Chapter 4
Properties of the metric projection
In the focus of this chapter are continuity and differentiability properties of the metric projections. We mainly restrict us to the case of a projection onto Chebyshev sets. First of all, we study geometric properties of Banach spaces which are helpful to analyze metric projections. In the focus of the geometric view on Banach spaces is the property of smoothness. For the material on geometric Banach space theory we refer to Borwein and Lewis [7], Chidume [10], Cioranescu [11], Deimling [14], Kabalo [24], Hiriart-Urruty and Lemarechal [22, 21], Megginson [30] and Werner [41].
4.1 Uniform convexity
Next we consider a class of Banach spaces which share with Hilbert spaces the following property: their unit ball B is rotund“ and touches“ convex sets just in one point. 1 ” ” We may say that the following considerations are devoted to the geometric properties of Banach spaces.
Definition 4.1. A Banach space is uniformly convex (or strongly rotund) if and X only if its modulus of convexityδ defined by X 1 δ (ε) := inf{1− x+y : x = y = 1, x−y =ε},ε [0, 2], X 2� � � � � � � � ∈ is positive for allε (0, 2]. ∈ � Clearly, a uniformly convex space is strictly convex. Let us collect some properties of the modulusδ . X Lemma 4.2. Let be a Banach space with modulus of convexityδ . Then we have: X X (1)δ (0) = 0,δ (ε) 1 ε,ε 0. X X ≤ 2 ≥ (2)(0, 2] ε δ (ε)ε−1 R is non-decreasing. � �− X ∈ (3) If is a Hilbert space thenδ (ε) =1− 1− 1 ε2,ε (0, 2]. X → H H 4 ∈ �
78 Proof: Ad(1) Evidentely,δ (0) =0. Letε [0, 2] and let x, y S with x−y =ε. Then X ∈ ∈ 1 � � 1 1 1 1 x+y = x+ (y−x) x − y−x 1− ε. 2� � � 2 � ≥ � � 2� � ≥ 2 Ad(2) Let0<η ε 2. Choose x, y S with x−y =ε. Set ≤ ≤ ∈ 1 � � η η x+y η η x+y u := x+(1− ) , w := y+(1− ) . ε ε x+y ε ε x+y � � � � Then η 1 1 η 1 η x+y u−w = x−y =η, (u+w) = (x+y) + (1− ) � � ε � � 2 2 ε 2 ε x+y � � and x+y 1 η 1 η η 1 η 1 − (u+w) = − x+y =1−(1− + x+y ) =1− u+v . � x+y 2 � ε 2 ε � � ε 2 ε � � 2� � � � Notice that x+y 1 1 − (x+y) =1− x+y . � x+y 2 � 2� � � � Therefore x+y 1 − (u+v) � x+y 2 � η 1 η � � = ( − x+y )/η u−v ε 2 ε � � � � x+y 1 − (x+y) 1 � x+y 2 � = (1− x+y )/ε= � � . 2� � x−y � � Now, we obtain
1 x+y 1 1− u+v − (u+v) δ (η) 2 � x+y 2 � X � � = � � η ≤ u−v u−v � � � � x+y 1 − (x+y) 1 1 � x+y 2 � 1− x+y 1− x+y = � � = 2� � = 2� � x−y x−y ε � � � � Taking the infimum with respect to the choice of x, y we obtain the assertion. Ad(3) Follows from the parallelogram identity. � Remark 4.3. Letδ be the modulus of convexity in the Banach space . There exists a X X constant 0 < L < 3.18 such that if0<η ε 2 then we have (see [16]) ≤ ≤ δ (η) δ (ε) X 4L X η2 ≤ ε2
�
79 Theorem 4.4. A uniformly convex Banach space is an E-space. X Proof: We know already that is strictly convex. To show the reflexivity of we may use the X X criterion of James (see the preliminaries in the preface). To apply this criterion we have to show that for each functionalλ ∗ with λ =1 there existsx S1 with λ, x =1. ∈X � � n ∈ � n� Letλ ∗ with λ =1. Then there exists a sequence(x )n N in S1 with limn λ, x =1. ∈X � � ∈ � � This implies lim xn =1 and from lim λ, xn +x m =1 we obtain n � � m,n� � 1 lim (xn +x m) =1. n,m �2 � n Due to the uniform convexity, the sequence(x )n N converges to somex S1 . Clearly, ∈ ∈ λ, x =1= λ . � � � � To prove the third property of an E-space, letx S1 and letx be the weak limit of the n ∈ sequence(x )n N in S1 . Assume thatx is not the strong limit of this sequence. Then ∈ nk nk there exists a subsequence(x )k N andε>0 such that x −x ε,k N. By the ∈ � � ≥ ∈ uniform convexity there exists ad>0 such that
1 n (x k +x) 1− d, k N. �2 � ≤ ∈ Letλ Σ(x). Then ∈ 1 1= λ, x = lim λ, (xnk +x) λ (1−d)<1 � � k � 2 � ≤ � � This is a contradiction. � Theorem 4.5. Let be a uniformly convex Banach space. Then every nonempty convex X closed subsetC of is a Chebyshev set. Moreover, the best approximation problem (2.1) X is strongly solvable. Proof: This follows from Theorem 4.4 and Theorem 2.24. � Remark 4.6. Notice that the property of uniform convexity is not stable under equivalent renorming the space. � Theorem 4.7 (Milman). A uniformly convex Banach space is reflexive. Proof: Follows from Theorem 4.4. � For concrete examples of uniformly convex spaces one can show the reflexivity ad hoc. For instance, the spacesl p, 1 < p < , are uniformly convex and reflexive. The modulus of convexity is given by ∞ 1 δ(ε) :=1− q 1−( ε)q , 0 <ε 2, � 2 ≤ 1 1 where p + q =1. To compute this result one uses a H¨older-type inequality. The reflexivity follows from the fact that the dual space ofl p is given byl q whereq is as above. Then we conclude thatl p∗∗ =l q∗ =l p . Obviously, the spacesl p, p= 1, are whether uniformly convex nor reflexive. ∞ 80 Remark 4.8. The class of reflexive Banach spaces is larger than the class of uniformly convex Banach spaces. This shows the famous result of M. Day (see [13]): There exists a separable reflexive strictly convex Banach space which is not isomorphic to a uniformly convex Banach space. On the other hand, a separable Banach space can be renormed in such a way that it X is strictly convex. This follows from the fact that there exists an injective linear mapping T: l 2 which leads to the new norm � : x x + Tx R. Now, X− �·� X� �− � � � � ∈ ( , �) is strictly convex; see Koethe [26]. X �·� � → → Remark 4.9. As we know from Theorem 4.7, each uniformly convex Banach space is reflexive. Obviously, the converse does not hold; consider for example the two-dimensional 2 spaceR endowed with thel 1 norm. A more sharp result concerning this question is the classical result of M. Day ([13]): There exist Banach spaces which are separable, reflexive, and strictly convex, but are not isomorphic to any uniformly convex space. This result has lead to the class of Banach spaces which are called superreflexive. Such spaces are characterized by the property that they admit an equivalent uniformly convex norm. We omit the question how a reflexive space which is not superreflexive looks like.�
Here is a result which connects uniform convexity to the quantitative behavior of the duality map.
Theorem 4.10 (Pr¨uss,1981). Let be a Banach space. Then is uniformly convex if X X and only if for eachR>0 there exists a gauge-functionγ R such that
J (x)−J (y), x−y γ R( x−y ) x−y for all x, y BR . (4.1) � X X � ≥ � � � � ∈ Proof: We follow [35]. Let be uniformly convex and chooseR>0. Letγ be defined byγ (0) =0, and X R R −1 γR(r) := inf{ λ−µ, x−y x−y : x, y BR, x−y r,λ J (x), µ J (y)}, r (0, 2R], � �� � ∈ � � ≥ ∈ X ∈ X ∈
We extendγ R by γ (r) :=γ (2R),r 2R . R R ≥ Obviously,γ R is nondecreasing. Let us prove thatγ R(r)>0 forr>0. Assume on the contrary thatγ R(r) =0 for somer (0, 2R]. Then there exist sequences n n n∈ n n n (x )n N,(y )n N belonging to B1 andλ n J (x ), µn J (y ) such that x −y r ∈ ∈ ∈ X ∈ X � � ≥ and lim λ −µ , xn −y n xn −y n −1 =0. Since we know (see (3.4)) n� n n �� � ( xn −y n )2 λ −µ , xn −y n � � � ≤ � n n � we may assume lim xn = lim y =a>0. Then n � � n � � lim inf | xn −1xn − y n −1yn|=a −1 lim inf xn −y n ra−1 . n � � � � n � � ≥ Therefore, by the uniform convexity of , X lim sup xn +y n =a lim sup | xn −1xn + y n yn| 2a(1−δ) n � � n � � � � ≤
81 for some0<δ=δ(r). On the other hand,
xn 2 + y n 2 − µ , xn − λ , yn = λ −µ , xn −y n . � � � � � n � � n � � n n � This implies lim λ , xn +y n = 2a 2 and leads to the contradiction n� n � 2 n n n n 2 2a = lim λn, x +y a lim sup x +y 2a (1−δ). n � � ≤ n � � ≤
Let us prove the sufficiency. Wefirst show that the range ofJ is closed. Let(λ n)n N n n ∈ be a sequence in ∗ withλ n J (x ), n N, and lim n λn =λ ∗ . Then(x )n N is nX ∈ X ∈ ∈X ∈ bounded, say x R, n N, due to the fact that(λ n)n N is bounded. Then wefindγ R � � ≤ ∈ ∈ with
n m n m n m n m γR( x −x ) x −y λn −λ m, x −x x −x λn −λ m , n, m N. � � � � ≤ � � ≤ � �� � ∈ n n Hence,(x )n N is a Cauchy sequence and therefore convergent. Let limn x =z. Then ∈ λ J (z) andJ has closed range. ∈ X ThereforeJ is surjective since ran(J) is dense in ∗ and is reflexive. This follows from X X−1 the Bishop-Phelps-theorem; see [6]. Now, we haveJ ∗ =J . Due to
γ ( x−y ) λ−µ forλ, µ ∗, λ R, µ R, x J ∗(λ), y J ∗(µ), R � � ≤� � ∈X � � ≤ � � ≤ ∈ ∈
henceJ ∗ is single-valued and uniformly continuous on bounded subsets of ∗ . Therefore X =( ∗)∗ is uniformly convex and the proof is complete. X X � 4.2 Uniform smoothness
Definition 4.11. A Banach space is called uniformly smooth if the the following X property holds:
ε δ>0 x, y ( x = 1, y δ= x+y + x−y −2<ε y ) ∀ ∃ ∀ ∈X � � � � ≤ � � � � � � ⇒ � To make the definition more applicable let us introduce a modulus which can be used to characterize uniformly smooth spaces.
