3. Calculus of variations3.1 Introduction

The calculus of variations deals with functionals mostly being formed as integrals involvingan unknown function and its derivatives. For example in many physical application suchfunctionals can be interpreted as the energy of the system. In the equilibrium such energyshould be minimal among all other states. This is the reason why we are interested onminimizers of such functionals One of the simplest case is the Dirichlet integral given by

F(v) :=12

∫Ω

|∇v|2 + f vdx, v ∈W 1,20 (Ω), f ∈ L2(Ω).

As we will see below there exists exactly one u ∈W 1,20 (Ω) such that

F(u) = minF(v).

It is easy to see that u satisfies the following Poisson equation

∆u = f in Ω, u|∂Ω = 0.

The above functional is of the form

F(v) =∫Ω

f (x,v,∇v)dx,

where F is convex and bounded with F(v)→+∞ as ‖v‖W 1,20 (Ω)→+∞. As we will see below

for such functionals there exists a minimizer.

3.2 Lower semi-continuous functionals

Let X be a normed space.

Definition 3.1 1) F : M⊆X→R∪+∞ is called sequentially weakly lower semi-continuous(short: sequ. w-lsc) in x ∈ X if

∀(uk)⊂ X ,uk u ⇒ F(u) ≤ liminfF(uk).

F is called sequentially weakly lower semi-continuous (short: sequ. w-lsc) on M if F is sequ.w-lsc in each point x ∈M.

2) M ⊂ X is called weakly sequentially closed [weakly sequentially compact] if

∀(uk)⊂M, uk u ⇒ u ∈M

[∀(uk)⊂M ∃(ukl),u ∈M : ukl u.]

Remark If M is closed and convex then M is weakly sequentially closed (cf. Th. Banach-Saksfor X Hilbert space).

1

Example 1 For C ⊆ X nonempty we set

IC(u) :=

0 for u ∈C

+∞ for u ∈ X \C.

Then IC : X → R∪+∞ is sequ. w-lsc iff C is weakly sequentially closed. Indeed, assumeIC is sequ. w-lsc. Let (uk) ⊂C,u ∈ X such that uk u in X . Then IC(u) ≤ liminfuk = 0.Thus, u ∈C and hence C is weakly sequentially closed.

On the other hand, assume C is weakly sequentially closed. Let (uk)⊂ X ,u ∈ X such thatuk u in X . In case liminf IC(uk) < +∞ by the definition of IC there exists a subsequence(ukl)⊂C. Thus, u ∈C and therefore IC(u) = liminf IC(uk). In case liminf IC(uk) = +∞ thenIC(u)≤ liminf IC(uk) is trivially fulfilled.

In particular, if C is closed and convex then IC is sequ. w-lsc.

Theorem 3.2 Let M ⊂ X be weakly sequentially compact, let F : M→ R∪+∞ be sequ.w-slc on M. Then

∃u0 ∈M : F(u0) = minu∈M

.

Proof The case infF = +∞ is trivial. Assume infF 6= +∞. Let (uk) ⊂ M minimizing se-quence for F , i.e. limF(uk) = infF . Then there exists (ukl) ⊂ M,u0 ∈ M : ukl u0.Thus

infF ≤ F(u0) ≤ liminfF(ukl) = limF(uk) = infF.

Theorem 3.3 Let X be a reflexive normed space. Let F : X → R∪+∞ sequ. w-slc on X.Assume that

(3.1) F(u)→+∞ as ‖u‖→+∞.

Then

∃u0 ∈ X : F(u0) = minu∈X

.

Proof The case infF = +∞ is trivial. Assume infF 6= +∞. Let (uk) ⊂ X be a minimizingsequence for F , i.e. limF(uk) = infF . By (3.1) the sequence (uk) is bounded. Since X isreflexive the exists (ukl)⊂ X ,u0 ∈ X such that

ukl u0 in X as l→+∞.

As in the proof of Th. 3.2 we obtain F(u0) = infF .

Example 2 (Weierstraß (1870)) X =C1([−1,1]),M := u∈X |u(−1) =−1,u(1) = 1. Con-sider

F(u) :=1∫−1

(tu′(t))2dt, u ∈ X .

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• M closed, convex, unbounded;

• F continuous;

• F(u)≥ 0 ∀u ∈ X ;

• infu∈M

F(u) = 0, but there is no u ∈M with F(u) = 0.

In order to verify infu∈M

F(u) = 0 we define

uε(t) :=arctan t

ε

arctan 1ε

, ε > 0, t ∈ [−1,1].

Clearly, uε ∈M. We calculate

u′ε =ε

ε2 + t21

arctan 1ε

and

F(uε) =1

arctan 1ε

1∫−1

ε2t2

(ε2 + t2)2 dt ≤ 1arctan 1

ε

1∫−1

ε2

ε2 + t2 dt = 2ε.

Thus, limε→0

F(uε) = 0;

Assume F(u) = 0 for u ∈ X . Then u = u0 = const and therefore u 6= M.

Example 3 X = C([0,1]),‖u‖X := maxt∈[0,1]

|u(t)|. M := u ∈ X |‖u‖ ≤ 1. Consider

F(u) :=1∫

0

u(t)dt−u(1), u ∈ X .

• M closed, convex, bounded;

• F linear; continuous;

• F(u)≥ (−1)1∫

0

dt−1 =−2 ∀u∈M. Thus, infu∈M

F(u)≥−2. In fact, we have infu∈M

F(u)=

−2. For, define

uk(t) :=

−1 for t ∈ [0,1−1/k]

2kt−2k +1 for t ∈ (1−1/k,1]

Clearly, uk ∈M and limk→∞

F(uk) = −2.

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On the other hand, assume u ∈ X with F(u) =−2. Then

1∫0

u(t)dt =−2+u(1) ≤ −1 =1∫

0

(−1)dt.

