MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3. (1) Let V be the vector space of all bounded or unbounded sequences of complex numbers.

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1 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 (1) Let V be the vector space of all bounded or unbounded sequences of complex numbers. (a) Define d : V V + {0} by d(x, y) = 1 ξ j η j 2 j 1 + ξ j η j. Show that d is a metric on V. Solution: Let x = (ξ j ), y = (η j ) V. For any j = 1, 2,, and j ξ j η j 1 + ξ j η j 1 2 j 1 2 j converges (geometric series). So by comparison test, d(x, y) < for all x, y V. Clearly then d(x, y) + {0}. Also clearly d(x, y) = d( y, x) for all x, y V i.e. (M2) is satisfied. If x = y then ξ j = η j for all j = 1, 2, and so d(x, y) = 0. Suppose that d(x, y) = 0. Then for any j = 1, 2,, 0 1 ξ j η j d(x, y) = 0 2 j 1 + ξ j η j 1 ξ j η j = 0, j = 1, 2, 2 j 1 + ξ j η j ξ j = η j, j = 1, 2, x = y. Thus, (M1) is satisfied. Now we show that the Triangle Inequality (M3). Let x = (ξ j ), y = (η j ) and z = (ζ j ). 1

2 2 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 Then for each j = 1, 2,, ξ j ζ j 1 + ξ j ζ j + ζ j η j ξ j ζ j 1 + ξ j ζ j, ζ j η j 1 + ζ j η j + ξ j ζ j ζ j η j 1 + ζ j η j. Recall that if 0 a b then (1) a 1 + a b 1 + b. n By triangle inequality with respect to, we have ξ j η j ξ j ζ j + ζ j η j for each j = 1, 2,. It then follows from the inequality (1) that ξ j η j 1 + ξ j η j ξ j ζ j + ζ j η j 1 + ξ j ζ j + ζ j η j ξ j ζ j 1 + ξ j ζ j + ζ j η j 1 + ζ j η j. So, for each n = 1, 2, we have 1 ξ j η j 2 j 1 + ξ j η j n 1 ξ j ζ j 2 j 1 + ξ j ζ j + n 1 ζ j η j 2 j 1 + ζ j η j. Finally by taking the limit n, we obtain the Triangle Inequality (M3). (b) Show that d is not translation invariant. Solution: Let a = (a j ) V. Then d(x + a, y + a) = = d(x, y). 1 ξ j + a j (η j + a j ) 2 j 1 + ξ j + a j (η j + a j )

3 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 3 Let α be a scalar. Then αξ j αη j d(αx, αy) = 1 + αξ j αη j = α ξ j η j 1 + α ξ j η j and this does not necessarily coincide with d(x, y). To see this, for example let x = (1, 1, 1, ), y = (0, 0, 0, ) and α = 2. Then d(αx, αy) = = j j 1 = = 2 3, while d(x, y) = 1 = 1. j+1 2 Hence, d is not translation invariant. Therefore, the metric d cannot be induced by a norm on a normed space. (2) Show that in a normed space X, vector addition and multiplication by scalars are continuous operations with respect to the norm. That is, the mappings defined by (x, y) x + y and (α, x) αx are continuous. Solution: Let a, b X. Let ε > 0 be given. Choose δ 1 = δ 2 = ε 2. Then whenever x a < δ 1 and y b < δ 2, x + y (a + b) x a + y b < ε.

4 4 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 Hence, the addition + is continuous. Let α be a scalar and a X. Let ε > 0 be given. If α = 0, then αx αa = 0 < ε. Now suppose α 0. Choose δ = ε > 0. Then α αx αa < α ε α = ε. Hence, the scalar multiplication is continuous. (3) Show that x n x and y n y implies x n + y n x + y. Show that α n α and x n x implies α n x n αx. Solution: Assume that x n x and y n y. Let ε > 0 be given. Then there exists positive integers N 1 > 0 and N 2 > 0 such that for all n N 1 and x n x < ε 2 y n y < ε 2 for all n N 2. Choose N = max{n 1, N 2 }. Then for all n N (x n + y n ) (x + y) x n x + y n y < ε 2 + ε 2 = ε. Assume that α n α and x n x. Let ε > 0 be given. Since (x n ) is convergent, there exists M > 0 such that x n < M. If α = 0, then there exists a positive integer N such that α n < ε for all n N. So, for all n N M α n x n αx = α n x n = x n α n < M ε M = ε. Now suppose that α 0. Then there exists a positive integer N 1 such that α n α < ε 2M

