Properties of the metric projection

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1 Chaper 4 Properies of he meric projecion In he focus of his chaper are coninuiy and differeniabiliy properies of he meric projecions. We mainly resric us o he case of a projecion ono Chebyshev ses. Firs of all, we sudy geomeric properies of Banach spaces which are helpful o analyze meric projecions. In he focus of he geomeric view on Banach spaces is he propery of smoohness. For he maerial on geomeric Banach space heory we refer o Borwein and Lewis [7], Chidume [10], Cioranescu [11], Deimling [14], Kabalo [4], Hiriar-Urruy and Lemarechal [, 1], Megginson [30] and Werner [41]. 4.1 Uniform convexiy Nex we consider a class of Banach spaces which share wih Hilber spaces he following propery: heir uni ball B 1 is round and ouches convex ses jus in one poin. We may say ha he following consideraions are devoed o he geomeric properies of Banach spaces. Definiion 4.1. A Banach space X is uniformly convex (or srongly round) if and only if is modulus of convexiy δ X defined by δ X (ε) := inf{1 1 x + y : x = y = 1, x y = ε}, ε [0, ], is posiive for all ε (0, ]. Clearly, a uniformly convex space is sricly convex. Le us collec some properies of he modulus δ X. Lemma 4.. Le X be a Banach space wih modulus of convexiy δ X. Then we have: (1) δ X (0) = 0, δ X (ε) 1 ε, ε 0. () (0, ] ε δ X (ε)ε 1 R is non-decreasing. (3) If X is a Hilber space H hen δ H (ε) = ε, ε (0, ]. 78

2 Ad (1) Evidenely, δ X (0) = 0. Le ε [0, ] and le x, y S 1 wih x y = ε. Then 1 x + y = x + 1 (y x) x 1 y x 1 1 ε. Ad () Le 0 < η ε. Choose x, y S 1 wih x y = ε. Se u := η ε x + (1 η ε ) x + y x + y, w := η ε y + (1 η ε ) x + y x + y. Then u w = η ε x y = η, 1 (u + w) = 1 η ε (x + y) + 1 (1 η ε ) x + y x + y and x + y x + y 1 (u + w) = η ε 1 η ε x + y = 1 (1 η ε + 1 η ε x + y ) = 1 1 u + v. Noice ha Therefore x + y x + y 1 (u + v) u v Now, we obain x + y x + y 1 (x + y) = 1 1 x + y. = ( η ε 1 η x + y )/η ε x + y = (1 1 x + y )/ε = x + y 1 (x + y). x y δ X (η) η = 1 1 u + v x + y = u v x + y x + y 1 (x + y) = x y x + y 1 (u + v) u v 1 1 x + y x y = 1 1 x + y ε Taking he infimum wih respec o he choice of x, y we obain he asserion. Ad (3) Follows from he parallelogram ideniy. Remark 4.3. Le δ X be he modulus of convexiy in he Banach space X. There exiss a consan 0 < L < 3.18 such ha if 0 < η ε hen we have (see [16]) δ X (η) η 4L δ X (ε) ε 79

3 Theorem 4.4. A uniformly convex Banach X space is an E-space. We know already ha X is sricly convex. To show he reflexiviy of X we may use he crierion of James (see he preliminaries in he preface). To apply his crierion we have o show ha for each funcional λ X wih λ = 1 here exiss x S 1 wih λ, x = 1. Le λ X wih λ = 1. Then here exiss a sequence (x n ) n N in S 1 wih lim n λ, x n = 1. This implies lim n x n = 1 and from lim m,n λ, x n + x m = 1 we obain lim 1 n,m (xn + x m ) = 1. Due o he uniform convexiy, he sequence (x n ) n N converges o some x S 1. Clearly, λ, x = 1 = λ. To prove he hird propery of an E-space, le x S 1 and le x be he weak limi of he sequence (x n ) n N in S 1. Assume ha x is no he srong limi of his sequence. Then here exiss a subsequence (x n k )k N and ε > 0 such ha x n k x ε, k N. By he uniform convexiy here exiss a d > 0 such ha Le λ Σ(x). Then This is a conradicion. 1 (xn k + x) 1 d, k N. 1 = λ, x = lim k λ, 1 (xn k + x) λ (1 d) < 1 Theorem 4.5. Le X be a uniformly convex Banach space. Then every nonempy convex closed subse C of X is a Chebyshev se. Moreover, he bes approximaion problem (.1) is srongly solvable. This follows from Theorem 4.4 and Theorem.4. Remark 4.6. Noice ha he propery of uniform convexiy is no sable under equivalen renorming he space. Theorem 4.7 (Milman). A uniformly convex Banach space is reflexive. Follows from Theorem 4.4. For concree examples of uniformly convex spaces one can show he reflexiviy ad hoc. For insance, he spaces l p, 1 < p <, are uniformly convex and reflexive. The modulus of convexiy is given by δ(ε) := 1 q 1 ( 1 ε)q, 0 < ε, where = 1. To compue his resul one uses a Hölder-ype inequaliy. The reflexiviy p q follows from he fac ha he dual space of l p is given by l q where q is as above. Then we conclude ha l p = l q = l p. Obviously, he spaces l p, p = 1, are wheher uniformly convex nor reflexive. 80

4 Remark 4.8. The class of reflexive Banach spaces is larger han he class of uniformly convex Banach spaces. This shows he famous resul of M. Day (see [13]): There exiss a separable reflexive sricly convex Banach space which is no isomorphic o a uniformly convex Banach space. On he oher hand, a separable Banach space X can be renormed in such a way ha i is sricly convex. This follows from he fac ha here exiss an injecive linear mapping T : X l which leads o he new norm : X x x + Tx R. Now, (X, ) is sricly convex; see Koehe [6]. Remark 4.9. As we know from Theorem 4.7, each uniformly convex Banach space is reflexive. Obviously, he converse does no hold; consider for example he wo-dimensional space R endowed wih he l 1 norm. A more sharp resul concerning his quesion is he classical resul of M. Day ([13]): There exis Banach spaces which are separable, reflexive, and sricly convex, bu are no isomorphic o any uniformly convex space. This resul has lead o he class of Banach spaces which are called superreflexive. Such spaces are characerized by he propery ha hey admi an equivalen uniformly convex norm. We omi he quesion how a reflexive space which is no superreflexive looks like. Here is a resul which connecs uniform convexiy o he quaniaive behavior of he dualiy map. Theorem 4.10 (Prüss, 1981). Le X be a Banach space. Then X is uniformly convex if and only if for each R > 0 here exiss a gauge-funcion γ R such ha J X (x) J X (y), x y γ R ( x y ) x y for all x, y B R. (4.1) We follow [35]. Le X be uniformly convex and choose R > 0. Le γ R be defined by γ R (0) = 0, and γ R (r) := inf{ λ µ, x y x y 1 : x, y B R, x y r, λ J X (x), µ J X (y)}, r (0, R], We exend γ R by γ R (r) := γ R (R), r R. Obviously, γ R is nondecreasing. Le us prove ha γ R (r) > 0 for r > 0. Assume on he conrary ha γ R (r) = 0 for some r (0, R]. Then here exis sequences (x n ) n N, (y n ) n N belonging o B 1 and λ n J X (x n ), µ n J X (y n ) such ha x n y n r and lim n λ n µ n, x n y n x n y n 1 = 0. Since we know (see (3.4)) ( x n y n ) λ n µ n, x n y n we may assume lim n x n = lim n y = a > 0. Then lim inf n x n 1 x n y n 1 y n = a 1 lim inf n Therefore, by he uniform convexiy of X, lim sup n x n y n ra 1. x n + y n = a lim sup x n 1 x n + y n y n a(1 δ) n 81