It follows1∫0(u(t)+1)dt ≤ 0. Taking into account u(t)+1 is continuous and nonnega-

tive implies u≡−1. But F(−1) = 0 which is a contradiction.

3.3 The epigraph and its properties

Definition 3.4 Let F : M ⊆ X → R∪+∞. Then the epigraph of F is defined by

epi(F) := (u, t) ∈ X×R | t ≥ F(u).

Remark For a given functional F on X we set

dom(F) := u ∈ X |F(u) ∈ R.

From the definition of the epigraph it follows that epi(F)⊂ dom(F)×R

Theorem 3.5 For F : M ⊆ X → R∪+∞ the two statements are equivalent

1 F is sequ. w-slc (slc) on M;

2 epi(F) is weakly sequentially closed (closed) in M×R.

Proof 1⇒ 2. Let (uk, tk) ∈ epi(F),(u, t) ∈M×R such that

(uk, tk) (u, t) in X×R as k→+∞.

Then uk u in X and tk→ t in R. Hence F(u)≤ liminfF(uk)≤ liminf tk = t, which proves2.

2⇒ 1. Let (uk)⊂M,u ∈ X with

uk u in X as k→+∞.

Set tk := maxF(uk),−m(m∈N). Without loss of generality we may assume that liminf tk <+∞. Thus, there exists a subsequence (tkl) and t ∈ R, such that tkl → t = liminf tk =maxliminfF(uk),−m. Consequently,

(uk, tk) (u, t) in X×R as k→+∞.

From 2 it follows that (u, t)∈ epi(F), i.e. F(u)≤ t = maxliminfF(uk),−m. Letting tend−m→−∞ gives F(u) ≤ liminfF(uk).Analogously, one proves F is lsc⇔ epi(F) is closed.

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3.4 Properties of convex functionals

Definition 3.6 Let M ⊆ X be convex. F : M ⊆ X → R∪+∞ is called convex if

F((1−λ)u+λv) ≤ (1−λ)F(u)+λF(v) ∀λ ∈ [0,1], ∀u,v ∈M.

Theorem 3.7 Let M⊆X be convex. For F : M→R∪+∞ the two statements are equivalent

1 F is convex; 2 epi(F) is convex.

Proof 1⇒ 2. Let (u, t),(v,s) ∈ epi(F). For λ ∈ [0,1] we have

F((1−λ)u+λv) ≤ (1−λ)F(u)+λF(v) ≤ (1−λ)t +λs.

Thus, (1−λ)(u, t)+λ(v,s) ∈ epi(F).2⇒ 1 Let u,v ∈M,λ ∈ [0,1]. Set t := F(u) and s := F(v). Then (u, t),(v,s) ∈ epi(F) andhence (1−λ)(u, t)+λ(v,s) ∈ epi(F), which is equivalent to

F((1−λ)u+λv) ≤ (1−λ)t +λs = (1−λ)F(u)+λF(v).

Corollary 3.8 Let M ⊆ X be convex. If F : M→R∪+∞ is convex and lsc, then F is sequ.w-lsc.

Proof From Th. 3.5 we see that epi(F) is closed. In addition, by Th. 3.7 epi(F) is convex.Thus, epi(F) is sequ. weakly closed. Once more applying Th. 3.5 implies that F is sequ.w-lsc.

Theorem 3.9 Let M ⊆ X be convex. Let F : M→R∪+∞ be convex with int(epi(F)) 6= /0.Then the restriction of F on U := int(dom(F)) is locally Lipschitz continuous, i.e. for everyu ∈U there exists an open set V ⊆U containing u such that F |V is Lipschitz continuous.

Before turning to the proof of Th. 3.9 we state the following

Lemma 3.10 Let U ⊆ X be open and convex. Let F : U→R be convex with int(epi(F)) 6= /0.Given (u, t) ∈ ∂(epi(F)), there exists w∗ ∈ X∗, such that

(3.2) 〈w∗,u− v〉 ≤ s− t ∀(v,s) ∈ epi(F).

Moreover there holds t ≤ F(u).

Proof From Th. 3.7 we deduce that epi(F) is convex. Let (u, t)∈ ∂epi(F). Since int(epi(F)) 6=/0 by the aid of a theorem of Mazur (cf. Theorem 7 below) there exists g ∈ (X ×R)∗ withg 6= 0 such that

g((u− v, t− s)) ≤ 0 ∀(v,s) ∈ epi(F).

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Define f (v) := g(v,0) for v ∈ X and α := g(0,1). Obviously, f ∈ X∗ and from the aboveinequality it follows

f (u− v) ≤ α(s− t) ∀(v,s) ∈ epi(F).

For v = u and s = |F(u)|+ |t|+1 one deducec α≥ 0. If we assume α = 0 then

f (u− v)≤ 0 ∀v ∈U.

Since U is open, there exists r > 0 such that Br(u) ⊆U . Hence, f (h) ≤ 0 for all h ∈ Br(0),which implies f ≡ 0. This together with α = 0 contradicts to g 6= 0. Thus, α > 0. Settingw∗ := α−1 f gives (3.2). In addition, in (3.2) setting v = u and s = F(u) yields t ≤ F(u).

Proof of Th. 3.9 Since M is convex we have dom(F) is convex. Thus, U = int(dom(M)) isconvex 1) . Let (u, t)∈ int(epi(F)). Then there exists r > 0, such that Br(u)× (t− r, t +r)⊆epi(F). This, shows that Br(u)⊆U . Therefore we may apply Lemma 3.10 to the restrictionof F on U , which will be denoted again by F .

