventually be removed as the Choquet capacitabilitytheorem will tell us the two competing definitions agree.

We start by checking that the smoothness requirement on admissible potentials can be relaxed.

Proposition 13.1. Let W0(K,Ω) = u ∈ W 1,p0 (Ω) ∩ C(Ω) ∶ u ≥ 1K for sets as above. Then we can

replace W (K,Ω) by W0(K,Ω) in the definition of the capacity without affecting its values.

Proof. As W0(K,Ω) ⊃W (K,Ω), we know

capp(K,Ω) ≥ infu∈W0(K,Ω)

∫Ω∣∇u∣p.

Take a sequence uj ∈W0(K,Ω) so that

∫Ω∣∇uj ∣p inf

u∈W0(K,Ω)∫

Ω∣∇u∣p

We take one of the se uj , denoted by u without subscript, and prove that it can be approximated byfunctions from W (K,Ω) in the W 1,p seminorm.

Let ε > 0. As u is continuous, the set U = u > 1 − ε ⊃K is open. Let v = min(u(1 − ε)−1,1) so that

∫Ω∣∇v∣p ≤ 1

1 − ε ∫Ω∣∇u∣p.

Let η be smooth with valuse in [0,1] and

η(x) =⎧⎪⎪⎨⎪⎪⎩

1, x ∈ Ω ∖U0, x ∈K

.

Let ϕj ∈ sico be such that ϕj → v in W 1,p(Ω) as j →∞. Set ψj = 1 − η(1 − ϕj) ∈W (K,Ω). Then

∫Ω∣∇(v − ψj)∣p = ∫

Ω∖U∣∇(v − ϕj)∣p + ∫

U∣∇(η(v − ϕj))∣p → 0.

Taking ε small and j large enough, we see that the energy of ψj can get arbitrarily close to that of u.

The following theorem contains most basic properties of the capacity.

Theorem 13.2. Let Ω be open and connected.

(i) Suppose E1 ⊂ E2 ⊂ Ω. Then capp(E1,Ω) ≤ capp(E2,Ω).(ii) Suppose Ω1 ⊂ Ω2 are open and E ⊂ Ω1. Then cap(E,Ω2) ≤ cap(E,Ω1).(iii) For K1,K2 ⊂ Ω compact, we have the strong subadditivity

capp(K1 ∪K2,Ω) + capp(K1 ∩K2,Ω) ≤ capp(K1,Ω) + capp(K2,Ω).

(iv) Let K1 ⊃K2 ⊃ ⋯ ⊃K = ∩∞i=1Ki be compact. Then

capp(K,Ω) = limi

capp(Ki,Ω).44

Proof. Items (i) and (ii) are direct consequences of the definition and the proofs are omitted. To provethe item (iii), take ui ∈W (Ki,Ω) for i = 1,2. Note that

∫ ∣∇max(u1, u2)∣p + ∫ ∣∇min(u1, u2)∣p = ∫ ∣∇u1∣p + ∫ ∣∇u2∣p

Also, max(u1, u2) ∈W0(K1 ∪K2) and min(u1, u2) ∈W0(K1 ∩K2). Hence

capp(K1 ∪K2,Ω) + capp(K1 ∩K2,Ω) ≤ ∫ ∣∇u1∣p + ∫ ∣∇u2∣p

and the claim follows by taking infimum over all admissible functions u1 and u2.To prove (iv), fix ε > 0 and take u ∈ W0(K,Ω) with value 1 in K and ∫ ∣∇u∣p < cap(K,Ω) + ε. For i

large enough, Ki ⊂ u ≥ 1 − ε. Hence

limi

capp(Ki,Ω) ≤ capp(u ≥ 1 − ε,Ω) ≤ (1 − ε)−p ∫ ∣∇u∣p.

Sending ε→ 0, the claim follows.

Lemma 13.3. Suppose E1, . . . ,Ek ⊂ Ω and Fi ⊂ Ei for each i. Then

capp(∪ki=1Ei,Ω) − capp(∪ki=1Fi,Ω) ≤k

∑i=1

(capp(Ei,Ω) − capp(Fi,Ω))

provided the first term on the left hand side is finite.

Proof. We first prove the claim for compact sets. In order to do so, we need an auxiliary inequality. Sup-pose C ⊂K and F are all compact. Then by monotonicity of the capacity as well as strong subadditivity(properties (i) and (iii))

capp(K ∪ F,Ω) + capp(C,Ω) ≤ capp(K ∪ (C ∪ F ),Ω) + capp(K ∩ (C ∪ F ),Ω)≤ capp(K,Ω) + capp(C ∪ F,Ω)

so that capp(K ∪ F,Ω) − capp(C ∪ F,Ω) ≤ capp(K,Ω) − capp(C,Ω).Next we proceed by induction. As k = 1, the claim is trivially true. Suppose then it is known for k −1

as k > 1. Then

capp(∪ki=1Ei,Ω) − capp(∪ki=1Fi,Ω) = capp(∪k−1i=1 Ei ∪Ek,Ω) − capp(∪k−1

i=1 Fi ∪Ek,Ω)+ capp(∪k−1

i=1 Fi ∪Ek,Ω) − capp(∪k−1i=1 Fi ∪ Fk,Ω)

≤ capp(∪k−1i=1 Ei,Ω) − capp(∪k−1

i=1 Fi,Ω) + capp(Ek,Ω) − capp(Fk,Ω)

≤k

∑i=1

(capp(Ei,Ω) − capp(Fi,Ω))

and the proof is complete for compact sets.To deduce the claim for open, fix ε > 0. For each i, take Ci ⊂ Fi ⊂ Ei compact so that capp(Fi,Ω) <

capp(Ci,Ω)+ε. Similarly, take K ⊂ ∪iEi compact with capp(∪Ei,Ω) < capp(K,Ω)+ε. Set K = K∪(∪iCi)and Ki = Ci ∪K ∖ ∪j≠iEj . Then K = ∪iKi and Ci ⊂Ki. Now

capp(∪ki=1Ei,Ω) − capp(∪ki=1Fi,Ω) ≤ capp(K,Ω) − capp(∪ki=1Ci,Ω) + ε= capp(∪ki=1Ki,Ω) − capp(∪ki=1Ci,Ω) + ε≤∑

i

(capp(Ki,Ω) − capp(Ci,Ω)) + ε

≤∑i

(capp(Ei,Ω) − capp(Fi,Ω) + ε) + ε.

Sending ε→ 0, the claim follows for compact sets. A similar argument can be used to conclude the prooffor general sets, but that one is omitted here.

Armed with the previous lemma, we are now in position to prove stability under unions as well ascountable subadditivity of the capacity.

Theorem 13.4. Let E1 ⊂ E2 ⊂ ⋯ ⊂ E = ⋃iEi ⊂ Ω where Ω is open. Then

capp(E,Ω) = limi

capp(Ei,Ω).45

Suppose that Fi ⊂ Ω are any sets. Then

capp(∪iFi,Ω) ≤∑i

capp(Fi,Ω)

as a consequence of the first item of the theorem.

