The Riesz–Thorin Interpolation TheoremOskar Nordenfors

Bachelor Thesis, 15hpBachelor in Mathematics, 180hp

Spring 2018Department of Mathematics and Mathematical Statistics

AbstractIn this essay we present some elementary measure theory and some theory of Lp–spaces

with the goal of proving the Riesz–Thorin interpolation theorem.

SammanfattningI denna uppsats presenteras grundläggande måtteori och något kring teorin om Lp–rum

med målet att bevisa Riesz–Thorins interpolationssats.

Contents

1. Introduction 32. Preliminaries 73. Lp–Spaces 174. The Riesz–Thorin Interpolation Theorem 255. The Hausdorff–Young Inequality 296. Acknowledgements 337. References 35

1. Introduction

In the world of analysis, one important tool at the modern mathematicians disposal isthe interpolation of operators. That is, if we know how the operator behaves in certainspaces, then we can find out how the operator behaves in spaces, which in some senselie in between the spaces in which the behaviour is known. The first theorem regardingthe interpolation of operators was proven by Marcel Riesz in 1927 [9]. The problemat that time was posed as finding out wether a particular function was convex or not,given certain assumptions. Convexity here means that the midpoint of any line segmentbetween two points on the graph of the function lies above, or on, the graph of thefunction. This is however not the formulation of the theorem that we will be using inthis essay. For a biography of Marcel Riesz see [4] p. 278-292. His student Olof Thorinproved a generalization of Riesz’s theorem in 1939 [13] which Thorin himself also laterexpanded on in 1948 [14]. For a biography of Olof Thorin see [1]. Though sometimesreferred to as the Riesz convexity theorem, due to the way he initially stated it in [9],this theorem is usually known as the Riesz–Thorin interpolation theorem.

The main goal of this essay is to provide the necessary background in measure theoryand the theory of Lp–spaces to be able to give a detailed proof of the Riesz–Thorininterpolation theorem which states the following:

Theorem 4.1 (The Riesz–Thorin Interpolation Theorem). Let (X,A, µ) and(Y,B, ν) be measure spaces. Let p0, p1, q0, q1 ∈ [1,∞], and if q0 = q1 = ∞, assumethat (Y,B, ν) is semifinite. For 0 < t < 1 define pt, qt by

1pt

= 1− tp0

+ t

p1and 1

qt= 1− t

q0+ t

q1.

If T : (Lp0 + Lp1)(X,A, µ) → (Lq0 + Lq1)(Y,B, ν) is a linear map such that for f0 ∈Lp0(X,A, µ) and f1 ∈ Lp1(X,A, µ) we have that

‖Tf0‖q0 ≤M0‖f0‖p0 and ‖Tf1‖q1 ≤M1‖f1‖p1 ,

where M0,M1 ∈ (0,∞). Then‖Tf‖qt ≤M1−t

0 M t1‖f‖pt ,

for f ∈ Lpt(X,A, µ).

Let us elucidate the meaning of this barrage of equalities and inequalities. One couldsay that the theorem gives a bound for the operator norm of some linear operator onsome Lp–space, given that we know the bounds for the norm on some Lp–spaces whichwe can interpolate between. To get a better understanding for why this might work, wecan compare this result to the case where we interpolate scalars between points insteadof operators between spaces. It is obviously true that for α0, α1, β0, β1 ∈ (0,∞), we havethat if

α0 ≤ β0 and α1 ≤ β1,

thenα1−t

0 αt1 ≤ β1−t0 βt1.

From this first theorem, the inquiry into the interpolation of operators has grown sub-stantially. With contributions from, among others, Józef Marcinkiewicz [8] and Elias M.

3

Stein [12]. The Marcinkiewicz interpolation theorem is sometimes seen as a generaliza-tion of the Riesz–Thorin interpolation theorem because it too gives bounds for operators,but with weaker assumptions. Nevertheless this is not so, for the Riesz-Thorin interpo-lation theorem provides a better estimate of the bound, should its assumption that T islinear hold. Over the years, the theorem has had some applications, such as, providingsimple proofs of bounds for the Hilbert Transform and for the Fourier transform. Thebound for the latter is known as the Hausdorff–Young inequality which states that:

Theorem 5.6 (The Hausdorff–Young Inequality). Suppose that 1 ≤ p ≤ 2 and q isthe conjugate exponent of p. If f ∈ Lp([0, 1],B([0, 1]), λ), then f̂ ∈ Lq(Z,P(Z), µ) and

‖f̂‖q ≤ ‖f‖p.

Where f̂ = Ff is the Fourier transform (Definition 5.2). So the norm of the transformoperator F is bounded by the norm of the function it is operating on. A less generalizedversion of this result was first proven by William Henry Young in 1913 [15], who provedit for certain values of q only. The general case was proven by Felix Hausdorff in 1923[5].

Now, for a brief overview of this essay. In Section 2, we provide some necessary def-initions and theorems from measure theory. In Section 3, we move closer to our maingoal and provide some necessary definitions and theorems from the theory of Lp–spaces.In Section 4, we prove the main goal of this essay, the Riesz–Thorin interpolation theo-rem. Lastly, in Section 5, we apply the Riesz–Thorin interpolation theorem to prove theHausdorff–Young inequality. In Figure 1.1 there is a simplified flowchart detailing whichresults depend on which, leading up to the Riesz–Thorin interpolation theorem.

This essay is based entirely on [3] with some help from [2]. The only exception to thisbeing the proof of Hölder’s inequality, which comes from [7].

The reader is expected to have some basic knowledge of analysis and linear algebra,perhaps from their undergraduate studies. The reader is also expected to be familiarwith complex analysis. If the reader is not familiar with complex analysis or needs torefresh their memory, the author recommends [11].

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Figure 1.1 – Flowchart of theorem dependency. Arrows point in the direction of impli-cation.

Lemma 3.7

Lemma 2.14

Lemma 3.6

Theorem 4.1

Lemma 2.18

Lemma 2.12

Theorem 3.4

Lemma 2.7

Theorem 3.3

Lemma 2.15 Theorem 3.10

Theorem 2.17

Corollary 3.11

Theorem 2.16

5

2. Preliminaries

In this section we will be reviewing some results from measure theory. This willhopefully help in understanding the main goal of this essay, the proof of the Riesz–Thorininterpolation theorem. We will talk about what σ–algebras, measures and measurablefunctions are, as well as some properties of these. We will also define the integration ofmeasurable functions. We shall begin by defining what a σ–algebra is.

Definition 2.1. Let X be a set. A σ–algebra on X is a collection A, of subsets of Xsuch that the following conditions hold;(i): X ∈ A;(ii): If A1, A2, . . . ∈ A, then ∪∞n=1An ∈ A;(iii): If A ∈ A, then Ac ∈ A.

Observe that the intersection of any nonempty collection of σ–algebras on X is itselfa σ–algebra on X. If P(X) is the power set of X, then we have that for any S ⊆ P(X),the intersection of all σ–algebras on X containing S is said to be the smallest σ–algebraon X containing S. This is often referred to as the σ–algebra generated by S. There is aspecial family of σ–algebras called the Borel σ–algebra on Rd which is useful in analysis,which we will now define.

Definition 2.2. The Borel σ–algebra, B(Rd), on Rd is the σ–algebra generated by theopen subsets of Rd.

We see that the Borel σ–algebra on Rd contains open and closed sets as well ascountable unions of closed sets and countable intersections of open sets, continuing inthis fashion ad infinitum. Next, we shall define what we mean by a measure.

Definition 2.3. Let (X,A) be a set X equipped with a σ–algebra A. A measure is aset function µ : P(X)→ [0,∞] on (X,A) such that the following conditions hold;(i): µ(∅) = 0;(ii): If {An} ∈ A is a countable sequence of disjoint sets, then

µ

( ∞⋃n=1

An

)=∞∑n=1

µ(An).

The measures we will be focusing on are non-negative and real-valued. In the followingtheorem are some of the properties of measures (for proof see [3] or [2]).

Theorem 2.4. Let µ be a measure on (X,A). Then the following statements are true;(i): If A,B ∈ A and A ⊆ B, then µ(A) ≤ µ(B);(ii): If {Ai} ⊆ A, then µ(∪∞i=0Ai) ≤

∑∞i=0 µ(Ai);

(iii): If {Ai} ⊆ A such that A1 ⊆ A2 ⊆ . . ., then µ(∪∞i=0Ai) = limi→∞ µ(Ai);(iv): If {Ai} ⊆ A such that . . . ⊆ A2 ⊆ A1, then µ(∩∞i=0Ai) = limi→∞ µ(Ai).