5 for some 0 < δ = δ(r). On he oher hand, x n + y n µ n, x n λ n, y n = λ n µ n, x n y n. This implies lim n λ n, x n + y n = a and leads o he conradicion a = lim λ n, x n + y n a lim sup x n + y n a (1 δ). n n Le us prove he sufficiency. We firs show ha he range of J is closed. Le (λ n ) n N be a sequence in X wih λ n J X (x n ), n N, and lim n λ n = λ X. Then (x n ) n N is bounded, say x n R, n N, due o he fac ha (λ n ) n N is bounded. Then we find γ R wih γ R ( x n x m ) x n y m λ n λ m, x n x m x n x m λ n λ m, n, m N. Hence, (x n ) n N is a Cauchy sequence and herefore convergen. Le lim n x n = z. Then λ J X (z) and J has closed range. Therefore J is surjecive since ran(j) is dense in X and X is reflexive. This follows from he Bishop-Phelps-heorem; see [6]. Now, we have J = J 1. Due o γ R ( x y ) λ µ for λ, µ X, λ R, µ R, x J (λ), y J (µ), hence J is single-valued and uniformly coninuous on bounded subses of X. Therefore X = (X ) is uniformly convex and he proof is complee. 4. Uniform smoohness Definiion A Banach space X is called uniformly smooh if he he following propery holds: ε δ > 0 x, y X ( x = 1, y δ = x + y + x y < ε y ) To make he definiion more applicable le us inroduce a modulus which can be used o characerize uniformly smooh spaces. Definiion 4.1. Le X be a Banach space. The modulus of smoohness of X is he funcion ρ X : [0, ) [0, ) defined by ρ X (τ) := sup{ 1 ( x + y + 1 x y ) 1 : x = 1, y = τ} = sup{ 1 ( x + τy + 1 x τy ) 1 : x = 1 = y } Theorem Le X be a Banach space. Then he following condiions are equivalen: (a) X is uniformly convex. 8

6 (b) lim τ 0 ρ X (τ) τ = 0. (a) = (b) Le ε > 0. Then here exiss δ > 0 such ha 1 ( x + y + x y ) 1 < 1 ε y for all x, y X wih x = 1, y = δ. Then ρ X (τ) < 1 ετ for all τ (0, δ). (b) = (a) Le ε > 0. Then here exiss δ > 0 such ha ρ X (τ) < 1 ετ for all τ (0, δ]. Le x, y X wih x = 1, y δ. Se τ := y. Then x + y + x y < ετ = ε y. Lemma Le X be a Banach space wih modulus of smoohness ρ X. Then we have: (1) ρ X (0) = 0. () max{0, τ 1} ρ X (τ) τ for all τ [0, ). (3) (0, ) τ ρ X (τ)τ 1 (0, ) is non-decreasing. (4) ρ X is a coninuous convex funcion from (0, ) o (0, ). Ad (1) Obviuosly rue. Ad () Le x, y S 1. For τ > 0 le x S 1 and se y := τx. Then 1 ( x+y + x y ) = τ. This shows ρ X (τ) τ. Le τ > 0. Since ρ X (τ) τ we have o show ρ X (τ) τ 1. Le x, y S 1. Then 1 ( x + τy + x τy ) 1 1 τy 1 = τ 1. Ad (3) Le x, y S 1. Le 0 < s. Applying Lemma 3.16 we obain 1 ( x + sy + x sy ) 1 s = 1 ( x + sy x x sy x ) s s 1 ( x + y x x y x ) 1 ( x + y + x y ) 1 = Therefore ρ X (s) ρ X (). s Ad (4) I is sufficien o show ha ρ X is convex since a convex funcion is coninuous on he inerior of is domain of definiion. Le x, y X wih x = y = 1. Consider f x,y () := x + y + x y 83 1.

7 Then we obain for a [0, 1] f x,y (a + (1 a)s) = x + (a + (1 a)s)y + x (a + (1 a)s)y 1 a x + y + (1 a) x + sy + a x y + (1 a) x sy 1 = af x,y () + (1 a)f x,y (s) Therefore f x,y is convex for all x, y. Le ε > 0. Then here exis x, y X wih ρ X (a+(1 a)s) ε f x,y (a+(1 a)s) af x,y ()+(1 a)f x,y (s) aρ X ()+(1 a)ρ X (s). Since ε > 0 was arbirary chosen, ρ X is convex. Theorem A uniformly smooh Banach space is smooh. Assume ha X is no smooh. Then here exis x 0 X and λ, µ X, such ha λ µ, λ = µ = 1 and λ, x 0 = x 0 = µ, x 0. We may assume x 0 = 1. Le y 0 X wih y 0 = 1 and λ µ, y 0 > 0. We obain for > 0 0 < λ µ, y 0 = 1 ( λ, x0 +y 0 + µ, x 0 y 0 ) 1 1 ( x0 +y 0 + x 0 y 0 ) 1. (4.) Hence 0 < λ µ, y 0 ρ X ()/. Since > 0 was arbirary chosen we conclude ha X is no uniformly smooh. Theorem 4.16 (Lindensrauss-Tzafiri, 1978). Le X be a Banach space. For every τ > 0 we have ρ X (τ) = sup{ 1 τε δ X (ε) : 0 < ε }, ρ X (τ) = sup{ 1 τε δ X (ε) : 0 < ε } (4.3) Le us proof he firs formula. Le τ > 0 < ε, x, y X wih x = y = 1. From he Hahn-Banach heorem we obain λ, µ X mi λ = µ = 1 such ha Then λ, x + y = x + y and µ, x y = x y. x + y + τ x y = λ, x + y + τ µ, x y = x y = λ + τµ, x + λ τµ, y λ + τµ + λ τµ sup{ λ + τµ + λ τµ : x = y = 1} = ρ X (τ) 84

8 If addiional x y ε hen we obain 1 τε ρ X (τ) 1 1 x + y. Therefore 1 τε ρ X (τ) δ X (ε) and since 0 < ε is arbirary chosen sup{ 1 τε δ X (ε) : 0 < ε } ρ X (τ). Choose λ, µ X wih λ = µ = 1 and δ > 0. For τ > 0 here exis x 0, y 0 X wih x 0 y 0 and x 0 = y 0 = 1 such ha λ + τµ λ + τµ, x 0 + δ, λ τµ λ τµ, y 0 + δ. Using hese inequaliies we obain λ + τµ + λ τµ λ + τµ, x 0 + λ τµ, y 0 + δ = λ, x 0 + y 0 + µ, x 0 τy 0 + δ x 0 + y 0 + τ µ, x 0 y 0 + δ. We se ε 0 := µ, x 0 y 0. Then we have 0 < ε 0 x 0 + y 0 and 1 ( x0 + τy 0 + x 0 τy 0 ) 1 1 τε 0 + δ δ X (ε 0 ) Since δ > 0 is arbirary chosen δ + sup{ 1 τε δ X (ε) : 0 < ε }. ρ X (τ) sup{ 1 τε δ X (ε) : 0 < ε } and he proof is complee. Le us prove he second formula. Le τ > 0 and le λ, µ X wih λ = µ = 1. For η > 0 here exis x 0, y 0 in X wih x 0 = y 0 = 1 such ha Then λ + µ η λ + µ, x 0, λ µ η λ µ, y 0. λ + µ + τ λ µ λ + µ, x 0 + τ λ µ, y 0 η(1 + τ) If 0 < ε λ µ hen we have sup{ x + τy + x τy : x = y = 1} + η(1 + τ) = ρ X (τ) + η(1 + τ). was 1 τε ρ X (τ) η(1 + τ) 1 1 (λ + µ), 1 τε ρ X (τ) η(1 + τ) δ X (ε). 85

