$d(T_n,T)\to 0$ if and only if $T_n\to T$ pointwise on the closed unit ball

Question

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.

Answer 1

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 $\leq 1$.