Convexity and Smoothness

Capter 4 Convexity and Smootness 4.1 Strict Convexity, Smootness, and Gateaux Differentiablity Definition 4.1.1. Let X be a Banac space wit a norm denoted by. A map f : X \{0} X \{0}, f f x is called a support mapping wenever. a) f(λx) =λf x, for λ> 0 and b) If x S X, ten f x = 1 and f x (x) = 1 (and tus f x (x) = x 2 for all x X). Often we only define f x for x S X and ten assume tat f x = x f x/ x, for all x X \{0}. For x X a support functional of x is an element x X, wit x = x and x,x = x 2. Tus a support map is a map f ( ) : X X, wic assigns to eac x X a support functional of x. We say tat X is smoot at x 0 S X if tere exists a unique f x S X, for wic f x (x) = 1, and we say tat X is smoot if it is smoot at eac point of S X. Te Banac space X is said to ave Gateaux differentiable norm at x 0 S X, if for all y S X ρ (x 0,y) = lim 0 x 0 + y x 0 exists, and we say tat is Gateaux differentiable if it is Gateaux differentiable norm at eac x 0 S X. 93

94 CHAPTER 4. CONVEXITY AND SMOOTHNESS Example 4.1.2. For X = L p [0, 1], 1 <p< te function f : L p [0, 1] L q [0, 1], f x (t) = sign(x(t)) x(t) p/q 1 = x p x p is a (and te only) support function for L p [0, 1]. p q x(t) p q In order establis a relation between Gateaux differentiability and smootness we observe te following equalities and inequalities for any x X, y S X, and >0: f x (y) x = f x(y) x = f x(x) x 2 + f x (y) x = f x(x + y) x 2 x f x(x + y) x 2 x f x x + y x 2 x x + y x = = x + y 2 x + y x x + y x + y 2 f x+y (x) x + y = f x+y(x + y) f x+y (x) x + y = f x+y(y)+f x+y (x) f x+y (x) x + y f x+y(y) x + y = f x+y(y) x + y and tus for any x X,y S X, and >0: (4.1) f x (y) x f x(x + y) x x + y x f x+y(y) x x + y.

4.1. STRICT CONVEXITY, SMOOTHNESS, AND GATEAUX DIFFERENTIABLITY95 Teorem 4.1.3. Assume X is a Banac space and x 0 S X Te following statements are equivalent: a) X is smoot at x 0. b) Every support mapping f : x f x is norm to w continuous from S X to S X at te point x 0. c) Tere exists a support mapping f ( ) : x f x wic is norm to w continuous from S X to S X at te point x 0. d) Te norm is Gateaux differentiable at x 0. In tat case f x (y) =ρ (x 0,y) = lim 0 x 0 + y x 0 for all y S X. Proof. (b) (a). Assume tat (x n ) S X is a net, wic converges in norm to x 0, but for wic f xn does not converge in w to f x0, were f ( ) : X X is te support map. We can assume tat tere is a w neigborood U of f x0, not containing any of te f xn, and by Alaoglu s Teorem 2.1.8 we can assume tat f xn as an accumulation point x U, wic cannot be equal to f x0. As x (x 0 ) 1 = x (x 0 ) f xn (x n ) x (x 0 ) f xn (x 0 ) + f xn (x 0 x n ) x (x 0 ) f xn (x 0 ) + x 0 x n n M,n 0, for some infinite M N it follows tat x (x 0 ) = 1, and since x 1 we must ave x = 1. Since x f x0, X cannot be smoot at x 0. (b) (c) is clear (since by Te Teorem of Han Banac tere is always at least one support map). (c) (d) Follows from (4.1), and from te fact tat (4.1) implies for x X, y X and >0, tat and x y x x x y x x x + ( y) x = f x( y) = f x (y) x x x + ( y) x = f x+( y)( y) x x + ( y) = f x+( y)(y) x + ( y).

96 CHAPTER 4. CONVEXITY AND SMOOTHNESS (d) (a) Let f S x be suc tat f(x) = x = 1. Since (4.1) is true for any support function it follows tat and f(y) x 0 + y x 0, for all y S X and >0, x 0 y x 0 = x 0 +( y) x 0 f( y) =f(y) for all y S X and <0. Tus, by assumption (d), ρ (x 0,y)=f(y), wic proves te uniqueness of f S X wit f(x 0 ) = 1. Definition 4.1.4. A Banac space X wit norm is called strictly convex wenever S(X) contains no non-trivial line segement, i.e. if for all x, y S X, x y it follows tat x + y < 2. Teorem 4.1.5. If X is strictly convex ten X is smoot, and if X is smoot te X is strictly convex. Proof. If X is not smoot ten tere exists an x 0 S X, and two functionals x y in S X wit x (x 0 )=y (x 0 ) = 1 but tis means tat x + y (x + y )(x 0 )=2, wic implies tat X is not strictly convex. If X is not strictly convex ten tere exist x y in S X so tat λx + (1 λ)y = 1, for all 0 λ 1. So let x S X suc tat x ( x + y ) =1. 2 But tis implies tat 1=x ( x + y ) = 1 2 2 x (x)+ 1 2 x (y) 1 2 + 1 2 =1, wic implies tat x (x) =x (y) = 1, wic by viewing x and y to be elements in X, implies tat X is not smoot. Exercises