A set X equipped with both a σ–algebra A and a measure µ is called a measure space.We write (X,A, µ). A measure space, (X,A, µ), is called finite if µ(X) <∞. Even if itis not finite it may be σ–finite or semifinite, which are defined as follows.

7

Definition 2.5. A measure space, (X,A, µ), is said to be σ–finite if there exists asequence of subsets {Ai}, Ai ∈ A for every i ∈ N, of X, such that

∞⋃i=0

Ai = X

andµ(Ai) <∞, for every i ∈ N.

Definition 2.6. A measure space, (X,A, µ), is said to be semifinite if for every subsetA ∈ A, of X, such that

µ(A) =∞,there exists a subset B ∈ A, of A, such that

0 < µ(B) <∞.

For semifinite measure spaces we have the following lemma, which will be of use later(see Figure 1.1).

Lemma 2.7. If (X,A, µ) is semifinite and µ(A) = ∞, then for every C ∈ R+ thereexists B ⊂ A such that

C < µ(B) <∞.

Proof. Assume that there exists C ∈ R+ such that for every subset B ⊂ A we have thateither µ(B) =∞ or µ(B) < C and let

M = sup{µ(B) : B ⊆ A and µ(B) <∞}.Then we have that 0 ≤M ≤ C (specifically, for n ∈ N, M − 1

n < M). So for every n ∈ Nthere exists An ⊆ A such that M − 1

n < µ(An) < ∞. Now, let Bn = ∪nj=1An for everyn ∈ N, and also let B = ∪∞n=1Bn. Then we have that

M − 1n≤ µ(An) ≤ µ(B)

and we also have that

µ(Bn) ≤n∑j=1

µ(An) <∞.

and Bn ⊆ A. Hence, µ(Bn) ≤M and therefore we get that

limn→∞

M − 1n≤ limn→∞

µ(B) = µ(B) = limn→∞

µ(Bn) ≤ limn→∞

M.

Thus, µ(B) = M and µ(A \ B) = ∞. But then, since (X,A, µ) is semifinite, we havethat there exists D ⊆ A \ B such that 0 < µ(D) < ∞. This is a contradiction of thedefinition of M as the supremum, since

M = µ(B) < µ(B) + µ(D) = µ(B ∪D) <∞,where B ∪D ⊆ A. Therefore, for every C ∈ R+ there exists B ⊂ A such that

C < µ(B) <∞.�

8

One of the most important measures is the Lebesgue measure, but before we candefine it we need to define what an outer measure is.

Definition 2.8. Let X be a set. An outer measure is a function µ∗ : P(X)→ [0,∞] forwhich the following holds;(i): µ∗(∅) = 0;(ii): If A ⊆ B ⊆ X, then µ∗(A) ≤ µ∗(B);(iii): If {An} is a sequence of subsets of X, then µ∗(∪∞n=1An) ≤

∑∞n=1 µ

∗(An).

Now, we define the Lebesgue outer measure as follows:

Definition 2.9. Let A ∈ P(R) and let {Ik} = {(ak, bk)} be a sequence of open intervals,with length L (Ik) = bk − ak. The Lebesgue outer measure is defined by

λ∗(A) = inf{ ∞∑k=1

L (Ik) : A ⊆∞⋃k=1

Ik

}.

The Lebesgue outer measure has the property of assigning to each interval on R itslength. Also, for r ∈ R, we define A+ r = {a+ r : a ∈ A} and rA = {ar : a ∈ A}. Thenit holds that λ∗(A+ r) = λ∗(A) and λ∗(rA) = |r|λ∗(A). The restriction of the Lebesgueouter measure to the set

C = {B ∈ P(R) : λ∗(A) = λ∗(A ∩B) + λ∗(A ∩Bc), for every A ∈ P(R)},

is a measure on (R, C). We write λ∗|C = λ, and call this measure the Lebesgue measure.Therefore, we have that the sets B ∈ C are Lebesgue measurable. For example, we havethat every B ∈ B(R) is Lebesgue measurable. Note that there exists non-Lebesguemeasurable sets, though the existence of such sets rely on the axiom of choice. Eventhough the Lebesgue measure is, as noted earlier, very important, we will not be seeingthis particular measure again until Section 5. Now, before we continue any further, weneed to define the concept of measurability of a function.

Definition 2.10. Let (X,A, µ) be a measure space. A function f : X → [−∞,∞] is A–measurable if f−1((r,∞)) ∈ A for every r ∈ R. A function f : X → C is A-measurableif Re(f) and Im(f) are A–measurable.

For example, if X = R, A = B(R) and µ = λ, then the continuous functions aremeasurable. Note that, if there exists some non-measurable set in a measure space, thenthere exists some non-measurable function defined on that measure space. We also havethe following properties (for proof see [3] or [2]).

Theorem 2.11. Let (X,A, µ) be a measure space. Then we have the following;(i): If f, g : X → C are A–measurable, then f + g and fg are A–measurable;(ii): If {fn} is a sequence of real-valued A–measurable functions, then supn∈Z fn, infn∈Z fn,lim supn→∞ fn and lim infn→∞ fn are all A–measurable;(iii): If f, g are real-valued A–measurable functions, then max(f, g) and min(f, g) areA–measurable.

Another useful property of measurable functions is that they can be approximatedfrom below by simple functions. The simple functions are functions φ : X → C, which

9

can be written as

φ(x) =N∑n=0

αnχAn(x),

where {An} is a sequence of pairwise disjoint sets in X, αn ∈ C nonzero and

χAn(x) ={

1 if x ∈ An0 otherwise

.

The function χAn is called the characteristic function of An.

Lemma 2.12. Let (X,A, µ) be a measure space. If f : X → C is A–measurable, thenthere exists a sequence {φn}, φn : X → C for every n ∈ N, of simple, A–measurablefunctions such that

|φ1(x)| < |φ2(x)| < . . . < |f(x)|and

f(x) = limn→∞

φn(x)

for every x ∈ X.

Proof. We will divide the proof into three parts. First we will demonstrate that a simlilarresult holds for f : X → [0,∞] and use this to prove that this similar result holds forf : X → [−∞,∞]. Finally we will prove this lemma in its full generality.

Part 1: Assume that we have f : X → [0,∞]. Then, for n ∈ N and 0 ≤ k ≤ 22n − 1,k ∈ N, let

Akn = f−1((k2−n, (k + 1)2−n

])and Bn = f−1

((2n,∞

]),

and define

φn =22n−1∑k=0

k2−nχAkn + 2nχBn .

It is clear that {φn} is a sequence of simple functions. Now we choose x ∈ Akn. Then bythe formula for an arithmetic series we have that

φn(x) = 2n−1(4n − 1) and φn+1(x) = 2n(4n+1 − 1).Hence, φn ≤ φn+1 on Akn clearly. Now choose x ∈ Bn. Then we similarly have that

φn(x) = 8n and φn+1(x) = 8n+1.

Hence, φn ≤ φn+1 for every x ∈ X, since Akn and Bn partition X for every n ∈ N. Now,for every x such that f(x) ≤ 2n, we have that

φn(x) ≤ f(x) ≤ φn(x) + 2−n

which is equivalent to0 ≤ |f(x)− φn(x)| ≤ 2−n

Now, by choosing

N > − lnεln2 ,

10

we have that for any given ε, there exists an N such that for N ≤ n, we get that|f(x)− φn(x)| ≤ 2−n ≤ 2−N < ε.

Thus, we have thatf(x) = lim

n→∞φn(x).

Part 2: Assume that we have f : X → [−∞,∞]. We divide f into its positive andnegative parts

f = f+ − f−

Since f+ : X → [0,∞] and f− : X → [0,∞] fulfill the assumptions of Part 1, we canapply the argument of Part 1 to both f+ and f−, yielding the functions φ+

n and φ−n ,such that

|φ+1 (x)| ≤ |φ+

2 (x)| ≤ . . . ≤ |f+| and |φ−1 (x)| ≤ |φ−2 (x)| ≤ . . . ≤ |f−|,as well as

limn→∞

φ+n (x) = f+(x) and lim

n→∞φ−n (x) = f−(x).

