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I admit this is an homework. However i'm quite unused to this kind of argument so i would like to receive a suggestion or a confirm about my guesses.. So..

Let $X$ be a separable Banach space with dual $X^*$. Let $B=\{x\in X:\|x\|\leq 1\}$ and let $(x_n)_{n\in\mathbb N}$ be a sequence dense in $B$. Setting $B^*=\{T\in X^*:\|T\|_{X^*}\leq 1\}$, let $d:B^*\times B^*\to\mathbb R$ be the distance defined by: $$d(S,T)=\sum_{n=1}^{+\infty}\;2^{-n}|S(x_n)-T(x_n)|,\qquad S,T\in B^*.$$

Prove that

  1. $d(T_n,T)\to 0$ if and only if $T_n\to T$ pointwise in B;

  2. $(B^*,d)$ is a metric and compact space.

Now my questions.. I can solve the implication $d(T_n,T)\to 0\Rightarrow T_n\to T$ pointwise, but what about the reverse implication? I was trying to find some uniform estimates on the norm $\|T_n-T\|$, maybe relying on Banach-Steinhaus theorem but I'm not sure whether it is possible to apply in this situation.

And for part 2. I can show that $d$ is a metric, but to conclude that $B^*$ is compact i was wondering if it were just a consequence of Banach Alaoglu or there is something more.

Last but not Least.. In my reflections it doesn't sound relevant the hypothesis of $X$ being separable. Was it used implicitly to assure the existence of a sequence $(x_n)_{n\in \mathbb N}$ dense in $B$ or am I missing something?

Thanks in advance to anybody who will answer or just share with me his thoughts.

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2  
Just a short comment: We have $\|T_n - T\| \leq 2$ by the triangle inequality, as $\|T_n\|,\|T\| \leq 1$ (this takes care of part a)). As for part b), you should rather do it using the Arzelà-Ascoli theorem instead of Banach-Alaoğlu. See also my answer here, (the lemma in the added bit). This could give some inspiration. –  t.b. Sep 6 '11 at 12:31
    
@Theo.. how would you finish part a)? –  uforoboa Sep 6 '11 at 12:41
    
Oh yes... what a stupid I am :).. so i'm left just with a finite number of terms of that sequence.. so i can say things like... Let $m$ be such $\max_{1\leq j\leq n_0} \{T_n(x_j)-T(x_j)\}\leq \frac{\varepsilon}{2(n_0-1)}$ for any $n\geq m$. right? –  uforoboa Sep 6 '11 at 13:01
    
Yes, that's the idea (if $\{\cdot\}$ means $\left|\cdot\right|$). –  Davide Giraudo Sep 6 '11 at 13:09
1  
@Davide: Okay, sure, then I'll leave it at a comment. The diagonal trick shows that the unit ball is compact with the metric topology (note that you only have this metric because $X$ is separable, otherwise you can't choose a dense sequence in the unit ball of $X$). In 1. you show that the the map $(B^{\ast},d) \to (B^{\ast}, \operatorname{weak}^{\ast})$ is continuous (you only need metrizability of the domain to conclude continuity from sequential continuity). Thus the usual "compact-Hausdorff-homeomorphism" argument lets you conclude that $d$ induces the weak$^{\ast}$ topology. –  t.b. Sep 6 '11 at 14:48

1 Answer 1

up vote 3 down vote accepted
  1. We assume that for all $x\in X$, $\lim_{n\to\infty}T_nx=Tx$. Let $\varepsilon>0$. We can find an integer $n_0$ such that $\displaystyle\sum_{n=n_0+1}^{+\infty}2^{-n+1}\leq \frac{\varepsilon}2$. Since for $k\in\{1,\ldots,n_0\}$, we have $\displaystyle\lim_{n\to\infty}T_n(x_k)=T(x_k)$, we can find, for $1\leq k\leq n_0$, an integer $N_k$ such that for $n\geq N_k$ we have $2^{-k}\left|T_n(x_k)-T(x_k)\right|\leq \frac{\varepsilon}{2n_0}$. We put $N:=\max_{1\leq k\leq n_0}N_k$. Then, for $n\geq N_k$, we have \begin{align*} d(T_n,T)&=\sum_{k=1}^{+\infty}2^{-k}|T_n(x_k)-T(x_k)|\\ &=\sum_{k=1}^{n_0-1}2^{-k}|T_n(x_k)-T(x_k)|+\sum_{k=n_0+1}^{+\infty} 2^{-k}|T_n(x_k)-T(x_k)|\\ &\leq \sum_{k=1}^{n_0-1}\frac{\varepsilon}{2n_0}+\sum_{k=n_0+1}^{+\infty} 2^{-k+1}\leq \varepsilon. \end{align*}
  2. Since the fact that $(B^*,d)$ is a metric space has been already proved, it suffice to establish that each sequence on $B^*$ admits a convergent subsequence to show that $(B^*,d)$ is compact. Let $\left\{T_n\right\}$ a sequence in $(B^*,d)$. For each $k\geq 1$ we can find an infinite subset $A_k$ of $\mathbb N^*$ such that the subsequence $\left\{T_n(x_k)\right\}_{n\in A_k}$ is convergent, since the sequence $\left\{T_n(x_k)\right\}$ is bounded. We can also assume that the sequence $\{A_k\}$ is strictly decreasing. Now let $\varphi(n)$ the $n$-th element of $A_n$. Then the sequence $\{T_{\varphi (n)}(x_k)\}$ is convergent for all $k\geq 1$. Now we show that the sequence $\{T_{\varphi(n)}(x)\}$ is convergent for all $x\in X$. Let $x\in X$ and $\varepsilon>0$. Let $k$ such that $\lVert x-x_k\rVert\leq \frac{\varepsilon}3$. We have for $m,n\in\mathbb N$: \begin{align*} \left|T_{\varphi(m)}(x)-T_{\varphi(n)}(x)\right|& \leq \left|T_{\varphi(m)}(x)-T_{\varphi(m)}(x_k)\right| +\left|T_{\varphi(m)}(x_k)-T_{\varphi(n)}(x_k)\right|\\ &+\left|T_{\varphi(n)}(x_k)-T_{\varphi(n)}(x)\right|\\ &\leq \lVert x-x_k\rVert+\left| T_{\varphi(m)}(x_k)-T_{\varphi(n)}(x_k)\right|+ \lVert x-x_k\rVert. \end{align*} Now, we pick $N\in\mathbb N$ such that for $m,n\geq N$ we have $\left| T_{\varphi(m)}(x_k)-T_{\varphi(n)}(x_k)\right|\leq \frac{\varepsilon}3$, and it shows that $\{T_{\varphi(n)}(x)\}\subset \mathbb R$ is a Cauchy sequence. Let $\displaystyle T(x):=\lim_{n\to\infty}T_{\varphi(n)}(x)$. We can conclude because $T$ is linear, continuous and its norm is $\leq 1$.
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1  
Very nice! Thanks for the effort (and saving me some work). Here are a few further remarks: 1. As I pointed out in a comment it's easy enough to see that $d$ in fact induces the weak$^{\ast}$ topology. 2. Metrizability of the unit ball doesn't imply metrizability of the dual space in the weak$^\ast$ topology (the weak$^{\ast}$ topology isn't even first countable!) 3. Separability of $X$ is necessary: a) the "coordinate functionals" on $\ell^{\infty}$ give a sequence in $(\ell^{\infty})^{\ast}$ without convergent subsequence. b) metrizability of $B^\ast$ in fact implies separability of $X$. –  t.b. Sep 6 '11 at 17:01

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