Proof. Let ε > 0. For each Ei, take open Ui ⊃ Ei with capp(Ui,Ω) ≤ capp(Ei,Ω) + 2−iε. By lemma 13.3

capp(∪ki=1Ui,Ω) − capp(∪ki=1Ei,Ω) ≤k

∑i=1

ε2−i < ε

for all finite k. Take any K ⊂ ⋃∞i=1Ui compact. By compactness, there is a finite cover so that K ⊂ ⋃ki=1Uifor some finite k. Then

capp(K,Ω) ≤ capp(∪ki=1Ui,Ω) ≤ capp(∪ki=1Ei,Ω) + ε.By the definition of capacity,

capp(E,Ω) ≤ capp(∪∞i=1,Ω) = supK⊂∪∞i=1UiK compact

capp(K,Ω) ≤ lim infk→∞

capp(∪ki=1Ei,Ω) + ε.

As ε > 0 was arbitrary, this concludes the proof of the first claim. The second claim is a simple consequenceof the first one and finite subadditivity (we proved it only for compact but extending to more general iseasy).

A function on sets satisfying monotonicity ((i) of Theorem 13.2), outer regularity by open sets andstability of increasing limits for nested sets (Theorem 13.4) is called a Choquet capacity. The previoustheorems establish that the capp is a Choquet capacity, and hence we can invoke Choquet’s capacitabilitytheorem. We do not prove it here.

Theorem 13.5 (Choquet’s theorem). Let C ∶ 2Rn → [0,∞] be a Choquet capacity. Then all Borel sets

E are capacitable in the sense thatC(E) = sup

K⊂EK compact

C(K).

This applies in particular to p-capacities.

Example 13.6. Let 0 < r < R <∞. Then

capp(Br,BR) ∼⎧⎪⎪⎨⎪⎪⎩

∣Rp−np−1 − r

p−np−1 ∣1−p, p ≠ n

(log Rr)1−n

, p = n.This actually holds as an equality with an constant but the comparability is enough for us

Proof. To prove a lower bound for the capacity, take any admissible function u. Then for any directiony ∈ ∂B(0,1) it holds

1 ≤ ∫R

r∣∂su(sy)∣ds ≤ ∫

R

r∣∇u(sy)∣ds ≤ (∫

R

rs

1−np−1 ds)

1/p′

(∫R

r∣∇u(sy)∣psn−1)

1/p

.

Raise to power p, set

A = ∫R

rs

1−np−1 ds =

⎧⎪⎪⎨⎪⎪⎩

∣Rp−np−1 − r

p−np−1 ∣1−p, p ≠ n

(log Rr)1−n

, p = n,and integrate over y ∈ ∂B(0,1) with respect to the surface measure to obtain

C(n, p) ≤ Ap−1 ∫∂B(0,1)

∫R

r∣∇u(sy)∣psn−1 dsdy = Ap−1 ∫

BR∣∇u∣p.

Taking infimum over all admissible u, we see that CA1−p ≤ capp(Br,BR). To prove the other inequality,one just has to test with

u(x) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

0, ∣x∣ ≥ R(∫

R∣x∣ t

1−np−1 dt) (∫

Rr t

1−np−1 dt)

−1, r < ∣x∣ < R

1, 0 ≤ ∣x∣ ≤ r.

Three special cases are the following:46

If p > 1, then capp(Br,B2r) ∼ rn−p. If p ∈ (1, n), then capp(Br,Rn) ∼ rn−p. If p > n, then capp(0,Br) ∼ rn−p. For the case p = n the capacitites of the last two condensers are infinite.

13.2. Hausdorff measure. Recall that for s, δ ∈ (0,∞) the s-Hausdorff δ-content of a set E is definedas

Λδs(E) = inf∑i

rsi

where the infimum is taken over all collections of balls Bi = B(xi, ri) that cover E and ri ≤ δ. TheHausdorff content fails to additive on disjoint compact sets so it is not useful as a measure. The Hausdorffmeasure is defined as

Λs(E) = supδ>0

Λδs(E) = limδ→0

Λδs(E).

A few facts whose proofs follow the lines of proving the basic properties of the Lebesgue measure (seeSection 2 in [6]) that we need are listed below.

Λs is Borel regular (all Borel sets are measurable and every set E has a Borel set B ⊃ E suchthat Λs(B) = Λs(E)).

Let t < s. If Λt(E) <∞, then Λs(E) = 0. We define that Hausdorff dimension as

dim(E) = inft > 0 ∶ Λt(E) = 0 = supt > 0 ∶ Λt(E) =∞

where we omit the proof of the later inequality.

Proposition 13.7. Suppose δ ∈ (0,∞]. Then Λs(E) = 0 if and only if Λδs(E) = 0.

Proof. If Λs(E) = 0, then clearly Λδs(E) ≤ Λs(E) ≤ 0 so it remains to establish the other implication.Assume Λδs(E) = 0. Take ε > 0. Take a cover of E with balls having radii ri < δ and ∑i rsi < εs. Thenri < ε and

Λεs(E) ≤∑i

rsi < εs.

Taking the limit ε→ 0 finishes the proof.

13.3. Sets of capacity zero. We start with the definition

Definition 13.8. A set E is said to be of capacity zero if capp(E,Ω) = 0 for all open Ω.

Proposition 13.9. Suppose E is bounded and there exists an open Ω0 ⊃ E such that capp(E,Ω0) = 0.Then E is of capacity zero.

Proof. By definition, there are open Ui with G = ∩iUi ⊃ E and capp(G,Ω0) = 0. Hence it suffices toassume E is Borel. Further, for any open Ω′ we know that even Ω′ ∩E is Borel so that we can invokeChoquet’s theorem to assume E is a compact subset of Ω′. Then let u ∈W (E,Ω0) and v ∈W (E,Ω′) sothat uv ∈W (E,Ω′). Now

capp(E,Ω′) ≤ ∫ ∣∇(uv)∣p ≤ 2p (∥v∥∞ ∫ ∣∇u∣p + ∥∇v∥p∞ ∫ ∣u∣p) ≤ C ∫ ∣∇u∣p → 0

as u approaches the minimizer of the energy. Here we used the Poincare inequality for functions withzero boundary value.

The estimate for the capacity of a ball suggested that the capacity should be n−p dimensional measurein some sense. Next we prove that the Hausdorff dimension n − p is the borderline case for p-capacitybeing zero or not. This result is split in two theorems.

Theorem 13.10. Let E ⊂ Rn be a compact set of p-capacity zero. Then dim(E) ≤ n − p provided thatp ∈ (1, n].

Remark 13.11. I stated this in the class for general sets, which I think is true, but to complete theproof in full generality seems to require more than what we did. We might return to this.

47

Proof. Suppose that E is compact. It suffices to show that for any s ∈ (n − p, p) it holds Λ∞s (E) = 0. To

this end, take u ∈W (E,BR) where R > 0 is a large number such that E ⊂ B2−1R ∶= B(0,2−1R). Then forany x ∈ BR we have the formula

∣u(x)∣ ≤ ∣u(x) − uBR ∣ + ∣uBR ∣ ≤ ∣uBR ∣ +∞∑i=1

∣uB21−kR

− uB2−kR

∣ ≲∞∑k=0

R2−k (⨏B

2−kR

∣∇u∣p)1/p

∼∞∑k=0

(R2−k)1−np +sp ( 1

(2−kR)s ∫B2−kR

∣∇u∣p)1/p

≲ R1+ s−np (Mn−s∣∇u∣p)1/p

because s − (n − p) > 0. Because u is an admissible function for the set E, we can further estimate

E ⊂ u ≥ 1/2 ⊂ Mn−s(∣∇u∣p) ≥ CR =∶ SNow for each x ∈ S we choose a ball B(x, r) so that

CR ≤ 1

rs∫B(x,r)

∣∇u∣p.