9 Since η > 0 is arbirary we obain for all ε (0, ] and finally 1 τε ρ X (τ) δ X (ε) sup{ 1 τε ρ X (τ) : 0 < ε } ρ X (τ). Le x, y X wih x = y = 1 and le τ > 0. From he Hahn-Banach heorem we obain λ, µ X wih λ = µ = 1, such ha Then λ, x + τy = x + τy and µ, x τy = x τy. x + τy + x τy = λ, x + τy + µ, x τy = x τy = λ + µ, x + τ λ µ, y λ + τµ + λ τµ We define ε 0 := λ µ, y. Then we have 0 < ε 0 x y and This implies 1 ( x + τy + x τy ) 1 1 ( λ + µ + τ λ µ, y ) 1 = 1 τε 0 (1 1 λ + µ ) 1 τε 0 δ X (ε 0 ) sup{ 1 τε δ X (ε) : 0 < ε }. sup{ 1 τε ρ X(τ) : 0 < ε } ρ X (τ). Le us add a few properies of he convexiy and smoohness modulus. Corollary Le X be a Banach space and le H be a Hilber space. (1) δ X is convex and coninuous. () δ X is sricly in increasing iff X is uniformly convex. (3) δ X (ε) δ H (ε), ε [0, ]. (4) ρ X (τ) ρ H (τ), τ (0, ). (5) ρ H (τ) = 1 + τ 1. 86

10 For he proofs of (1) (4) we refer o he lieraure ([11, 30, 41]). Ad (5) his follows from he second formula in Theorem Theorem 4.18 (Shmulian, 1940). Le X be a Banach space. We have: (a) (b) X is uniformly smooh if and only if X is uniformly convex. X is uniformly convex if and only if X is uniformly smooh. Ad (a) Le X be uniformly smooh. Assume ha X is no uniformly convex. Then here exiss ε 0 (0, ] wih δ X (ε 0 ) = 0 and we obain from Theorem 4.16 for all τ > 0 0 < 1ε 0 ρ X (τ)/τ. This implies ha X is no uniformly smooh. Le X be uniformly convex. Assume ha X is no uniformly smooh. Then lim τ 0 ρ X (τ)/τ 0. Hence, here exiss ε > 0 such ha for all δ > 0 here exiss wih 0 < < δ and ε ρ X (). Then here exiss a sequence (τ n ) n N wih 0 < τ n < 1, ρ X (τ n ) > 1ετ n, n N, lim n τ n = 0. Due o Theorem 4.16 we obain for each n N ε n (0, ] such ha 1ετ n 1τ nε n δ X (ε n ). This implies 0 < δ X (ε n ) 1τ n(ε n ε). Morreover, ε < ε n und lim n δ X (ε n ) = 0. Since δ X is nondecreasing we obain δ X (ε) δ X (ε n ), n N. This shows δ X (ε) 0. Hence X is no uniformly convex. Ad (b) The proof folows along he lines above by exchanging he roles of X and X. Lemma Le X be a uniformly smooh Banach space. Then is dualiy map J is single-valued and uniformly coninuous on bounded ses. Since X is uniformly smooh X is uniformly convex; see Theorem Then we know ha J is single-valued; see Lemma I is enough o prove he uniformly boundeness propery for he bounded se S 1. Le ε > 0 and le x, y S 1 wih x y < δ X (ε). Then J X (x) J X (y) J X (x) + J X (y), x = J X (x), x + J X (y), y + J X (y), x y This shows J X (x) J X (y) < ε. x + y x y J X (y) > (1 δ X ).(ε) Le H be a Hilber space. Then wih he resuls in Theorem 4.16 we obain δ H (ε) = ε = 1 8 ε + O(ε 4 ), ρ H (τ) = 1 + τ 1 = 1 τ + O(τ 4 ). Hilber spaces are he mos uniformly convex and mos uniformly smooh Banach spaces in he sense ha if any space whose dimension is a leas wo, hen δ H (ε) δ X (ε) and ρ H (τ) ρ X (τ) ; see for insance [34], Day [8], [30]. Quaniaive versions of sric convexiy and smoohness are of paricular imporance in Banach space heory. Here hey can be used o quanify he coninuiy properies of he meric projecion; see below. Remark 4.0. For uniformly convex smooh Banach spaces here exis crieria for convexiy of heir Chebyshev ses: if C be a Chebyschev subse of he uniformly convex smooh Banach space X hen C is convex iff C is approximaely compac. 87

11 4.3 Srongly smoohness Le X be a Banach space wih norm and norm mapping ν : X x x R. If he norm is Gaeaux differeniable in x X \{θ} hen he limi x + y x lim 0 (4.4) exiss and is equal o he value ν (x, y) of he uniquely deermined funcional ν (x, ) X ; see Secion 3.3. Definiion 4.1. Le X be a Banach space. (a) The norm in X is called Fréche-differeniable a x X \{θ} if (b) x + y x lim 0 exiss uniformly in y S 1. The limi is called he Fréche derivaive a x in direcion y. The norm in X is called Fréche-differeniable if he norm is Fréche-differeniable a each x X \{θ}. (c) X is called srongly smooh if he norm is Fréche-differeniable a each x X \{θ}. (d) The norm in X is called uniformly Fréche-differeniable if he limi in exiss uniformly for (x, y) S 1 S 1. x + y x lim 0 Le X be a Banach space wih norm and norm-mapping ν. I is clear ha a a poin x where he norm is Fréche differeniable he norm is Gaeaux differeniable and he Fréche derivaive in x in direcion y is equal o ν (x, y). Since he norm is homogeneous one has o check he Fréche differeniabiliy a x S 1 only. In he general, Gaeaux differeniabiliy of he norm does no imply Fréche differeniabiliy. An example is he norm in l 1 ; see [30]. Lemma 4.. Le X be a Banach space wih norm and norm-mapping ν. Le x X \{θ}. Then he following saemens are equivalen: (a) The norm is Fréche differeniable in x. (b) x + v x ν (x, v) lim v θ v Observe ha v = y wih = v, y = v v 1 S 1. = 0. 88

12 Lemma 4.3 (Shmulian, 1939). Le X be a Banach space wih norm-mapping ν. Then for x S 1 he following condiions are equivalen: (a) The norm is Fréche-differeniable in x. (b) If (λ n ) n N is a sequence in X wih λ n 1, n N, and lim n λ n, x = 1 hen his sequence is convergen. (a) = (b) Le λ := ν (x, ) and le (λ n ) n N be a sequence in X wih λ n 1, n N, and lim n λ n, x = 1. Since 1 λ n λ n, x, n N, and lim n λ n, x = 1 we may suppose ha λ n = 1 for all n N. Assume ha (λ n ) n N does no converge o λ. Then here exiss ε > 0 and (z n ) n N in X wih z n = 1, n N, wih λ n λ, z n ε, n N. We se x n := z n (1 λ n, x )ε 1. Then lim n x n = θ and (see 4.) ν(x + x n ) ν(x) λ, x n x n λ n, x + x n 1 λ, x n x n = λ n, x 1 + λ n λ, z n (1 λ n, x )ε 1 (1 λ n, x )ε 1 = λ n λ, z n ε ε This conradics he Fréche-differeniabiliy of ν in x. (b) = (a) Se λ := ν +(x, ). Assume ha here exiss ε > 0 and (x n ) n N wih lim n x n = θ such ha Then ν(x + x n ) ν(x) λ, x n x n ε, n N. x + x n λ, x + x n ε x n, n N. Choose λ n S 1 such ha λ n, x + x n = x + x n, n N. Since lim n x n = θ we have lim n λ n, x = x = 1. Bu λ n λ λ n λ, x n x n 1 λ, x λ n, x x n + ε ε since λ, x = x λ n, x. This means ha (λ n ) n N does no converge o λ. Corollary 4.4. Le X be a Banach space and le he norm in he dual space X be Fréche differeniable. Then X is reflexive. We use he resul of James which says ha X is reflexive if and only if each nonzero funcional λ X aains is norm a some x S 1 ; see he preliminaries in he preface. Le λ X, λ = 1, and le (x n ) n N be a sequence in S 1 such ha lim n λ, x n = 1. By applying Smulian s lemma, lim n x n = x S 1. Therefore λ, x = lim n λ, x n = 1 = λ. If now µ X \{θ} apply he consideraion above o λ := µ µ 1. 89

