Let $X$ be a complex Banach space. We say a sequence $x^{i} \in X$ converges to $x \in X$ weakly if $\phi\left(x^{i}\right) \rightarrow \phi(x)$ for all $\phi \in X^{*}$. Let $T: X \rightarrow Y$ be bounded and linear. Show that if $x^{i}$ converges to $x$ weakly, then $T x^{i}$ converges to $T x$ weakly.

Now let $X=\ell_{2}$. Show that for a sequence $x^{i} \in X, i=1,2, \ldots$, with $\left\|x^{i}\right\| \leqslant 1$, there exists a subsequence $x^{i_{k}}$ such that $x^{i_{k}}$ converges weakly to some $x \in X$ with $\|x\| \leqslant 1$.

Now let $Y=\ell_{1}$, and show that $y^{i} \in Y$ converges to $y \in Y$ weakly if and only if $y^{i} \rightarrow y$ in the usual sense.

Define what it means for a linear operator $T: X \rightarrow Y$ to be compact, and deduce from the above that any bounded linear $T: \ell_{2} \rightarrow \ell_{1}$ is compact.