By the usual Vitali covering argument we can find a countable pairwise subfamily Bi so that 5Bi are acover of S. Now

Λ∞s (E) ≤ Λ∞

s (S) ≤ 5s∑i

ri ≲∑i∫Bi

∣∇u∣p ≤ ∫ ∣∇u∣p.

Taking infimum over all admissible u, we see that Λ∞s (E) = 0 so that Λs(E) = 0 and we conclude

dim(E) ≤ s for all s > n − p. In other words dim(E) ≤ n − p.

To prove a partial converse, we need two lemmas. The first one connects the finiteness of the Hausdorffmeasure to the capacity, and the second one gives a suffiecient condition on zero capacity through auniformity condition over condenser.

Lemma 13.12. If Λn−p(E) < ∞, then there is M ∈ (0,∞) such that for all open Ω it holds capp(E ∩Ω,Ω) ≤M .

Proof. Take Ω to be open. As Λn−p is Borel regular, we find a Borel set with finite measure containingE. By monotonicity of capacity, it suffices to estimate the capacity of this set. Its intersection with Ω isstill Borel, and by Choquet we can prove a uniform bound for capacities of its compact subset. Hencewe can assume E is compact.

Let δ = d(E,Ωc) and let 4ε < min(1, δ). Take a cover Bi of E so that the radii ri < ε. Note that if2Bi ∩Ωc ≠ ∅, then Bi ∩E = ∅ and we can discard such a ball from the covering. Now

capp(E,Ω) ≤∑i

capp(Bi,Ω) ≤∑i

cap(Bi,2Bi) ≲∑i

rn−pi → Λn−p(E)ε → Λn−p(E)

as we take the infimum over all coverings and then ε→ 0.

Forcing the condencer to be very thin, we make the admissible functions very steep which causes thegradients to blow up eventually. However, if the energies still stay bounded, this is a manifestation ofthe condenser being very small. This is the content of the next lemma.

Lemma 13.13. Suppose E is compact and there is M ∈ (0,∞) such that for all Ω it holds capp(E∩Ω,Ω) ≤M . Then E is of p-capacity zero.

Proof. As E is compact, we can take Ωi = d(x,E) < 2−i so that they are open, Ωi+1 ⊂ Ωi and

⋂∞i=1 Ωi = E. Let ϕi ∈ C∞c (Ωi) satisfy 0 ≤ ϕi ≤ 1 and ϕi = 1 in E and

∫ ∣∇ϕi∣p <M + 1

where the inequality is from the definition of the capacity and the assumption. Now ϕi(x), ∣∇ϕi(x)∣→ 0for every x ∉ E and ∣E∣ = 0 so by the dominated convergence theorem

capp(E,Ω1) ≤ ∫ ∣∇ϕi∣p → 0

as i→∞.

Putting the lemmas together, we have proved the following theorem

Theorem 13.14. Let E be a set with Λn−p(E) <∞. Then E is of p capacity zero.

The two theorems together tell that capacity zero implies dim(E) ≤ n − p and dim(E) < n − p impliescapacity zero.

48

13.4. Exceptional sets for Sobolev spaces. Recall we characterized the spaces W 1,p(Ω) as closures ofC∞(Ω) with respect to the Sobolev norms. This characterizetion is the most intuitive way to understandthe difference between spaces defined on Ω and Ω ∖ E when E is a closed set of measure zero. Eventhough Lebesgue measure does not distinguish the domains of definition here, the smooth functions andin many cases also the Sobolev norms do. Expressions of the type W 1,p

0 (Ω) = W 1,p0 (Ω ∖ E) mean that

every function from W 1,p0 (Ω) can be approximated in the Sobolev norm by the elements of the other

space and vice versa. A necessary condition for this is ∣E∣ = 0. It is, however, not sufficient. A sharpcondition of removability of a set can be formulated in terms of the capacity. This section follows [14]and [6].

Theorem 13.15. Let Ω be open and E ⊂ Ω relatively closed. Then W 1,p(Ω) =W 1,p0 (Ω ∖E) if and only

if capp(E) = 0.

Proof. Assume first that E has capacity zero. Take any ϕ ∈ C∞c (Ω). First note that ∣E∣ = 0 so that

both Sobolev spaces have the same norm. Hence it suffices to show ϕ can be approximated in W 1,p byfunctions in W 1,p

0 (Ω ∖E).As E has capacity zero, we can find sequence of functions uj ∈ W 1,p

0 (Ω) ∩ C(E) with uj = 1 in aneighbourhood of E and uj → 0 in the norm of W 1,p. Set vj = (1 − uj)ϕ so that it has compact supportin Ω ∖E and

∫ ∣∇(vj − ϕ)∣p =≤ 2p ∫ ∣∇ϕ∣p∣uj ∣p + 2p ∫ ∣∇uj ∣p∣ϕ∣p → 0

∫ ∣vj − ϕ∣p = ∫ ∣ujϕ∣p → 0.

Hence W 1,p0 (Ω) ⊂W 1,p

0 (Ω ∖E). The other inclusion is trivial as C∞c (Ω ∖E) ⊂ C∞

c (Ω).To prove the other implication, assume that W 1,p

0 (Ω) ⊂ W 1,p0 (Ω ∖ E). By inner regularity of the

capacity we can assume that E is compact and hence proper subset of Ω. Take ϕ ∈W 1,p0 (Ω) ∩C(Ω) so

that ϕ = 1 in a neighbourhood of K. By the assumption, there is a sequence ϕi ∈W 1,p0 (Ω∖E)∩C(Ω∖E)

such that ϕ → ϕi in the Sobolev norm. Then ϕ − ϕj is an admissible function for the condenser (K,Ω)and

capp(K,Ω) ≤ ∫ ∣∇ϕi −∇ϕ∣p → 0.

The zero capacity condition is enough also without zero boundary values but in this case it need notbe necessary.

Proposition 13.16. Let E ⊂ Ω be relatively closed. If E has capacity zero, then W 1,p(Ω) =W 1,p(Ω∖E).

Proof. By zero capacity ∣E∣ = 0 so that the function spaces have the same norm. Since C∞c (Ω) ⊂

C∞0 (Ω∖E), it is also clear that the closures share the same inclusion. To prove the other inclusion, take

any bounded u ∈ W 1,p(Ω ∖E). Again, we choose smooth functions 1E ≤ vj ≤ 1Ω realizing the capacity(that is, converging to zero in Sobolev norm). As in the proof of the previous theorem uj = u(1−vj)→ uin the Sobolev norm. As bounded functions are dense, the proof is complete.

The interest of the previous proposition is that is helps to characterize the domains where zero bound-ary values follow automatically from the definition without boundary conditions.

Theorem 13.17. W 1,p(Ω) =W 1,p0 (Ω) if and only if capp(Ωc) = 0.

Proof. Assume first capp(Ω) = 0. Then

W 1,p(Ω) =W 1,p(Rn ∖Ωc) =W 1,p(Rn) =W 1,p0 (Rn) =W 1,p

0 (Rn ∖Ωc) =W 1,p0 (Ω).