Let u ∈ U . First, we claim (u,F(u)) ∈ ∂epi(F). Indeed, every ball Br(u)× (F(u)−r,F(u) + r) contains (u,F(u)) ∈ epi(F) and (u,F(u)− r/2) 6∈ epi(F), which shows that(u,F(u)) ∈ ∂epi(F). Hence, by the aid of Lemma 3.10 setting t = F(u) therein there ex-ist w∗ ∈ X∗:

(3.2) 〈w∗,u− v〉 ≤ s−F(u) ∀(v,s) ∈ epi(F).

In particular, setting s = F(v) therein yields

(3.3) F(v)≥ 〈w∗,u− v〉+F(u) ∀v ∈U.

Next, we claim (u,F(u)+ 1) ∈ int(epi(F)). Otherwise, observing (u,F(u)+ 1) ∈ ∂epi(F)by Lemma 3.10 one deduces F(u)+1≤F(u), which cannot be true. Hence there exists r > 0such that

Br(u)× (F(u)+1− r,F(u)+1+ r)⊆ epi(F),

which implies together with (3.3)

(3.4) C ≤ F(v) ≤ F(u)+1 ∀v ∈ Br(u),

where C =−r‖w∗‖+F(u).Now, let v,w ∈ Br/2(u) with F(w) > F(v). Set h := w−v

‖w−v‖ and a := ‖w− v‖. Thus,w = v+ah. Since φ : τ 7→ ‖v+ τh−u‖ is continuous there exists b > 0, such that

‖v+bh−u‖ = r.

1) Let K be convex. Let x,y ∈ int(K). For λ ∈ (0,1) we set z0 := (1−λ)x+λy. There exists r > 0 such thatBr(x) ⊆ K and Br(y) ⊆ K. Set R := minλr,(1−λ)r. Then for z ∈ BR(z0) we write z = (1−λ)x′+ λy′ with

x′ :=z+z0

2 −λy1−λ

and y′ :=z+z0

2 −(1−λ)yλ

. One easily verifies ‖x− x′‖ < R2(1−λ) ≤ r and ‖y− y′‖ < R

2λ≤ r. Thus,

x′,y′ ∈ K and therefore z ∈ K. Hence, z0 ∈ int(K).

6

Using the triangular inequality gives

b−a = ‖v+bh−u− (v+ah−u)‖ ≥ ‖v+bh−u‖−‖v+ah−u‖

= r−‖w−u‖ ≥ r2.

Observing w = v+ah =(

1− ab

)v+

ab(v+bh) by means of the convexity of F one obtains

F(w) ≤(

1− ab

)F(v)+

ab

F(v+bh).

Observing (3.4) gives

F(w)−F(v) ≤ ab(F(v+bh)−F(v))≤ 2

r(F(u)+1−C)‖w− v‖.

Theorem 3.11 Let M ⊆ X be convex. Let F : M→ R∪+∞ be convex. Then the followingstatements are equivalent

1 int(epi(F)) 6= /0;

2 ∃u ∈ int(dom(F)) such that F is continuous in u;

3 ∃u ∈ intM and r > 0 such that Br(u)⊆M and supv∈Br(u)

F(v) < +∞.

Proof 1⇒ 2 follows from T h.3.9 and 2⇒ 3 follows from the definition of the continuity.It remains to show that 3⇒ 1. Indeed, from 3 it follows that t ≥C≥ F(v) for all v∈ Br(u)and t ∈ (C,C +2). Thus,

Br(u)× (C,C +2)⊂ epi(F),

which shows that (u,C +1) is an inner point of epi(F).

Remark According to Th. 3.11 the assumption int(dom(F)) in Th. 3.9 can be replaced by theassumption that there exists u ∈ int(dom(F)), such that F is continuous in u. On the otherhand, if dim(X) = +∞ this condition is necessary since there exists a convex functionalbeing everywhere discontinuous. Namely, if aii∈I denotes a Hamel basis of X , the set I isinfinite. Thus, there exists a countable set A = i1, i2, . . . ⊆ I. Then we define

f (ain) = n‖ain‖, n ∈ N, f (ai) = 0, i ∈ I \A.

For every x∈ X there exists λi ∈R(i∈ I) with λi = 0 except finite i∈ I, such that x = ∑i∈I

λiai.

In addition, the numbers λi are unique. Then we define

f (x) = ∑i∈I

λi f (ai).

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Then f is an unbounded linear functional, which is also convex and is discontinuous in everypoint.

Theorem 3.12 Let X be reflexive. Let M ⊆ X be closed and convex. Let F : M→ R∪+∞be convex, such that∀(uk)⊂ dom(F) with ‖uk‖→+∞ or dist(uk,∂(dom(F))→ 0

there holds F(uk)→+∞.

If int(epi(F)) 6= /0, then there exists u ∈M, such that F(u) = minv∈M

F(v). In addition, if F is

strictly convex, then the minimal element u is unique.

Proof Without loss of generality we may assume that F 6≡ +∞. Then let (uk) ⊂ dom(F),such that lim

k→∞F(uk) = inf

v∈MF(v). From our assumption it follows that (uk) is bounded. By

virtue of the reflexivity eventually passing to a subsequence there exists u ∈M such that

uk u in X as k→+∞.

Set K := convu1,u2, . . .. Then u ∈ K. Since K ⊂ dom(F) we have u ∈ dom(F). We claimK ⊆ int(dom(F)). Assume this is not true. Let w ∈ K \ int(dom(F)). Then w ∈ ∂dom(F)and there exists a sequence (vn) ⊂ K such that vn → w in X . By the assumption of thetheorem this implies F(vn)→ +∞, which contradicts to F |K is bounded from above 2) .

Hence, by Th. 3.9 G := F |K is continuous. Thus epi(G) is closed and convex and byTh. 3.7 epi(G) is sequ. weakly closed. Therefore, G is sequ. w-lsc on K. This provesF(u)≤ liminfF(uk) = inf

v∈MF(v)≤ F(u).