13 Theorem 4.5. Le X be a Banach space wih dualiy map J and norm mapping ν. Then we have he equivalence of he following condiions: (a) (b) (c) (d) J is single-valued and uniformly norm-o-norm coninuous an bounded subses. The norm in X is uniformly Fréche differeniable. X is uniformly smooh. X is uniformly convex. (a) = (b) Le x S 1, y X. Le > 0 be such ha x + y > 0. Then Hence J X (x), y = J X (x), x x + J X (x), y = J X (x), x + y x x + y x = x + y x x + y x + y J X (x + y), x + y J X (x + y), x x + y = J X (x + y), y + J X (x + y), x J X (x + y), x x + y J X (x + y), y x + y J X (x), y x x + y x J X (x + y), y x + y Using he propery (a) we read off ha he norm is uniformly Fréche differeniable. (b) = (c) Assuming (b) we can choose for ε > 0 a δ > 0 so ha for all (0, δ) and all x, y S 1 x + y x ν (x, y) < 1 ε. Bu his implies ha for all (0, δ) and all x, y S 1 ( ) x + y x x + y x y = + ν (x, y) ( ) x + ( y) x + ν (x, y) < + ε. This implies he asserion in (c) (c) = (d) Le ε > 0. By (c) we can find δ > 0 such ha for all x S 1 and z X wih z δ we have x + z + x z < ε z.. 90

14 Le λ, µ X wih λ = µ = 1 and λ µ ε. Then here is a z X wih z 1δ and λ µ, z 1 εδ. This implies λ + µ = sup x S 1 λ + µ, z = sup x S 1 λ, x + z + µ, x z λ µ, z sup( x + z + x z ) 1 x S 1 εδ ε z 1 εδ < 1 4 εδ. (c) = (a) J is single-valued due o Lemma Le ε > 0. Then here exiss δ > 0 such ha for all λ, µ S 1 X we have λ + µ > δ implies λ µ < ε. Choose x, y S 1 X wih x +µ = 1 and x y = δ. Then wih λ J X (x), µ J X (y) we obain λ + µ λ + µ, y = λ, x + µ, y λ, x y > x y > δ. Therefore J is norm-o-norm coninuous on uni sphere. Since J is homogeneous, i.e. J X (ax) = aj X (x) for all x X, a R, J is norm-o-norm coninuous on he uni ball. This implies ha J is uniformly coninuous on each bounded se. Remark 4.6. If he he norm in a Banach space is Fréche differeniable hen he dualiy map is uniformly coninuous on bounded subses from he norm-opology o he weak - opology; see [10]. Theorem 4.7. Le X be a Banach space. Then he following saemens are equivalen: (a) (b) X is an E-space. X is srongly smooh. (a) = (b) We know by definiion ha X is reflexive and sricly convex and herefore X smooh by Theorem 3.7. We wan o use Lemma 4.3 o verify ha he norm in X is Fréche differeniable a each λ X. Le λ X wih λ = 1 and le (x n ) n N be a sequence in B 1 wih lim n λ, x n = 1. Now, 1 x n λ, x n, n N, which implies lim n x n = 1. Le z n := x n x n 1, n N, and le z be a weak cluser poin of he sequence (z n ) n N in S 1 ; i exiss due o he reflexiviy of X. Then z 1, bu λ, z = lim n λ, x n x n 1 = 1. So z = 1. Since X is an E- space we obain ha z is cluser poin of (x n ) n N wih respec o he norm convergence. Because X is smooh, z is uniquely deermined by he condiions z S 1, λ, z = 1. Therefore, lim n z n = z, and so lim n x n = x also. (b) = (a) This is proved similarly. We omi he proof. 91

15 4.4 Radial projecion and norm inequaliies Here we wan o sudy he projecion ono he uni ball in a Banach space X. Such a projecion is he mapping { x if x 1 P rad : X x x x if x 1 B 1 X. (4.5) Lemma 4.8. Le X be a Banach space and le x X \{θ}. Then P rad P B1, especially B 1 is a proximal subse. We wan o use he characerizaion resul 3.3. Le λ J X (x x x wih µ J X (x). We wan o show ha λ N( x x, B 1). This follows from ). Then λ = (1 1 x )µ λ, u x x = (1 1 x 1 )( µ, u µ, ) (1 x x x )( µ u x ) 0 for all u B 1. As a consequence of Lemma 4.8, P B1 (x) = P rad (x) for all x X \{θ} if X is sricly convex; see Theorem.15. Thus in a Hilber space he radial projecion P rad (x) is he bes approximaion of x in he uni ball. Thus, in his Hilber space case we know ha P rad is nonexpansive. As we see in he following, he nonexpansiviy propery of P rad does no hold in he Banach space case in general. To sudy his quesion, le us begin wih a few famous inequaliies concerning he riangle inequaliy in Hilber and Banach spaces which may be used o compue he Lipschiz consan of he projecion P rad. Theorem 4.9 (Dunkl and Williams, 1964). Le H be a Hilber space and le x, y H, x θ, y θ. Then This implies x x x x y y x + y y = x y x y y = x x ( x + y ) x y x x y = y = x y (4.6) x x y y y y 1 ( x y x y ) x y ( x ) y ) (( x + y ) x y 4 x y The righhand side in he las equaliy is nonnegaive as one easily can check. 9

16 Theorem 4.30 (Maligranda, 006). Le X be a Banach space and le x, y X, x θ, y θ. (a) x + y x + y ( x + y ) min( x, y ) x y (b) x + y x + y ( x + y ) max( x, y ) x y If eiher x = y = 1 or y = ax wih a > 0, hen equaliy holds in boh (a) or (b). We follow Maligranda [9]. Wihou loss of generaliy assume y x, i.e. min( x, y ) = x. Ad (a) x + y = x x x + x x y + (1 y y )y x x x + y + y x y ( x = y + x + y 1 ) x y ( x = x + y + x + y ) x. y Ad (b) x + y = y y x + y ( x x + 1 y ) x x y y y + x x y x = y y y + x y + x x ( = x + y x x + y ) y y The addiional asserion concerning he equaliy can easily be verified. Corollary Dunkl and Williamson, 1964] Le X be a Banach space and le x, y X, x θ, y θ. Then x x From (b) in Theorem 4.30 we obain x x y y y y 4 x + y x y + x y max( x, y ) x y (4.7) which implies (4.7). 93