Definition 4.12. Let be a Banach space. The modulus of smoothness of is the X X functionρ :[0, ) [0, ) defined by X − 1 1 ρ (∞τ) :=→ sup∞{ ( x+y + x−y )−1: x = 1, y =τ} X 2 � � 2� � � � � � 1 1 = sup{ ( x+τy + x−τy )−1: x =1= y } 2 � � 2� � � � � � � Theorem 4.13. Let be a Banach space. Then the following conditions are equivalent: X (a) is uniformly convex. X 82 ρ (τ) (b) limτ 0 Xτ =0.
Proof: ↓ (a)= (b) Letε>0. Then there existsδ>0 such that 1 1 ( x+y + x−y )−1< ε y ⇒ 2 � � � � 2 � � for all x, y with x = 1, y =δ. Thenρ (τ)< 1 ετ for allτ (0,δ). ∈X � � � � X 2 ∈ (b)= (a) Letε>0. Then there existsδ>0 such thatρ (τ)< 1 ετ for allτ (0,δ]. X 2 ∈ Let x, y with x = 1, y δ. Setτ := y . Then ∈X � � � � ≤ � � ⇒ x+y + x−y −2< ετ=ε y . � � � � � � � Lemma 4.14. Let be a Banach space with modulus of smoothnessρ . Then we have: X X (1)ρ (0) =0. X (2) max{0,τ−1} ρ (τ) τ for allτ [0, ). ≤ X ≤ ∈ (3)(0, ) τ ρ (τ)τ−1 (0, ) is non-decreasing. � �− X ∈ ∞ (4)ρ is a continuous convex function from(0, ) to(0, ). X∞ → ∞ Proof: Ad(1) Obviuosly true. ∞ ∞ Ad(2) Let x, y S . Forτ>0 letx S and sety :=τx . Then 1 ( x+y + x−y ) =τ. ∈ 1 ∈ 1 2 � � � � This showsρ (τ) τ. X ≤ Letτ>0. Sinceρ (τ) τ we have to showρ (τ) τ−1. Let x, y S1 . Then X ≤ X ≥ ∈ 1 1 ( x+τy + x−τy )−1 2τy −1=τ−1. 2 � � � � ≥ 2� � Ad(3) Let x, y S . Let0
83 Then we obtain fora [0, 1] ∈ x+(at+(1−a)s)y + x−(at+(1−a)s)y f (at+(1−a)s) = � � � � −1 x,y 2 a x+ ty +(1−a) x+ sy +a x− ty +(1−a) x− sy � � � � � � � � −1 ≤ 2 = afx,y(t) + (1−a)f x,y(s)
Thereforef x,y is convex for all x, y . Letε>0. Then there exist x, y X with ∈
ρ (at+(1−a)s)−ε f x,y(at+(1−a)s) af x,y(t)+(1−a)fx,y(s) aρ (t)+(1−a)ρ (s). X ≤ ≤ ≤ X X Sinceε>0 was arbitrary chosen,ρ is convex. X � Theorem 4.15. A uniformly smooth Banach space is smooth.
Proof: 0 Assume that is not smooth. Then there existx andλ, µ X ∗, such that X ∈X ∈ λ= µ, λ = µ =1 and λ, x 0 = x 0 = µ, x 0 . We may assume x 0 =1. Let � � � � � � � � � � � � � y0 X with y 0 =1 and λ− µ, y 0 >0. We obtain fort>0 ∈ � � � � 1 1 0 < t λ−µ, y0 = ( λ, x0+ty0 + µ, x0−ty0 )−1 ( x0+ty0 + x0−ty0 )−1. (4.2) � � 2 � � � � ≤ 2 � � � � Hence0< λ− µ, y 0 ρ (t)/t . Sincet>0 was arbitrary chosen we conclude that � � ≤ X X is not uniformly smooth. � Theorem 4.16 (Lindenstrauss-Tzafiri, 1978). Let be a Banach space. For everyτ>0 X we have 1 1 ρ (τ) = sup{ τε−δ (ε):0<ε 2},ρ (τ) = sup{ τε−δ (ε):0<ε 2} (4.3) X ∗ 2 X ≤ X 2 X ∗ ≤ Proof: Let us proof thefirst formula. Letτ>0<ε 2, x, y with x = y =1. From the Hahn-Banach theorem we ≤ ∈X � � � � obtainλ, µ ∗ mit λ = µ =1 such that ∈X � � � � λ, x+y = x+y and µ, x−y = x−y . � � � � � � � � Then
x+y +τ x−y −2= λ, x+y +τ µ, x−y = x−y −2 � � � � � � � � � � = λ+τµ, x + λ−τµ, y −2 � � � � λ+τµ + λ−τµ −2 ≤ � � � � sup{ λ+τµ + λ−τµ −2: x = y =1} ≤ � � � � � � � � =2ρ (τ) X ∗
84 If additional x−y ε then we obtain � � ≥ 1 1 τε−ρ (τ) 1− x+y . 2 X ∗ ≤ 2� � Therefore 1 τε−ρ (τ) δ (ε) and since0<ε 2 is arbitrary chosen 2 X ∗ ≤ X ≤ 1 sup{ τε−δ (ε):0<ε 2} ρ (τ). 2 X ≤ ≤ X ∗ 0 0 Chooseλ, µ X ∗ with λ = µ =1 andδ>0. Forτ>0 there existx , y with ∈ � � � � ∈X x0 =y 0 and x 0 = y 0 =1 such that � � � � � λ+τµ λ+τµ, x 0 +δ, λ−τµ λ−τµ, y 0 +δ. � � ≤ � � � � ≤ � � Using these inequalities we obtain
λ+τµ + λ−τµ −2 λ+τµ, x 0 + λ−τµ, y 0 −2+2δ � � � � ≤� � � � = λ, x 0 +y 0 + µ, x 0 −τy 0 −2+2δ � � � � x0 +y 0 −2+τ| µ, x 0 −y 0 |+2δ. ≤ � � � � We setε :=| µ, x 0 −y 0 |. Then we have0<ε x0 +y 0 2 and 0 � � 0 ≤ � � ≤ 1 1 ( x0 +τy 0 + x 0 −τy 0 )−1 τε +δ−δ (ε ) 2 � � � � ≤ 2 0 X 0 1 δ+ sup{ τε−δ (ε):0<ε 2}. ≤ 2 X ≤ Sinceδ>0 is arbitrary chosen 1 ρ (τ) sup{ τε−δ (ε):0<ε 2} X ∗ ≤ 2 X ≤ and the proof is complete. Let us prove the second formula. 0 0 Letτ>0 and letλ, µ ∗ with λ = µ =1. Forη>0 there existx , y in with ∈X � � � � X x0 = y 0 =1 such that � � � � λ+µ −η λ+ µ, x 0 , λ−µ −η λ− µ, y 0 . � � ≤� � � � ≤� � Then
λ+µ +τ λ−µ −2 λ+ µ, x 0 +τ λ− µ, y 0 −2η(1+τ) � � � � ≤� � � � sup{ x+τy + x−τy −2: x = y =1}+η(1+τ) ≤ � � � � � � � � =2ρ (τ) +η(1+τ). X If0<ε λ−µ 2 then we have ≤� � ≤ 1 1 τε−ρ (τ)−η(1+τ) 1− (λ+µ) , 2 X ≤ � 2 � was 1 τε−ρ (τ)−η(1+τ) δ (ε). 2 X ≤ X ∗ 85 Sinceη>0 is arbitrary we obtain 1 τε−ρ (τ) δ (ε) 2 X ≤ X ∗ for allε (0, 2] andfinally ∈ 1 sup{ τε−ρ (τ):0<ε 2} ρ (τ). 2 X ≤ ≤ X Let x, y with x = y =1 and letτ>0. From the Hahn-Banach theorem we ∈X � � � � obtainλ, µ ∗ with λ = µ = 1, such that ∈X � � � � λ, x+τy = x+τy and µ, x−τy = x−τy . � � � � � � � � Then
x+τy + x−τy −2= λ, x+τy + µ, x−τy = x−τy −2 � � � � � � � � � � = λ+ µ, x +τ λ− µ, y −2 � � � � λ+τµ + λ−τµ −2 ≤ � � � � We defineε :=| λ− µ, y |. Then we have0<ε x−y 2 and 0 � � 0 ≤ � � ≤ 1 1 ( x+τy + x−τy )−1 ( λ+µ +τ| λ− µ, y |)−1 2 � � � � ≤ 2 � � � � 1 1 = τε −(1− λ+µ ) 2 0 2� � 1 τε0 −δ (ε0) ≤ 2 X ∗ 1 sup{ τε−δ (ε):0<ε 2}. ≤ 2 X ∗ ≤ This implies 1 sup{ τε−ρ X(τ):0<ε 2} ρ (τ). 2 ≤ ≥ X � Let us add a few properties of the convexity and smoothness modulus.
Corollary 4.17. Let be a Banach space and let be a Hilbert space. X H (1)δ is convex and continuous. X (2)δ is strictly in increasing iff is uniformly convex. X X (3)δ (ε) δ (ε),ε [0, 2]. X ≤ H ∈ (4)ρ (τ) ρ (τ),τ (0, ). X ≥ H ∈ (5)ρ (τ) = √1+τ 2 −1. H ∞
86 Proof: For the proofs of(1)−(4) we refer to the literature ([11, 30, 41]). Ad(5) this follows from the second formula in Theorem 4.16.