Conversely, if W 1,p(Ω) =W 1,p0 (Ω), then

W 1,p0 (Rn) ⊂W 1,p(Rn) ⊂W 1,p(Ω) =W 1,p

0 (Ω) ⊂W 1,p0 (Rn)

and hence

W 1,p0 (Rn) =W 1,p

0 (Ω) =W 1,p0 (Rn ∖Ωc)

so that by the characterization for the zero boundary value spaces capp(Ωc) = 0. 49

13.5. Precise representatives. The observations of the previous section hint that a Sobolev functionmight be defined in a set larger than almost everywhere. In this subsection, we show that the set ofLebesgue points of a Sobolev function is everything except a set of capacity zero. In particular, theneglected set has Hausdorff dimension at most n − p < n. For this section, we define all function spacesover Rn and leave them out from the notation most of the time. We begin with a useful lemma.

Lemma 13.18. Suppose that u ∈ Lp with p ∈ (1,∞) and suppose that ∥∆hu∥Lp ≤ A∣h∣ for all ∣h∣ < ε andsome ε > 0 and A <∞. Then u ∈W 1,p and ∥∇u∥Lp ≤ A.

Proof. Denote Dhu = ∣h∣−1∆hu. Take ϕ ∈ C∞c . Then by a change of variable

∫ uDhϕ = −∫ (D−hu)ϕ.

Fix a coordinate direction i and assume h = ∣h∣ei. By assumption and weak compactness, there is asequence hj → 0 so that D−hju v ∈ Lp. Then

∫ u∂iϕ = limj∫ uDhjϕ = − lim

j∫ D−hjuϕ = −∫ vϕ

so that v is the weak derivative.

This lemma can be used to extend boundedness of the Hardy–Littlewood maximal operator from Lp

spaces to Sobolev spaces.

Lemma 13.19. Let T be a sublinear operator that commutes with translations, is bounded Lp → Lp forsome p > 1 and maps all functions to non-negative ones. Then T ∶W 1,p →W 1,p is bounded.

Proof. By sublinearity and commutation with translations, for all x,h ∈ Rn

Tf(x) = T (f − f(⋅ + h) + f(⋅ + h)) ≤ T (f − f(⋅ + h)) + Tf(⋅ + h)so that (∆hTf)(x) ≤ T (∆hf)(x). Then by the Lp bound

∥∆hTf∥Lp ≤ ∥T∆hf∥Lp ≤ ∥∆hf∥Lp .The right hand side is bounded by the norm of the gradient by a simple approximation argument andhence the claim follows from the previous lemma.

Remark 13.20. Note in particular that the Hardy–Littlewood maximal operator

Mf(x) = supr>0

⨏B(x,r)

∣f(y)∣dy

satisfies the hypotheses of the lemma above so that it is bounded in W 1,p with p > 1. This is the casethat is needed in the following capacitary weak type estimate.

Lemma 13.21. Let p > 1 nd suppose u ∈W 1,p(Rn). Then there is a constant C such that for all ε > 0it holds

capp(Mu > ε,Rn) ≤ C

εp∫ ∣∇u∣p

Proof. By lower semicontinuity of the maximal function of a locally finite Borel measure, the level set onthe left is open. It suffices to study the capacities of its compact subsets. Let K ⊂ Mu > ε be compact.Take R > 0 so large that K ⊂ B(0,R) and choose 1B(0,R) ≤ ϕ1B(0,2R) with ∣∇ϕ∣ ≤ R−1. Then ϕMu/ε isadmissible for the condenser (K,Rn) so that

capp(K) ≤ 2p

εp(∫ ∣∇Mu∣pϕp + ∫ ∣Mu∣p∣∇ϕ∣p) R→∞Ð→ 2p

εp∫ ∣∇Mu∣pϕp ≲ ∫ ∣∇u∣p

where the last step used the previous lemma for the Hardy–Littlewood maximal function.

To state the refined Lusin and Lebesgue theorem for Soboelv functions, we need the following defini-tions.

Definition 13.22. Let p > 1.

Suppose E is of p-capacity zero and a property holds in Ec. Then it is said to hold p-quasieverywhere (q.e.).

Suppose that for every ε > 0 there is an open U such that capp(U) < ε and the restriction f ∣U iscontinuous. Then f is said to be p-quasicontinuous.

50

Let f be locally integrable. We define the precise representative at every point as

fpr(x) = lim supr→0

⨏B(x,r)

f dy.

Theorem 13.23. Let p ∈ (1, n) and f ∈W 1,p(Rn).

(i) The limit limr→0 fB(x,r) exists p-quasieverywhere. In particular, the precise representative equalsf p-quasieverywhere.

(ii) For p-q.e. x it holds

limr→0

⨏B(x,r)

∣f(y) − fpr(x)∣dy.

(iii) fpr is p-quasicontinuous.

Proof. Consider first the sets

Ak = x ∶ lim supr→0

rp−n ∫B(x,r)

∣∇f ∣p > 1

k, k = 1,2, . . .

These are the sets where gradient blows up. They should be small. Indeed, by the usual coveringargument (compare to the subsection about capacity and Hausdorff dimension), we see that Λn−p(Ak) <∞ so that by Theorem 13.14 capp(Ak) = 0 for each k and by countable subadditivity also A = ∪∞i=1Aiis of capacity zero. Outside of this set, the gradient behaves in a tame way and Poincare’s inequalityallows us to infer vanishing oscillation: If x ∉ A, then

(13.1) limr→0

(⨏B(x,r)

∣f − fB(x,r)∣p∗

)1/p∗

≲ lim supr→0

(rp−n ∫B(x,2r)

∣∇f ∣p)1/p

= 0.

This is one thing we need.Second, towards the quasicontinuoity, we attempt uniform approximation by continuous functions off

a small set. Let fi ∈ C∞ ∩W 1,p be such that

∫ ∣∇fi −∇f ∣p ≤1

2i(p+1) .

Define Ci = M(fi − f) > 2−i. By the capacitary weak type estimate 13.21

capp(Ci) ≲ 2ip ∫ ∣∇fi −∇f ∣p ≤ 2−i.

Letting then Ek = A ∪ (∪∞i=kCi), we see that

capp(Ek) ≤ capp(A) +∑j≥k

capp(Cj) ≲ ∑j≥k

2−j ≲ 2−k.

Hence these sets are of arbitrarily small capacity.For any x ∉ Ek

∣fB(x,r) − fi(x)∣ ≤ ⨏B(x,r)

∣f − fB(x,r)∣ + ⨏B(x,r)

∣f − fi∣ + ⨏B(x,r)

∣fi − fi(x)∣→ cn,p2−i

as r → 0 because the first term vanishes by x ∉ A, the second term admits the estimate because x ∉ Ciand the third term vanishes by continuity of fi. Further, for any j, i ≥ k, we can compare fi and fj tothe mean value in a small ball so that the above inequality implies

∣fi(x) − fj(y)∣ ≲ 2−i + 2−j .

Hence fi is Cauchy in the supremum norm in Ek. Hence there exists a continuous limit functiong ∈ C(Rn ∖Ek). However,

∣g(x) − fpr(x)∣ ≤ lim supr→0

∣g(x) − fB(x,r)∣ ≤ lim supr→0

(∣g(x) − fi(x)∣ + ∣fi(x) − fB(x,r)∣) = 0

so that g(x) = fpr(x) and the limsup in the definition of the precise representative actually exists as alimit (the second expression converging to zero).

The proof of (i) is complete. The item (ii) is immediate from cap(A) = 0. To finalize the proof of thequasicontinity (iii), we still have to make the exceptional sets Ek open. But this follows directly fromthe outer approximation by open sets in the definition of the capacity.