Therefore, by letting φn = φ+n − φ−n , we have that a sequence of simple functions {φn},

φn : X → [0,∞] for every n ∈ N, such that|φ1(x)| ≤ |φ2(x)| ≤ . . . ≤ |f(x)| and lim

n→∞φn(x) = f(x)

exists for f : X → [−∞,∞].Part 3: Assume now that f : X → C and write

f = g + ih,then by Part 2, we can choose sequences of simple functions, {ϕn} and {ψn}, ϕn, ψn :X → R for every n ∈ N, such that

|ϕ1(x)| < |ϕ2(x)| < . . . < |g(x)| and |ψ1(x)| < |ψ2(x)| < . . . < |h(x)|,as well as

limn→∞

ϕn(x) = g(x) and limn→∞

ψn(x) = h(x),

If we now defineφn = ϕn + iψn,

then {φn} is a sequence of simple functions such that|φ1(x)| < |φ2(x)| < . . . < |f(x)|,

andf(x) = lim

n→∞φn(x),

for every x ∈ X. This concludes the proof of the lemma. �

Let (X,A, µ) be a measure space. We will now define the integral for A–measurablefunctions. To do this we will first define for a simple, A–measurable function, φ : X →[0,∞], such that φ(x) =

∑ni=0 αiχAi , the integral

ˆφdµ =

n∑i=1

αiµ(Ai).

11

Where {Ai} is a sequence of pairwise disjoint sets in X and αi ∈ (0,∞]. Then we definefor an A–measurable function f : X → [0,∞] the integralˆ

fdµ = sup{ˆ

φdµ : 0 ≤ φ ≤ f, φ simple and A–measurable}.

For A–measurable functions f : X → [0,∞] we have the following important results,known as the monotone convergence theorem and Fatou’s lemma (for proofs see [3] or[2]).

Theorem 2.13 (The Monotone Convergence Theorem). Let (X,A, µ) be a mea-sure space. If {fn} is a sequence of A–measurable functions fn : X → [0,∞] such thatfn ≤ fn+1, for every n ∈ N, and f = limn→∞ fn. Thenˆ

fdµ = limn→∞

ˆfndµ.

Lemma 2.14 (Fatou’s Lemma). Let (X,A, µ) be a measure space. If {fn} is a se-quence of A–measurable functions fn : X → [0,∞], thenˆ

lim infn→∞

fndµ ≤ lim infn→∞

ˆfndµ.

We also have the following basic result, which we are going to use in Theorem 3.10.

Lemma 2.15. Let (X,A, µ) be a measure space and f : X → [0,∞] a A-measurablefunction. If t ∈ (0,∞), and

At = {x ∈ X : f(x) ≥ t},

thenµ(At) ≤

1t

ˆAt

fdµ ≤ 1t

ˆfdµ.

Proof. By the definition of At we have that

0 ≤ tχAt ≤ fχAt ≤ f,

which yields ˆtχAtdµ ≤

ˆAt

fdµ ≤ˆfdµ.

Now, division by t gives us that

µ(At) ≤1t

ˆAt

fdµ ≤ 1t

ˆfdµ.

�

Now, take f+ and f− as the positive and negative parts of f respectively. Then wedefine for f : X → R A–measurable, the integralˆ

fdµ =ˆf+dµ−

ˆf−dµ

12

and note that since |f | = f+ + f− we say that f is integrable if´|f |dµ <∞.

Now, for f : X → C A–measurable we define the integralˆfdµ =

ˆRe(f)dµ+ i

ˆIm(f)dµ.

Again, we say that f is integrable if´|f |dµ < ∞. For integrable functions we have

the following theorem, known as the dominated convergence theorem (for proof see [3]).Note that µ–a.e. means for every x ∈ X except for possibly a set A ∈ A with µ(A) = 0and is read as "µ almost everywhere".

Theorem 2.16 (The Dominated Convergence Theorem). Let (X,A, µ) be a mea-sure space and let {fn} be a sequence of integrable functions fn : X → C such that:(i): limn→∞ fn = f µ–a.e;(ii): There exists an integrable function g : X → [0,∞] such that |fn| ≤ g µ–a.e. forevery n ∈ N.Then f : X → C, is integrable andˆ

fdµ = limn→∞

ˆfndµ.

This is quite an important result in analysis, which we will use, for example, in theproofs of the Riesz–Thorin interpolation theorem (Theorem 4.1) and Hölder’s inequality(Theorem 3.4).

Next, we shall prove a convergence theorem that we will make use of in Section 3.Before we can state this theorem, we must define that what we mean when we say thata sequence of functions, {fn}, converges to a function, f , in measure, is that

limn→∞

µ

({x ∈ X : |fn(x)− f(x)| > ε

})= 0, for every ε > 0.

Theorem 2.17. Let (X,A, µ) be a measure space and f, fn : X → C be A–measurablefunctions, n ∈ N. If {fn} converges to f in measure, then a subsequence of {fn} con-verges to f µ–a.e.

Proof. Our assumption that {fn} converges to f in measure means that

limn→∞

µ

({x ∈ X : |fn(x)− f(x)| > ε

})= 0, for every ε > 0.

Now define a sequence {nk}, nk ∈ N for every k ∈ N. First define n1 by

µ

({x ∈ X : |fn1(x)− f(x)| > 1

})≤ 1

2 ,

and define nk inductively such that nk > nk−1 and

µ

({x ∈ X : |fnk(x)− f(x)| > 1

k

})≤ 1

2k ,

where k = 2, 3, .... For k ∈ N let

Ak ={x ∈ X : |fnk(x)− f(x)| > 1

k

}.

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If x /∈ ∩∞i=1 ∪∞k=i Ak, then there exists i ∈ N such that x /∈ ∪∞k=iAk. Thus,

|fnk(x)− f(x)| < 1k,

for k ≥ i. That is, {fnk} converges to f whenever x /∈ ∩∞i=1 ∪∞k=i Ak. Now, since

µ

( ∞⋃k=i

Ak

)≤∞∑k=i

µ(Ak) ≤∞∑k=i

12k = 1

2i−1 ,

for i ∈ N and {∪∞k=iAk} is a decreasing sequence with respect to i, it follows that

0 = limi→∞

12i−1 ≥ lim

i→∞µ

( ∞⋃k=i

Ak

)= µ

( ∞⋂i=k

∞⋃k=i

Ak

).

Therefore, {fn} has a subsequnce, {fnk}, that converges to f µ–a.e. �

This concludes the section on measure theory. However, before we continue withthe next section, we will present the proof of a result from complex analysis, knownas the three lines lemma. We are going to use this result directly in the proof of theRiesz–Thorin interpolation theorem in Section 4. Note here that we are interpolatinga function across a strip in the complex plane. This is crucial to the technique we aregoing to employ in later proofs (see Theorem 3.4 and Theorem 4.1), where we are goingto fit the points of functions into certain complex curves.

Lemma 2.18. Let φ be a bounded continuous function on the strip 0 ≤ Re(z) ≤ 1 whichis holomorphic on the interior of the strip and let M0,M1 > 0. If

|φ(z)| ≤M0,

for Re(z) = 0 and|φ(z)| ≤M1,

for Re(z) = 1, then|φ(z)| ≤M1−t

0 M t1,

for Re(z) = t, t ∈ (0, 1).

Proof. For ε > 0 defineφε(z) = φ(z)Mz−1

0 M−z1 eεz(z−1).

For y ∈ R we then have that

|φε(iy)| ≤M0M−10 |eεiy(iy−1)| = e−εy

2≤ 1,

and|φε(1 + iy)| ≤M1M

−11 |eεiy(iy+1)| = e−εy

2≤ 1.

Moreover, we have that limIm(z)→∞ |φε(z)| = 0. Thus we can choose A ∈ R, sufficientlylarge, such that |φε(z)| ≤ 1 on the boundary of the rectangle given by

0 ≤ Re(z) ≤ 1 and −A ≤ Im(z) ≤ A.The maximum modulus principle (see [11] p. 217) then yields that |φε(z)| ≤ 1 on 0 ≤Rez ≤ 1. Then we have that

|φ(z)|M t−10 M−t1 = lim

ε→0|φε(z)| ≤ 1,

14

for Re(z) = t. Which for Re(z) = t is equivalent to|φ(z)| ≤M1−t

0 M t1.

�

15

3. Lp–Spaces

This section is a direct continuation of the previous section. In this section we shalldefine what an Lp–space is and then go over some of its properties, such as the importantHölder’s inequality (Theorem 3.4). We shall also prove a few results which we will makeuse of in Section 4. We will begin by defining what an Lp–space is.

