Compact operators and weak convergence Let $X$ and $Y$ be Banach spaces. 
(a) Let $T \in \mathcal{L}(X, Y )$. For each sequence $(x_n)_{n \geq 1}$ in $X$ and each $x \in X$, show that
$x_n →x$ weakly, as $n \rightarrow  \infty$ ,implies that $Tx_n \rightarrow Tx$ weakly, as $n\rightarrow \infty$.  
(b)  Let $T \in \mathcal{K}(X, Y )$. For each sequence $(x_n)_{n \geq 1}$ in $X$ and each $x \in X$, show that
$x_n →x$ weakly, as $n \rightarrow  \infty$ ,implies that $||Tx_n -Tx|| \rightarrow 0$ , as $n\rightarrow \infty$.. 
(c) Conversely, if $X$ is reflexive and separable, and $T \in \mathcal{L}(X,Y)$ satisfies that $∥Tx_n − Tx∥ \rightarrow 0$, as $n \rightarrow \infty$, whenever $(x_n)_{n\geq 1}$ is a sequence in $X$ converging weakly to $x \in X$, then $T \in \mathcal{K}(X, Y )$.
 
(d) Show that each $T \in \mathcal{L}(X,l_1(\mathbb{N}))$ is compact, whenever $X$ is reflexive and separable.  
(e) Let $Y$ be infinite dimensional. Show that no $T \in \mathcal{K}(X, Y )$ is open.  
(f) Show that there is no reflexive separable Banach space $X$ such that $l_1(\mathbb{N}) = T(X)$, for some $T \in \mathcal{L}(X,l_1(\mathbb{N}))$.
My attempt:
(a) We have that $x_n \rightarrow x$ weakly if and only if $f(x_n) \rightarrow f(x)$ weakly for every $ f \in X^*$. Now $Tx_n \rightarrow Tx$ weakly if and only if $g(Tx_n) \rightarrow g(Tx)$ weakly for every $g \in Y^*$. But for every $g \in Y^*$ we have $gT \in X^*$. Therefore  $Tx_n \rightarrow Tx \Leftrightarrow g(Tx_n) \rightarrow g(Tx) \Leftrightarrow  (gT)x_n \rightarrow (gT)x \Leftrightarrow x_n \rightarrow x$ weakly. 
b) Since $T$ is compact I know that every sequence is sent to a sequence that has a convergent subsequence, but then I don't know how to proceed.
c) I have a hint for this problem:
Suppose that $T$ is not compact. Show that there exists $\delta > 0$ and a sequence $(x_n)_{n\geq 1}$ in the unit ball of $X$ such that $∥Tx_n −Tx_m∥ \geq \delta$, for all $n \neq m$.. Show next that $(x_n)_{n\geq 1}$ has a weakly convergent subsequence.
(d) Now let $(x_n)_{n \geq 1}$ be a sequence in $ X$  and $x \in X$ such that $ x_n \rightarrow x$ weakly, as $n \rightarrow \infty$. By part (a) since $T \in \mathcal{L}(X,l_1(\mathbb{N}))$ we get $Tx_n \rightarrow Tx$ weakly, as $n \rightarrow \infty$. But weak convergence is the same as norm convergence in $l_1(\mathbb{N})$. Therefore $||Tx_n-Tx|| \rightarrow 0$, as $n \rightarrow \infty$. So we can use part $c)$ to deduce that $T \in \mathcal{K}(X,l_1(\mathbb{N}))$, as requested.
(e) I know that if $Y$ is infinite dimensional the unit ball in $Y$ is not compact.
(f) Since $X$ is reflexive and separable, by part (d) we get that $T$ is compact. Suppose now that $l_1(\mathbb{N})=T(X)$, i.e. $T$ is surjective. By the Open Mapping Theorem $T$ is open and this contradicts part (e).
 A: (a) Looks good!
(b) Using this fact and (a), you should be set. If $Tx_n$ fails to converge to $Tx$ in norm, then there must be a subsequence $(Tx_{n_k})_k$ that maintains at least some $\varepsilon > 0$ distance from $Tx$. Take a norm-convergent subsequence of this subsequence, and it will be a subsequence that fails to weakly converge to $Tx$ (as it norm-converges to another point). This contradicts (a).
(c) Not sure where separability comes in here, but if $T$ is not compact, then $T(B_X)$ is not totally bounded. That is, there exists a $\delta > 0$ such that no finite number of balls radius $\delta$ that cover $T(B_X)$. Hence, we may choose some $Tx_1 \in T(B_X)$, and for each $n$,
$$Tx_{n+1} \in T(B_X) \setminus \bigcup_{k=1}^n B[x_k; \delta],$$
which is never empty by definition of $\delta$. Thus, we have a sequence $(x_n) \in B_X$ such that
$$\|Tx_n - Tx_m\| > \delta$$
for all $n, m$. Using reflexivity, as well as Eberlein-Smulian, the sequence $(x_n)$ must have a weakly convergent subsequence. Note that this weakly convergent subsequence must also have its terms separated by at least $\delta$, which means it cannot be mapped to a norm-convergent sequence (as such a sequence would have to be Cauchy). This shows the contrapositive of (c).
(d) Good.
(e) If $T$ is compact and open, then $T(B[0; 1])$ contains $T(B(0; 1))$, which is open, since $T$ is an open map. Thus, $T(B[0; 1])$ contains some ball $B[y; r]$. Since $T$ is compact, the closure of $T(B[0; 1])$ is compact, hence so is the closed subset $B[y; r]$. A scaling and translation argument yields that $Y$ has a compact unit ball, and hence is finite-dimensional.
(f) Also good.