� Theorem 4.18 (Shmulian, 1940). Let be a Banach space. We have: X (a) is uniformly smooth if and only if ∗ is uniformly convex. X X (b) is uniformly convex if and only if ∗ is uniformly smooth. X X Proof: Ad(a) Let be uniformly smooth. Assume that ∗ is not uniformly convex. X X Then there existsε 0 (0, 2] withδ ∗ (ε0) =0 and we obtain from Theorem 4.16 for all 1 ∈ X τ>00< ε0 ρ (τ)/τ. This implies that is not uniformly smooth. 2 ≤ X X Let ∗ be uniformly convex. Assume that is not uniformly smooth. X X Then limτ 0 ρ (τ)/τ=0. Hence, there existsε>0 such that for allδ>0 there exists X � t with0
Let be a Banach space with norm and norm mappingν: x x R. If X �·� X� �− � � ∈ the norm is Gateaux differentiable inx \{θ} then the limit ∈X → x+ ty − x lim � � � � (4.4) t 0 t
exists and is equal to the valueν �(x,→ y) of the uniquely determined functionalν �(x, ) · ∈ ∗ ; see Section 3.3. X Definition 4.21. Let be a Banach space. X (a) The norm in is called Fr´echet-differentiable atx \{θ} if X ∈X x+ ty − x lim � � � � t 0 t
exists uniformly iny S . The→ limit is called the Fr´echet derivative atx in ∈ 1 directiony. (b) The norm in is called Fr´echet-differentiable if the norm is Fr´echet-differentiable X at eachx \{θ}. ∈X (c) is called strongly smooth if the norm is Fr´echet-differentiable at eachx \{θ}. X ∈X (d) The norm in is called uniformly Fr´echet-differentiable if the limit in X x+ ty − x lim � � � � t 0 t
exists uniformly for(x, y) S →S . ∈ 1 × 1 �
Let be a Banach space with norm and norm-mappingν. It is clear that at a X �·� pointx where the norm is Fr´echet differentiable the norm is Gateaux differentiable and the Fr´echet derivative inx in directiony is equal toν �(x, y). Since the norm is homogeneous one has to check the Fr´echet differentiability atx S only. In the general, Gateaux ∈ 1 differentiability of the norm does not imply Fr´echet differentiability. An example is the norm inl 1 ; see [30]. Lemma 4.22. Let be a Banach space with norm and norm-mappingν. Let X �·� x \{θ}. Then the following statements are equivalent: ∈X (a) The norm is Fr´echetdifferentiable inx. (b) x+v − x −ν �(x, v) lim � � � � =0. v θ v � � Proof: → Observe thatv= ty witht= v , y=v v −1 S . � � � � ∈ 1 � 88 Lemma 4.23 (Shmulian, 1939). Let be a Banach space with norm-mappingν. Then X forx S the following conditions are equivalent: ∈ 1 (a) The norm is Fr´echet-differentiable inx.
(b) If(λ n)n N is a sequence in ∗ with λ n 1, n N, and lim n λn, x =1 then this ∈ X � � ≤ ∈ � � sequence is convergent. Proof:
(a)= (b) Letλ :=ν �(x, ) and let(λ n)n N be a sequence in ∗ with λ n 1, n N, · ∈ X � � ≤ ∈ and lim λ , x =1. Since n� n � ⇒ 1 λ n λn, x , n N, and lim λn, x =1 ≥� � ≥ � � ∈ n � �
we may suppose that λ n =1 for alln N. � � ∈ n Assume that(λ n)n N does not converge toλ. Then there existsε>0 and(z )n N in n ∈ n n n ∈ −X1 with z = 1, n N, with λ n −λ, z 2ε,n N. We setx :=z (1− λ n, x )ε . � � n ∈ � � ≥ ∈ � � Then limn x =θ and (see 4.22) ν(x+x n)−ν(x)− λ, x n λ , x+x n −1− λ, x n � � � n � � � xn ≥ xn � � � � λ , x −1+ λ −λ, z n(1− λ , x )ε−1 = � n � � n � n � � (1− λ , x )ε−1 � n � = λ −λ, z n −ε ε � n � ≥ This contradicts the Fr´echet-differentiability ofν inx. n (b)= (a) Setλ :=ν +� (x, ). Assume that there existsε>0 and(x )n N with n · ∈ limn x =θ such that ⇒ ν(x+x n)−ν(x)− λ, x n � � ε,n N. xn ≥ ∈ � � � � � � Then � � � n n � n x+x − λ, x+x ε x ,n N. � � � � ≥ � � ∈ n n n Chooseλ n S1 such that λ n, x+x = x+x , n N. Since lim n x =θ we have ∈ � � � � ∈ lim λ , x = x =1. But n� n � � � λ, x − λ , x λ −λ λ −λ, x n xn −1 � � � n � +ε ε � n � ≥ � n �� � ≥ xn ≥ � � since λ, x = x λ n, x . This means that(λ n)n N does not converge toλ.� � � � � ≥ � � ∈ Corollary 4.24. Let be a Banach space and let the norm in the dual space ∗ be X X Fr´echetdifferentiable. Then is reflexive. X Proof: We use the result of James which says that is reflexive if and only if each nonzero X functionalλ ∗ attains its norm at somex S1 ; see the preliminaries in the preface. ∈X n ∈ n Letλ ∗, λ = 1, and let(x )n N be a sequence in S1 such that limn λ, x =1. By ∈X � � ∈ � � applying Smulian’s lemma, lim xn =x S . Therefore n ∈ 1 λ, x = lim λ, xn =1= λ . � � n � � � � −1 If nowµ ∗\{θ} apply the consideration above toλ :=µ µ . ∈X � � � 89 Theorem 4.25. Let be a Banach space with duality mapJ and norm mappingν. Then X we have the equivalence of the following conditions:
(a)J is single-valued and uniformly norm-to-norm continuous an bounded subsets. (b) The norm in is uniformly Fr´echetdifferentiable. X (c) is uniformly smooth. X (d) ∗ is uniformly convex. X Proof: (a)= (b) Letx S , y . Lett>0 be such that x+ ty >0. Then ∈ 1 ∈X � � J (x), x − x + J (x), ty ⇒ J (x), y = � X � � � � X � � X � t J (x), x+ ty − x = � X � � � t x+ ty − x � � � � ≤ t x+ ty 2 − x x+ ty = � � � �� � t x+ ty � � J (x+ ty), x+ ty −| J (x+ ty), x | � X � � X � ≤ t x+ ty � � t J (x+ ty), y + J (x+ ty), x −| J (x+ ty), x | = � X � � X � � X � t x+ ty � � J (x+ ty), y � X � ≤ x+ ty � � Hence J (x), y x+ ty − x J (x+ ty), y � X � � � � � � X � . x ≤ t ≤ x+ ty � � � � Using the property(a) we read off that the norm is uniformly Fr´echet differentiable. (b)= (c) Assuming(b) we can choose forε>0aδ>0 so that for allt (0,δ) and ∈ all x, y S1 ∈ x+ ty − x 1 ⇒ � � � � −ν �(x, y) < ε. t 2 � � But this implies that for all�t (0,δ) and all x, y S� � ∈ ∈ 1� x+ ty − x x+ ty − x− ty =2+t � � � � −ν �(x, y) � � � � t � � x+t(−y) − x +t � � � � −ν �(x,−y) t � � < 2+tε.
This implies the assertion in(c) (c)= (d) Letε>0. By(c) we canfindδ>0 such that for allx S andz with ∈ 1 ∈X z δ we have x+z + x−z <2+ 1 ε z . � � ≤ � � � � 4 � � ⇒
90 1 Letλ, µ ∗ withλ = µ =1 and λ−µ ε. Then there is az with z δ ∈X � � � � � ≥ ∈X � � ≤ 2 and λ− µ, z 1 εδ. This implies � � ≥ 2 λ+µ = sup | λ+ µ, z � � x S � � ∈ 1 = sup | λ, x+z + µ, x−z − λ− µ, z | x S � � � � � � ∈ 1 1 sup( x+z + x−z )− εδ ≤ x S � � � � 2 ∈ 1 1 1 1 2+ ε z − εδ<2− εδ. ≤ 4 � � 2 4 (c)= (a)J is single-valued due to Lemma 3.14. Letε>0. Then there existsδ>0 such that for allλ, µ S ∗ we have ∈ 1 ⊂X ⇒ λ+µ >2−δ implies λ−µ <ε. � � � �
Choose x, y S1 with x +µ =1 and x−y =δ. Then withλ J (x), µ J (y) ∈ ⊂X � � � � � ∈ X ∈ X we obtain
λ+µ λ+ µ, y = λ, x + µ, y − λ, x−y >2− x−y >2−δ. � � ≥ � � � � � � � � � � ThereforeJ is norm-to-norm continuous on unit sphere. SinceJ is homogeneous, i.e. J (ax) = aJ (x) for allx ,a R,J is norm-to-norm continuous on the unit ball. X X ∈X ∈ This implies thatJ is uniformly continuous on each bounded set. � Remark 4.26. If the the norm in a Banach space is Fr´echetdifferentiable then the duality map is uniformly continuous on bounded subsets from the norm-topology to the weak∗- topology; see [10]. � Theorem 4.27. Let be a Banach space. Then the following statements are equivalent: X (a) is an E-space. X (b) ∗ is strongly smooth. X Proof: (a)= (b) We know by definition that is reflexive and strictly convex and therefore X ∗ smooth by Theorem 3.27. We want to use Lemma 4.23 to verify that the norm in ∗ X X is Fr´echet⇒ differentiable at eachλ ∗ . ∈Xn n Letλ ∗ with λ =1 and let(x )n N be a sequence in B1 with limn λ, x =1. Now, ∈Xn �n � ∈ n n n �n −1 � 1 x λ, x , n N, which implies lim n x =1. Letz :=x x , n N, and ≥� � ≥ � � ∈ �n � � � ∈ letz be a weak cluster point of the sequence(z )n N in S1 ; it exists due to the reflexivity n n ∈−1 of . Then z 1, but λ, z = lim n λ, x x =1. So z =1. Since is an E- X � � ≤ � � � n�� � � � X space we obtain thatz is cluster point of(x )n N with respect to the norm convergence. ∈ Because ∗ is smooth,z is uniquely determined by the conditionsz S1, λ, z =1. X n n ∈ � � Therefore, limn z = z, and so limn x =x also. (b)= (a) This is proved similarly. We omit the proof. �
⇒ 91 4.4 Radial projection and norm inequalities
Here we want to study the projection onto the unit ball in a Banach space . Such a X projection is the mapping
x if x 1 x � � ≥ Prad : x � � B1 . (4.5) X� �− �x if x 1 ∈ ⊂X � � ≤ → Lemma 4.28. Let be a Banach space and letx \{θ}. ThenP rad P , especially X ∈X ∈ B1 B1 is a proximal subset. Proof: We want to use the characterization result 3.32. Letλ J (x− x ). Thenλ=(1− 1 )µ ∈ X x x x � � � � withµ J (x). We want to show thatλ N( , B1). This follows from ∈ X ∈ x � � x 1 x 1 λ, u− =(1− )( µ, u − µ, ) (1− )( µ u − x ) 0 for allu B . � x � x � � � x � ≤ x � �� � � � ≤ ∈ 1 � � � � � � � � � As a consequence of Lemma 4.28,P (x) =P rad(x) for allx \{θ} if is strictly B1 ∈X X convex; see Theorem 2.15. Thus in a Hilbert space the radial projectionP rad(x) is the best approximation ofx in the unit ball. Thus, in this Hilbert space case we know that Prad is nonexpansive. As we see in the following, the nonexpansivity property ofP rad does not hold in the Banach space case in general. To study this question, let us begin with a few famous inequalities concerning the triangle inequality in Hilbert and Banach spaces which may be used to compute the Lipschitz constant of the projectionP rad . Theorem 4.29 (Dunkl and Williams, 1964). Let be a Hilbert space and let x, y H ∈ ,x=θ, y=θ. Then H � � x y 2 − x−y (4.6) x y ≤ x + y � � �� � � �� � � � � Proof: � �
x y 2 x y x y − = − − x y � x y x y � �� � � �� � � x� ��y� � � � � � =2−2 � � x y � � � � � 1 � = (2�x y − x|y ) x y � �� � � � � �� � This implies
2 x + y 2 x y 2 ( x − y ) x−y 2 − � � � � − = � � � � ( x + y ) 2 − x−y 2 � � 2 x y 4 x y � � � � � � � � �� � � �� � �� � � � The righthand side in the last� equality is� nonnegative as one easily can check. �
92 Theorem 4.30 (Maligranda, 2006). Let be a Banach space and let x, y ,x= X ∈X � θ, y=θ. � y (a) x+y x + y −(2− x + ) min( x , y ) � � ≤ � � � � x y � � � � � � � � � x y � (b) x+y x + y −(2− � + �) max( x , y ) � � ≥ � � � � x y � � � � � � � � � � If either x = y =1 ory= ax with� a > 0, then� equality holds in both(a) or(b). � � � � Proof: We follow Maligranda [29]. Without loss of generality assume y x , i.e. min( x , y ) = x . Ad(a) � � ≥ � � � � � � � � x x x x+y = � �x+ � � y+(1− � � )y � � x y y �� � x � �y � � � �x + + y − x � ≤ � � x y � � � � � �� � x� ��y = y +� + � −1 x � � x y � � ��� � �x �� y � = x + y +� +� −2 x . � � � � x y � � ��� � � �� � Ad(b) � �
y y y x+y = � �x+ � �x+ 1− � � x � � y x x �� � y � �x � � � � � �y + −| x − y |� ≥ � � y x � � � � � ��y� �x�� = y � + � − y + x � � y x � � � � �� � � �� x y = x + �y − 2−� + y � � � � x y � � � �� � � ��� The additional assertion concerning the equality can� easily be veri� fied.� Corollary 4.31. Dunkl and Williamson, 1964] Let be a Banach space and let x, y X ∈ ,x=θ, y=θ. Then X � � x y 4 − x−y (4.7) x y ≤ x + y � � �� � � �� � � � � Proof: � � From(b) in Theorem 4.30 we obtain
x y x−y +| x − y | − � � � � � � x y ≤ max( x , y ) �� � � �� � � � � � � which implies (4.7). �
93 2 Example 4.32. ConsiderR endowed with thel 1-norm. Takex=(1,ε) andy := (1, 0) whereε is small. Then x y x + y 4+2ε − � � � � = x y x−y 1+ε �� � � �� � � where � � 4+2ε lim =4. ε 0 1+ε 2 If the norm inR is given by thel ↓ -norm then we obtain forx := (1, 1) andy := (1−ε, 1+ε), whereε>0 is small, ∞ x y x + y 4+2ε lim − � � � � = lim =4. ε 0 x y x−y ε 0 1+ε �� � � �� � � ↓ � � ↓ �
Let be a Banach space and let x, y \{θ}. The quantity X ∈X x y α(x, y) := − x y �� � � �� is called the angular distance or Clarkson� distance �of x, y . As we see from the results in Theorem 4.30 this quantity controls the triangle inequality. In [12] a detailed analysis of the triangle inequality bas been made for uniformly convex Banach spaces. From(b) in Theorem 4.30 we obtain (see the proof of 4.31
2 x−y α(x, y) � � . (4.8) ≤ max( x , y � � � � This a sligthly strengthening of the Dunkl-Williams inequality (4.6) in Theorem 4.29. Kirk and Smiley showed in [25] that this inequality characterizes inner product spaces: If in a Banach space the inequality (4.8) holds for all x, y \{θ} then is actually a X ∈X X Hilbert space; see also [32, 37]. This observation may be seen a motivation to introduce the so called Dunkl-Williams- constant: x + y DW( ) := sup{α(x, y)� � � � : x, y \{θ}, x=y}. (4.9) X x−y ∈X � � � 2 Observe that2 DW( ) 4. We know from example 4.32 that DW(R , 1) = ≤ X ≤ �·� DW(R2, ) =4. The Dunkl-Williams-constant may be considered as a measure how �·� much a Banach space is close to a Hilbert space. ∞
Let us come back to the radial projectionP rad . It is obvious that the inequalities above play an important role for studying the Lipschitz continuity ofP rad . Theorem 4.33. Let be a Banach space. Then X 4 P (x)−P (y) x−y for all x, y . (4.10) � rad rad � ≤ x + y � � ∈X � � � �
94 Proof: Let x, y . ∈X First case: x, y / B . Then (4.10) follows from Corollary 4.31. ∈ 1 Second case:x B , y / B . Then ∈ 1 ∈ 1 y y P (x)−P (y) x− x−y + y− � rad rad � ≤ � y ≤ � � � y � � � � � = x−y + y −1 x−y + y − x 2 x−y . � � � � ≤� � � � � � ≤ � � � From Theorem 4.33 we obtain a Lipschitz constant 2 for the radial projection. This constant cannot improved in general as the following example shows.
2 Example 4.34. ConsiderR endowed with thel 1-norm. Takex=(0, 2) andy := (ε, 2) whereε is small. ThenP rad(x) = (0, 1),Prad(y) = (ε, 1−ε) and
2ε= P (x)−P (y) =2ε=2 x−y . � rad rad � � � �
4.5 Continuity of a single-valued metric projection
The considerations in the last chapters show that the metric projection operators have extremely good properties in Hilbert spaces, especially metric projections onto subspaces: they are linear orthogonal projectors, non-expansive, uniformly continuous, self-adjoint and idempotent. In the last section we have seen a big difference concerning metric projections in Hilbert spaces and Banach spaces: metric projections in a Banach space are not nonexpansive in general. Additionally, we have to realize that the metric projection is set-valued in general. What should be a definition of continuity of such a map?
Before we investigate the general case in the next section let us study continuity prop- erties of single-valued metric projections in this section.
Theorem 4.35. Let be a a Banach space and letC be an approximately compact X Chebyshev subset of . ThenP is continuous. X C Proof: n n Letx and let(x )n N withx= lim n x . SinceC is a Chebyshev set there exist ∈X ∈ for everyx n au n C with u n −x n = dist(x n,C);u n =P (xn). Then lim un − ∈ � � C n � x = dist(x, C) since dist( ,C) is continuous; see Lemma lem:nonexpdist. From the fact � · nk thatC is approximately compact we obtain a subsequence(u )k N andu with ∈ nk ∈X w= lim k x . SinceC as a Chebyshev set is closed we haveu C. Due to u−x = nk ∈ � � limk u −x = dist(x, C) we haveu P C(x). SinceC is a Chebyshev set we have � � ∈ n {u}=P C(x). From this uniqueness we conclude that(u )n N converges toP C(x). ∈ � Corollary 4.36. Let be a Banach space and letC be a compact Chebyshev subset of X . ThenP is continuous. X C
95 Proof: Follows from Theorem 4.35 since a compact set is approximately compact.� Corollary 4.37. Let be afinite-dimensional Banach space and letC be a Chebyshev X subset of . ThenP is continuous. X C Proof: Letx . ThenP (x) C B (x) withr> dist(x, C). Now,C B (x) is closed and ∈X C ∈ ∩ r ∩ r bounded, therefore compact. Apply now Corollary 4.36. � Now, we come to continuity results which use geometric properties of the space under consideration. In the light of Corollary 4.36 it is natural to ask whether every Chebyshev set has a continuous metric projection mapping. But it is surprising that this question may fail even in a reflexive space. Unfortunately,finding an example of a Chebyshev set with a discontinuous metric projection mapping is not as straightforward as one might first imagine. Brown gave an example of cheyshevian subspace in a smooth reflexive Banach space with a discontinuous metric projection; see [8].1
Theorem 4.38. Let be a strictly convex Banach space and letC be a nonempty compact X convex subset of . ThenC is a Chebyshev set andP :X C is continuous. X C − Proof: → Clearly,C is closed and a Chebyshev set. n n Letx and let(x )n N be a sequence in with lim n x =x. Then the sequence ∈ ∈nX X nk (PC(x ))n N belongs to the compact setC. Therefore, there exists a subsequence(P C(x ))k N ∈ ∈ andz C with lim P(xnk ) =z. ∈ k Letu C. Then x nk −P (xnk ) xnk −u and therefore x−z x−u since ∈ � C � ≤ � � � � ≤ � � lim (xnk ,P (xnk )) = (x, z). This impliesz P (x) andC is a Chebyshev setz=P (x). k C ∈ C C � Theorem 4.39. Let be an E-space and letC be a nonempty closed convex subset. Then X the metric pojectionP :X C is continuous. C − Proof: Letx and letx= lim xn . Then ∈X → n P (xn)−x P (xn)−x n + x n −x P (x)−x n + x n −x . � C � ≤ � C � � � ≤ � C � � � Therefore, lim P (xn)−x = dist(x, C). Then lim P (xn) =P (x); see Corollary 2.20. n � C � n C C � There is no hope that strengthening the E-property to uniformly convexity would result in the uniform continuity of the metric projection since this fails even in thefinite- dimensional case.
Corollary 4.40. Let be a uniformly convex Banach space and letC be a nonempty X closed convex subset. Then the metric pojectionP :X C is continuous. C − Proof: → This is a consequence of Theorem 4.39 and Theorem 4.4. � 1Another example of a discontinuous metric projection is due to Kripke (unpublished).