Definition 3.1. Let (X,A, µ) be a measure space and let 0 < p < ∞. For an A–measurable function f : X → C, we define

‖f‖p =(ˆ

|f |pdµ) 1p

.

Furthermore, for p =∞ define

‖f‖∞ = infα>0

{α ∈ R : µ({x ∈ X : |f(x)| > α}) = 0

}.

Note that we may have ‖f‖p =∞. Using ‖ · ‖p we will now define what an Lp–spaceis.

Definition 3.2. Let (X,A, µ) be a measure space. For 0 < p <∞, we defineLp(X,A, µ) = {f : X → C : f isA−measurable and ‖f‖p <∞}.

For p =∞, defineL∞(X,A, µ) = {f : X → C : f is bounded,A−measurable and ‖f‖∞ <∞}.

It is easy to see that if f, g ∈ Lp(X,A, µ), then f + g ∈ Lp(X,A, µ) and βf ∈Lp(X,A, µ) for every β ∈ C. Thus Lp(X,A, µ) is a vector space. Furthermore, if weconsider two arbitrary functions f, g ∈ Lp(X,A, µ) to be the same element in Lp(X,A, µ)whenever f = g µ–a.e. we shall see that Lp(X,A, µ) is a normed vector space with ‖ · ‖pas its norm (see Theorem 3.5).

The first theorem of this section is fundamental to our proof the Riesz–Thorin inter-polation theorem. Recall that if a set A ⊂ B is dense in B, then for every point b ∈ B,we have that either b ∈ A or b = limn→∞ an for some sequence of points, {an} ⊂ A.

Theorem 3.3. Let (X,A, µ) be a measure space and let 0 < p <∞. The set of simplefunctions φ : X → C, such that

φ =n∑i=1

αiχAi ,

where {Ai} is a sequence of pariwise disjoint sets in X, αi ∈ C nonzero andµ(Ai) <∞, for every i ∈ N,

is dense in Lp(X,A, µ).

Proof. To see that such functions, φ, are in Lp(X,A, µ) simply observe that we havethat

‖φ‖pp =ˆ|φ|pdµ =

n∑i=1|αi|pµ(Ai)p <∞,

17

since µ(Ai) < ∞ for every i. If f ∈ Lp(X,A, µ), then, by Lemma 2.12, we can choosea sequence of simple, A–measurable functions {φn}, φn : X → C for every n ∈ N, suchthat

|φn| < |f | and limn→∞

φn = f µ−a.e.

Then φn ∈ Lp(X,A, µ) for every n, and also|φn − f |p ≤ 2p|f |p ∈ L1(X,A, µ).

By the dominated convergence theorem (Theorem 2.16) we then have thatlimn→∞

‖f − φn‖p = 0.

Moreover, since the {φn} are of the form

φn =n∑i=1

αiχAi ,

where the αi are nonzero and the Ai are pairwise disjoint, we have thatn∑i=1

αiµ(Ai) =ˆ|φ|dµ = ‖φn‖pp <∞.

From this it follows thatµ(Ai) <∞, for every i ∈ N.

Thus, the set of simple functions φ : X → C, such that

φ =n∑i=1

αiχAi ,

where αi ∈ C, Ai ⊆ X andµ(Ai) <∞, for every i ∈ N,

is dense in Lp(X,A, µ). �

We shall henceforth refer to the set of simple, A–measurable functions which satisfythe conditions of Theorem 3.3 as ΣX .

Next, we will present what is the most important theorem in the theory of Lp–spaces,Hölder’s inequality. Despite its name, the theorem was first proven by Leonard JamesRogers in 1888 [10] and independently by Otto Hölder in 1889 [6]. The proof we willpresent is from [7] and the reader may observe a similarity between this proof and theproof of the Riesz–Thorin interpolation theorem later on.

Theorem 3.4 (Hölder’s Inequality). Let (X,A, µ) be a measure space, 1 ≤ p ≤ ∞and let q satisfy 1/p+ 1/q = 1. If f ∈ Lp(X,A, µ) and g ∈ Lq(X,A, µ), then

‖fg‖1 ≤ ‖f‖p‖g‖q.

Proof. Assume that f, g : X → C are simple functions. Now, let z = 1/q. Then we wantto prove the inequality ˆ

|f |p(1−z)|g|qzdµ ≤ ‖f‖p(1−z)p ‖g‖qzq .

18

Now, since f, g are simple functions we define

φ(z) =ˆ|f |p(1−z)|g|qzdµ =

n∑j=1

m∑k=1|αj ||βk|eλjp(1−z)eδkqz.

This function is bounded and continuous on the strip {z ∈ C : 0 ≤ Re(z) ≤ 1} andholomorphic on the interior of the strip. Furthermore, we have that φ(σ + it) ≤ φ(σ),σ, t ∈ R. Whereby,

φ(0 + it) ≤ φ(0) ≤ˆ|f |pdµ and φ(1 + it) ≤ φ(1) ≤

ˆ|g|qdµ.

Thus, by Lemma 2.18, we have thatˆ|f |p(1−σ)|g|qσdµ ≤ ‖f‖p(1−σ)

p ‖g‖qσq .

By letting σ = 1/q we obtain Hölder’s inequality.Now assume that f ∈ Lp(X,A, µ) and g ∈ Lq(X,A, µ), are not necessarily simple.

By Theorem 3.3 we have that there exists simple, A–measurable functions fn, gn suchthat |fn| ≤ |f |, limn→∞ fn = f , |gn| ≤ |g| and limn→∞ gn = g. Thus,ˆ

|fn||gn|dµ ≤ ‖fn‖p‖gn‖q.

Now we let n→∞, and by the dominated convergence theorem (Theorem 2.16) we havethat

‖fg‖1 =ˆ|f ||g|dµ ≤ ‖f‖p‖g‖q.

�

Numbers, p and q, that fulfill the equation 1/p+ 1/q = 1 are sometimes referred to asHölder conjugates, or conjugate exponents; we will be using the latter in this text. Thissimple-looking result is used everywhere in modern analysis to prove all sorts of theorems.Among others, the following result is proved using Hölder’s inequality (Theorem 3.4).

Theorem 3.5 (Minkowski’s Inequality). Let (X,A, µ) be a measure space and 1 ≤p ≤ ∞. If f, g ∈ Lp(X,A, µ), then

‖f + g‖p ≤ ‖f‖p + ‖g‖p.

Proof. The result clearly holds if f + g = 0 µ–a.e. The case where p = 1 follows fromthe regular triangle inequality. If p = ∞, then we have that |f | ≤ ‖f‖∞ µ–a.e. and|g| ≤ ‖g‖∞ µ–a.e. This means that

|f + g| ≤ |f |+ |g| ≤ ‖f‖∞ + ‖g‖∞ µ–a.e.Thus, we have that ‖f‖∞ + ‖g‖∞ is an upper bound for |f + g| µ-a.e. But, by thedefinition of the L∞–norm ‖f + g‖∞ is the infimum over these upper bounds. Therefore

‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞.Now, assume that 1 < p < ∞ and f + g 6= 0 µ–a.e. We observe that if q is the

conjugate exponent of p, then we have that(|f + g|p−1)q = |f + g|p.

19

We also observe that|f + g|p ≤

(|f |+ |g|

)|f + g|p−1.

Now, using what we noted and applying Hölder’s inequality (Theorem 3.4) to thefunctions f and |f + g|p−1 and to the functions f and |f + g|p−1, we get thatˆ|f + g|pdµ ≤

ˆ (|f |+ |g|

)|f + g|p−1dµ =

ˆ|f ||f + g|p−1dµ+

ˆ|g||f + g|p−1dµ

≤ ‖f‖p‖|f + g|p−1‖q + ‖g‖p‖|f + g|p−1‖q =(‖f‖p + ‖g‖p

)(ˆ|f + g|pdµ

) 1q

.

Dividing by (´|f + g|pdµ)1/q yields(ˆ

|f + g|pdµ)1− 1

q

≤ ‖f‖p + ‖g‖p.

Which is equivalent to‖f + g‖p ≤ ‖f‖p + ‖g‖p.