51

14. Modulus of a curve family

We mostly follow [7, 11, 13, 23], the notes by Vuorinen being the most substantial source for modulus,the paper by Haj lasz for Sobolev functions. The previous section gives refinement of Lebesgue differ-entiation theorem. Sobolev functions may fail to be absolutely continuous on lines in general, but thefailure is not so dramatic. In this section we quantify the lack of absolute continuity.

Definition 14.1. We recall the standard terminology about curves.

A curve is a continuous map γ ∶ I → Rn where I is an interval. If I is closed, we define the length of γ as

`(γ) = sup(ti)i

k

∑i=1

∣γ(ti) − γ(ti−1)∣

where the supremum is taken over all partitions t0 = a < ti < ⋯ < tk = b where I = [a, b]. If I isnot closed, then we take supremum of lengths of its closed subcurves.

A curve is rectifiable if it has finite length. It is localy rectifiable if its compact subcurves havefinite length.

Arclength parametrization is the unique s ∶ I → [0, `(γ)] so that γ = γ s with γ 1-Lipschitz.We say that γ is parametrized along arclength.

For γ parametrized along arclength and f ≥ 0 Borel, we define

∫γf d` = ∫

`(γ)

0f(γ(s))ds

Next we define the modulus of curve family.

Definition 14.2. Let Γ be a family of curves and p ≥ 1. The functions admissible for Γ are

F(Γ) = f ≥ 0, Borel ∶ ∫γf d` ≥ 1 for all locally rectifiable γ ∈ Γ.

The p-modulus is

modp(Γ) = inff∈F(Γ)

∫Rnfp

with the convention inf ∅ =∞.

Remark 14.3. A few remarks are in order

The definition of admissible function is restricted to locally rectifiable ones as for the non-rectifiable any non-negative function meeting them would be admissible so they would notcontribute to the value of modulus anyway.

The same definition can be done with a family of measures instead of curves. This is whatFuglede does in his paper. Hence it would be equally easy to study modulus of surface families.

The rough electrostatic analogue here is the following: Suppose that Γ is the family of curvesjoining the two parts of a condenser. The admissible functions are roughly the electric fields andthe admissibility means the work done on a unit charge traveling along that curve must be atleast one. Hence the modulus in this case is the least energy needed to have potential differenceone. It will turn out that modulus of such a curve family, indeed, coincides with the capacityof such a condenser.

The reciprocal of the modulus of a curve family is sometimes referred to as the extremal length.

It is natural to think the modulus as a measure. If we add more curves to a family, it becomes moredifficult for Borel functions to be admissible. Consequently their set is at most the same and the infimumof Lp integrals is at least the same. Next we prove that the modulus is an outer measure. We allowexceptions in sets of p-modulus zero by saying something holds for p-almost every curve.

Theorem 14.4. Let p ≥ 1. Then

(i) modp(∅) = 0,(ii) Suppose Γ1 ⊂ Γ2. Then modp(Γ1) ≤ modp(Γ2). This holds even in the case that for each γ ∈ Γ1

there is γ ∈ Γ2 such that γ ⊂ γ.(iii) modp(∪∞i=1γi) ≤ ∑∞

i=1 modp(Γi).52

Proof. If there are no curves, every Borel function is admissile. Testing with 0 gives (i). If Γ1 ⊂ Γ2, thenF(Γ2) ⊂ F(Γ1), and taking infima over quantities indexed along these sets reverses the order once more.Hence (ii) is clear.

To prove the countable subadditivity, let ε > 0. For each i, let fi ∈ F(Γi) be such that ∫ fpi ≤ε2−i + modp(Γi). Then g = supi fi ∈ F(∪∞i=1Γi) and it is increasing limit of gk = max1≤j≤k fj . LetXi = i smallest s.t. fi = gk and notice that these sests are disjoint. Then

∫ gpk ≤k

∑i=1∫Xifpi ≤∑

i

modp(Γi) + ε

and by monotone convergence theorem

modp(∪∞i=1Γi) ≤ ∫ gp = limk∫ gpk ≤∑

i

modp(Γi) + ε

and the proof is complete as ε was arbitrary.

Notation. For two closed sets E and F with E ∩F = ∅ we let Γ(E,F ) be the set of curves γ that meetboth sets.

Example 14.5. One can re-examine the computation of the capacity of the spherical condenser to verifythat for 0 < r < R <∞ it holds

modp(B(0, r),B(0,R)c) = capp(B(0, r),B(0,R)) ∼ rn−p.This gives an example of family with p-modulus zero: The curves inside a ball meeting both the center

point and the exterior. The following simple proposition tells that modulus does not give much weighton long curves. Compare this to the fact that the shortest curve possible, the constant, has infinitemodulus.

Proposition 14.6. Let G be a Borel set of finite measure. Let r > 0 and define the curve familyΓ = γ ⊂ G ∶ `(γ) ≥ r. Then modp(Γ) ≤ ∣G∣r−p.

Proof. The function f(x) = 1G(x)r−1 is admissible and give the right upper bound.

Corollary 14.7. Let G be a Borel set of measure zero. If Γ is a family of curves contained in G, thenmodp(Γ) = 0.

Proof. Let Γj = γ ∈ Γ ∶ `(γ) ≥ 1/j. Then the claim follows from the previous proposition and countablesub-additivity.

Proposition 14.8. Let E ⊂ Rn−1 be Borel, I a finite interval and Γ = x×I ∶ x ∈ E. Then modp(Γ) = 0if and only if ∣E∣ = 0 where the measure is the n − 1 one dimensional Lebesgue measure.

Proof. Suppose first the modulus is zero. Take ε > 0 and an admissible function f with p-integral lessthan ε. Then by montonicity and Holder as well as admissibility

ε > ∫Rnfp ≥ ∫

E∫Ifp ≥ ∫

E∣I ∣−p/p

′

(∫If)

p

≥ ∣E∣∣I ∣−p/p′

.

As I is finite and ε > 0 arbitrary, we conclude that ∣E∣ = 0.To prove the other direction, just note that if the measure is zero, then 1E×I ∣I ∣−1 is an admissible test

function with zero Lp norm.

The following test is useful in checking whether the modulus is zero in more general cases.

Lemma 14.9. The p-modulus of a family Γ is zero if and only if there exists an admissible function fsuch that

∫Rnfp <∞, ∫

γf =∞

where the second equality is asked to hold for all locally rectifiable curves in Γ.

Proof. Assume first the existence of the function f . Then we see that every fk−1 with k > 0 is anadmissible function so that modp(Γ) ≤ ∫ (fk−1)p ≲ k−1 → 0 as k →∞. Hence the modulus must be zero.

Conversely, if the modulus is zero, we find a sequence fk ∈ F(Γ) such that ∫ fpk ≤ 4−k. Define

f = (∑∞k=1 f

pk2k). Then

∫Rnfp =

∞∑k=1

2k ∫Rnfpk ≤

∞∑k=1

2−k = 1

53

and for every γ ∈ Γ and every k

∫γf ≥ 2k/pfk

whence the claim follows by sending k →∞.

As an application, we can show that all non-rectifiable curves in the whole Rn are of p-modulus zeroprovided p ≥ n. Note that for p < n we could conclude this only for bounded curves.

Proposition 14.10. Let Γ be a family of curves in Rn. Let Γr be the subfamily of rectifiable curves.Then modp(Γ) = modp(Γr).

Proof. Let f(x) = (∣x∣ log ∣x∣)−11B(0,2) + 1B(0,1). Then this function has finite Lp norm for p ≥ n. It alsohas non-zero lower bound cB in any bounded set B so for any non-rectifiable curve γ ⊂ B it holds

∫γf ≥ cB ∫

γ=∞.