96 Theorem 4.41. Let be a strictly convex and reflexive Banach space and letC be a X nonempty convex closed subset of . ThenC is a Chebyshev set and the metric projection X P : is continuous. C X− X Proof: → See [27]. � Theorem 4.42. Let be a reflexive Banach space with a Kadek-Klee norm and letC be X a weakly closed Chebyshev subset of . Then the metric projectionP is continuous. X C Proof: n n Letx . Let(x )n N be a sequence with limn x =x. SinceP C is locally bounded ∈ ∈X nk (see Lemma 4.63) and is reflexive, there exists a subsequence(x )k N and a point ∈ nXk y C such that(P C(x ))k N converges toy with respect to the weak topology. SinceC ∈ ∈ is weakly closed,y C. Therefeore, ∈
nk nk dist(x, C) x−y lim inf x −P C(x ) ≤� � ≤ k � � nk nk nk lim sup x −P C(x ) = lim dist(x ,C) = dist(x, C) ≤ k � � k
nk nk since(x −PC(x ))n N converges tox−y with respect to the weak topology on and the ∈ X norm in is weakly lower semicontinuous. Now, becauseC is a Chebyshev set it follows X nk nk thaty=P C(x). Furthermore, due to Kadek-Klee property(x −P C(x ))n N converges ∈ nk in norm tox−y=x−P C(x). Therefore,P C(x ))n N converges in norm toP C(x). Due to n ∈ n this uniqueness of the cluster point ofP C(x ))n N we conclude that(P C(x ))n N converges ∈ ∈ in norm toP C(x). � Corollary 4.43. Let be a uniformly convex Banach space and letC be a weakly closed X Chebyshev subset of . Then the metric projectionP is continuous. X C Proof: A uniformly convex Banach space has a Kadek-Klee norm. � A famous problem in approximation theory is the convexity of Chebyshev sets. We have seen that a Banach space has the property that every closed convex subsetC X is a Chebyshev set if and only if is reflexive and strictly convex. It is natural to ask X what are the Banach spaces in which the converse property holds, i.e. in which every X Chebyshev setC is convex. This problem has been solved for 2- and 3-dimensional spaces . For Banach spaces offinite dimension 4 only sufficient conditions are known. For X ≥ infinite-dimensional Banach spaces the problem is considerable more difficult, even the X answer to the following problem beeing unknown:
In a Hilbert space, is every Chebyshev set necessarily convex?
At the beginning of the analysis of this problem the approach was to add a condition on the set itself in order to derive convexity. An important step was the idea of Klee of imposing continuity conditions on the metric projection. It was quickly realized that continuity properties of the metric projection play a key role in determining the convexity of Chebyshev sets. Here is the most important result in this concept.
97 Theorem 4.44 (Vlasov). Let be a Banach space and letC be a Chebyshev set. X ⊂X IfX ∗ is strictly convex andP C is continuous thenC is convex. Proof: [40]. � Corollary 4.45. IfC is a Chebyshev set in a Hilbert space with a continuous metric projection thenC is convex.
Proof: Follows from Theorem 4.44. � Corollary 4.46. IfC is a weakly closed Chebyshev set in a smooth uniformly convex Banach space thenC is convex. X Proof: Follows from � Lemma 4.47. LetC be a nonempty weakly closed subset of a Hilbert space . Then the H metric projectionP C is single-valued and continuous. Proof:
� Theorem 4.48. LetC be a nonempty weakly closed subset of a Hilbert space . Then H the following conditions are equivalent:
(a)C is a Chebyshev set. (b)C is a convex set.
Proof: Ad(a)= (b) Consider the function
1 2 ⇒ f: x x +δ C(x) R { }. H� �− 2� � ∈ ∪
Notice thatf ∗ is convex. Then fory → ∈H ∞
1 2 f∗(y) = sup( y|u − y ) u C � � 2� � ∈ 1 1 1 = y|y + (− y|y +2 y|u − u|u ) 2� � 2 u C 2� � � � � � ∈ 1 1 1 1 = y 2 − inf y−u 2 = y 2 − dist(y, C)2 2 2 u C 2 2 � � ∈ � � � � 1 1 1 = y 2 − y−P (y) 2 = y|P (y) − P (y) 2 2� � 2� C � � C � 2� C � = y|P (y) −f(P (y)) � C � C
98 This shows dom(f∗) = andf ∗(y) = y|P C(y) −f(P C(y)), y . We concludeP C(y) H � � ∈H 1 ∈ ∂f∗(y) for ally . Now supposez ∂f ∗(y). Sety :=y+ (z−P (y)). Then ∈H ∈ n n C limn yn =y and hence lim n PC(yn) =P C(y) by Lemma 4.47. From 1 0 y n −y|P C(yn)−z = z−P C(y)|PC(yn)−z ,n N, ≤� � n� � ∈ which is a consequence of the defining property of a subdifferential we conclude
2 0 lim inf z−P C(y)|PC(yn)−z =− z−P C(y) . ≤ n � � � �
Consequentely,z=P (y). This shows∂f ∗(y) ={P (y)} for ally . Sincef ∗ is C C ∈H continuous and∂f ∗(y) is a singletonf ∗ is Gateaux differentiable. Now,− p := inf (f(z) +g(z)) =d := max (−f∗(y)−g ∗(−y))< . z y ∈H ∈H Sincef ∗(y) = x+y , g ∗(y) =δ (y), y , and sinceC is closed, we∞ obtain � � C ∈H d= max − x+y −δ C(−y)) =−dist(x, y). y ∈H � � Choose anyu C. Observe that0 dist(x, C) x+y . Therefore,P (x) must be ∈ ≤ ≤� � C nonempty. Uniqueness follows from the convexity ofC. � Remark 4.49. IfC is a weakly closed Chebyshev set in a smooth uniformly convex Banach space then the metric projection is continuous. As a consequence, a nonempty weakly X closed subset is then convex if and only ifC is a Chebyshev set.� Corollary 4.50. IfC is a weakly closed Chebyshev set in a Hilbert space thenC is convex. Proof: � Theorem 4.51 (Asplund, ). Let be a Hilbert space and letC be a Chebyshev set. X ⊂X IfP : is continuous when is endowed with the norm-topology in the domain C H− H H and is endowed with the weak topology in the range of the mapping thenC is convex. H → Proof: [?]. � 99 Remark 4.52. The metric projectionP :=P U onto a closed linear subspaceU of a Hilbert space is continuous and it has the the propertyP P=P (Idempotencity). Moreover, H ◦ it is continuous and nonexpansive. In a Banach space it is not clear that a projection X P onto a closed subspace is continuous. A necessary und sufficient condition is that the subspaceU is complemented. This means that there exists a closed subspaceV such that X=U V. But it is known that not each closed subspace is complemented. ⊕ � Let us add a surprising result concerning the linearity of a metric projection. We need two definitions. Definition 4.53. Let be a Banach space and letU be a proximal subspace of . X X (1)U 0 :={x :θ P (x)} is called the kernel ofP . ∈X ∈ U U (2) P := sup{ y :y P (x), x S }. � U� � � ∈ U ∈ 1 � Let , as in Definition 4.53. Obviously,U 0 ={x : x = dist(x, U). Moreover, X ∈X � � P [1, 2] sinceP | =I, and P (x) P (x)−x + x 2 x , x . � U� ∈ U U � U � ≤ � U � � � ≤ � � ∈X Lemma 4.54 (Holmes, Kripke, 1968). Let be a Banach space and letU be a Chebyshev X subspace of . X (1) IfP is continuous thenI+P is a homoemorphism of into itself. U U X (2) IfP is linear, then P =1 if and only ifU 0 is a proximal subspace andx−P (x) U � U� U ∈ P 0 (x) for eachx . U ∈X 0 (3) IfP U is linear andU is a Chebyshev subspace, then the following statements are equivalent: (i) P =1. � U� (ii)P U0 =I−P U . (iii)(U 0)0 =U. Proof: See [23] for a proof. � As we know, in a Hilbert space the metric projection onto a closed linear subspace is always linear and has norm one. In general normed spaces however, with the exception of Chebyshev subspaces of codimension one, Chebyshev subspaces having linear metric projections are relatively scarce. For example, the spaceC[0, 1] of continuous functions endowed with the supremum norm has none withfinite dimension; see [9]. Deutsch and Lambert (see [15]) gave for eachr [0, 2] an example of a one-dimensional subspaceU in 2 ∈ 0 R , endowed with thel 1-norm which has the properties:U,U are Chebyshev subspaces, P ,P 0 are linear, P =r. U U � U� 100 4.6 Continuity properties of a multi-valued metric projections Let be a Banach space and letC be a proximal subset of . Then the metric projection X X is a mappingP : with nonempty closed convex subsetsP (x), x ; clearly, C X⇒X C ∈X #P (x) =1 forx C. To study continuity propertis of this mapping one has to introduce C ∈ concepts of neighborhoods and/or convergence of families of sets. LetX,Y be a topological spaces with the system , of open sets. LetG:X⇒Y. T X T Y The most important continuity concept for set-valued mapping is that of lower semi- continuity since in this concept under some additional circumstances Michael’s selection theorem may be used to obtain a continuous mappingS:X Y such thatS(x) G(x) − ∈ for allx X. For example, the radial projection is a continuous