17 Example 4.3. Consider R endowed wih he l 1 -norm. Take x = (1, ε) and y := (1, 0) where ε is small. Then where x x y y x + y x y lim ε ε 1 + ε = 4. = 4 + ε 1 + ε If he norm in R is given by he l -norm hen we obain for x := (1, 1) and y := (1 ε, 1 + ε), where ε > 0 is small, lim ε 0 x x y y x + y x y = lim ε ε 1 + ε = 4. Le X be a Banach space and le x, y X \{θ}. The quaniy α(x, y) := x x y y is called he angular disance or Clarkson disance of x, y. As we see from he resuls in Theorem 4.30 his quaniy conrols he riangle inequaliy. In [1] a deailed analysis of he riangle inequaliy bas been made for uniformly convex Banach spaces. From (b) in Theorem 4.30 we obain (see he proof of 4.31 α(x, y) x y max( x, y. (4.8) This a slighly srenghening of he Dunkl-Williams inequaliy (4.6) in Theorem 4.9. Kirk and Smiley showed in [5] ha his inequaliy characerizes inner produc spaces: If in a Banach space X he inequaliy (4.8) holds for all x, y X \{θ} hen X is acually a Hilber space; see also [3, 37]. This observaion may be seen a moivaion o inroduce he so called Dunkl-Williamsconsan: x + y DW(X ) := sup{α(x, y) : x, y X \{θ}, x y}. (4.9) x y Observe ha DW(X ) 4. We know from example 4.3 ha DW(R, 1 ) = DW(R, ) = 4. The Dunkl-Williams-consan may be considered as a measure how much a Banach space is close o a Hilber space. Le us come back o he radial projecion P rad. I is obvious ha he inequaliies above play an imporan role for sudying he Lipschiz coninuiy of P rad. Theorem Le X be a Banach space. Then P rad (x) P rad (y) 4 x + y x y for all x, y X. (4.10) 94

18 Le x, y X. Firs case: x, y / B 1. Then (4.10) follows from Corollary Second case: x B 1, y / B 1. Then P rad (x) P rad (y) x y x y + y y y y = x y + y 1 x y + y x x y. From Theorem 4.33 we obain a Lipschiz consan for he radial projecion. This consan canno improved in general as he following example shows. Example Consider R endowed wih he l 1 -norm. Take x = (0, ) and y := (ε, ) where ε is small. Then P rad (x) = (0, 1), P rad (y) = (ε, 1 ε) and ε = P rad (x) P rad (y) = ε = x y. 4.5 Coninuiy of a single-valued meric projecion The consideraions in he las chapers show ha he meric projecion operaors have exremely good properies in Hilber spaces, especially meric projecions ono subspaces: hey are linear orhogonal projecors, non-expansive, uniformly coninuous, self-adjoin and idempoen. In he las secion we have seen a big difference concerning meric projecions in Hilber spaces and Banach spaces: meric projecions in a Banach space are no nonexpansive in general. Addiionally, we have o realize ha he meric projecion is se-valued in general. Wha should be a definiion of coninuiy of such a map? Before we invesigae he general case in he nex secion le us sudy coninuiy properies of single-valued meric projecions in his secion. Theorem Le X be a a Banach space and le C be an approximaely compac Chebyshev subse of X. Then P C is coninuous. Le x X and le (x n ) n N wih x = lim n x n. Since C is a Chebyshev se here exis for every x n a u n C wih u n x n = dis(x n, C) ; u n = P C (x n ). Then lim n u n x = dis(x, C) since dis(, C) is coninuous; see Lemma lem:nonexpdis. From he fac ha C is approximaely compac we obain a subsequence (u n k )k N and u X wih w = lim k x n k. Since C as a Chebyshev se is closed we have u C. Due o u x = lim k u n k x = dis(x, C) we have u PC (x). Since C is a Chebyshev se we have {u} = P C (x). From his uniqueness we conclude ha (u n ) n N converges o P C (x). Corollary Le X be a Banach space and le C be a compac Chebyshev subse of X. Then P C is coninuous. 95

19 Follows from Theorem 4.35 since a compac se is approximaely compac. Corollary Le X be a finie-dimensional Banach space and le C be a Chebyshev subse of X. Then P C is coninuous. Le x X. Then P C (x) C B r (x) wih r > dis(x, C). Now, C B r (x) is closed and bounded, herefore compac. Apply now Corollary Now, we come o coninuiy resuls which use geomeric properies of he space under consideraion. In he ligh of Corollary 4.36 i is naural o ask wheher every Chebyshev se has a coninuous meric projecion mapping. Bu i is surprising ha his quesion may fail even in a reflexive space. Unforunaely, finding an example of a Chebyshev se wih a disconinuous meric projecion mapping is no as sraighforward as one migh firs imagine. Brown gave an example of cheyshevian subspace in a smooh reflexive Banach space wih a disconinuous meric projecion; see [8]. 1 Theorem Le X be a sricly convex Banach space and le C be a nonempy compac convex subse of X. Then C is a Chebyshev se and P C : X C is coninuous. Clearly, C is closed and a Chebyshev se. Le x X and le (x n ) n N be a sequence in X wih lim n x n = x. Then he sequence (P C (x n )) n N belongs o he compac se C. Therefore, here exiss a subsequence (P C (x n k ))k N and z C wih lim k P(x n k ) = z. Le u C. Then x n k PC (x n k ) x n k u and herefore x z x u since lim k (x n k, PC (x n k )) = (x, z). This implies z PC (x) and C is a Chebyshev se z = P C (x). Theorem Le X be an E-space and le C be a nonempy closed convex subse. Then he meric pojecion P C : X C is coninuous. Le x X and le x = lim n x n. Then P C (x n ) x P C (x n ) x n + x n x P C (x) x n + x n x. Therefore, lim n P C (x n ) x = dis(x, C). Then lim n P C (x n ) = P C (x) ; see Corollary.0. There is no hope ha srenghening he E-propery o uniformly convexiy would resul in he uniform coninuiy of he meric projecion since his fails even in he finiedimensional case. Corollary Le X be a uniformly convex Banach space and le C be a nonempy closed convex subse. Then he meric pojecion P C : X C is coninuous. This is a consequence of Theorem 4.39 and Theorem Anoher example of a disconinuous meric projecion is due o Kripke (unpublished). 96

20 Theorem Le X be a sricly convex and reflexive Banach space and le C be a nonempy convex closed subse of X. Then C is a Chebyshev se and he meric projecion P C : X X is coninuous. See [7]. Theorem 4.4. Le X be a reflexive Banach space wih a Kadek-Klee norm and le C be a weakly closed Chebyshev subse of X. Then he meric projecion P C is coninuous. Le x X. Le (x n ) n N be a sequence wih lim n x n = x. Since P C is locally bounded (see Lemma 4.63) and X is reflexive, here exiss a subsequence (x n k )k N and a poin y C such ha (P C (x n k ))k N converges o y wih respec o he weak opology. Since C is weakly closed, y C. Therefeore, dis(x, C) x y lim inf x n k P C (x n k ) k lim sup x n k P C (x n k ) = lim dis(x n k, C) = dis(x, C) k k since (x n k PC (x n k ))n N converges o x y wih respec o he weak opology on X and he norm in X is weakly lower semiconinuous. Now, because C is a Chebyshev se i follows ha y = P C (x). Furhermore, due o Kadek-Klee propery (x n k PC (x n k ))n N converges in norm o x y = x P C (x). Therefore, P C (x n k ))n N converges in norm o P C (x). Due o his uniqueness of he cluser poin of P C (x n )) n N we conclude ha (P C (x n )) n N converges in norm o P C (x). Corollary Le X be a uniformly convex Banach space and le C be a weakly closed Chebyshev subse of X. Then he meric projecion P C is coninuous. A uniformly convex Banach space has a Kadek-Klee norm. A famous problem in approximaion heory is he convexiy of Chebyshev ses. We have seen ha a Banach space X has he propery ha every closed convex subse C is a Chebyshev se if and only if X is reflexive and sricly convex. I is naural o ask wha are he Banach spaces X in which he converse propery holds, i.e. in which every Chebyshev se C is convex. This problem has been solved for - and 3-dimensional spaces X. For Banach spaces of finie dimension 4 only sufficien condiions are known. For infinie-dimensional Banach spaces X he problem is considerable more difficul, even he answer o he following problem beeing unknown: In a Hilber space, is every Chebyshev se necessarily convex? A he beginning of he analysis of his problem he approach was o add a condiion on he se iself in order o derive convexiy. An imporan sep was he idea of Klee of imposing coninuiy condiions on he meric projecion. I was quickly realized ha coninuiy properies of he meric projecion play a key role in deermining he convexiy of Chebyshev ses. Here is he mos imporan resul in his concep. 97