Let F be strictly convex. Assume there are u1,u2 ∈ M,u1 6= u2 with F(u1) = F(u2) =

infv∈M

F(v). Then F(u1 +u2

2

)<

F(u1)+F(u2)2

= infv∈M

F(v) which is a contradiction.

3.5 Differentiability

Definition 3.13 F : X → R is called in u ∈ X directional differentiable if for all h ∈ X thereexists

DF(u;h) := limt→0

F(u+ th)−F(u)t

in R.

Remark Clearly, D(u;λh) = λD(u;h) ∀λ ∈ R, however it is possible that

D(u;h1 +h2) 6= D(u;h1) + D(u;h2).

2) Notice, c0 := supF(uk) < +∞. Let v ∈ K with v =m∑

i=1λiuki (λi ≥ 0,

m∑

i=1λi = 1). By the convexity of F one

estimates F(v)≤m∑

i=1λiF(uki) ≤ c0.

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Definition 3.14 F : X → R is called GATEAUX differentiable in u ∈ X if F is in u direc-tional differentiable and DF(u; ·) is linear and bounded, i.e. DF(u; ·) ∈ X∗. Then we writegradF(u) instead of DF(u; ·).

F : X → R is called GATEAUX differentiable on M ⊆ X if F is GATEAUX differentiablein every u ∈M. Then gradF : u ∈M 7→ gradF(u) is called the GATEAUX derivative or thegradient of F on M.

Example 4 Let X = H be a Hilbert space with scalar product (·, ·). Let T ∈L (H,H). Con-sider, F(u) := (Tu,u),u ∈ H. Then F is Gateaux differentiable on X . Indeed, for u,h ∈ Hone gets

F(u+ th)−F(u)t

=(T (u+ th),u+ th)− (Tu,u)

t

=t(Tu,h)+ t(T h,u)+ t2(T h,h)

t= (Tu,h)+(T h,u)+ t(T h,h)→ (Tu,h)+(T h,u) as t→ 0.

Thus, DF(u;h) = (Tu,h)+(T h,u) = (Tu+T ∗u,h). Since h 7→ (Tu+T ∗u,h) belongs to X∗

we have 〈gradF(u),h〉 = (Tu+T ∗u,h) for all u,h ∈ H.Especially, for T = id we have F(u) = ‖u‖2 and gradF(u) = 2u.

Next, Consider G(u) := ‖u‖,u ∈ H. Then, ‖ · ‖2 = G2 and

2(u,h) = limt→0

G2(u+ th)−G2(u)t

= limt→0

(G(u+ th)+G(u))G(u+ th)−G(u)

t

For ‖u‖ 6= 0 the limit DG(u; ·) exists, and there holds

2(u,h) = 2‖u‖DG(u; ·) ⇐⇒ DG(u;h) =( u‖u‖

,h)∀h ∈ H.

Example 5 Let Ω be an open set. Let X = Lp(Ω)(1 < p < ∞). We consider F( f ) = ‖ f‖pp, f ∈

Lp(Ω). Then there holds

〈gradF( f ),h〉 = p∫Ω

sign( f )| f |p−1hdx, u,h ∈ Lp(Ω).

Proof Set φ(s) = |s|p,s ∈ R. Then φ ∈C1(R) with

φ′(s) = psign(s)|s|p−1 ∀s ∈ R.

Let f ,h ∈ Lp(Ω). For a. e. x ∈Ω we calculate

limt→0

| f (x)+ th(x)|p−| f (x)|p

t= φ

′( f (x))h(x)

= psign( f (x))| f (x)|p−1h(x).

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On the other hand, using the mean value theorem there exists θ ∈ [0,1], such that∣∣∣ | f (x)+ th(x)|p−| f (x)|p

t

∣∣∣ ≤ |φ′( f (x)+θth(x))||h(x)|

= p| f (x)+θth(x)|p−1|h(x)|≤ p2p−1(| f (x)|p−1|h(x)|+ |h(x)|p)

for a. e. x ∈Ω, for all t ∈ (0,1). Since | f |p−1|h|+ |h|p ∈ L1(Ω) we are in a position to applyLebegue’s theorem of dominated convergence to get

limt→0

F( f + th)−F( f )t

= limt→0

1t

∫Ω

| f + th|p−| f |pdx = p∫Ω

sign( f )| f |p−1hdx.

We consider G( f ) := ‖ f‖p. Let f ,h ∈ Lp(Ω) with f 6= 0. Then having F( f ) = Gp( f ) yields

〈grad( f ),h〉 = limt→0

Gp( f + th)−Gp( f )t

= limt→0

Gp( f + th)−Gp( f )G( f + th)−G( f )

· G( f + th)−G( f )t

.

Verifying that

limt→0

Gp( f + th)−Gp( f )G( f + th)−G( f )

= p‖ f‖p−1p 6= 0

it follows that limt→0

G( f + th)−G( f )t

= DG( f ;h) exists and there holds

〈grad( f ),h〉 = p‖ f‖p−1p DG(u;h).

This shows that

DG( f ;h) = 〈gradG( f ),h〉 = ‖ f‖1−p∫Ω

sign( f )| f |p−1hdx.

Analogously, one proves

〈J( f ),h〉 = 〈grad12‖ f‖2,h〉 = ‖ f‖2−p

p

∫Ω

sign( f )| f |p−1hdx.

Theorem 3.15 Let F : X → R be directional differentiable on X. Let U ⊆ X be open andu ∈U such that

F(u) = minv∈U

F(v).

Then DF(u; ·) = 0.

10

Proof Let r > 0 such that Br(u) ⊆ U . Let h ∈ Br(0). Set φ(t) := F(u + th), t ∈ (−1,1).Clearly, F(u) = ϕ(0) = min

t∈(−1,1)ϕ(t). Thus, DF(u;h) = ϕ′(0) = 0. Whence, DF(u; ·) = 0.

