PROBLEMS Problem 0.1. Problem 0.2. Let V := (Ln1 l(n) 1 )2 with the norm k(xn)n1k2 := (X1 n=1 kxnk2l(n) 1 )1/2, (xn)n1 2 V. Prove that l1(N) is finitely representable in V. Let V0 l1(N) be a finite dimensional subspace with dimV0 = n. Since V0 is finite dimensional it is isometrically isomorphic to Fn,F = R or C. Now, we want to embed V0 in V. Let {˜ei, i = 1, 2, ...n} be a basis for V0 having k˜eikl1 = 1. Define Wm = {0}, for m 6= n and Wn = ln1 Set W0 := (LmnWm)2, a subspace of V. Define T : V0 ! W0 by T(˜ei) = (0, 0, ..., ei, 0, ..., ) = eni, say . where ei, the ith unit vector in the natural basis of ln1 , is the nth coordinate. kenik2 = (keik2l(n) 1 )1/2 = 1 Let x = Pni=1 xi˜ei 2 V0. kxkl(n) 1 = Xn i=1 |xi| T(x) = Xn i=1 xiT(˜ei) = Xn i=1 xieni Date: 6-April-2010. 12 FUNCTIONAL ANALYSIS kT(x)k =Xn i=1 |xi| = kxkl(n) 1 So, T is isometric. It is enough to prove that T is injective. T(x) = 0 ) Xn i=1 xieni= 0 ) xi = 0, 8i ) x = 0 kerT = (0) and hence T is injective. Thus, T is an isometric isomorphism of V0 onto T(V0) W0, a subspace of V. T−1 exists since T is bijective. Hence the conclusion. Problem 0.3. (1) Show that any Hilbert space H is a uniformly convex Banach space and compute its modulus of convexity H(), > 0. (2) Show that the Banach space V := (Ln1 l(n) 1 )2 is not uniformly convex. (1) If > 0, x, y 2 H, kxk = kyk = 1, kx − yk , then we have to show that 9> 0 : kx + y 2 k 1 − Since H is a Hilbert space, kx + yk2 + kx − yk2 = 2(kxk2 + kyk2) = 4 kx + yk2 = 4 − kx − yk2 4 − 2 kx + y 2 k2 1 − ( 2)2 kx + y 2 k r1 − ( 2)2 = 1 − , where = 1 −r1 − ( 2)2 So, H is uniformly convex.FUNCTIONAL ANALYSIS 3 Now, the modulus of convexity, () := inf{1 − kx + y 2 k : kxk = kyk = 1, kx − yk } kx + y 2 k 1 − ) 1 − kx + y 2 k So, () = = 1 −r1 − ( 2)2 (2) By problems 5 and 2, l1(N) is finitely representable in V. l1(N) is uniformly convex. l1(N) is not reflexive) l1(N) is not uniformly convex ) V is not uniformly convex. Problem 0.4. Prove that if (V, k.k) is a uniformly convex Banach space,then any closed subspace and any quotient space of V are uniformly convex. (1) Let M be a closed subspace of V. Let > 0, x, y 2 M such that , kxk = kyk = 1, kx− yk . Since M V, x, y 2 V, which is uniformly convex and therefore kx + y 2 k 1 − for some > 0 So, W is a uniformly convex. (2) Lemma If V is a uniformly convex, then given 1 > 0, x, y 2 V such that 1 kxk 1 + 1, 1 kyk 1 + 1, kx − yk 1 then kx + y 2 k 1 − ˜2 where ˜is such that 8˜x, ˜y 2 V, k˜xk 1, k˜yk 1, k˜x − ˜yk 24 FUNCTIONAL ANALYSIS we have k ˜x + ˜y 2 k 1 − ˜Solution Let M be a closed subspace of V. The quotient space V/M = {x +M : x 2 V } is a Banach space with the norm defined as, for [x] := x +M, k[x]k1 = inf{kx + mk : m 2 M} Given > 0, let [x], [y] 2 V/M, k[x]k1 = k[y]k1 = 1, k[x] − [y]k1 k[x]k1 = k[y]k1 = 1 ) 9(xn) 2 [x], in (yn) in [y] such that kxnk ! 1, kyn ! 1k kxn − ynk inf{kx − y + mk : m 2 M} By lemma, given 1 > 0, 1 kxnk 1 + 1, 1 kynk 1 + 1, kxn − ynk we have kxn + yn 2 k 1 − ˜2 Since kxnk ! 1, kyn ! 1, 9N 2 N such that 1 kxnk 1 + 1, 1 kynk 1 + 1, 8n 2 N. Also, inf{kx + y 2 +mk : m 2 M} kxn + yn 2 k 1−˜2 = 1−, d := ˜2 > 0FUNCTIONAL ANALYSIS 5 Finally, k[x] + [y] 2 k1 = kx +M + y +M 2 k1 = kx + y 2 +Mk1 and inf{kx + y 2 + mk : m 2 M} 1 − Therefore k[x] + [y] 2 k1 1 − Therefore, V/M is uniformly convex. Problem 0.5. Show that if V is a uniformly convex Banach space, and W is finitely representable in V , then W is uniformly convex as well. Lemma 1 IfW is finitely representable in V , then given 0 > 0, and W0 a finite dimensional subspace of W, there exists a finite dimensional subspace V0 of V with dimW0 = dimV0 and 9T : W0 ! V0 such that 1 (1 + 0)2kxk kTxk kxk Lemma 2 Given > 0, x, y 2 V, kxk 1, kyk 1, kx − yk ", 9v > 0 : kx + y 2 k 1 − v Solution Given > 0, let x, y 2 W, kxk = 1, kyk = 1, kx − yk " Chose W0 as a finite dimensional subspace of W containing x, y. Then,by Lemma 1, 9T : W0 ! V0 such that 1 (1 + 0)2kxk kTxk kxk6 FUNCTIONAL ANALYSIS Also, kTxk kxk = 1, kTyk kyk = 1 Now, Tx, Ty 2 V0 V kTx−Tyk = kT(x−y) kx − yk (1 + "0)2 (1 + "0)2 = 1, say By Lemma 2, 9v > 0 : kTx + Ty 2 k 1 − v kT(x + y 2 )k 1 − v 1 (1 + 0)2kx + y 2 k kT(x + y 2 )k 1 − v kx + y 2 k (1 − v)(1 + 0)2 = 1 − , say , where = v(1 + 0)2 − 20(1 + 0) Choose 0 sufficiently small to make > 0 Hence the conclusion.