21 Theorem 4.44 (Vlasov). Le X be a Banach space and le C X be a Chebyshev se. If X is sricly convex and P C is coninuous hen C is convex. [40]. Corollary If C is a Chebyshev se in a Hilber space wih a coninuous meric projecion hen C is convex. Follows from Theorem Corollary If C is a weakly closed Chebyshev se in a smooh uniformly convex Banach space X hen C is convex. Follows from Lemma Le C be a nonempy weakly closed subse of a Hilber space H. Then he meric projecion P C is single-valued and coninuous. Theorem Le C be a nonempy weakly closed subse of a Hilber space H. Then he following condiions are equivalen: (a) (b) C is a Chebyshev se. C is a convex se. Ad (a) = (b) Consider he funcion Noice ha f is convex. Then for y H f : H x 1 x + δ C (x) R { }. f (y) = sup( y u 1 u C y ) = 1 y y + 1 u C( 1 y y + y u u u ) = 1 y 1 inf y u C u = 1 y 1 dis(y, C) = 1 y 1 y P C(y) = y P C (y) 1 P C(y) = y P C (y) f(p C (y)) 98

22 This shows dom(f ) = H and f (y) = y P C (y) f(p C (y)), y H. We conclude P C (y) f (y) for all y H. Now suppose z f (y). Se y n := y + 1 (z P n C(y)). Then lim n y n = y and hence lim n P C (y n ) = P C (y) by Lemma From 0 y n y P C (y n ) z = 1 n z P C(y) P C (y n ) z, n N, which is a consequence of he defining propery of a subdifferenial we conclude 0 lim inf z P C (y) P C (y n ) z = z P C (y). n Consequenely, z = P C (y). This shows f (y) = {P C (y)} for all y H. Since f is coninuous and f (y) is a singleon f is Gaeaux differeniable. Now, < f (x) f(x), x Hs. Moreover, f is weakly lower semiconinuous since C is weakly closed. This implies ha f is convex and hence dom(f) = C is convex. Ad (b) = (a) We may assume wihou loss of generaliy, θ C. Choose any x H and consider he funcions f : H z x z + δ B1 (z) R { }, g : H z sup(z) R { }. u C Since f is coninuous a θ dom(f) dom(g) we have no dualiy gap and we know p := inf (f(z) + g(z)) = d := max z H y H ( f (y) g ( y)) <. Since f (y) = x + y, g (y) = δ C (y), y H, and since C is closed, we obain d = max y H x + y δ C( y)) = dis(x, y). Choose any u C. Observe ha 0 dis(x, C) x + y. Therefore, P C (x) mus be nonempy. Uniqueness follows from he convexiy of C. Remark If C is a weakly closed Chebyshev se in a smooh uniformly convex Banach space X hen he meric projecion is coninuous. As a consequence, a nonempy weakly closed subse is hen convex if and only if C is a Chebyshev se. Corollary If C is a weakly closed Chebyshev se in a Hilber space hen C is convex. Theorem 4.51 (Asplund, ). Le X be a Hilber space and le C X be a Chebyshev se. If P C : H H is coninuous when H is endowed wih he norm-opology in he domain and H is endowed wih he weak opology in he range of he mapping hen C is convex. [?]. 99

23 Remark 4.5. The meric projecion P := P U ono a closed linear subspace U of a Hilber space H is coninuous and i has he he propery P P = P (Idempoenciy). Moreover, i is coninuous and nonexpansive. In a Banach space X i is no clear ha a projecion P ono a closed subspace is coninuous. A necessary und sufficien condiion is ha he subspace U is complemened. This means ha here exiss a closed subspace V such ha X = U V. Bu i is known ha no each closed subspace is complemened. Le us add a surprising resul concerning he lineariy of a meric projecion. We need wo definiions. Definiion Le X be a Banach space and le U be a proximal subspace of X. (1) U 0 := {x X : θ P U (x)} is called he kernel of P U. () P U := sup{ y : y P U (x), x S 1 }. Le X, as in Definiion Obviously, U 0 = {x X : x = dis(x, U). Moreover, P U [1, ] since P U U = I, and P U (x) P U (x) x + x x, x X. Lemma 4.54 (Holmes, Kripke, 1968). Le X be a Banach space and le U be a Chebyshev subspace of X. (1) If P U is coninuous hen I + P U is a homoemorphism of X ino iself. () If P U is linear, hen P U = 1 if and only if U 0 is a proximal subspace and x P U (x) P U 0(x) for each x X. (3) If P U is linear and U 0 is a Chebyshev subspace, hen he following saemens are equivalen: (i) P U = 1. (ii) P U 0 = I P U. (iii) (U 0 ) 0 = U. See [3] for a proof. As we know, in a Hilber space he meric projecion ono a closed linear subspace is always linear and has norm one. In general normed spaces however, wih he excepion of Chebyshev subspaces of codimension one, Chebyshev subspaces having linear meric projecions are relaively scarce. For example, he space C[0, 1] of coninuous funcions endowed wih he supremum norm has none wih finie dimension; see [9]. Deusch and Lamber (see [15]) gave for each r [0, ] an example of a one-dimensional subspace U in R, endowed wih he l 1 -norm which has he properies: U, U 0 are Chebyshev subspaces, P U, P U 0 are linear, P U = r. 100

24 4.6 Coninuiy properies of a muli-valued meric projecions Le X be a Banach space and le C be a proximal subse of X. Then he meric projecion is a mapping P C : X X wih nonempy closed convex subses P C (x), x X ; clearly, #P C (x) = 1 for x C. To sudy coninuiy properis of his mapping one has o inroduce conceps of neighborhoods and/or convergence of families of ses. Le X, Y be a opological spaces wih he sysem T X, T Y of open ses. Le G : X Y. The mos imporan coninuiy concep for se-valued mapping is ha of lower semiconinuiy since in his concep under some addiional circumsances Michael s selecion heorem may be used o obain a coninuous mapping S : X Y such ha S(x) G(x) for all x X. For example, he radial projecion is a coninuous selecion for he meric projecion ono he closed uni ball of Banach space; see Secion 4.4. Definiion Le X, Y be a opological spaces wih he sysem T X, T Y of open ses. Le G : X Y. (a) G is called upper semiconinuous if for every open se W in Y he se {x X : G(x) W} is open in X. (b) G is called lower semiconinuous if for every open se W in Y he se {x X : G(x) W} is open in X. (c) G is coninuous if i is boh upper semiconinuous and lower semiconinuous. (d) dom(g) := {x X : G(x) }. (e) graph(g) := {(x, y) X Y : y G(x)} is called he graph of G. A paracompac opological space X is defined as follows: Every open covering of X has an open refinemen U = i=1 U i where each U i is a locally finie collecion of open subses of X. Every compac opological space, every meric space is paracompac. Theorem 4.56 (Michael, 1953). Le X be a paracompac opological Hausdorff space and le F : X X be a lower semiconinuous mapping wih nonempy closed convex subses F(x), x X. Then here exiss a coninuous mapping f : X X wih f(x) F(x) for all x X. Le us collec some opological conceps for he convergence of families of ses. Definiion Le X be a Banach space and le (C n ) n N be sequence of ses. (a) (C n ) n N is said o be Hausdorff-convergen o a se C if (i) for each ε > 0 here exiss N N wih C n {x X : dis(x, A) ε} for all n N ; 101