5 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 5 for all n N 1 and there exists a positive integer N 2 such that x n x < ε 2 α for all n N 2. Choose N = max{n 1, N 2 }. Then for all n N α n x n αx = α n x n αx n + αx n αx α n α x n + α x n x < ε 2M M + α ε 2 α = ε. (4) Show that the closure Ȳ of a subspace Y of a normed space X is again a vector subspace. Solution: Let y 1, y 2 Ȳ. Then there exist sequences (ξ n ), (η n ) Y such that ξ n y 1 and η n y 2. Then by #3, ξ n + η n y 1 + y 2. Since (ξ n + η n ) Y, y 1 + y 2 Ȳ. Also by #3, αξ n αy 1. Since (αξ n ) Y, αy 1 Ȳ. (5) In a normed space, convergence of a series implies absolute convergence of that series. But the converse need not be true. In fact, we see, throughout the questions (a)-(c), that in a normed space X absolute convergence of a series implies convergence of that series if and only if X is complete. (a) Show that in a normed space absolute convergence of a series does not necessarily imply convergence of that series. Solution: Let Y be the set of all sequences with only finitely many nonzero complex terms. Then clearly Y is a subset of l. Let (y n ) be a sequence such that y 1 = (1, 0, 0, ), 0, 12, 0, 2 y 2 =, 0, 0, 13, 0, 2 y 3 =.,

6 6 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 Then ( y n ) Y. y 1 + y 2 + y 3 + = = n=1 1 n 2 <. So, y n=1 n converges absolutely. However, 1, 12, 13, 2 2 y 1 + y 2 + y 3 + = Y, i.e. n=1 y n does not converge. (b) If in a normed space X, absolute convergence of any series always implies convergence of that series, show that X is complete. Solution: Let (x n ) X be a Cauchy sequence. Let a 1 = x 2 x 1, a 2 = x 3 x 2,. a n = x n+1 x n,. and let b n = a n, n = 1, 2,. Also let s denote the n n-th partial sum of the b n s. For m < n s n s m = b n + b n b m+1 = x n+1 x n + x n x n x m+2 x m+1. Since m < n, n = m + l for some positive integer l. Let ε > 0 be given. Then there exists a positive integer N such that x n x m < ε l for all m, n N. So, for all m.n N, s n s m < ε l + + ε l = ε,

7 k=1 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 7 i.e. (s ) is a Cauchy sequence in. Since is complete, (s ) is a convergent sequence: n n k=1 a k = lim n s n <. That is, k=1 a k is absolutely convergent. By assumption, it is convergent. Let us denote the n-th partial sum of the a n s by s n. Then a k = lim n s n = lim n [(x 2 x 1 ) + (x 3 x 2 ) + + (x n+1 x n )] = lim n (x n+1 x 1 ) <. Hence, (x n ) is a convergent sequence in X. Now we have shown that any Cauchy sequence in X is a convergent sequence and therefore X is complete. (c) Show that in a Banach space, an absolutely convergent series is convergent. Solution: Let x n=1 n be an absolutely convergent series. For m < n, s n s m = x m x n x m x n x m+1 + = x n ( x x m ) n=1 0 as m. So, (s n ) is a Cauchy sequence in the Banach space and hence it is convergent. (6) Show that (e n ), where e n = (δ nj ), is a Schauder basis for l p, where 1 p <.

8 8 MAT 771 FUNCTIONAL ANALYSIS HOMEWORK 3 Solution: Let x = (ξ j ) l p. Then x = (ξ 1, ξ 2, ξ 3, ) = ξ 1 (1, 0, 0, ) + ξ 2 (0, 1, 0, ) + ξ 3 (0, 0, 1, 0, ) + = ξe 1 + ξ 2 e 2 + ξ 3 e 3 + and e 1, e 2, e 3, l p. Hence, (e n ) is a Schauder basis for l p.