�

This is the triangle inequality for the Lp(X,A, µ)–norm, known as Minkowski’s in-equality. What about the other properties of the norm? Indeed ‖f‖p = 0 if, and onlyif, f = 0 µ–a.e. and ‖βf‖p = |β|‖f‖p for every β ∈ C. Thus Lp(X,A, µ) is a normedvector space with ‖ · ‖p as its norm. Furthermore, it can be shown that Lp(X,A, µ) is aBanach space, i.e. a complete normed vector space (see [3], p. 183 for a proof).

Next, we will prove two lemmas which we are also necessary for Section 4 (see Fig-ure 1.1). They both deal with finding bounds for the Lp(X,A, µ)–norm. The first ofthese lemmas should be compared to the Riesz–Thorin interpolation theorem (Theo-rem 4.1), as it is essentially a special case of that theorem. We are giving a bound forthe norm of a function (as opposed to a linear map) by interpolating between Lp–spaces.

Lemma 3.6. Let (X,A, µ) be a measure space. If 1 ≤ p < q < r ≤ ∞. Then

Lp(X,A, µ) ∩ Lr(X,A, µ) ⊂ Lq(X,A, µ),

and‖f‖q ≤ ‖f‖tp‖f‖1−t

r ,

where t ∈ (0, 1) fulfills the equation1q

= t

p+ (1− t)

r.

Proof. If r =∞, then we have that t = p/q and

|f |q = |f |q−p|f |p ≤ ‖f‖q−p∞ |f |p,

which yields‖f‖q ≤ ‖f‖p/qp ‖f‖1−p/q

∞ = ‖f‖tp‖f‖1−t∞ .

20

If r <∞, then we have that

1 = 1ptq

+ 1r

(1−t)q.

Hölder’s inequality (Theorem 3.4) then yields thatˆ|f |qdµ =

ˆ|f |tq|f |(1−t)qdµ ≤ ‖|f |tq‖p/tq‖|f |(1−t)q‖r/(1−t)q

=(ˆ

|f |pdµ)tq/p(ˆ

|f |rdµ)(1−t)q/r

= ‖f‖tqp ‖f‖(1−t)qr .

Now taking the qth root gives the wanted result. �

Lemma 3.7. Let (X,A, µ) be a measure space, 0 ≤ p ≤ ∞ and q defined such that1/p + 1/q = 1. Assume that f : X → C is a A–measurable function such that fφ ∈L1(X,A, µ), where φ ∈ ΣX . Assume also that

Mq(f) = sup{∣∣∣∣ ˆ fφdµ

∣∣∣∣ : φ ∈ ΣX , ‖φ‖p = 1}<∞.

Assume further that Sf = {x ∈ X : f(x) 6= 0} is σ–finite or that (X,A, µ) is semifinite.Then f ∈ Lq(X,A, µ) and Mq(f) = ‖f‖q.

Proof. Let φ be a bounded A–measurable function which vanishes outside a set, A, offinite measure and with ‖φ‖p = 1, then, by Lemma 2.12, we have that there exists asequence of simple functions {φn}, φn : X → C for every n ∈ N, such that |φn| ≤ |φ| andlimn→∞ φn = φ µ–a.e. These φn also vanish outside A, that is |φn| ≤ |φ|χA which meansthat |φn| ≤ ‖φ‖∞χA µ–a.e. Now, since fχA ∈ L1(X,A, µ), we have that the dominatedconvergence theorem (Theorem 2.16) yields that∣∣∣∣ˆ φfdµ

∣∣∣∣ = limn→∞

∣∣∣∣ ˆ φnfdµ∣∣∣∣ ≤ lim

n→∞Mq(f) = Mq(f).

Assume now that q <∞. Then we can assume that Sf is σ-finite. We know that either(X,A, µ) is semifinite or Sf is σ–finite. Assume that (X,A, µ) is semifinite, choose ε > 0and let B = {x ∈ X : |f(x)| > ε}. If µ(B) = ∞, then by Lemma 2.7, there existsBn ⊂ B such that n < µ(Bn) <∞ for every n ∈ N. But since

µ(Bn) ≤ 1ε

ˆ|f(x)|χBndµ = (µ(Bn))

1p

ε

ˆf(x) f(x)

|f(x)|(µ(Bn))1p

χBndµ

≤ (µ(Bn))1p

εMq(f)

we have that the subsets Bn are uniformly bounded. This contradicts the fact that wecan find Bn ⊆ B for every n, such that n < µ(Bn) < ∞ (since it fails for n sufficientlylarge). Therefore µ(B) <∞ for every ε > 0. Thus, Sf is σ-finite.

Now, since Sf is σ–finite we can choose a sequence of sets {An}, An ⊆ X for every n ∈N, with finite measure such that Sf = ∪nAn. By Lemma 2.12, we can choose a sequence

21

{ϕn} of simple functions such that |ϕn| ≤ |f | and ϕn → f . Now let fn = ϕnχAn . Thenwe have that |fn| ≤ |f |, limn→∞ fn = f , and that fn vanishes outside An. Now, let

Φn = |fn|q−1sgn(f)‖fn‖q−1

q

.

Where sgn is the familliar sign function, sgn(z) = z/|z|, for z 6= 0. We have thatp(q − 1) = q and that |

´φfdµ| ≤Mq(f). Whereby,

‖Φn‖pp =

´ ∣∣∣∣|fn|q−1sgn(f)∣∣∣∣pdµ

‖fn‖p(q−1)q

=´|fn|qdµ‖fn‖qq

=´|fn|qdµ´|fn|qdµ

= 1.

From Fatou’s lemma (Lemma 2.14) it follows that

‖f‖q ≤ lim infn→∞

‖fn‖q = lim infn→∞

ˆ|Φnfn|dµ

≤ lim infn→∞

ˆ|Φnf |dµ = lim inf

n→∞

ˆΦnfdµ ≤Mq(f).

By Hölder’s inequality (Theorem 3.4) we have thatMq(f) ≤ ‖Φn‖p‖f‖q = ‖f‖q.

Therefore, when q <∞, we have that ‖f‖q = Mq(f).Now assume q =∞. Given ε > 0, let

A = {x ∈ X : |f(x)| ≥M∞(f) + ε}.Now, assume that µ(A) > 0. Then we can choose a subset, B, of A such that 0 < µ(B) <∞, because either (X,A, µ) is semifinite or A ⊂ Sf and Sf is σ–finite. Let

Φ = µ(B)−1χBsgn(f).Then

‖Φ‖1 =ˆ|Φ|dµ = µ(B)−1µ(B) = 1,

andˆΦfdµ = µ(B)−1

ˆB

sgn(f)dµ = µ(B)−1ˆB

|f |dµ

≥ µ(B)−1µ(B)(M∞(f) + ε) = M∞(f) + ε.

But as we noted earlier,´

Φfdµ ≤Mq(f), which is a contradiction. Therefore, µ(A) = 0and by the definition of the L∞(X,A, µ)–norm we have that

‖f‖∞ ≤M∞(f).In addition, we have that∣∣∣∣ˆ Φfdµ

∣∣∣∣ ≤ ˆ |Φf |dµ ≤ ‖Φ‖1‖f‖∞ = ‖f‖∞,

which implies thatM∞(f) ≤ ‖f‖∞,

22

and soMq(f) = ‖f‖q for 0 ≤ q ≤ ∞.

�

Before we end this section we need to define what we mean by (Lp + Lq)(X,A, µ),and what makes this set special. Without this one cannot understand the statement ofthe Riesz–Thorin interpolation theorem (Theorem 4.1).

Definition 3.8. Let (X,A, µ) be a measure space, and let Lp0(X,A, µ) and Lp1(X,A, µ)be Lp–spaces. Then

(Lp0 + Lp1)(X,A, µ) ={f : f = f0 + f1, f0 ∈ Lp0(X,A, µ) and f1 ∈ Lp1(X,A, µ)

}.

That is, it is the space of functions f = f0 + f1 where f0 ∈ Lp0(X,A, µ) and f1 ∈Lp1(X,A, µ).

It is easy to see from this definition that if we have a linear map T , such that

T : Lp0(X,A, µ)→ Lq0(Y,B, ν) and T : Lp1(X,A, µ)→ Lq1(Y,B, ν),

thenT : (Lp0 + Lp1)(X,A, µ)→ (Lq0 + Lq1)(Y,B, ν),

since, for f ∈ (Lp0 + Lp1)(X,A, µ), we have that

Tf(x) = T (f0 + f1)(x) = Tf0(x) + Tf1(x) = g0(y) + g1(y) = g(y).