Theorem 3.16 Let F : X→R be Gateaux differentiable on X. Then the following statementsare equivalent.

1 F is convex;

2 gradF : X → X∗ is monotone;

3 F(v)≥ F(u)+ 〈gradF(u),v−u〉 ∀u,v ∈ X.

Proof 1⇒ 2. Let u,v∈X . Set h := u−v. Define ϕ(t) := F(v+th), t ∈R. Since F is convexϕ is differentiable and convex on R. This implies ϕ′ is non decreasing. Consequently,

ϕ′(1) = 〈gradF(u),h〉 ≥ ϕ

′(0) = 〈gradF(v),h〉

which implies

〈gradF(u)−gradF(v),u− v〉 ≥ 0 ∀u,v ∈ X .

2⇒ 3. Let u,v∈ X . Set h := v−u and ϕ(t) := F(u+ th), t ∈R. Then we have from gradFbeing monotone one gets

ϕ(1) = ϕ(0) +1∫

0

ϕ′(t)dt = F(v) +

1∫0

〈gradF(u+ th),h〉dt

= F(u)+ 〈gradF(u),v−u〉 +1∫

0

〈gradF(u+ th)−gradF(u),h〉dt

≥ F(u)+ 〈gradF(u),v−u〉.

Whence 3.3⇒ 1. Let w,v ∈ X and λ ∈ [0,1]. Set h := v−w. Then applying 3 with u = w + λh =v+(λ−1)h, first setting v = w and then v = v yields

F(w)≥ F(w+λh)−λ〈gradF(w+λh),h〉,F(v)≥ F(w+λh)+(1−λ)〈gradF(w+λh),h〉.

Multiplying the first inequality by 1−λ, the second by λ and taking the sum of both yields

(1−λ)F(w)+λF(v)≥ F(w+λh) = F((1−λ)w+λv).

Definiiton 3.17 Let F : X→R∪+∞. Let u∈ dom(F). Then w∗ ∈ X∗ is called subgradientof F in u, if

F(v)≥ F(u)+ 〈w∗,v−u〉 ∀v ∈ X ;

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By ∂F(u) we denote the set of all subgradients of F in u and the mapping u 7→ ∂F(u) iscalled the subdifferential of F .

F is called subdifferentiable in u ∈ dom(F) if ∂F(u) 6= /0.

Example 6 X = R, F(x) = |x|,x ∈ R. Then

∂F(x) =

1∗ for x > 0

α∗ | −1≤ α≤ 1 for x = 0

−1∗ for x < 0

where 〈α∗,x〉 := α · x,α,x ∈ R.

Example 7 Let C ⊂ X be a convex set with int(C) 6= /0. For the indicator functional IC wehave dom(F) = C. For u ∈C there holds

∂IC(u) = w∗ ∈ X∗ | 〈w∗,v−u〉 ≤ 0 ∀v ∈C.

In particular ∂IC(u) = 0 for every u ∈ int(C). If u ∈ ∂C by a well known theorem ofMazur (cf. Th. 7 below) there exists w∗ ∈ X∗ such that 〈w∗,u− v〉 ≤ 0 for all v ∈C. Thus,w∗ ∈ ∂IC(u). This shows that IC is subdifferentiable on C.

Example 8 Let F : X → R be convex and GATEAUX differentiable in u ∈ X . Then

∂F(u) = gradF(u).

Clearly, by Th. 3.16; 3 we have gradF(u) ∈ ∂F(u). Let w∗ ∈ ∂F(u). Let h ∈ X . Then wehave

F(u+ th)−F(u)≥ t〈w∗,h〉 ∀ t ∈ R.

Now, dividing both sides by t > 0 and letting tend t→ 0 gives

〈gradF(u),h〉 ≥ 〈w∗,h〉.

Replacing h by −h one gets the opposite inequality, which proves the assertion.

Lemma 3.18 (Properties of ∂F) Let F : X → R∪+∞. Then

1. ∂F(u) is convex and sequ. weakly*- closed,

2. F subdifferential in u ∈ X ⇒ F is sequ. w-lsc. in u,

3. F subdifferential in u ∈ X ⇒ F has global minimum in u iff 0 ∈ ∂F(u),

4. 〈w∗1−w∗2,u1−u2〉 ≥ 0 ∀w∗i ∈ ∂F(ui) (i = 1,2),

i.e. ∂F : X → 2X∗ is a monotone map.

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Proof 1. The convexity of ∂F(u) is obvious. Let (w∗k)⊂ ∂F(u) such that

w∗k∗→ w∗ in X∗ as k→+∞.

Then for v ∈ X we have

〈w∗,v−u〉 = limk→∞〈w∗k ,v−u〉 ≤ F(v)−F(u).

Consequently, w∗ ∈ ∂F(u).

2. Let w∗ ∈ ∂F(u). Let (uk)⊂ X with uk u.

F(u)≤ F(uk)+ 〈w∗,u−uk〉 ∀k ∈ N.

Since liminf〈w∗,u−uk〉= 0 one gets F(u)≤ liminfF(uk).

3. Let w∗ ∈ ∂F(u). Clearly, F(u) = minF implies 0 ∈ ∂F(u). Assume 0 ∈ ∂F(u). Then bythe definition of ∂F(u) it follows F(v)≥ F(u) for all v ∈ X which proves F(u) = minF .

4. Let w∗i ∈ ∂F(ui)(i = 1,2). Then from the definition of ∂F one gets

F(u2)−F(u1)≥ 〈w∗1,u2−u1〉F(u1)−F(u2)≥ 〈w∗2,u1−u2〉.

Taking the sum of both inequalities implies 0 ≥ 〈w∗1−w∗2,u2− u1〉, which is equivalent tothe assertion.