For any unbounded γ, we see that there is x ∈ γ with ∣x∣ > 6 and

∫γf ≥ ∫

∞

∣x∣

1

r log rdr =∞.

Hence f satisfies the conditions given in the previous lemma and we can conclude Γ ∖ Γr has moduluszero. The exact form given in the statement follows by monotonicity and sub-additivity.

14.1. Curves and condensers. Next we relate the modulus of curve family to capacity of a condencer.This follows mostly [24]. We start with an auxiliary lemma from measure theory. The proof can befound in many textbooks. It is a direct consequence of the fact that every measurable set is trappedbetween open and closed sets up to arbitrarily small measure and that characteristic functions of openand closed sets are lower and upper semicontinuous, respectively.

Lemma 14.11 (Vitali–Caratheodory). Let f ∈ L1 and ε > 0. Then there exists an upper semicontinuousa and lower semicontinuous b such that a ≤ f ≤ b and ∥b − a∥L1 < ε.

Corollary 14.12. The admissible function in the definition of modulus can always be replaced by lowersemicontinuous ones.

Proof. Take a sequence of admissible Borel functions fi with finite Lp norms so that ∥fi∥Lp → c for someconstant c. Then for every i there is a lower semicontinuous bi such that

∫Rnbpi ≤ ∫Rn

+2−i, b ≥ f.

Now

lim infi

∫Rnbpi ≤ lim inf

i(∫

Rnfpi + 2−i) = lim inf

i∫Rnfpi .

In particular, if the sequence fi realizes the value of the modulus of a curve family so does bi.

To motivate the following set of lemmas, we give an overview of the setup of the main theorem.Suppose E is compact and Ω ⊃ E open and bounded. Let Γ be the curves meeting both E and Ωc. Toestimate the capacity by modulus, we have to construct a function admissible for capacity starting froma function admissible for modulus. By the Vitali–Caratheodory lemma, we can assume that the functiong admissible for modulus is lower semicontinuous. To define a “primitive”, we let

(14.1) f(x) = infγ∫γg d`

where the infimum is taken over all curves meeting both x and Ωc. Then the work to be done consistsof proving that this function is admissible for capacity, both in terms of boundary values and regularity.The following lemma gives existence of suitable minmizing curve in many situations.

Lemma 14.13. Let B be a ball, M > 0 and βi curves in B so that `(βi) ≤ M for all i. Assume thatthere are xi, yi ∈ βi with xi → x and yi → y with x ≠ y.

Then there exists a curve β such that x, y ∈ β and

lim infi

∫βif d` ≥ ∫

βf d`

for all lower semicontinuous f .54

Proof. First let γi = βi([xi, yi]) as a set. Denote ai = `(γi) ≤M and assume it is parametrized along thearclength. Hence γi ∶ [0, ai] → B is 1-Lipschitz and it can be extended to a 1-Lipschitz function on thewhole interval [0,M]. Now the family γi is equicontinuous and uniformly bounded. By Arzela–Ascolitheorem there is a subsequence, still denoted by γi, converging uniformly to a function γ which is still1-Lipschitz though not necessarily curve parametrized along arclength. This is the existence of the curve.

By passing to a subsequence once more, we get γi(0) = xi → x = γ(0) and similarly

∣y − γi(a)∣ ≤ ∣y − γi(ai)∣ + ∣γi(ai) − γi(a)∣ ≤ ∣y − γi(ai)∣ + ∣ai − a∣→ 0

so that γ(a) = limi γi(a) = y. This tells that the limit points sit on the curve whose existence we justproved.

Finally, take ε > 0. For i large enough ai > a − ε. Then

lim infi

∫βif d` ≥ lim inf

i∫γif d` ≥ lim inf

i∫

a−ε

0f(γi(t))dt

≥ ∫a−ε

0f(γ(t))dt ≥ ∫

a−ε

0f(γ(t))∣γ′(t)∣dt→ ∫

γf d`

as ε→ 0 by Fatou and lower semicontinuity.

The next lemma is concerned a about regularity of functions of type (14.1). It is relatively easy toprove they are Sobolev, but the continuity requirement asked by the definition of the capacity remainsunknwon. However, this problem can be overcome by different means, and the regularity lemma belowcan be upgraded to one more characterization of Sobolev spaces.

Definition 14.14. Let f ∶ Rn → R be a function. A Borel function g ≥ 0 is called its upper gradient iffor all x, y ∈ Rn and almost all curves γ ∋ x, y

(14.2) ∣f(x) − f(y)∣ ≤ ∫γg d`.

If the above inequality only holds for almost every x, y and p-almost every curve γ ∋ x, y, then we say gis a p-weak upper gradient.

Lemma 14.15. Let f ∈ Lp and assume g ∈ Lp is a p-weak upper gradient of f . Then f ∈ W 1,p and∣∇f ∣ ≤ g almost everywhere.

Proof. Assume first g is lower semicontinuous upper gradient. Take any curve γ parametrized along itsarclength. As γ is continuous and g lower semicontinuous, the function g(x+γ(t)) is lower semicontinuousin both variables. In particular, it is measurable both in x and t. Taking γxy to be the line segmentjoining x to y and using the definition of the upper gradient on it, we see that for any ball B

⨏ ∣f − fB ∣ ≤ ⨏B⨏B∣f(x) − f(y)∣dxdy ≤ ⨏

B⨏B∫

1

0g(x + t(y − x))dtdxdy ≤ r(B)⨏

Bg

Hence f ∈W 1,p and g ≥ ∣∇f ∣ almost everywhere by a refined version of Theorem 3.5.To remove the additional assumptions, we start with the upper gradient property. Suppose g is p-weak

upper gradient. Let Γ be the set of curves on which (14.2) does not holds. Since modp(Γ) = 0, we knowby there exists a Borel function ρ ∈ Lp with ∫γ ρd` for all γ ∈ Γ. For each i, we let gi = g + ρ

i∥ρ∥Lp. Then

each gi is an upper gradient. To make gi lower semicontinous, we can apply Vitali–Caratheodory lemmato find gii ≥ gi lower semicontinuous with ∥(gii)p − g

pi ∥L1 ≤ 1/i. Now gii is lower semicontinuous upper

gradient in Lp, and the remaining claims follow by taking the limit as i→∞.

Now we are in position to prove the equality of capacity of condenser and the modulus of the curvestraveling from one plate to another.

Theorem 14.16. Let Ω be open and K ⊂ Ω compact and p > 1. Then capp(K,Ω) = modp(Γ(K,Ωc)).Recall Γ(K,Ωc) are the curves meeting both K and Ωc.

Proof. One direction is easy. Suppose u is admissible function for the p-capacity. Then ∣∇u∣ is a contin-uous function such that for any γ ∈ Γ(K,Ωc)

1 ≤ infx∈K∩γ

u(x) − supx∈Ωc∩γ

u(x) ≤ ∫γ∣∇u∣d`

so that ∣∇u∣ is admissible for modulus. Then

modp(Γ(K,Ωc)) ≤ ∫Rn

∣∇u∣p dx55

whence the inequality modp(Γ(K,Ωc)) ≤ capp(K,Ω) follows by taking infimum over all u ∈ F(K,Ω).For the other direction, we start with f ∈ F(Γ(K,Ω)). By Vitali–Caratheodory, we can assume f

is lower semicontinuous. By the truncation f i = 1Ω max(f, i−1) we can assume that f is bounded frombelow (in Ω, we do not keep repeating we only care about what happens in Ω from this point on). Indeed,f i ≥ f so it is admissible and f i → f in Lp as i →∞ by dominated convergence as Ω is bounded so therestriction to f bounded from below does not affect the value of the modulus.

