# 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).

(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).
(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.