25 (b) (ii) for each ε > 0 here exiss N N wih C {x X : dis(x, A n ) ε} for all n N. We will wrie lim H C n = C. (C n ) n N is said o converge Wijsman o a se C if lim n dis(x, C n ) = dis(x, C) for all x X. (c) (d) We will wrie lim W C n = C. (C n ) n N is said o converge Mosco o a se C if (i) for each x C here exis N N such ha for each n N a x C n exiss such ha (x n ) n N converges o x ; (ii) if here exiss a subsequence (n k ) k N and an associaed sequence (x n k )k N wih x n k Cnk, k N, such ha (x n k )k N converges weakly o a poin x X hen x C. We will wrie lim M C n = C. (C n ) n N is said o be slice-convergen o a se C if dis(b, C n ) converges o dis(b, C) for every bounded convex se B. We will wrie lim S C n = C. I is appearen ha Wijsman convergence depends on he precise norm used, while Mosco convergence is preserved by equivalen renorming. Lemma Le X be a Banach space and le (C n ) n N be sequence of ses. (1) If lim W C n = hen lim M C n =. () If X is reflexive and if lim M C n = hen lim W C n =. (3) If X is reflexive and if lim M C n = C hen lim W C n = C. Remark If X is n reflexive hen here exiss a sequence (C n ) n N of compac convex ses such ha lim M C n exiss and is a nonempy compac convex se, bu lim W C n fails o exis. Thus, a Banach space is reflexive if and only if Mosco convergence implies Wijsman convergence of closed convex subses of X ; see [?]. Lemma Le X be a Banach space and le (C n ) n N be sequence of closed ses and le (r k ) k N an increasing sequence of posiive real numbers. Assume ha D k := lim M (C n B rk ) exiss for all k N. Then C := lim M C n exiss and C = k N D k. Moreover, C is closed. We use he convenion dis(x, ) =. 10

26 There is a connecion o he Vieoris-opology and he opology induced by he Hausdorff meric in he case of closed-valued/compac-valued mappings. Lemma Le X be a Banach space and le C be a nonempy closed subse of X. If (x n, y n ) n N is a sequence in X X wih y n P C (x n ), n N, and lim n (x n, y n ) = (x, y) hen y P C (x). Consider he sequence as above. Firsly, y C since C is closed. Furhermore, x y = lim n x n y n = lim n dis(x n, C) = d(x, C) since dis(, C) is coninuous due o Lemma.1. Thus, y P C (x). Definiion 4.6. Le X, Y be Banach spaces and le Φ : X Y. Φ is called locally bounded if for every x X here exis r > 0, R > 0 such ha Φ(B r (x)) B R. Lemma Le X be a Banach space and le C be a nonempy subse of X. Then he merix projecion mapping P C : X X is locally bounded. Le x X and le r := 1, R := dis(x, C)+ x +. Suppose y P C (B r (x)), i.e. y P C (u) for some u B r (x). Using he nonexpansiviy of he disance funcion we obain y y u + u x + x = dis(u, C) + u x + x dis(x, C) + u x + u x + x dis(x, C) + + x = R. Hence, P C (B r (x)) B R. Definiion Le X, Y be Banach spaces and le Φ : X Y. Φ is called coninuous a x X if Φ(x) is a singleon and for any sequence (x n, y n ) n N in he graph of Φ wih lim n x n = x we have lim n y n = z wih z Φ(x). Theorem Le X be a Banach space and le C be a boundedly compac subse of X. If P C (x) is a singleon for x X, hen he meric projecion mapping P C is coninuous a x. Le x X and le P C (x) = {z}. Le (x n, y n ) n N be a sequence in he graph of P C wih lim n x n = x. Since P C is locally bounded and since C is boundedly compac here exiss a subsequence (x n k, y n k)k N such ha lim k y n k = y for some y C. Since limk (x n k, y n k) = (x, y) we conclude by Lemma 4.61 y P C (x). Thus, y = z as required. 103

27 4.7 Differeniabiliy properies of he disance mapping Le X be a Banach space and le C be a nonempy closed subse of X. We know ha he disance mapping X x dis(x, C) := inf x u R u C is nonexpansive. As a consequence, he differenial quoiens dis(x + h, C) dis(x, C), 0, x X \{θ}, h X, are bounded by 1. Therfore, one should ask for differeniabiliy properies of he disance mapping. Theorem Le H be a Hilber space and le C is Chebyshev subse of H. Then he disance funcion dis(, C) is Fréche differeniable and Ddis(, C)(x) = 1. Le f : X R be a Lipschiz coninuous funcion. Clarke s generalized direcional derivaive of f a a poin x in he direcion h X, denoed by f (x; h), is given by : f f(z + h) f(z) (x; h) := lim sup. (4.11) z x, 0 Clarke s generalized subdifferenial of f a x is given by f(x) := {λ X : f (x; h) f(y) for all y X }. (4.1) Theorem Le X be a Banach space such ha X is smooh. Then a nonempy closed subse C of X is convex if is disance funcion dis(, C) saisfies lim sup y θ dis(x + y, C) dis(x, C) y = 1 for all x X \C. (4.13) In paricular, his differeniabiliy condiion is saisfied if dis(, C) is Gaeaux-differeniable a x and dis(x, C) S 1 of he dual space X. See [5]. Theorem Le X be a Banach space such ha X is sricly convex. Le C X be a Chebyshev se, le x C such ha dis(x, C) is a singleon. Then he following asserions are equivalen: (a) (b) C is convex. dis(, C) is convex. (c) dis(, C) is Gaeaux differeniable a x. 104

28 (d) (e) dis(x + h) dis(x) There exiss h S 1 such ha lim 0 dis(x + y) dis(x) lim sup y θ = 1. y = 1. (a) = (b) Since C is convex i can easily be seen ha dis(, C) is convex. (b) = (c) Since dis(, C) is convex and coninuous a x and dis(, C) is a singleon, dis(, C) is Gaeaux differeniable a x and dis(, C)(x) = {d C (x)}. (c) = (d) Since C is a chebyshev se we have x P C (x) = dis(x, C). I follows from Gaeaux differeniabiliy of dis(, C) ha lim inf 0 exiss for every y X. For each > 0 we have Hence for y := x P C (x) we se lim inf 0 dis(x + h, C) dis(x, C) dis(x + (x P C (x), C) dis(x, C) dis(x, C). dis(x + (x P C (x), C) dis(x, C) = dis(x, C). Since x X \C, dis(x, C) > 0 and if = /dis(x, C) as 0. Then by he above lim inf 0 dis(x + (x P C (x), C) dis(x, C) If now z = (x P C (x)) x P C (x) 1 hen we have lim inf 0 dis(x + z, C) dis(x, C) On he oher hand, dis(, C)) is a Lipschiz funcion and so lim sup 0 dis(x + y, C) dis(x, C) y (d) = (e) Since dis(, C) is a Lipschiz funcion = dis(x, C). = 1. = 1. lim sup 0 dis(x + y, C) dis(x, C) y 1. On he oher hand, for each v S 1 lim 0 dis(x + v, C) dis(x, C) in paricular for v := z in (d), we have 1 lim sup y θ lim sup y θ dis(x + y, C) dis(x, C) y dis(x + y, C) dis(x, C) y., (e) = (a) Follows from 105