4.1. STRICT CONVEXITY, SMOOTHNESS, AND GATEAUX DIFFERENTIABLITY97 1. Sow tat l 1 admits an equivalent norm wic is strictly convex and (l 1, ) is (isometrically) te dual of c 0 wit some equivalent norm. 2. Assume tat T : X Y is a linear, bounded, and injective operator between two Banac spaces and assume tat Y is strictly convex. Sow tat X admits an equivalent norm for wic X is strictly convex.

98 CHAPTER 4. CONVEXITY AND SMOOTHNESS 4.2 Uniform Convexity and Uniform Smootness Definition 4.2.1. Let X be a Banac space wit norm. We say tat te norm of X is Frécet differentiable at x 0 S X if x 0 + y x 0 lim 0 exists uniformly in y S X. We say tat te norm of X is Frécet differentiable if te norm of X is Frécet differentiable at eac x 0 S X. Remark. By Teorem 4.1.3 it follows from te Frecét differentiability of te norm at x 0 tat tere a unique support functional f x0 S X and lim x 0 + y x 0 f x0 (y) =0, 0 uniformly in y and tus tat (put z = y) x 0 + z x 0 f x0 (z) lim =0. z 0 z In particular, if X as a Frécet differentiable norm it follows from Teorem 4.1.3 tat tere is a unique support map x f x. Proposition 4.2.2. Let X be a Banac space wit norm. Ten te norm is Frécet differentiable if and only if te support map is norm-norm continuous. Proof. (We assume tat K = R) Assume tat (x n ) S X converges to x 0 and put x n = f xn, n N, and x 0 = f x 0. It follows from Teorem 4.1.3 tat x n(x 0 ) 1, for n. Assume tat our claim were not true, and we can assume tat for some ε> 0 we ave x n x 0 > 2ε, and terefore we can coose vectors z n S X, for eac n N so tat (x n x 0 )(z n) > 2ε. But ten x 0(x 0 ) x n(x 0 ) ( x 0(x 0 ) x n(x 0 ) )( 1( x ε n (z n ) x ) ) 0(z n ) 1 }{{} >2ε ( x 0 (z n ) x n(z n ) )( x n(x 0 ) x 0(x 0 ) ) = ( x n(x 0 ) x 0(x 0 ) ) + 1 ε = ( x n x 0 ) ( 1( x 0 + z n x ε 0 (x 0 ) x n(x 0 ) ))

4.2. UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS 99 ( x 1( n x 0 + z n x ε 0 (x 0 ) x n(x 0 ) )) ( x 1( 0 x 0 + z n x ε 0 (x 0 ) x n(x 0 ) )) 1( x 0 + z n x ε 0 (x 0 ) x n(x 0 ) ) ( x 0 x 1( 0 z n x ε 0 (x 0 ) x n(x 0 ) )). Tus if we put 1( y n = z n x ε 0 (x 0 ) x n(x 0 ) ), it follows tat y n 0, if n, and, using te Frécet differentiability of te norm tat (note tat ( x 0 (x 0) x n(x 0 ) ) / y n = ε) we deduce tat 0 <ε = x 0 (x 0) x n(x 0 ) y n x 0 + y n x 0 x 0 (y n) y n n 0, wic is a contradiction. From (4.1) it follows tat for x, y S X, and R x + y x f x (y) f x+y (y) x + y f x(y) f x+y (y) f x (y) f x+y (y) + x + y f x+y(y) 1 f x+y f x + 1+ 1 f x+y, wic converges uniformly in y to 0 and proves our claim. Definition 4.2.3. Let X be a Banac space wit norm. We say tat te norm is uniformly Frécet differentiable on S X if x + y x lim f x (y), 0 uniformly in x S X and y S X. In oter words if for all ε> 0 tere is a δ> 0 so tat for all x, y S X and all R, 0< <δ x + y x f x (y) < ε.