selection for the metric ∈ projection onto the closed unit ball of Banach space; see Section→ 4.4. Definition 4.55. LetX,Y be a topological spaces with the system , of open sets. Let T X T Y G:X⇒Y. (a)G is called upper semicontinuous if for every open setW inY the set{x X: ∈ G(x) W} is open inX. ⊂ (b)G is called lower semicontinuous if for every open setW inY the set{x X: ∈ G(x) W} is open inX. ∩ (c)G is continuous if it is both upper semicontinuous and lower semicontinuous. (d) dom(G) :={x X:G(x)= }. ∈ � ∅ (e) graph(G) :={(x, y) X Y:y G(x)} is called the graph ofG. ∈ × ∈ � A paracompact topological spaceX is defined as follows: Every open covering of X has an open refinement = U where eachU is a locallyfinite collection of open U ∪ i=1 i i subsets ofX. Every compact topological∞ space, every metric space is paracompact. Theorem 4.56 (Michael, 1953). LetX be a paracompact topological Hausdorff space and letF:X⇒X be a lower semicontinuous mapping with nonempty closed convex subsets F(x), x X. Then there exists a continuous mappingf:X X withf(x) F(x) for ∈ − ∈ allx X. ∈ → Let us collect some topological concepts for the convergence of families of sets. Definition 4.57. Let be a Banach space and let(C n)n N be sequence of sets. X ∈ (a)(C n)n N is said to be Hausdorff-convergent to a setC if ∈ (i) for eachε>0 there existsN N withC n {x : dist(x, A) ε} for all ∈ ⊂ ∈X ≤ n N; ≥ 101 (ii) for eachε>0 there existsN N withC {x : dist(x, A n) ε} for all n N. ∈ ⊂ ∈X ≤ ≥ We will write limH Cn =C. (b)(C n)n N is said to converge Wijsman to a setC if ∈ lim dist(x, Cn) = dist(x, C) for allx . n ∈X 2 We will write limW Cn =C. (c)(C n)n N is said to converge Mosco to a setC if ∈ (i) for eachx C there existN N such that for eachn Nax C n exists ∈ n ∈ ≥ ∈ such that(x )n N converges tox; ∈ nk (ii) if there exists a subsequence(n k)k N and an associated sequence(x )k N with ∈ ∈ nk nk x C nk , k N, such that(x )k N converges weakly to a pointx then x C.∈ ∈ ∈ ∈X ∈ We will write limM Cn =C. (d)(C n)n N is said to be slice-convergent to a setC if ∈ dist(B, Cn) converges to dist(B, C) for every bounded convex setB. We will write limS Cn =C. � It is appearent that Wijsman convergence depends on the precise norm used, while Mosco convergence is preserved by equivalent renorming. Lemma 4.58. Let be a Banach space and let(C n)n N be sequence of sets. X ∈ (1) If lim C = then lim C = . W n ∅ M n ∅ (2) If is reflexive and if lim C = then lim C = . X M n ∅ W n ∅ (3) If is reflexive and if lim C =C then lim C =C. X M n W n Proof: � Remark 4.59. If is nt reflexive then there exists a sequence(C n)n N of compact convex X ∈ sets such that limM Cn exists and is a nonempty compact convex set, but limW Cn fails to exist. Thus, a Banach space is reflexive if and only if Mosco convergence implies Wijsman convergence of closed convex subsets of ; see [?]. X � Lemma 4.60. Let be a Banach space and let(C n)n N be sequence of closed sets and let X ∈ (rk)k N an increasing sequence of positive real numbers. Assume thatD k := limM(Cn Brk ) ∈ ∩ exists for allk N. ThenC := lim M Cn exists andC= k NDk . Moreover,C is closed. ∈ ∪ ∈ 2We use the convention dist(x, ) = . ∅ ∞ 102 Proof: � There is a connection to the Vietoris-topology and the topology induced by the Haus- dorff metric in the case of closed-valued/compact-valued mappings. Lemma 4.61. Let be a Banach space and letC be a nonempty closed subset of . If n n X n n n n X (x , y )n N is a sequence in withy P C(x ), n N, and lim n(x , y ) = (x, y) ∈ X×X ∈ ∈ theny P (x). ∈ C Proof: Consider the sequence as above. Firstly,y C sinceC is closed. Furthermore, ∈ x−y = lim xn −y n = lim dist(xn,C) =d(x, C) � � n � � n since dist( ,C) is continuous due to Lemma 2.1. Thus,y P (x). · ∈ C � Definition 4.62. Let , be Banach spaces and letΦ: .Φ is called locally X Y X⇒Y bounded if for everyx X there exist r > 0, R > 0 such thatΦ(B (x)) B . ∈ r ⊂ R � Lemma 4.63. Let be a Banach space and letC be a nonempty subset of . Then the X X metrix projection mappingP : is locally bounded. C X⇒X Proof: Letx and letr := 1, R := dist(x, C)+ x +2. Supposey P (B (x)), i.e.y P (u) ∈X � � ∈ C r ∈ C for someu B (x). Using the nonexpansivity of the distance function we obtain ∈ r y y−u + u−x + x � � ≤ � � � � � � = dist(u, C) + u−x + x � � � � dist(x, C) + u−x + u−x + x ≤ � � � � � � dist(x, C) +2+ x =R. ≤ � � Hence,P (B (x)) B . C r ⊂ R � Definition 4.64. Let , be Banach spaces and letΦ: ⇒ .Φ is called continuous X Y X nY n atx ifΦ(x) is a singleton and for any sequence(x , y )n N in the graph ofΦ with ∈X ∈ lim xn =x we have lim yn =z withz Φ(x). n n ∈ � Theorem 4.65. Let be a Banach space and letC be a boundedly compact subset of . X X IfP C(x) is a singleton forx , then the metric projection mappingP C is continuous at x . ∈X Proof: n n Letx and letP C(x) ={z}. Let(x , y )n N be a sequence in the graph ofP C with ∈nX ∈ limn x =x. SinceP C is locally bounded and sinceC is boundedly compact there exists a nk nk nk nk nk subsequence(x , y )k N such that limk y =y for somey C. Since lim k(x , y ) = ∈ ∈ (x, y) we conclude by Lemma 4.61y P (x). Thus,y=z as required. ∈ C � 103 4.7 Differentiability properties of the distance map- ping Let be a Banach space and letC be a nonempty closed subset of . We know that the X X distance mapping x dist(x, C) := inf x−u R u C X� �− ∈ � � ∈ is nonexpansive. As a consequence, the differential quotients → dist(x+ th, C)− dist(x, C) , t= 0, x \{θ}, h , t � ∈X ∈X are bounded by1. Therfore, one should ask for differentiability properties of the distance mapping. Theorem 4.66. Let be a Hilbert space and letC is Chebyshev subset of . Then the H H distance function dist( ,C) is Fr´echetdifferentiable and Ddist( ,C)(x) =1. · � · � Proof: � Letf: R be a Lipschitz continuous function. Clarke’s generalized direc- X− tional derivative off at a pointx in the directionh , denoted byf ◦(x;h), is given ∈X by : → f(z+ th)−f(z) f◦(x;h) := lim sup . (4.11) z x,t 0 t Clarke’s generalized subdifferential→ of↓f atx is given by ∂f(x) :={λ X ∗ :f ◦(x;h) f(y) for ally }. (4.12) ∈ ≥ ∈X Theorem 4.67. Let be a Banach space such thatX ∗ is smooth. Then a nonempty X closed subsetC of is convex if its distance function dist( ,C) satisfies X · dist(x+ y, C)− dist(x, C) lim sup =1 for allx \C. (4.13) y ∈X y θ � � In particular, this di→fferentiability condition is satisfied if dist( ,C) is Gateaux-differentiable · atx and dist(x, C) S of the dual spaceX ∗ . ∈ 1 Proof: See [5]. � Theorem 4.68. Let be a Banach space such thatX ∗ is strictly convex. LetC X ⊂X be a Chebyshev set, letx C such that∂dist(x, C) is a singleton. Then the following ∈ assertions are equivalent: (a)C is convex. (b) dist( ,C) is convex. · (c) dist( ,C) is Gateaux differentiable atx. · 104 dist(x+ th)− dist(x) (d) There existsh S such that lim =1. ∈ 1 t 0 t dist(x+y)− dist(x) (e) lim sup =↓1. y θ y � � Proof: → (a)= (b) SinceC is convex it can easily be seen that dist( ,C) is convex. · (b)= (c) Since dist( ,C) is convex and continuous atx and∂dist( ,C) is a singleton, · · dist( ,C) is Gateaux differentiable atx and∂dist( ,C)(x) ={d � (x)}. ·⇒ · C (c)= (d) SinceC is a chebyshev set we have x−P (x) = dist(x, C). It follows from ⇒ � C � Gateaux differentiability of dist( ,C) that · ⇒ dist(x+ th, C)− dist(x, C) lim inf t 0 t exists for everyy . For eacht>0↓ we have ∈X dist(x+t(x−P (x),C)− dist(x, C) tdist(x, C). C ≤ Hence fory :=x−P C(x) we set dist(x+t(x−P (x),C)− dist(x, C) lim inf C = dist(x, C). t 0 t Sincex \C, dist(x,↓ C)>0 and ift � = t/dist(x, C) ast 0. Then by the above ∈X dist(x+t �(x−P C(x),C)− dist(x, C) lim inf ↓ = dist(x, C). t � 0 t If nowz=(x−P ↓(x)) x−P (x) −1 then we have C � C � dist(x+ tz, C)− dist(x, C) lim inf =1. t 0 t On the other hand, dist( ,C↓)) is a Lipschitz function and so · dist(x+ y, C)− dist(x, C) lim sup =1. y t 0 � � (d)= (e) Since dist( ,C) is↓ a Lipschitz function · dist(x+ y, C)− dist(x, C) ⇒ lim sup 1. y ≤ t 0 � � On the other hand, for each↓v S ∈ 1 dist(x+ tv, C)− dist(x, C) dist(x+ y, C)− dist(x, C) lim lim sup , t 0 t ≤ y y θ � � in particular↓ forv :=z in(d), we have → dist(x+ y, C)− dist(x, C) 1 lim sup . ≤ y y θ � � (e)= (a) Follows from → � ⇒ 105 