Next we set

fk = min(k, f), uk = infβ∫βfk d`

where the infimum is over all the curves in the family β ∋ x ∶ β ∩Ωc ≠ ∅. Note that the uk(x) <∞ forevery x since both fk and Ω are bounded.

Suppose βi are curves such that

uk(x) = limi∫βifk d`.

Without loss of generality, we assume `(βi ∩Ω) = `(βi) and that x is an enpoint of the curve. For eachi, there is yi ∈ ∂Ω. As ∂Ω is compact, we can pass to a subsequence to assume that there is y ∈ ∂Ω sothat yi → y. Moreover, as fk ≥ c inside Ω, uk(x) ≥ ∫βi fk d` ≥ `(βi)c. Hence the assumptions of Lemma

14.13 are fulfilled and we infere there exists a curve β ∋ x, y such that

uk(x) = limi∫βifk d` ≥ ∫

βfk d` ≥ uk(x)

where the last step follows from β being admissible in the definition of uk.Next we show uk is k-Lipschitz and ∣∇uk ∣ ≤ fk. For x, y ∈ Ω, we let β be the curve as above and γxy

the lline segment joining x to y. Then

uk(y) ≤ ∫βfk d` + ∫

γxyfk d` ≤ uk(x) + k∣x − y∣

and the same argument applies with the roles of x and y reversed so we conclude uk is k-Lipschitz. Thefact that ∣∇uk ∣ ≤ fk is the content of Lemma 14.15. In particular, every uk is in W 1,p ∩C.

Next, it is clear uk = 0 in Ωc. Unfortunately it is not quite clear what is the value of uk in K. Letmk = minuk(x) ∶ x ∈K. We claim that

(14.3) lim infk

mk ≥ 1.

Let xk ∈ K be such that uk(xk) = mk. Let βk be the curve as constructed above. Suppose, forcontradiction, we had lim infk uk(xk) = lim infkmk < 1. Take a subsequence such that

∫βkfk d` < 1

for all k. By compactness we can pass to subsequence twice to find a limit point of xk as well as a limitpoint from βk ∩ Ωc. Again, as fk is bounded from below, we see that `(βk) < c−1 uniformly in k. ByLemma 14.13 we find a curve β meeting both K and Ωc so that the lower semicontinuity estimate

lim infk

∫βkρd` ≥ ∫

βρd`

holds for all lower semicontinuous functions ρ. Let then ε > 0 and find an index m so large that

∫βfm d` ≥ ∫

βf d` − ε ≥ 1 − ε

provided the quantity in the middle is finite. In case it is not, take m large enough so that the left handside is larger than the right hand side. As uk is increasing in k, we see that for all k ≥m

mk = ∫βkfk d` ≥ ∫

βkfm d`.

Then

lim infk

mk ≥ lim infk

∫βkfm d` ≥ ∫

βfm d` ≥ 1 − ε→ 1

which is a contradiction. Hence (14.3) holds.In conclusion, given f lower semicontinuous and bounded from below, we can take uk as constructed

above. Then uk = 0 off Ω, ukm−1k ≥ 1 in K, uk ∈W 1,p ∩C and ∣∇uk ∣ ≤ fk ≤ f . Hence the function ukm

−1k

is admissible for the capacity of the condenser and

capp(K,Ω) ≤ lim infk

1

mpk∫ ∣∇uk ∣p dx ≤ ∫ fp dx.

56

Optimizing over all f admissible for modulus conclude the proof.

14.2. Sobolev spaces using curves. Recall that in one dimension Sobolev functions are absolutelycontinuous. Similar property holds in higher dimensions for almost every line parallel to coordinateaxes. Using the notion of modulus, we can extract similar information for more general families ofcurves. Moreover, we can use the notion of upper gradient as in Definition 14.14 to give one morecharacterization of Sobolev spaces. To make all this precise, we start with a lemma of Fuglede.

Lemma 14.17. Let fi ∈ Lp be a sequence such that fi → f in Lp. Then there is a subsequence, stilldenoted fi, such that for p-almost every curve γ

∫γ∣f − fi∣d`→ 0.

Proof. Let fi be a subsequence such that ∥f − fi∥pLp ≤ 2−i(p+1). Denote gi = ∣f − fi∣ and let

Ai = γ ∶ ∫γgi d` > 2−i, Bj =

∞⋃i=jAi, E =

∞⋂j=1

Bj .

Now 2igi is admissible function for every γ ∈ Ai so modp(Ai) ≤ 2−i. Hence for all k

modp(E) ≤ modp(Bk) ≤∞∑j=k

modp(Aj) ≤ 2−k → 0.

On the other hand, for all γ ∉ E there is i such that γ ∉ Bi and hence for all j > i

∫γgj d` = ∫

γ∣fj − f ∣d` ≤ 2−j

and as the right hand side goes to 0 as j →∞, the proof is complete.

Now we can collect the observations about Sobolev functions and their line integrals.

Theorem 14.18. Let f ∈ Lp with p ≥ 1.

i. If f ∈W 1,p, then f has a p-weak upper gradient.ii. If there is a p-weak upper gradient g ∈ Lp, then f ∈W 1,p and g ≥ ∣∇f ∣.

iii. f ∈ W 1,p if and only if it is absolutely continuous on almost every line parallel to coordinateaxes and the partial derivatives are controlled as ∣∂if ∣ ≤ gi ∈ Lp.

Proof. Suppose f is a Sobolev function. Let ϕi ∈ C∞ ∩W 1,p be an approximation ϕi → f in Sobolevnorm and pointwise (e.g. the usual convolution approximation). Pass to the subsequence given by theprevious lemma. Take any pair of Lebesgue points x, y and any curve γ so that the convergence promisedin the lemma takes place. Then

f(x) − f(y) = limi(ϕi(x) − ϕi(y)) ≤ lim sup

i∫γ∣∇ϕi∣d` = ∫

γ∣∇f ∣d`.

This conclude the proof of the first item.The second item was already proved in Lemma 14.15. The third item is left as an exercise (or see

[6]).

14.3. Remark on metric spaces. Let (X,d,µ) be a metric measure space. With some additionalassumptions on the structure of the space (at least µ being doubling measure and metric balls havingnon-zero and finite mass) one can develop surprisingly rich theory of Sobolev spaces. In that context,different characterizations seen so far go with different names and they need not always coincide. As asmall dictionary, one should remember the following two:

Haj lasz-Sobolev space M1,p is defined as the functions f ∈ Lp satisfying

∣f(x) − f(y)∣ ≤ d(x, y)(g(x) + g(y))for almost every x, y and some g ∈ Lp. As was seen, this conincides with Hardy–Sobolev space,which we denoted by H1,p, in Rn. This is very robust definition, and it works under minimalassumptions on the underlying space. For instance, one automatically has Poincare inequality.

Newtonian spaceN1,p is defined as the functions satisfying the condition on weak upper gradientsas in theorem 14.18. This definition of course is useful only in spaces that contain enough curves.Moreover, validity of Poincare inequality for Newton–Sobolev functions is very much dependenton the geometry of the space. This should be though as W 1,p of metric spaces.