100 CHAPTER 4. CONVEXITY AND SMOOTHNESS X is uniformly convex if for all ε> 0 tere is a δ> 0 so tat for all x, y S X wit x y ε it follows tat (x + y)/2 < 1 δ. We call δ X (ε) = inf { 1 x + y 2 : x, y S X, x y ε }, for ε [0, 2] te modulus of uniform convexity of X. X is called uniform smoot if for all ε> 0 tere exists a δ> 0 so tat for all x, y S X and all (0,δ] x + y + x y < 2+ε. Te modulus of uniform smootness of X is te map ρ : [0, ) [0, ) { } x + z x z ρ X (τ) = sup + 1:x, z X, x =1, z τ. 2 2 Remark. X is uniformly convex if and only if δ X (ε) > 0 for all ε> 0. X is uniformly smoot if and only if lim τ 0 ρ X (τ)/τ = 0. Teorem 4.2.4. For a Banac space X te following statements are equivalent. a) Tere exists a support map x f x wic uniformly continuous on S X wit respect to te norms. b) Te norm on X is uniformly Frécet differentiable on S X. c) X is uniformly smoot. d) X is uniformly convex. e) Every support map x f x is uniformly continuous on S X wit respect to te norms. Proof. (a) (b) We proceed as in te proof of Proposition 4.2.2. From (4.1) it follows tat for x, y S X, and R x + y x f x (y) f x+y (y) x + y f x(y)

4.2. UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS 101 f x+y (y) f x (y) f x+y (y) + x + y f x+y(y) 1 f x+y f x + 1+ 1 f x+y wic converges by (a) uniformly in x and y, to 0. (b) (c). Assuming (b) we can coose for ε> 0aδ> 0 so tat for all (0,δ) and all x, y S X x + y x f x (y) < ε/2. But tis implies tat for all (0,δ) and all x, y S X we ave x + y + x y ( x + y x = 2 + 2+ε, ( x + ( y) x ) ) f x (y)+ f x ( y) wic implies our claim. (c) (d). Let ε> 0. By (c) we can find δ> 0 suc tat for all x S X and z X, wit z δ, we ave x + z + x z 2+ε z /4. Let x,y S X wit x y ε. Tere is a z X, z δ/2 so tat (x y )(z) εδ/2. Tis implies x + y = sup x S X (x + y )(x) = sup x S X x (x + z)+y (x z) (x y )(z) sup x S X x + z + x z εδ/2 2+ε z /4 εδ/2 < 2 εδ/4. (d) (e). Let x f x be a support functional. By (d) we can coose for ε> a δ so tat for all x,y S X we ave x y <ε, wenever x + y > 2 δ. Assume now tat x, y S X wit x y <δ. Ten f x + f y 1 2 (f x + f y )(x + y)

102 CHAPTER 4. CONVEXITY AND SMOOTHNESS = f x (x)+f y (y)+ 1 2 f x(y x)+ 1 2 f y(x y) 2 x y 2 δ, wic implies tat f x f y <ε, wic proves our claim. (e) (a). Clear. Teorem 4.2.5. Every uniformly convex and every uniformly smoot Banac space is reflexive. Proof. Assume tat X is uniformly convex, and let x S X. Since B X is w -dense in B X we can find a net (x i ) i I wic converges wit respect to w to x. Since for every η > 0 tere is a x S X wit lim i I x (x i )=x (x ) > 1 η, it follows tat lim i I x i = 1 and we can terefore assume tat x i = 1, i I. We claim tat χ(x i ) is a Caucy net wit respect to te norm to x, wic would finis our proof. So let ε> 0 and coose δ so tat x+y > 2 δ implies tat x y <ε, for any x, y S X. Ten coose x S X, so tat x (x ) > 1 δ/4, and finally let i 0 I so tat x (x i ) 1 δ/2, for all i i 0. It follows tat x i + x j x (x i + x j ) 2 δ wenever i, j i 0, and tus x i x j <ε, wic verifies our claim. If X is uniformly smoot it follows from Teorem 4.2.4 tat X is uniformly convex. Te first part yields tat X is reflexive, wic implies tat X is reflexive. Exercises 1) Sow tat for tere is a constant c>0 so tat for all ε> 0, δ l2 (vp) cε 2. (Here δ l2 is te modulus of uniform convexity of l 2 ). 2) Prove tat for every ε> 0, C>1and any n N tere is an N = (n, ε, C) so tat te following olds: If X is an N dimensional space wic is C-isomorpic to l N 1 ten X as an n-dimensional subspace Y wic is (1 + ε) ismorpic to l n 1.

4.2. UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS 103 Hint: prove first te following: Assume tat is a norm on l n2 1 so tat 1 C x 1 x x < for all x l n2 1, ten tere is a -normalized block sequence (x 1,x 2,... x n ) so tat : 1 C n b i i=1 3) Sow tat ( ) n=1 ln 1 convex. l 2 n b i x i i=1 n b i. i=1 does not admit a norm wic is uniformly