To fully appreciate the usefulness of the set (Lp +Lq)(X,A, µ), we have the followingresult. The reader is advised to pay attention to the proof, as we will utilize the sametechnique in one part of the proof of the Riesz–Thorin interpolation theorem (Theo-rem 4.1).

Theorem 3.9. Let (X,A, µ) be a measure space. If 0 < p < r < q ≤ ∞, thenLr(X,A, µ) ⊆ (Lp + Lq)(X,A, µ).

Proof. Let A = {x ∈ X : |f(x)| > 1}. For f ∈ Lr(X,A, µ) set g = fχA and seth = fχAC . Then

|g(x)|p = |f(x)|pχA(x) ≤ |f(x)|rχA(x) <∞,since |f(x)| > 1, r > p and f ∈ Lr(X,A, µ). Similarly,

|h(x)|q = |f(x)|qχAC(x) ≤ |f(x)|rχAC <∞,

since |f(x)| ≤ 1, r < q and f ∈ Lr(X,A, µ). This means that g ∈ Lp(X,A, µ) andh ∈ Lq(X,A, µ), and since f ∈ Lr(X,A, µ) was chosen arbitrarily, this proves thetheorem for q <∞.

Now, assume q =∞. Then

‖h‖∞ = infα>0

{α ∈ R : µ({x ∈ X : |h(x)| > α}) = 0

}≤ 1,

so h ∈ L∞(X,A, µ), and the theorem is proved. �

23

What this means is that if 0 < p < r < q ≤ ∞, then every function f ∈ Lr(X,A, µ)can be decomposed into a sum of functions g ∈ Lp(X,A, µ) and h ∈ Lq(X,A, µ).

We will end this section by proving a convergence result in Lp(X,A, µ)–norm. Wewill use this result (Corollary 3.11) in the last part of the proof of the Riesz–Thorininterpolation theorem.

Theorem 3.10. Let (X,A, µ) be a measure space, let 1 ≤ p ≤ ∞ and let f, fn : X → Cbe A–measurable functions, n ∈ N. If {fn} converges to f in Lp(X,A, µ)–norm then{fn} converges to f in measure for every p.

Proof. Assume first that 1 ≤ p < ∞ and let {fn} be a sequence of A–measurablefunctions which converges to f in Lp(X,A, µ)–norm. This means that

limn→∞

‖fn − f‖p = limn→∞

(ˆ|fn − f |pdµ

)1/p= 0.

Now, let ε > 0. What we want to show is thatlimn→∞

µ({x ∈ X : |fn(x)− f(x)|p > ε}) = 0.

From Lemma 2.15 we have that

µ({x ∈ X : |fn(x)− f(x)|p > ε}) ≤ 1ε

ˆ|fn(x)− f(x)|pdµ = 1

ε‖fn − f‖pp.

Taking the limit of this inequality, as n goes to infinity, yields

0 ≤ limn→∞

µ({x ∈ X : |fn(x)−f(x)|p > ε}) ≤ limn→∞

1ε

ˆ|fn(x)−f(x)|pdµ = lim

n→∞

1ε‖fn−f‖pp = 0,

because limn→∞ ‖fn − f‖p = 0. Since ε > 0 was chosen arbitrarily, this means thatlimn→∞

µ({x ∈ X : |fn(x)− f(x)| > ε}) = 0, for every ε > 0,

and so {fn} converges to f in measure.Now assume that p =∞. Then, for every α > 0, we have that

limn→∞

infα

{α ∈ R : µ({x ∈ X : |fn(x)− f(x)| > α}) = 0

}= 0,

by our assumption that {fn} is a sequence of A–measurable functions which convergesto f in Lp(X,A, µ)–norm. This means that

limn→∞

µ({x ∈ X : |fn(x)− f(x)| > ε}) = 0, for every ε > 0.

By definition this means that {fn} converges to f in measure. �

From this, we immediately get the following corollary.

Corollary 3.11. Let (X,A, µ) be a measure space, let 1 ≤ p ≤ ∞ and let f, fn : X → Cbe A–measurable functions, n ∈ N. If {fn} converges to f in Lp(X,A, µ)–norm thenthere exists a subsequence of {fn} that converges to f µ–a.e.

Proof. Since {fn} converges in norm it converges in measure by Theorem 3.10. Then byTheorem 2.17, since it converges in measure there exists some subsequence {fnk} whichconverges to f µ–a.e. �

24

4. The Riesz–Thorin Interpolation Theorem

We are now going to prove the Riesz–Thorin interpolation theorem. The method weshall employ is a standard limiting technique. We are going to start by using Lemma 3.6to show that the result holds if we interpolate between a space and itself, i.e. if thereno interpolation at all. Then we assume that the functions we are operating with arenot just functions in some Lp(X,A, µ), but are in fact simple functions. We are goingto use Lemma 3.7, Lemma 2.18 and Hölder’s inequality (Theorem 3.4) to show that theresult holds for the simple functions. Then by Theorem 3.3 we know that every functionin Lp(X,A, µ) is the limit point of some sequence of simple functions. We will then usethe dominated convergence theorem (Theorem 2.16), Corollary 3.11 and Fatou’s lemma(Lemma 2.14) to show that the result holds in the limit as well. Thus proving that itholds for every function in some Lp(X,A, µ).

Theorem 4.1 (The Riesz–Thorin Interpolation Theorem). Let (X,A, µ) and(Y,B, ν) be measure spaces. Let p0, p1, q0, q1 ∈ [1,∞], and if q0 = q1 = ∞, assumethat (Y,B, ν) is semifinite. For 0 < t < 1 define pt, qt by

1pt

= 1− tp0

+ t

p1and 1

qt= 1− t

q0+ t

q1.

If T : (Lp0 + Lp1)(X,A, µ) → (Lq0 + Lq1)(Y,B, ν) is a linear map such that for f0 ∈Lp0(X,A, µ) and f1 ∈ Lp1(X,A, µ) we have that

‖Tf0‖q0 ≤M0‖f0‖p0 and ‖Tf1‖q1 ≤M1‖f1‖p1 ,

where M0,M1 ∈ (0,∞). Then‖Tf‖qt ≤M1−t

0 M t1‖f‖pt ,

for f ∈ Lpt(X,A, µ).

Proof. If p0 = p1 then pt = p0 = p1 = p. Then we have that‖Tf‖qt ≤ ‖Tf‖1−t

q0‖Tf‖tq1

≤M1−t0 M t

1‖f‖1−tp ‖f‖tp = M1−t

0 M t1‖f‖p.

by Lemma 3.6.Now, assume that p0 6= p1, which means that pt < ∞ when 0 < t < 1. Let ΣX and

ΣY be the sets of all simple functions on X and Y respectively, which vanish outsidesets of finite measure. Then by Theorem 3.3, ΣX and ΣY are dense in Lp(X,A, µ) andLp(Y,B, ν) respectively, for 1 ≤ p < ∞. Note that, for f ∈ ΣX , Tf ∈ Lq0(Y,B, ν) ∩Lq1(Y,B, ν), so {y ∈ Y : Tf(y) 6= 0} must be σ–finite as long as we do not haveq0 = q1 = ∞. But if q0 = q1 = ∞, then (Y,B, ν) is semifinite, and the hypothesis ofLemma 3.7 is fulfilled. Therefore, we have that

‖Tf‖qt = sup{∣∣∣∣ˆ (Tf)gdν

∣∣∣∣ : g ∈ ΣY , ‖g‖q′t = 1},

where q′t is the conjugate exponent to qt. We may assume without loss of generality thatf 6= 0, then we rescale f such that ‖f‖pt = 1. Thus, we want to show that for f ∈ ΣXsuch that ‖f‖pt = 1, we have∣∣∣∣ˆ (Tf)gdν

∣∣∣∣ ≤M1−t0 M t

1, for every g ∈ ΣY such that ‖g‖q′t = 1.

25

Let

f =n∑j=1

αjχAj and g =m∑k=1

βkχBk ,

where {Aj} and {Bk} are sequences of pairwise disjoint sets in X and Y respectivelyand αj , βk ∈ C nonzero. We write αj and βk in polar coordinates,

αj = |αj |eiθj , βk = |βk|eiψk .Now, define

γ(z) = 1− zp0

+ z

p1and δ(z) = 1− z

q0+ z

q1.

