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Spring 2015 - Problem 6

Banach spaces

When $B_1$ and $B_2$ are Banach spaces, we say that a linear operator $\func{T}{B_1}{B_2}$ is compact if for any bounded sequence $\set{x_n}_n$ is $B_1$ , the sequence $\set{Tx_n}_n$ has a convergent subsequence. Show that if $T$ is compact then $\im{T}$ has a dense countable subset.

Solution.

Let $n \geq 1$ and observe that $K_n = T\p{B\p{0, n}}$ is precompact. Indeed, since $T$ is a compact operator, it follows that $\cl{K_n}$ is sequentially compact since $B_2$ is complete, hence compact since we are in a metric space. For each $r = \frac{1}{k}$ , notice that

\cl{K_n} \subseteq \bigcup_{x \in K_n} B\p{x, \frac{1}{k}},

so by compactness, there exist finitely many $x^{\p{k}}_1, \ldots, x_{m_k}^{\p{k}} \in K$ such that $\set{B\p{x^{\p{k}}_j, \frac{1}{k}}}_j$ cover $\cl{K_n}$ . Notice then that $S_n = \bigcup_k \set{x^{\p{k}}_j}_j$ is countable and dense in $K_n$ . Indeed, it is certainly countable as a countable union, and given $y \in K_n$ , for each $k \geq 1$ , there exists $x_k \in S_n$ such that $y \in B\p{x_k, \frac{1}{k}}$ by construction. Thus, we get a sequence such that $\abs{y - x_k} < \frac{1}{k}$ , i.e., $x_k \to y$ .

Let $S = \bigcup_n S_n$ . As a countable union of countable sets, $S$ remains countable, and it is dense in $\im{T}$ . Indeed, given any $Tx \in \im{T}$ , there exists $n \geq 1$ such that $x \in B\p{0, n}$ , i.e., $Tx \in K_n$ . Since $S_n$ is dense in $K_n$ , it follows that $S$ is still dense in $K_n$ , which completes the proof.