Theorem 3.19 Let F : X → R∪+∞ be convex with int(epi(F)) 6= /0. Then

∂F(u) 6= /0 ∀u ∈ int(dom(F)).

Proof Let u ∈ int(dom(F)). As we have seen in the proof of Th. 3.9, (u,F(u)) ∈ ∂epi(F).Applying Lemma 3.10 with (u, t) = (u,F(u)) one gets a functional w∗ ∈ X∗ such that

F(v)≥ F(u) + 〈w∗,v−u〉 ∀v ∈ int(dom(F)).

Since u ∈ int(dom(F)) there exists r > 0 such that Br(u)⊆ int(dom(F)), which implies

F(u+h)≥ F(u) + 〈w∗,h〉 ∀h ∈ Br(0).

For h ∈ Br(0) we define, Φ(t) := F(u + th)−F(u)− t〈w∗,h〉, t ∈ R. Clearly, Φ : R→ R∪+∞ is convex with

Φ(t)≥ 0 ∀ t ∈ (−1,1), Φ(0) = 0.

Assume Φ(t) < 0 for t ∈ R. Then by the convexity of Φ we have for every λ ∈ (0,1)

Φ(λt) = Φ((1−λ)0+λt) ≤ (1−λ)Φ(0)+λφ(t) < 0,

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which is a contradiction. Thus, Φ(t)≥ 0 for all t ∈ R and therefore

F(u+ th)≥ F(u) + 〈w∗, t h〉 ∀ t ∈ R.

Whence, w∗ ∈ ∂F(u).

Definition 3.20 (Duality map) Let u ∈ X . We define the duality map J : X → 2X∗ by

J(u) := w∗ ∈ X∗ |‖w∗‖2∗ = ‖u‖2 = 〈w∗,u〉.

Remark Thanks to Hahn-Banach’s theorem we have J(u) 6= /0. It is also readily seen that

〈w∗1−w∗2,u1−u2〉 = ‖u1‖2 +‖u2‖2−〈w∗1,u2〉−〈w∗2,u1〉 ≥ (‖u1‖−‖u2‖)2

for all ui ∈ X ,w∗i ∈ J(ui)(i = 1,2). Hence, J is strictly monotone.

Theorem 3.21 For every u ∈ X the set J(u) equals to the subgradient of the FunctionalF : v 7→ 1

2‖v‖2 in u.

Proof Let w∗ ∈ ∂F(u), i.e.

(1)12‖v‖2 ≥ 1

2‖u‖2 + 〈w∗,v−u〉, ∀v ∈ X .

Let h ∈ X ,‖h‖= 1. Into (1) inserting v = u+ th, t > 0 one obtains

〈w∗, th〉 ≤ 12(‖u+ th‖2−‖u‖2) =

12(‖u+ th‖+‖u‖)(‖u+ th‖−‖u‖)

≤ t2(‖u+ th‖+‖u‖).(2)

Dividing both sides by t and letting tend t→ 0 gives

〈w∗,h〉 ≤ ‖u‖ =⇒ ‖w∗‖∗ = suph∈X ,‖h=1‖

〈w∗,h〉 ≤ ‖u‖.

Next, in (1) setting v = tu(t ∈ R) one gets

(t−1)〈w∗,u〉 ≤ t2−12‖u‖2.

For t < 1 dividing both sides by t−1 gives 〈w∗,u〉 ≥ t +12‖u‖2. Letting tend t → 1 shows

that

〈w∗,u〉 ≥ ‖u‖2 =⇒ ‖w∗‖∗ ≥ ‖u‖.

This implies, that ‖w∗‖∗ = ‖u‖. Furthermore, estimating 〈w∗,u〉 ≤ ‖w∗‖∗‖u‖= ‖u‖2 yields〈w∗,u〉= ‖u‖2 = ‖w∗‖2

∗, i.e. w∗ ∈ J(u).

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On the other hand, let w∗ ∈ J(u). Then for every v ∈ X

〈w∗,v−u〉 ≤ ‖u‖‖v‖−‖u‖2 = −12(‖v‖−‖u‖)2 +

‖v‖2

2− ‖u‖

2

2

≤ ‖v‖2

2− ‖u‖

2

2.

Thus, w∗ ∈ ∂F(u).

3.6 Functionals on W 1, p(Ω)(1 < p < ∞)

Let Ω⊂ Rn open, bounded.

Theorem (SCORZA-DRAGONI) The following statements are equivalent

1 f : Ω×R×Rn→ R is a Caratheodory function;

2 ∀K ⊂Ω compact, ∀ε > 0 ∃Kε ⊂ K, such that

|K \Kε| ≤ ε, f |Kε×Rm×RN is continuous.

(Ekeland/Temam; S. 218-219)

Let f : Ω×RN×RnN be a Caratheodory function. Set

F(u) :=

∫Ω

f (x,u,∇u)dx if f (·,u,∇u) ∈ L1(Ω)

+∞ else.

For each α ∈ R the set

Fα =

u ∈W 1, p(Ω) |F(u)≤ α

is closed.

Theorem 2.1 (SERRIN) Let f : Rn→ R be convex, continuous, f (ξ)≥ 0 for all ξ ∈ Rn. Letu,uk ∈W 1,1(Ω)(k ∈ N) with

uk u in L1(Ω) as k→ ∞.

Then there holds

I(u;Ω) :=∫Ω

f (∇u)dx ≤ liminf I(uk;Ω).

Proof Let ωρ ∈C∞(Rn)(ρ > 0) denote the usual mollifying kernel satisfying

(i) supp(ωρ)⊆ Bρ,

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(ii) 0≤ ωρ(x) ≤ c0ρ−n (c0 = const > 0),

(iii)∫Rn

ωρ(x)dx = 1.