Whereby it holds that

γ(t) = 1pt

and δ(t) = 1qt, for every t ∈ (0, 1).

Fix t ∈ (0, 1). In this case we have assumed that pt <∞, and thus γ(t) > 0. Therefore,we can define

fz =n∑j=1|αj |

γ(z)γ(t) eiθjχAj

We define

gz ={∑m

k=1 |βk|1−δ(z)1−δ(t) eiψkχBk , if δ(t) < 1

g, if δ(t) = 1.

Now, let

Φ(z) =ˆ

(Tfz)gzdν.

Then we have that

Φ(z) ={∑

j,k Ej,k|αj |γ(z)γ(t) |βk|

1−δ(z)1−δ(t) , if δ(t) < 1∑

j,k Ej,k|αj |γ(z)γ(t) |βk|, if δ(t) = 1

,

whereEj,k = ei(θj+ψk)

ˆ(TχAj )χBkdν.

Now, Φ(z) is an entire function, that is bounded on 0 ≤ Re(z) ≤ 1, with´

(Tf)gdν =Φ(t) for t ∈ (0, 1). We have that by Lemma 2.18 it suffices to show that |Φ(z)| ≤M0 forRe(z) = 0 and |Φ(z)| ≤M1 for Re(z) = 1, since then

|Φ(z)| ≤M1−t0 M t

1, for Re(z) = t, t ∈ (0, 1),and Re(z) = t when z = t, t ∈ (0, 1).

We begin with the case δ(t) < 1. For r ∈ R we have that

γ(ir) = 1− irp0

+ ir

p1= 1p0

+ ir( 1p1− 1p0

),

and1− δ(ir) = 1− 1− ir

q0− ir

q1= (1− 1

q0)− ir( 1

q1− 1q0

).

26

Whereby

|fir| = |f |Re(γ(ir)γ(t)

)= |f |

ptp0 ,

and

|gir| = |g|Re(

1−δ(ir)1−δ(t)

)= |g|

q′tq′0 .

Therefore, by Hölder’s inequality (Theorem 3.4), we have that|Φ(ir)| ≤ ‖Tfir‖q0‖gir‖q′0 ≤M0‖fir‖p0‖gir‖q′0 = M0‖f‖pt‖g‖q′t = M0.

Likewise, we have that

γ(1 + ir) = 1− (1 + ir)p0

+ 1 + ir

p1= 1p1

+ ir( 1p1− 1p0

),

and1− δ(1 + ir) = 1− 1− (1 + ir)

q0− 1 + ir

q1= (1− 1

q1)− ir( 1

q1− 1q0

).

Whereby

|f1+ir| = |f |Re(γ(1+ir)γ(t)

)= |f |

ptp1 ,

and

|g1+ir| = |g|Re(

1−δ(1+ir)1−δ(t)

)= |g|

q′tq′1 .

By Hölder’s inequality (Theorem 3.4) we then have that|Φ(1 + ir)| ≤ ‖Tf1+ir‖q1‖g1+ir‖q′1 ≤M1‖f1+ir‖p1‖g1+ir‖q′1 = M1‖f‖pt‖g‖q′t = M1.

Now, for the case δ(t) = 1, observe that since δ(t) = 1/qt we have that qt = 1 forevery t ∈ (0, 1), which means that qt = q0 = q1 = 1. As above, we have that, for r ∈ R,

|Φ(ir)| ≤M0‖fir‖p0‖gir‖q′0 = M0,

since ‖fir‖p0 = ‖f‖pt = 1, gir = g, ‖g‖q′t = 1 and q0 = qt. Likewise,

|Φ(1 + ir)| ≤M1‖f1+ir‖p1‖g1+ir‖q′1 = M1,

since ‖fir‖p1 = ‖f‖pt = 1, g1+ir = g, ‖g‖q′t = 1 and q1 = qt. Thus, by Lemma 2.18 wehave that

‖Tf‖qt = M1−t0 M t

1‖f‖pt , for every f ∈ ΣX .Now assume that f ∈ Lpt(X,A, µ) is an arbitrary (not necessarily simple) function.

Then, by Theorem 3.3, we can choose a sequence of functions {fn}, fn ∈ ΣX for everyn, such that |fn| ≤ |f | and limn→∞ fn = f . Now, let

A = {x ∈ X : |f(x)| > 1}.Also, let g = fχA, gn = fnχA, h = fχAC , hn = fnχAC . Then we have that f = g+h andfn = gn+hn. Now, assume that 1 ≤ p0 < pt < p1 ≤ ∞, for if not we may simply relabelp0, p1. Then we have that g ∈ Lp0(X,A, µ) and h ∈ Lp1(X,A, µ), since ‖g‖p0 ≤ ‖f‖ptand ‖h‖p1 ≤ ‖f‖pt , by the same argument as in the proof of Theorem 3.9. By thedominated convergence theorem (Theorem 2.16) we have that

limn→∞

‖fn − f‖pt = 0, limn→∞

‖gn − g‖p0 = 0 and limn→∞

‖hn − h‖p1 = 0,

27

which means thatlimn→∞

‖Tgn − Tg‖q0 = 0 and limn→∞

‖Thn − Th‖q1 = 0,

since‖T (gn − g)‖q0 ≤M0‖gn − g‖p0 and ‖T (hn − h)‖q1 ≤M1‖hn − h‖p1 .

Then, by Corollary 3.11, we have that there exists subsequences {Tgnk} and {Thnk}, of{Tgn} and {Thn} respectively, such that

limn→∞

Tgn = Tg ν–a.e. and limn→∞

Thn = Th ν–a.e.

But this means thatlimn→∞

Tfn = Tf ν–a.e.

Thus, by Fatou’s lemma (Lemma 2.14) we have that‖Tf‖qt ≤ lim inf

n→∞‖Tfn‖qt ≤ lim inf

n→∞M1−t

0 M t1‖fn‖pt = M1−t

0 M t1‖f‖pt .

�

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5. The Hausdorff–Young Inequality

In this section we are going to apply the Riesz–Thorin interpolation theorem (The-orem 4.1) to prove a result known as the Hausdorff–Young inequality. But first we aregoing to provide some necessary definitions and proofs that have to do with the topic ofFourier analysis, since the inequality we are to prove is the following:

‖f̂‖q ≤ ‖f‖p,

where f̂ is the Fourier transform as defined in Definition 5.2, and the norms are definedas in Definition 5.1. We are going to use some results from Fourier analysis to show thatwe have bounds for some Lp–spaces and then we will interpolate between them to provethe inequality.

First we will define the norm we are going to be working with.

Definition 5.1. For Lp([0, 2π],B([0, 2π]), λ), 1 ≤ p < ∞, where λ is the Lebesguemeasure and B([0, 2π]) is the Borel σ–algebra, define the norm

‖f‖p =(

12π

ˆ 2π

0|f |pdλ

) 1p

.

For Lp(Z,P(Z), µ), 0 ≤ p <∞, where µ is the counting measure, define the norm

‖f‖p =(∑n∈Z|f(n)|p

) 1p

and for p =∞ define‖f‖∞ = sup

n∈Z|f(n)|.

Note that for Lp(Z,P(Z), µ) this is just a special case of the regular Lp–space wedefined in an earlier section. Next, we define what we mean by the Fourier transform.

Definition 5.2. For f ∈ L1([0, 2π],B([0, 2π]), λ), we define

f̂(k) = (Ff)(k) = 12π

ˆ 2π

0f(x)e−ikxdλ(x).

This function f̂ = (Ff)(k) is, as mentioned earlier, known as the Fourier transformof f . Note that since

12π

∣∣∣∣ˆ 2π

0f(x)e−ikxdλ(x)

∣∣∣∣ ≤ 12π

ˆ 2π

0|f(x)|dλ(x) = ‖f‖1 <∞,

we have that this function is well-defined, and since f̂(k) <∞, for every k ∈ Z, we havethat ‖f̂‖∞ < ∞ so that f̂ ∈ L∞(Z,P(Z), µ). Now, for a finite measure space, such as([0, 2π],B([0, 2π]), λ), we have the following useful result.

Theorem 5.3. Let (X,A, µ) be a measure space. If µ(X) < ∞, and 1 ≤ p < q ≤ ∞,then Lq(X,A, µ) ⊆ Lp(X,A, µ), and

‖f‖p ≤ ‖f‖qµ(X)1p−

1q , for every f ∈ Lq(X,A, µ).