29 Theorem Le X be a Banach space such ha X is smooh. Then a nonempy closed subse C of X is convex if for some r > 0 he se P C,r := {x 0 rz : x 0 P C (C), p(x 0 ) C, x 0 p(x 0 ) = dis(x 0, C) > r, z = (x 0 p(x 0 )) x 0 p(x 0 ) 1 } is dense in X \C. See [18, 19]. Lemma Le X be a Banach space and le C be a nonempy subse of X. If he disance mapping dis(, C) is Gaeaux-differeniable in x X \C and w P C (x) hen ddis(x)(x w) = x w and ddis(x) = 1. (4.14) See [17] 4.8 Differeniabiliy properies of he meric projecions Le H be a Hilber space and le C be a nonempy closed convex subse o H. The he meric projecion C is welldefined and one may ask for differeniabiliy properies of P C. We say ha P C is direcionally differeniable a a poin x H if d + P C (x)(h) := lim 0 P C (x + h) P C (x) (4.15) exiss for each h H. If he mapping h d + P C (x) is linear and coninuous hen P C is called Gaeaux-differeniable a x H. This mapping, when i exiss, is denoed by dp C (x). If he above limi exiss uniformly for h B 1 and if h d + P C (x) is linear and coninuous we say ha P C is Fréche-differeniable a x and denoe he corresponding linear operaor by DP C (x). Lemma Le H be a Hilber space and le C is Chebyshev subse of H. Then we define W : H\C x 1 ( x dis(x, C) R. (4.16) Then W is Fréche differeniable and DW(x) = P C (x), x H. Moreover, W is convex and 1 W(x) = sup u C ( x u u ), x H\C. (4.17) By he commens above, he meric projecion P T(x,C) exiss. Zarananello [4] and Haraux [0] (see also [39]) have proved he following resul. 106

30 Theorem 4.7. Le H be a Hilber space and le C be a nonempy closed convex subse of H. Then for any x C we have P C (x + h) = x + P T(x,C) (h) + o(), > 0, h H. (4.18) Le x C and h H. We have o prove In a firs sep we prove for > 0 d + P C (x)(h) = lim 0 P C (x + h) P C (x) = P T(x,C) (h). (4.19) We have for > 0 and y C P C (x + h) P C (x) = P (C x)/ (h). (4.0) where ^y = (C x)/. In a second sep we verify 0 (x + h) P C (x + h) y P C (x + h) = (x + h) P C (x + h) y (P C (x + h) + x x) = x P C(x + h) + h y x P C(x + h) x = h P C(x + h) x ^y P C(x + h) x lim 0 inf{ h u : u (C x)/} = inf{ h v : v T(x, C)} (4.1) which means lim 0 h P (C x)/ (h) = h P T(x,C) (h). Noice ha he meric projecions P (C x)/ (h), P T(x,C) (h) exis since (C x)/, T(x, C) are closed convex ses. If 0 1 hen (C x)/ (C x)/ 1 due o he fac ha C is convex. This shows ha q := lim 0 h P (C x)/ (h) exis. Moreover since (C x)/ T(x, C) for all > 0 we have q h P T(x,C) (h). Since P T(x,C) (h) T(x, C) we obain q = h P T(x,C) (h). Now, we obain P (C x)/ P T(x,C) (h) = h P (C x)/ + h P T(x,C) 4 h 1 (P (C x)/ + P T(x,C) ) h P (C x)/ + h P T(x,C) 4 h P T(x,C) and we conclude ha lim 0 P (C x)/ P T(x,C) (h) = 0 which implies finally lim 0 P C (x + h) P C (x) = P T(x,C) (h). We conclude from he heorem above ha a each poin x C, he map P C has a direcional derivaive in every direcion h H given by P T(x,c) (h). Since he meric 107

31 projecion P T(x,C) will be linear only in he very special case when T(X, C) is a linear subspace, i is clear ha dp C (x) does no generally exis a poins x of C. In paricular, if C H and C has nonempy inerior and x C, hen T(x, C) will no be a subspace, so dp C (x) will no exis. Rademacher s heorem (see [36] and [43]) saes ha a Lipschiz coninuous funcion from R n ino R m is dffereniable almos everywhere. The heorem of Alexandrov (see [1] and [31]) says ha each coninuous convex funcion on R n is almos everywhere wice differeniable. For n = 1 his reduces o Lebesgue s heorem abou he differeniabiliy almos everywhere of a monoone funcion. Alexandrov s heorem explois Asplund (see []) as follows: Theorem 4.73 (Asplund, 1973). Le C be a nonempy closed convex subse he inner produc space H := R n and le P C be he associaed meric projecion from H ino C. Then P C is differenable almos everywhere. Define f : H R, x sup( x y 1 y C y ). Obviously, f is convex, acually f is he Fenchel-conjugae of j : H u 1 u R. One can show ha f(x) = 1 1 x inf y C x y = 1 x 1 x P C(x), x H. From his we conclude ha P C (x) is in he subdifferenial of f a x ; see he chaper concerning he fundamenals facs on convex analysis. Since we know from Alexandrov s heorem ha f is wice (Fréche-)differeniable almos everywhere he meric projecion is almos everywhere (Fréche-)differeniable. Le C be a nonempy closed convex subse he inner produc space H := R n and le P C be he associaed meric projecion from H ino C. I is known ha if x C, hen P C is direcionally differeniable a x and he corresponding direcional derivaives d + P C (x)(h) = lim 0 P C (x + h) P C (x) are given by he meric projecion of h ono he suppor cone S C (x) o C a x. In he case x / C direcional differeniabiliy of C a x is guaraneed only under cerain condiions on smoohness of he boundary of C. Here is an example in R 3 (consruced by Shapiro in [38]) whose meric projecion fails o be direcionally differeniable; see also [33]. Example Consider a sequence (α n ) n N of posiive numbers defined by λ 1 := π/ and α n+1 := c n α 1, n N, wih c (0, 1). Then his sequence is monoone decreasing and lim n α n = 0. Le C be defined as he convex hull of {(cos(α n ), sin(α n )) : n N} {(0, 0), (1, 0)}. Le x := (, 0), d := (0, 1). We will show ha P C (x) is no direcionally differeniable in x in direcion d. Consider for n =, 3,... n := sin(α n )+( cos(α n )) an( 1 (α +α n 1 )), s n := sin(α n )+( cos(α n )) an( 1 (α +α n+1 )). 108

32 Then we have P C (x) = (1, 0) and P C (x + n d) = P C (x + s n d) = (cos(α n ), sin(α n )), n =, 3,.... Since we obain lim n lim n sin(α n ) n = 3 + c 1, lim n sin(α n ) s n = 3 + c π (P C (x + n d) π (P C (x)) n lim n π (P C (x + s n d) π (P C (x)) s n, where π (u) denoes he second coordinae u of u = (u 1, u ). 4.9 Appendix: Subdifferenial Le X be a Banach space and le f : X ^R be proper, convex and closed. We consider he epsilon-subdifferenial: ε f(x 0 ) := {λ X f(y) f(x 0 )+ λ, y x 0 ε for all y X}, ε 0, x 0 dom(f). (4.) For ε = 0 we wrie f(x 0 ) := 0 f(x 0 ). The epsilon-subdifferenial defines a poin-o-se operaor ε f : X x ε f(x) POT(X ) ; we use he noaion from he heory of se-valued mappings: ε f : X X I can easily proved ha for all x X, ε 0 he se ε f(x) is convex and weak -closed (possible empy). Moreover, he following properies are useful: x X ε 1, ε 0(ε 1 ε = ε1 f(x) ε f(x)) (4.3) x X ε 0 ( ε f(x) = η>ε η f(x)) (4.4) Theorem Le X be a Banach space, le f : X ^R be convex and suppose ha f is coninous in x 0 dom(f). Then (1) f(x 0 ). () f(x 0 ) is convex and σ(x, X)-closed. (3) f(x 0 ) is bounded (as a subse of X ). Noice ha he assumpion ha f is coninuous in x 0 dom(f) canno be replaced by he assumpion ha f is lower semiconinuous in x 0 dom(f) as wes see in he example { 1 x f : R x if x 1 R { }. else Corollary Le X be a Banach space, le f : X ^R be convex, U dom(f) convex and open, and suppose ha f is bounded from above in a neighborhood V of a poin x U. Then f(x 0 ) for all x 0 U. 109