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.