57

A good starting point for learning more about Sobolev spaces on metric spaces is the article by Haj lasz[11], which has been written in a very reader–friendly manner.

15. Riesz capacity and Frostman’s lemma

In the final section we discuss an improtant characterization of sets with positive Hausdorff measure.We mostly follow [19]. To motivate the discussion, we introduce one more notion of capacity which istightly related to but still different from the p-capacity we studied in detail. We consider positive Radonmeasures µ on Rn. For a set A, we letM(A) be the sets of positive Radon measures supported inside theset A. With more work than what we were ready to do, it would have been be possible to characterizethe p-capacity as follows: for K compact and Ω ⊃K open, it holds

capp(K,Ω) = supµ(Rn) ∶ µ ∈M(A), 0 ≤ uµ ≤ 1where uµ is function in W 1,p−1 solving

div(∣∇u∣p−2∇u) = µin the weak sense. For compact set A, there exists an extremizing measure (capacitary distribution)and a corresponding solution (capacitary potential). A set of capacity zero is one that does not supportmeasures with finite potential. Even though nonlinear potentials (and their estimates through moreexplicit potential operators) are difficult to study, that can be done, see the last edition in the book[14]. For the purpose of studying dimension however, the approach through potentials is available in asimplified form.

The case of 2-Laplacian is simpler. The 2-Poisson equation with measure data is linear and can besolved through a linear potential operator with convolution kernel cn∣x∣−n+2 (in full space). Attemptingto study Hausdorff dimensions, the problem we now encounter is that we are not allowed to move thegradient integrability parameter p. However, passing to 2-Laplacians of fractional order of smoothnessallows us to stay linear while gaining a parameter to vary. As we saw earlier, the equations

(−∆)s/2u = µadmit a linear solution operator (up to constants, as always with these equations). Recall that for t > 0the Riesz potential is defined

In−tµ(x) = ∫ ∣x − y∣−t dµ(y).We define the s-Riesz energy as

Esµ = ∫ In−tµ(x)dµ(x)and the capacity of a set A

Cs(A) = supEsµ−1 ∶ µ ∈M(A), µ(Rn) = 1.Using the capacity, we define the capacitary dimension as

dimcA ∶= sups ∶ ∃µ ∈M(A), µ(Br) ≤ rs ∀r > 0 = inft ∶ ∃µ ∈M(A),Etµ <∞= sups ∶ Cs(A) > 0 = infs ∶ Cs(A) = 0.

The first equality is the definition and the remaining ones are simple exercises to verify.

Proposition 15.1. If s > 0 and Λs(A) <∞, then Cs(A) = 0.

Proof. Suppose that Cs(A) > 0. Then there is µ ∈M(A) such that Esµ <∞. In particular, In−sµ(x) <∞for µ-almost every x. Consequently, µ(x) = 0 for any such x and by dominated convergence theorem

limr→0

∫B(x,r)

∣x − y∣−s dµ(y) = 0.

Choose compact B ⊂ A so that µ(B) > 1/2. Let ε > 0. Then there is δ so that for every x ∈ B andr ∈ (0, δ] it holds

µ(B(x, r)) ≤ ∫ ( r

∣x − y∣ )s

dµ(y) ≤ εrs.

Finally, fix a covering Bi of B by balls with radii ri ≤ δ/2 so that Λs(A) + 1 ≥ ∑i rsi . Then for each ithere is xi ∈ B ∩Bi so that B(xi, ri) ⊃ B ∩Bi and

1

2< µ(B) ≤∑

i

µ(B(xi, ri)) ≲ ε∑i

rsi ≲ ε(Λs(A) + 1).

58

As ε can be taken arbitrarily small, this can be true only if Λs(A) =∞. This concludes the proof.

To prove a converse of the previous proposition, one has to find a measure with finite energy startingfrom information about Hausdorff measure. Frostman’s lemma is a way to achieve this. Note that thislemma has nothing to do with Riesz capacities so it is equally useful when studying other capacities orproblems as well.

Theorem 15.2 (Frostman’s lemma). Let B be closed. Then Λs(B) > 0 if and only if there existsµ ∈M(B) such that µ(B(x, r)) ≤ rs for all x ∈ Rn and r > 0.

Proof. Suppose µ exists. Let Bi be a covering of B. Then

0 < µ(B) ≤∑i

µ(Bi) ≤∑i

rsi

so that Λ∞s (B) ≥ µ(B) > 0. Hence the same holds for the Hausdorff measure.

Conversely, assume B has positive measure. By countable subadditivity, we can assume B is compact.Let Q0 be a cube that contains B. Let Di be the collection of cubes obtained by partitioning Q0 into2ni cubes with side length 2−il(Q0). By assumption, there is b > 0 so that for any i

∑j

l(Qj)s ≥ b

whenever the family Qj consists of cubes that cover B.We construct a sequence of measures indexed along the smallest non-trivial scale. Let m > 0. For

Q ∈ Dm set

µmm∣Q =⎧⎪⎪⎨⎪⎪⎩

2−ms dx∣Q∣ , if B ∩Q ≠ ∅0, otherwise.

Next we force the correct behaviour on the parent cubes through a recursion. For k ∈ [1,m], if µmk hasbeen defined, we let for Q ∈ Dk−1

µmk−1∣Q =⎧⎪⎪⎨⎪⎪⎩

µmk ∣Q, if µmk (Q) ≤ 2−(k−1)s

2−(k−1)s µmk ∣Qµmk

(Q) .

Denote µm = µm0 .

For any k ≤m and Q ∈ Dm−k it clearly holds µm(Q) ≤ 2−(m−k)s. Moreover, for each x ∈ B we can find

k and Q ∈ Dm−k so that µm(Q) = 2−(m−k)s. Of all such cubes collect a subfamily maximal with respectto inclusion and call it Qi. By the lower bound on the Hausdorff content of B

µm(Rn) =∑i

µm(Qi) ∼∑i

l(Qi)s ≳ b.

As the lower bound is non-zero and independent of m, we can normalizev νm = µm(Rn)−1µm so that νis a probability measure.

The sequence νm is bounded in total mass so there is a subsequence converging weakly* to a Radonmeasure ν (to see it is non-zero apply on a continuous function equal to one in Q0). Moreover, for B(x, r)there are 2n cubes of from the dyadic system generated from Q0 with side length ∼ r so that for theirunion U we get

ν(B(x, r)) ≤ ν(U) ≤ lim infi

νi(U) ≲ rs.

This concludes the proof.

The more general case of Borel or analytic sets can be dealt with using Choquet capacitability on theHausdorff content. But now, the Frostman’s lemma having been proved, we can see that capacitary andHausdorff dimensions are equal:

Theorem 15.3. Let A be closed. If Λs(A) < ∞, then Cs(A) = 0. If Cs(A) = 0, then Λt(A) = 0 for allt > s. In particular, dimc(A) = dim(A).

59

Proof. The first claim was done in the beginning of this section. For the second one, suppose Λt(A) > 0.By Frostman, there is a measure µ ∈M(A) with µ(Br) ≤ rs for all balls with radius r. Now

Esµ = ∫ (∫∞

0µ(y ∶ ∣x − y∣−s > λ)dλ) dµ(x)

= ∫ (∫∞

0µ(B(x,λ−1/s))dλ) dµ(x)

≤ ∫ (µ(Rn) + ∫∞

1λ−t/s dλ) dµ(x) <∞

so that Cs(A) > 0.

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