Theorem 4.69. Let be a Banach space such thatX ∗ is smooth. Then a nonempty X closed subsetC of is convex if for somer>0 the set X P :={x 0−rz:x 0 P (C), p(x0) C, x 0−p(x0) = dist(x 0,C) > r, z=(x 0−p(x0)) x0−p(x0) −1} C,r ∈ C ∈ � � � � is dense in \C. X Proof: See [18, 19]. � Lemma 4.70. Let be a Banach space and letC be a nonempty subset of . If the X X distance mapping dist( ,C) is Gateaux-differentiable inx \ C andw P (x) then · ∈X ∈ C ddist(x)(x−w) = x−w and ddist(x) =1. (4.14) � � � � Proof: See [17] � 4.8 Differentiability properties of the metric projec- tions Let be a Hilbert space and letC be a nonempty closed convex subset o . The the H H metric projection C is welldefined and one may ask for differentiability properties ofP C . We say thatP is directionally differentiable at a pointx if C ∈H PC(x+ th)−P C(x) d+PC(x)(h) := lim (4.15) t 0 t exists for eachh . If the mappingh ↓ d P (x) is linear and continuous thenP ∈H �− + C C is called Gateaux-differentiable atx . This mapping, when it exists, is denoted by ∈H dP (x). If the above limit exists uniformly forh B and ifh d P (x) is linear and C → ∈ 1 �− + C continuous we say thatP C is Fr´echet-differentiable atx and denote the corresponding linear operator byDP C(x). → Lemma 4.71. Let be a Hilbert space and letC is Chebyshev subset of . Then we H H define 1 2 2 W: \C x ( x − dist(x, C) R. (4.16) H � �− 2 � � ∈ ThenW is Fr´echetdifferentiable andDW(x) =P (x), x . Moreover,W is convex → C ∈H and 1 W(x) = sup (2 x|u − u 2),x \C. (4.17) u C 2 � � � � ∈H ∈ Proof: � By the comments above, the metric projectionP T(x,C) exists. Zarantanello [42] and Haraux [20] (see also [39]) have proved the following result. 106 Theorem 4.72. Let be a Hilbert space and letC be a nonempty closed convex subset H of . Then for anyx C we have H ∈ P (x+ th) =x+ tP (h) +o(t) , t > 0, h . (4.18) C T(x,C) ∈H Proof: Letx C andh . We have to prove ∈ ∈H PC(x+ th)−P C(x) d+PC(x)(h) = lim =P T(x,C) (h). (4.19) t 0 t In afirst step we prove fort>0 ↓ P (x+ th)−P (x) C C =P (h). (4.20) t (C−x)/t We have fort>0 andy C ∈ 0 (x+ th)−P (x+ th)|y−P (x+ th) ≥� C C � = (x+ th)−P (x+ th)|y−(P (x+ th) +x−x) � C C � x−P (x+ th) y−x P (x+ th)−x =t 2 C +h| − C � t t t � P (x+ th)−x P (x+ th)−x =t 2 h− C |^y− C � t t � where^y=(C−x)/t . In a second step we verify lim inf{ h−u :u (C−x)/t}= inf{ h−v :v T(x, C)} (4.21) t 0 � � ∈ � � ∈ which means lim↓ h−P (h) = h−P (h) . Notice that the metric projections t 0 � (C−x)/t � � T(x,C) � P(C−x)/t(h),PT(x,C) (h) exist since(C−x)/t, T(x, C) are closed convex sets. If0=t t then↓ (C−x)/t (C−x)/t due to the fact thatC is convex. This shows � 1 ≤ 2 2 ⊂ 1 thatq := lim h−P (h) exist. Moreover since(C−x)/t T(x, C) for allt>0 t 0 � (C−x)/t � ⊂ we haveq h−P T(x,C) (h) . SinceP T(x,C) (h) T(x, C) we obtainq= h−P T(x,C) (h) . ≥� ↓ � ∈ � � Now, we obtain P −P (h) 2 =2 h−P 2 +2 h−P 2 −4 h � (C−x)/t T(x,C) � � (C−x)/t� � T(x,C) � � 1 − (P +P ) 2 2 (C−x)/t T(x,C) � 2 h−P 2 +2 h−P 2 −4 h−P 2 ≤ � (C−x)/t� � T(x,C) � � T(x,C) � and we conclude that lim P −P (h) 2 =0 which impliesfinally t 0 � (C−x)/t T(x,C) � ↓ PC(x+ th)−P C(x) lim =P T(x,C) (h). t 0 t ↓ � We conclude from the theorem above that at each pointx C, the mapP has ∈ C a directional derivative in every directionh given byP (h). Since the metric ∈H T(x,c) 107 projectionP T(x,C) will be linear only in the very special case whenT(X, C) is a linear subspace, it is clear that dPC(x) does not generally exist at pointsx ofC. In particular, ifC= andC has nonempty interior andx ∂C, thenT(x, C) will not be a subspace, � H ∈ so dPC(x) will not exist. Rademacher’s theorem (see [36] and [43]) states that a Lipschitz continuous function fromR n intoR m is dfferentiable almost everywhere. The theorem of Alexandrov (see [1] and [31]) says that each continuous convex function onR n is almost everywhere twice differentiable. Forn=1 this reduces to Lebesgue’s theorem about the differentiability almost everywhere of a monotone function. Alexandrov’s theorem exploits Asplund (see [2]) as follows: Theorem 4.73 (Asplund, 1973). LetC be a nonempty closed convex subset the inner n product space :=R and letP C be the associated metric projection from intoC. H H ThenP C is differentable almost everywhere. Proof: Define 1 2 f: R,x sup ( x|y − y ). H− �− y C � � 2� � ∈ Obviously,f is convex, actuallyf is the Fenchel-conjugate ofj: u 1 u 2 R. → → H� �− 2 � � ∈ One can show that 1 2 1 2 1 2 1 2 → f(x) = x − inf x−y = x − x−P C(x) , x . 2 y C 2 2 2 � � ∈ � � � � � � ∈H From this we conclude thatP C(x) is in the subdifferential off atx; see the chapter concerning the fundamentals facts on convex analysis. Since we know from Alexandrov’s theorem thatf is twice (Fr´echet-)differentiable almost everywhere the metric projection is almost everywhere (Fr´echet-)differentiable. � LetC be a nonempty closed convex subset the inner product space :=R n and let H P be the associated metric projection from intoC. It is known that ifx C, thenP C H ∈ C is directionally differentiable atx and the corresponding directional derivatives PC(x+ th)−P C(x) d+PC(x)(h) = lim t 0 t are given by the metric projection ofh onto↑ the support coneS C(x) toC atx. In the case x / C directional differentiability of atx is guaranteed only under certain conditions ∈ C on smoothness of the boundary ofC. Here is an example inR 3 (constructed by Shapiro in [38]) whose metric projection fails to be directionally differentiable; see also [33]. Example 4.74. Consider a sequence(α n)n N of positive numbers defined byλ 1 :=π/2 n ∈ andα n+1 :=c α1, n N, withc (0, 1). Then this sequence is monotone decreasing ∈ ∈ and limn αn =0. LetC be defined as the convex hull of{(cos(α n), sin(αn)) :n N} ∈ ∪ {(0, 0),(1, 0)}. Letx := (2, 0), d := (0, 1). We will show thatP C(x) is not directionally differentiable inx in directiond. Consider forn= 2, 3, . . . 1 1 t := sin(α )+(2−cos(α )) tan( (α α )),s := sin(α )+(2−cos(α )) tan( (α α )). n n n 2 + n−1 n n n 2 + n+1 108 Then we haveP C(x) = (1, 0) and PC(x+t nd) =P C(x+s nd) = (cos(αn), sin(αn)),n= 2, 3, . . . . Since sin(αn) 2 sin(αn) 2 lim = −1 , lim = n tn 3+c n sn 3+c we obtain π (P (x+t d)−π (P (x)) π (P (x+s d)−π (P (x)) lim 2 C n 2 C = lim 2 C n 2 C , n tn � n sn whereπ 2(u) denotes the second coordinateu 2 ofu=(u 1, u2). � 4.9 Appendix: Subdifferential LetX be a Banach space and letf:X R^ be proper, convex and closed. We consider − the epsilon-subdifferential: 0 0 → 0 0 ∂ f(x ) :={λ X ∗|f(y) f(x )+ λ, y−x −ε for ally X},ε 0, x dom(f). (4.22) ε ∈ ≥ � � ∈ ≥ ∈ 0 0 Forε=0 we write∂f(x ) :=∂ 0f(x ). The epsilon-subdifferential defines a point-to-set operator∂ f:X x ∂ f(x) POT(X ∗); we use the notation from the theory of ε � �− ε ∈ set-valued mappings: → ∂εf:X⇒X ∗ It can easily proved that for allx X,ε 0 the set∂ f(x) is convex and weak∗-closed ∈ ≥ ε (possible empty). Moreover, the following properties are useful: x X ε ,ε 0(ε ε = ∂ f(x) ∂ f(x)) (4.23) ∀ ∈ ∀ 1 2 ≥ 1 ≤ 2 ε1 ⊂ ε2 x X ε 0(∂ f(x) = ∂ f(x)) (4.24) ∀ ∈ ∀ ≥ ε ⇒ η η>ε � Theorem 4.75. LetX be a Banach space, letf:X R^ be convex and suppose thatf − is continous inx 0 dom(f). Then ∈ (1)∂f(x 0)= . → � ∅ 0 (2)∂f(x ) is convex andσ(X ∗,X)-closed. 