29

Proof. If q =∞, then

‖f‖pp =ˆ|f |pdµ =

ˆ|f |pχXdµ ≤ ‖f‖p∞

ˆχXdµ = ‖f‖p∞µ(X).

Taking the pth root yields the result.If q < ∞, then by Hölder’s inequality (Theorem 3.4) applied to the conjugate expo-

nents q/p and q/(q − p) we have that

‖f‖pp =ˆ|f |pdµ =

ˆ ∣∣|f |p∣∣dµ = ‖|f |p‖1 ≤ ‖|f |p‖ qp‖1‖ q

q−p

=( ˆ ∣∣|f |p∣∣ qp dµ

) pq(ˆ

χXdµ) q−p

q

= ‖f‖pqµ(X)q−pq .

Taking the pth root of both sides yields

‖f‖p = ‖f‖ppq µ(X)

q−pqp = ‖f‖qµ(X)

1p−

1q .

�

An immediate consequence of this is thatLp([0, 2π],B([0, 2π]), λ) ⊆ L1([0, 2π],B([0, 2π]), λ) for every p ≥ 1.

Therefore, f̂ = Ff is well-defined for every f ∈ Lp([0, 2π],B([0, 2π]), λ), p ≥ 1. Forf, g ∈ L2([0, 2π],B([0, 2π]), λ) we have that there exists an inner product defined as

〈f, g〉 = 12π

ˆ 2π

0f(x)g(x)dλ(x).

Thus, we have that

f̂(k) = 〈f, eikx〉 = 12π

ˆ 2π

0f(x)e−ikxdλ(x).

An important property of the set {eikx} is that it is an orthonormal set in L2([0, 2π],B([0, 2π]), λ).What this means is that the L2([0, 2π],B([0, 2π]), λ)–norm of eikx is 1, for every k, andthe inner product of einx and eimx is 0, whenever n 6= m.

Lemma 5.4. The set {eikx} is orthonormal in L2([0, 2π],B([0, 2π]), λ).

Proof. Firstly, we have that

‖eikx‖2 =(

12π

ˆ 2π

0|eikx|2dλ(x)

) 12

=(

12π

ˆ 2π

01dλ(x)

) 12

=(2π

2π

) 12 = 1.

Secondly, we have that for n 6= m, n,m ∈ Z,

〈eimx, einx〉 = 12π

ˆ 2π

0eimxe−inxdλ(x) = 1

2π

ˆ 2π

0ei(m−n)xdλ(x) = 1

2π (1− 1) = 0.

�

Before we prove the Hausdorff–Young inequality, there is one last theorem we need toprove. This theorem is known as Bessel’s Inequality.

30

Theorem 5.5 (Bessel’s Inequality). Let {φn} be an orthonormal set of functions suchthat φn ∈ L2([0, 2π],B([0, 2π]), λ) for every n ∈ Z. Then∑

n∈Z|〈f, φn〉|2 ≤ ‖f‖2

2.

Proof. First, note that⟨f, 〈f, φn〉φn

⟩= 〈f, φn〉〈f, φn〉 = |〈f, φn〉|2,

and that we have, for Φn = 〈f, φn〉φn,∥∥∥∥ N∑−N〈f, φn〉φn

∥∥∥∥2

2=∥∥∥∥ N∑−N

Φn∥∥∥∥2

2=∥∥Φ−N + . . .+ Φ0 + . . .+ ΦN‖2

2

=⟨Φ−N + . . .+ Φ0 + . . .+ ΦN ,Φ−N + . . .+ Φ0 + . . .+ ΦN

⟩= 〈Φ−N ,Φ−N 〉+ . . .+ 〈Φ0,Φ0〉+ . . .+ 〈ΦN ,ΦN 〉

= ‖Φ−N‖22 + . . .+ ‖Φ0‖2

2 + . . .+ ‖ΦN‖22 =

N∑−N|〈f, φn〉|2.

Thus, for N ∈ Z, we have that

0 ≤∥∥∥∥f − N∑

−N〈f, φn〉φn

∥∥∥∥2

2= ‖f‖2

2 − 2Re(⟨

f,

N∑−N〈f, φn〉φn

⟩)+∥∥∥∥ N∑−N〈f, φn〉φn

∥∥∥∥2

2

= ‖f‖22 − 2

N∑−N|〈f, φn〉|2 +

N∑−N|〈f, φn〉|2 = ‖f‖2

2 −N∑−N|〈f, φn〉|2.

Which is equivalent toN∑−N|〈f, φn〉|2 ≤ ‖f‖2

2

and taking the limit as N →∞ we have the desired result. �

In our case this inequality happens to also be an equality known as Parseval’s Identity,but we shall not prove that here, as it is not required for the following proof.

Now we have everything that we need in order to prove the Hausdorff–Young inequal-ity. The proof is fairly straightforward; we simply interpolate between L1([0, 2π],B([0, 2π]), λ)and L2([0, 2π],B([0, 2π]), λ).

Theorem 5.6 (The Hausdorff–Young Inequality). Suppose that 1 ≤ p ≤ 2 and qis the conjugate exponent of p. If f ∈ Lp([0, 2π],B([0, 2π]), λ), then f̂ ∈ Lq(Z,P(Z), µ)and

‖f̂‖q ≤ ‖f‖p.

Proof. We have that

‖f̂‖∞ = ‖Ff‖∞ = supk∈Z|f̂(k)| =

∣∣∣∣ˆ 2π

0f(x)e−ikxdλ(x)

∣∣∣∣ ≤ ˆ 2π

0|f(x)|dλ(x) = ‖f‖1.

31

That is, F : L1([0, 2π],B([0, 2π]), λ)→ L∞(Z,P(Z), µ) such that ‖f̂‖∞ ≤ ‖f‖1.By Lemma 5.4 and Bessel’s Inequality (Theorem 5.5), we have that

‖f̂‖22 = ‖Ff‖2

2 =∑k∈Z|f̂(k)|2 ≤ ‖f‖2

2.

Taking the square root yields‖f̂‖2 ≤ ‖f‖2.

That is, F : L2([0, 2π],B([0, 2π]), λ)→ L2(Z,P(Z), µ) such that ‖f̂‖2 ≤ ‖f‖2.By the Riesz-Thorin interpolation theorem (Theorem 4.1) we then have that

‖f̂‖q ≤ ‖f‖p,where 1 ≤ p ≤ 2 and q is the conjugate exponent of p. �

This means that F : Lp([0, 2π],B([0, 2π]), λ) → Lq(Z,P(Z), µ) is bounded, where1 ≤ p ≤ 2 and q is the conjugate exponent of p. This can, for example, be used to show,via the open mapping theorem (see [3] p. 162), that F : Lp([0, 2π],B([0, 2π]), λ) →Lq(Z,P(Z), µ) is not surjective.

32

6. Acknowledgements

I offer thanks to my advisor Per Åhag, who helped me get started on this essay, andwithout whose help I would never have finished it.

I would like to thank my examiner Olow Sande, for providing valuable criticism; Iwould also like to thank Aron Persson, for providing valuable criticism, even though hewas not required to.

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1999.[4] L. Gårding, Matematik och matematiker: matematiken i Sverige före 1950, Lunds Univ. Press,

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163–169.[6] O. Hölder, Über einen mittelwertsatz, Gött. Nachr. (1889), 38–47.[7] M. Lewko, Hölder’s inequality via complex analysis, Lewko’s Blog (2009). [Available at https:

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[8] J. Marcinkiewicz, Sur l’interpolation d’opérateurs, C.R. Acad. Sci. Paris 208 (1939), 1272–1273.[9] M. Riesz, Sur les maxima des formes bilinéaires et sur les fonctionnelles linéaires, Acta Math. 49

(1927), 465–497.[10] L. J. Rogers, An extension of a certain theorem in inequalities, Messenger Math. 17 (1888), 145–

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and science–third edition, Pearson Education, 2003.[12] E. M. Stein, Interpolation of linear operators, Trans. Amer. Math. Soc. 83 (1956), 482–492.[13] O. Thorin, An extension of a convexity theorem due to M. Riesz, K. Fysiogr. Sällskap. i Lund Förh.

8 (1939), no. 14.[14] , Convexity theorems generalizing those of M. Riesz and Hadamard with some applications,

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The author of this essay can be reached via e-mail at:oskar.nordenfors@gmail.com

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