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MA3422 (Functional Analysis 2) Tutorial sheet 8 [March 20, 2015] Name: Solutions 1. Let E be a normed space and (xn )∞ n=1 a sequence in E, and x ∈ E. Show that the sequence converges to x in the weak topology if and only if limn→∞ |α(xn −x)| = 0 for each α ∈ E ∗ . [Hint: an open set for the weak topology that contains x must contain Bq (x, 1) = {y ∈ E : q(x − y) < 1} for some continuous seminorm q on E.] Solution: Recall first that xn → x as n → ∞ means that for each open U with x ∈ U there is N such that xn ∈ U for all n ≥ N . If the sequence converges to x in the weak topology and α ∈ E ∗ then Bpα (x, ε) = {y ∈ E : pα (x − y) < ε} = {y ∈ E : |α(x − y)| < ε} is an open set for the weak topology that contains x. So there is N so that |α(x − xn )| < ε ∀n > N That means lim |α(xn − x)| = 0 n→∞ Conversely, assume that this holds for each α ∈ E ∗ . Choose U ⊂ E open in the weak topology with x ∈ U . Then, as we showed (and according to the hint) there is a continuous seminorm q (continuous for the weak topology) with Bq (x, 1) ⊆ U . However we also know that such a seminorm must satisfy q(x) ≤ C1 pα1 (x) + · · · + Cm pαm (x) for some m and some α1 , . . . , αm ∈ E ∗ , some constants Cj ≥ 0. So q(x − xn ) ≤ C1 pα1 (x − xn ) + · · · + Cm pαm (x − xn ) → 0 as n → ∞ Thus there is N with q(x − xn ) < 1 ∀n > N (so xn ∈ Bq (x, 1) ⇒ xn ∈ U ∀n > N ). Thus xn → x (as n → ∞) in the weak topology. 2. Show that a weakly convergent sequence in a normed space E must be a bounded sequence. ∗ [Hint: If (xn )∞ n=1 is a sequence in E, consider the sequence of operators Tn : E → K where Tn (α) = α(xn ) and use the uniform boundedness principle. What is the norm of Tn ∈ E ∗∗ ?] Solution: Suppose the sequence (xn )∞ n=1 in E, converges to x ∈ E in the weak topology. By the previous question limn→∞ α(xn −x) = limn→∞ α(xn )−α(x) = 0, or limn→∞ α(xn ) = ∗ α(x). So (α(xn ))∞ n=1 is a convergent sequence and hence bounded (for each α ∈ E ). Defining (as in the hint) Tn : E ∗ → K by Tn (α) = α(xn ) we actually have Tn = J(xn ) where J : E → E ∗∗ is the canonical embedding. So kTn k = kJ(xn )k = kxn k (and in particular Tn is bounded for each n). The sequence Tn is pointwise bounded since we have just observed that Tn (α) = α(xn ) is a bounded sequence for each α ∈ E ∗ . By the uniform boundedness theorem (which applies since E ∗ is a Banach space), supn kTn k < ∞. So supn kxn k < ∞. 3. Let E = c0 and consider the weak topology on E. Using the fact that E ∗ = `1 , show that a sequence (xn )∞ n=1 in c0 converges weakly to zero if and only if supn kxn k∞ < ∞ and limn→∞ xn,j = 0 for each j, where xn = (xn,1 , xn,2 , . . .) (and 0 ∈ c0 is the constant sequence of zeroes). Solution: By the previous question, if a sequence (xn )∞ n=1 in c0 converges weakly to zero (that is converges in the weak topology to 0), then supn kxn k∞ < ∞. If we define e∗j ∈ c∗0 by e∗j (y) = e∗j (y1 , y2 , . . .) = yj we have limn→∞ e∗j (xn ) = limn→∞ xn,j = e∗j (0) = 0 (using question 1). For the converse, asssume supn kxn k∞ = M < ∞ and limn→∞ xn,j = 0. Take α ∈ c∗0 . By the identification c∗0 = `1 there is z = (z1 , z2 , . . .) ∈ `1 with α(y) = α(y1 , y2 , . . .) = ∞ X yj zj j=1 We can estimate ∞ X |α(xn )| = xn,j zj ≤ j=1 ∞ X |xn,j ||zj | j=1 = ≤ N X |xn,j ||zj | + j=1 j=N +1 N X ∞ X |xn,j ||zj | + j=1 ≤ ∞ X N X |xn,j ||zj | kxn k∞ |zj | j=N +1 |xn,j ||zj | + M j=1 ∞ X j=N +1 2 |zj | P P∞ Since ∞ |z | < ∞, we can choose N so big that j j=1 j=N +1 |zj | is small (say < ε/(2M + PN 1) for ε > 0 given). Then limn→∞ j=1 |xn,j ||zj | = 0, and so for all n big we will have PN j=1 |xn,j ||zj | < ε/2 also. Thus limn→∞ α(xn ) = 0. As this holds for each α ∈ E ∗ , it shows (by question 1) that the sequence (xn )∞ n=1 converges to 0 in the weak topology. Richard M. Timoney 3