0 (3)∂f(x ) is bounded (as a subset ofX ∗). Notice hat the assumption thatf is continuous inx 0 dom(f) cannot be replaced by ∈ the assumption thatf is lower semicontinuous inx 0 dom(f) as wes see in the example ∈ −√1−x 2 if|x| 1 f:R x ≤ R { }. � �− � else ∈ ∪ → ∞ Corollary 4.76. LetX be a Banach space, letf:X R^ be convex,U dom(f) ∞ − ⊂ convex and open, and suppose thatf is bounded from above in a neighborhoodV of a point x U. Then∂f(x 0)= for allx 0 U. ∈ � ∅ ∈ → 109 Theorem 4.77. LetX be a Banach space and letf:X R^ be proper, convex and − closed. Ifx 0 dom(f),ε>0 then∂ f(x0) is nonempty. ∈ ε Proof: → The setA := epi(f) is nonempty, convex and closed and we have: (x0, f(x0)−ε)/ epi(f),B :={(x 0, f(x0)−ε)} convex, compact ∈ Therefore we may apply a separation theorem (see for instance [3]) to the setsA,B and obtainµ X ∗,α R such that ∈ ∈ µ, x0 +α(f(x 0)−ε)< µ, y +αt for ally dom(f), t f(y). � � � � ∈ ≥ If we choosey=x 0, t=f(x 0) we obtain−αε<0 which impliesα>0. Withλ :=−1/αµ andt=f(y) we have − λ, x0 +f(x 0)−ε<− λ, y +f(y) for ally X. � � � � ∈ This showsλ ∂ f(x0). ∈ ε � LetX be a Banach space. We setN:X x x R. Then the subfifferential of � �− � � ∈ the norm functionN has the form x → x ifx=θ ∂N(x) = � � � (4.25) �B1 ifx=θ Theorem 4.78 (Bronsted and Rockafellar, 1965). LetX be a Banach space and letf: 0 0 X R^ be proper, convex and closed. Ifx dom(f),ε > 0,γ>0 andξ 0 ∂ εf(x ) − ∈ ∈ then there existsx dom(f),ξ X ∗ with ∈ ∈ → ξ ∂f(x 0), x−x 0 ε/γ, ξ−ξ γ. (4.26) ∈ � � ≤ � 0� ≤ In particular, the domain dom(∂f) is dense in dom(f). Proof: We have ξ , y−x 0 f(y)−f(x 0) +ε for ally X. � 0 � ≤ ∈ We consider the mappingg:X y f(y)− ξ 0, y (− , ]. Theng is proper, � �− 0 � � ∈ convex and closed and we haveg(x ) inf x X g(x) +ε. Now, we may apply Ekeland’s ≤ ∈ variational principle. This givesz dom(g) = dom(f) such that ∈ → ∞ ∞ z−x 0 ε/γ, g(z) g(y) +γ y−z for ally X. � � ≤ ≤ � � ∈ With the mappingh:X y γ y−z R we havez dom(g) dom(h) and � �− � � ∈ ∈ ∩ θ ∂(g+h)(z) =∂g(z) +∂h(z) →∈ sinceg is convex and closed andh is continuous. Therefore, there existµ∂g(z),η ∂h(z) ∈ withθ=µ+η. Clearly,∂g(z) =∂f(z)−ξ 0 . Moreover, ∂h(z) ={ζ X ∗| ζ γ} ∈ � � ≤ 110 which is proved as follows. Ifζ ∂h(z) then ∈ ζ, y−z h(y)−h(z) =γ y−z for ally X � � ≤ � � ∈ and therefore ζ γ. � � ≤ Ifζ X ∗ with ζ γ thenζ ∂h(z) since ∈ � � ≤ ∈ ζ, y−z ζ y−z γ y−z =h(y)−h(z) for ally X. � � ≤ � �� � ≤ � � ∈ Thusµ=ξ−ξ withξ ∂f(z) andµ=−η, hence ξ−ξ γ. Now, we have the 0 ∈ � 0� ≤ result by choosingx :=z. The assertion concerning the domain dom(∂f) is an easy consequence.� The connection of subdifferentiability and conjugation is content of the following the- orem. Theorem 4.79. LetX be a Banach space, letf:X R^ be convex and letx 0 − ∈ dom(f),λ X ∗ . Then the following conditions are equivalent: ∈ (a)λ ∂f(x 0). → ∈ 0 0 (b)f(x ) +f ∗(λ) = λ, x . � � (c)f(x 0)− λ, x 0 f(x)− λ, x for allx X. � � ≤ � � ∈ Theorem 4.80 (Moreau-Rockafellar, 1965). LetX be a Banach space, let f, g:X R^ − convex and suppose thatf org is continuous in a pointz dom(f) dom(g). Then we 0 ∈ ∩ have for allx X → ∈ 0 0 0 ∂(f+g)(x ) =∂f(x ) +∂g(x ) (4.27) Corollary 4.81. LetX be a Banach space and letF i :X R^ be convex,i= 1, . . . , m . − m We assume thatF 1,...,Fm−1 are continuous in a point inz i=1 dom(Fi). Then we have 0 for allx X → ∈ m m � 0 0 ∂( Fi)(x ) = ∂Fi(x ) i 1 i 1 �= �= Lemma 4.82. LetX be a Banach space and letf:X R^ be convex. Then − ∂(tf)(x0) =t∂f(x 0) for allx 0 X, t > 0 . →∈ Theorem 4.83. LetX,Y be Banach spaces, letA:X Y be a linear and continuous − mapping ans letf:Y R^ be convex. Suppose that there exists a pointz X such that − ∈ f is continuous in Az . Then → → 0 0 0 ∂(f A)(x ) =A ∗∂f(Ax ) for allx X. (4.28) ◦ ∈ Definition 4.84. LetX be a Banach space, letf:X R^ and letx 0 dom(f), v X. − ∈ ∈ The limit 0 1 0 0 f+� (x , v) := lim (f(x + hv→)−f(x )) h 0 h ¯ 0 (in R) is called the (onesided) directional↓ derivative off inx in directionv.� 111 0 ¯ Notice thatf +� (x , ):X R is convex and positively homogeneous. · − Theorem 4.85 (Moreau-Psenichnii, 1965). LetX be a Banach space, letf:X R^ → − be proper, convex and suppose thatf is continuous inx 0 dom(f). Then we have for all ∈ v X: ∈ → 0 (1)f +� (x , v) = maxλ ∂f(x0) λ, v . ∈ � � 0 (2)−f +� (x ,−v) = min λ ∂f(x0) λ, v . ∈ � � Remark 4.86. If we in Theorem 10.54 ommit the assumption thatf is continuous one can prove the following result (see [4]): 0 f+� (x , v) = inf sup λ, v . (4.29) ε 0 0 λ ∂εf(x )� � ∈ ↓ � Definition 4.87. LetX be a Banach space, letf:X R^ and letx 0 dom(f). If the − ∈ limit df(x0)(h) := lim(f(x0 + th)−f(x 0))/t , h X t 0 → ∈ 0 0 exists inR for allh X and if df(x ):X R is linear and bounded (i.e. df(x ) X ∗) ∈ → − ∈ thenf is called Gateaux-differentiable inx 0 , and df(x0) is the Gateaux derivative 0 inx . → � Corollary 4.88. LetX be a Banach space, letf:X R^ be convex and letx 0 dom(f). − ∈ Then the following conditions areequivalent: → (a)f is Gateaux-differentiable inx 0 . (b)∂f(x 0) is a singleton. 1 p Lemma 4.89. Let be a Banach space and letj p :X x x R with X � �− p � � ∈ 1 < p < . Then for eachx Xs holds: ∈� → p−1 J (x) :=∂j (x) ={λ X ∗ : λ, x = x λ , λ = x . (4.30) ∞ p p ∈ � � � �� � � � � � Proof: Siehe Kien � In the casep=2 we use the short notationJ=J 2 . Let us prove the sandwich theorem. Here we need Definition 4.90. Let be a Banach space and letS be a subset of . We sayx core(S) X X ∈ if = r(S−x). X ∪ r>0 � 112 Theorem 4.91 (Sandwich Theorem). Let be a Banach space and let f,−g Γ ( ) X ∈ 0 X and suppose f(x) g(x) for allx . ≥ ∈X Assume that the following constraint qualification (CQ) holds: θ core(dom(f)− dom(−g)). (4.31) ∈ Then there is n affine functiona such that f(x) a(x) g(x) for allx . ≥ ≥ ∈X Proof: Considerh: z inf u (f(u)−g(u−z)) R^. Thenh is convex by Lemma?? X� �− ∈X ∈ and continuous atz=θ. Hence there is some−λ ∂h(θ), and this provides the linear ∈ part of the assured→ affine function. Indeed, we have f(x)−g(z−u) h(z)−h(θ) λ, z . ≥ ≥� � � Corollary 4.92. Let be a Banach space and let f, g Γ ( ) and let X ∈ 0 X θ core(dom(f) + dom(g)). (4.32) ∈ Then ∂(f+g) =∂f+∂g . Proof: � Corollary 4.93. Let be a Banach space with duality mappingJ , letf Γ 0( ), let X X ∈ X j := 1 2, and leth :=f j . Then 2 �·� � (a) dom(h) = andh is continuous at eachx . X ∈X (b)λ ∂f(v) +J (v) if and only ifh ∗(λ) +h(v)− λ, v 0. ∈ X � � ≤ Proof: � 4.10 Conclusions and comments 4.11 Exercises 1.) Let be a normed space. Then the following conditions are equivalent: X (a) is uniformly convex. X 113 n n n n (b) If(x )n N,(y )n N are sequences in with x = y = 1, n N, then we ∈ ∈ X � � � � ∈ have: 1 From lim (xn +y n) =1 we conclude lim (xn −y n) =θ. n �2 � n n n n n (c) If(x )n N,(y )n N are sequences in with lim sup n x 1, lim supn y ∈ ∈ X � � ≤ � � ≤ 1, n N, we have: ∈ 1 From lim( (xn +y n) ) =1 we conclude lim (xn −y n) =θ. n �2 � n n 2.) Let be a uniformly convex Banach space. If(x )n N is a sequence in with X ∈ X 1 α := lim xn = lim (xn +x m) , n � � n,m �2 � Then this sequence is convergent. n 3.) Let be a uniformly convex Banach space and let(x )n N be a sequence inX. X ∈ Then the following conditions are equivalent: n (a) limn x =x. (b)w− lim xn =x, lim xn = x . n n � � � � 4.) We say that the Banach space has a quadratic modulus of smoothness if with a X positive constantk we have x+y + x−y 2+k y 2 for all x, y Xs . � � � � ≤ � � ∈ Show that the Banach spacesl , 2 p< , has a quadratic modulus of smooth- p ≤ ness. 5.) Show: If a Banach space has a quadratic∞ modulus of smoothness then the associ- ated duality map is Lipschitz continuous. 2 2 1 6.) Let :=C[0, 1] endowed with the norm f := ( f + f ) 2 . Then is strictly X � � � � � � 2 X convex but nor uniformly convex. ∞ 7.) Let be a Hilbert space and letC be nonempty closed convex subset of . Let H H P be the metric projection. Show for x, y : C ∈H P (x)−P (y) < x−y orP (x)−P (y) =x−y. � C C � � � C C 8.) Let be a Banach space and letU,V closed subspaces of . X X (a) Show that ifU is afinite dimensional subspace thenU+V is closed. (b) Consider now the space :=c of real sequences converging to zero endowed X 0 with the supremum norm. Let n n n−1 U :={(x )n N c 0 :x = nx , n even}, ∈ n ∈ n V :={(x )n N c 0 :x = 0, n odd}. ∈ ∈ Show thatU+V is dense inc 0 but not closed. 114 2 9.) LetR endowed with the euclidean norm. Consider the sequence(C n)n N with ∈ Cn :={x, x/n):x R}, n N. Show tahtC := lim M Cn exists. ∈ ∈ 1 n 10.) Let Hs be the Hilbert spacel 2 and consider the setsC n := span({e , e }), n N. ∈ What can be said about limM Cn, limW Cn, limS Cn? 11.) Show that the spacesc 0 andl 1 are isomorphic. Herec 0 is the space of real sequences converging to zero endowed with the supremum norm. 12.) Show: (a) The closed unit ball inc 0 has no extreme point. (b) The closed unit ball inc has an extreme point. Herec is the space of converging real sequences andc 0 is the space of real sequences converging to zero; both spaces are endowed with the supremum norm. Is it possible thatc 0, c are isometric isomorph? 13.) Let be a Banach space and letF: be lower semicontinuous. Moreover X X⇒X letF(x) not empty for allx . 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