Fix Ω′ ⊂⊂Ω. Set d0 = d(Ω′,∂Ω) > 0. Then for 0 < ρ < d0 we define

Iρ(v;Ω′) :=

∫Ω′

f (∇vρ)dx, v ∈ L1(Ω).

Since

| f ((∇v)ρ(x)| =∣∣∣∣∫Rn

ωρ(x− y)∇v(y)dy∣∣∣∣ ≤ c1ρ

−n−1‖v‖L1

the functional Iρ : L1(Ω)→ R is convex and bounded. By Th. 3.9 and Th. 3.7 this functionalis sequ. w-lsc, which shows that

Iρ(u;Ω′) ≤ liminf Iρ(uk;Ω

′).

On the other hand, by Jensen’s inequality one finds

f ((∇uk)ρ(x)) = f(∫

Rn

ωρ(x− y)∇uk(y)dy)

= f(

1µρ,x(Rn)

∫Rn

∇uk(y)dµρ,x

)≤ 1

µρ,x(Rn)

∫Rn

f (∇uk)dµρ,x =∫Ω

ωρ(x− y) f (∇uk)dy,

for a. e. x ∈Ω′, where

µρ,x(A) :=∫A

ωρ(x− y)dy, A⊆ Rn (A Lebesgue mesurable).

Thus, applying Fubini’s theorem gives∫Ω′

f ((∇uk)ρ(x))dx ≤∫Ω′

(∫Ω

ωρ(x− y) f (∇uk(y))dy)

dx

=∫Ω

∫Ω′

ωρ(x− y)dx f (∇uk(y))dy

≤∫Ω

∫Rn

ωρ(x)dx f (∇uk(y))dy =∫Ω

f (∇uk)dy.

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Consequently,

Iρ(u;Ω′) ≤ liminf

k→∞

∫Ω′

f ((∇uk)ρ(x))dx

≤ liminfk→∞

∫Ω

f (∇uk)dx = liminfk→∞

I(uk;Ω).

By Fatou’s Lemma one gets

I(u;Ω′) =

∫Ω′

liminfρ→0

f ((∇u)ρ)dx

≤ liminfρ→0

∫Ω′

f ((∇u)ρ)dx ≤ liminfk→∞

I(uk;Ω).

Finally, by virtue of monotone convergence (Ω′→Ω) one arrives at

I(u;Ω) ≤ liminfk→∞

I(uk;Ω).

Theorem 2.2 Let f : Ω×R×Rn→ R be a Caratheodory function, such that

(1) −c0 ≤ f (x,u,ξ) ≤ c1(1+ |ξ|m) for all (x,u,ξ) ∈Ω×R×Rn

(m > 1;c0,c1 = const≥ 0),

(2) | f (x,u,ξ)− f (x,v,ξ)| ≤ ω(|u− v|)(1+ |ξ|)m for all x ∈Ω,u,v ∈ R,ξ ∈ Rn,

where ω : [0,∞)→ [0,∞) with ω(τ)→ 0 as τ→ 0

(3) ξ 7→ f (x,u,ξ) is convex for all (x,u) ∈Ω×R,

(4) I(u;Ω)→+∞ as ‖u‖W 1,m(Ω)→+∞.

Let W 1,m0 (Ω)⊆V ⊆W 1,m(Ω) be a closed subspace. Then there exists u ∈V such that

I(u;Ω) = minv∈V

I(v;Ω),

where

I(v;Ω) =∫Ω

f (x,v,∇v)dx, v ∈W 1,m0 (Ω).

Proof Let α := infv∈V F(v). By (1) we have α ∈ R. Let uk ∈V such that

F(uk)→ α .

From (4) one deduces that (uk) is bounded in W 1,m(Ω). By means of the reflexivity of V andRiesz-Fischer’s theorem there exists u ∈V such that (eventually passing to a subsequence)

uk u in V,

uk→ u a. e. in Ω as k→+∞.

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Let ε > 0 be arbitrarily chosen. By Egorov’s theorem, there exists a measurable set Eε ⊂ Ω

with |Ω\Eε| ≤ ε and

uk|Eε→ u|Eε uniformly in as k→+∞.

We define

F(v) :=∫Eε

f (x,u,∇v)dx, v ∈V.

Then F is convex and bounded. In particular, F is sequ. w-slc. Hence,

I(u;Eε) = F(u;Eε) ≤ liminfk→∞

F(uk;Eε) = liminfk→∞

∫Eε

f (x,u,∇uk)dx

≤ liminfk→∞

∫Eε

f (x,uk,(∇uk)ρ)dx.

By (3) we get

f (x,uk,∇uk) = f (x,uk,∇uk)− f (x,u,∇uk)+ f (x,u,∇uk)≥ f (x,u,∇uk)−ω(|uk−u|)(1+ |∇uk|m).

In addition,

limk→∞

∫Eε

ω(|uk−u|)(1+ |∇uk|m)dx = 0

Thus,

α≥ liminfk→∞

∫Eε

f (x,uk,∇uk)dx≥ liminfk→∞

∫Eε

f (x,u,∇uk)dx≥ I(u;Eε).

Hence

I(u;Ω) = I(u;Eε)+ I(u;Ω\Eε) ≤ α+ I(u;Ω\Eε)→ α as ε→ 0.

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Geometric version of Hahn-Banachs theorem

Theorem 1 Let p :→ [0,∞) be a functional with

(i) p(x+ y) ≤ p(x)+ p(y) ∀x,y ∈ X,

(ii) p(tx) = t p(x) ∀x ∈ X ,∀ t ≥ 0.

Let X0 ⊂ X a linear subspace of X and f0 : X0→R a linear functional with f0(x) ≤ p(x) forall x ∈ X0. Then there exists a linear functional f : X → R such that

f |X0 = f0, f (x) ≤ p(x) ∀x ∈ X .

(For the proof cf. K. Yosida; Functional Analysis).

Definition 2 Let C ⊆ X be a convex, open set with 0 ∈C. Define

pC(x) := inf1

t

∣∣∣ tx ∈C, t > 0

The functional pC : X → [0,∞) is called the Minkowski functional for C.

Lemma 3 Let C ⊂ X be convex and open. Then the functional pC satisfies the properties (i)and (ii) of Theorem 1.

Proof First we prove that pC(x) < +∞ for all x∈X . Indeed, letting x∈X because 0∈C and Cis open there exists t > 0 such that tx∈C, which shows that pC(x)≤ 1

t . Next, let ε > 0 and letx,y∈ X . Take s > 0 such that sx∈C, 1

s ≤ pC(x)+ε and t > 0 such that ty∈C, 1t ≤ pC(y)+ε.

Then by means of the convexity of C there holds

11s + 1

t

(x+ y) =t

s+ tsx +

ss+ t

ty ∈C.

Thus, pC(x + y) ≤ 1s + 1

t ≤ pC(x)+ pC(y)+ 2ε. Whence (i). Next, let ε > 0 and let x ∈ Xand t > 0. Let s > 0 such that stx ∈C and pC(tx)1

s ≤ pC(tx)+ ε. Then

p(x) ≤ 1st

= (pC(tx)+ ε)1t.

which implies t p(x) ≤ p(tx). Replacing x by y = tx and t by τ = 1t yields

p(tx) = p(y) ≤ 1τ

p(τy) = t p(x).

This implies (ii).

Lemma 4 Let C ⊂ X be convex and open. Let f : X →R be linear with f (x) ≤ pC(x) for allx ∈ X. Then f ∈ X∗.

19

Proof Since 0 ∈C and Cis open there exists r > 0 such that Br(0) ⊆C. Therefore for everyx ∈ X with ‖x‖ ≤ 1 there holds r

2x ∈C. Thus,

f (x) ≤ pC(x) ≤ 2r∀x ∈ X with ‖x‖ ≤ 1.

This shows that f is bounded and therefore f ∈ X∗.

Theorem 5 Let A,B ⊆ X convex with A∩B = /0, where A is open. Then there exists f ∈ X∗

such that

f (a) < f (b) ∀a ∈ A, ∀b ∈ B.

Proof Let a0 ∈ A and b0 ∈ B. Set x0 := b0−a0. Define

C := A−B+ x0 = a−b+ x0 |a ∈ A,b ∈ B.

Clearly, C is open and convex with 0 ∈C. Set X0 = linx0 and f0 : X0→ R by

f0(tx0) := t, t ∈ R.

It is easy to see that f0(tx0)≤ pC(tx0) for all t ∈R. Indeed, for t ≤ 0 the inequality is triviallyfulfilled since pC ≥ 0. On the other hand, observing x0 6∈C we have pC(x0) > 1. Thus, forall t > 0

f0(tx0) = t < t pC(x0) = pC(tx0).

According to Theorem 1 along with Lemma 4 there exists f ∈ X∗ such that

f (x) ≤ pC(x) ∀x ∈ X .

Thus,

f (a−b+ x0) = f (a)− f (b)+1 ≤ p(a−b+ x0)≤ 1 ∀a ∈ A, ∀b ∈ B.

which proves the assertion.

Theorem 6 Let C⊂ X be convex set. Let u ∈ X \C. Then there exists f ∈ X∗ with f 6= 0 suchthat

〈 f ,u− v〉 ≤ 0 ∀v ∈C.

Proof There exists an open ball B = Br(u) ⊆ X \C. Let v0 ∈ C. Set U := B−C + v0− u.Clearly, U is an open and convex set with 0 ∈U . Set X0 = t(v0−u) | t ∈ R. Define

〈 f0, t(v0−u)〉 = t, t ∈ R.

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According to Theorem 1 along with Lemma 4 there exists f ∈ X∗, such that 〈 f ,x〉 ≤ pU(x)for all x ∈ X . Thus,

〈 f ,b− v+ v0−u〉 = 〈 f ,b− v〉+1 ≤ pU(b− v+ v0−u) ≤ 1 ∀v ∈C,b ∈ B.

In particular, setting b = u therein one gets

〈 f ,u− v〉 ≤ 0 ∀v ∈C.

Theorem 7 Let C ⊂ X be a closed convex set with int(C) 6= /0. Then for every u ∈ ∂C thereexists f ∈ X∗, f 6= 0, such that

〈 f ,u− v〉 ≤ 0 ∀v ∈C.

Proof Let u ∈ ∂C. There exists a sequence (uk) ⊂ X \C with uk → u as k→ ∞. ApplyingTheorem 6 for every k ∈ N one gets a functional fk ∈ X∗ with ‖ fk‖∗ = 1, such that

〈 fk,u− v〉 ≤ 0 ∀v ∈C.

By Tychonov’s theorem the ball w∗|‖w∗‖∗ ≤ 1 is weakly* compact in X∗. Thus, thereexists a subsequence ( fk j)⊂ X∗, f ∈ X∗, such that

fk j∗ f in X∗ as j→+∞.

This shows that for every v ∈C there holds

〈 f ,u− v〉 = limj→∞〈 fk j ,uk j − v〉 ≤ 0.

Thus, it only remains to verify that f 6= 0. By our assumption there exists w ∈C and r > 0such that Br(w)⊂C. Let h ∈ X with ‖h‖= 1. Then w+ rh ∈C. For every j ∈ N we have

〈 fk j ,w−uk j〉 ≥ −r〈 fk j ,h〉 =⇒ 〈 fk j ,w−uk j〉 ≥ r ∀ j ∈ N.

This yields 〈 f ,w−u〉 ≥ r and therefore f 6= 0.

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