As an exercise I tried to prove the following theorem:
If $X,Y$ are Banach spaces and $u \in B(X,Y)$ is a bounded linear
operator then the following are equivalent:(1) $u$ is compact
(2) for every bounded set $S \subseteq X$ the image $u(S)$ is
relatively compact(3) if $x_n$ is a bounded sequence in $X$ then $u(x_n)$ admits a
convergent subsequence in $Y$
Please could someone check my proof?
Proof:
Recall the definition of compact operator: $u$ is compact iff if the image of the unit ball is relatively compact.
$(1) \iff (2)$:
Since multiplication by $n$ is a linear homeomorphism, $u$ is compact iff $u(B(0,1))$ is relatively compact iff $u(B(0,n))$ is relatively compact. From this it is obvious that $(1) \iff (2)$.
$(2) \implies (3)$: Let $x_n$ be a bounded sequence. Then $S=\{x_n\}_{n \in \mathbb N}$ is a bounded set hence $u(S)$ is relatively compact hence $u(x_n)$ admits a subsequence that converges in $Y$.
$(2) \Longleftarrow (3)$: Let $S$ be a bounded set. Let $x_n$ be any sequence in $\overline{u(S)}$. Then $x_n=u(s_n)$ for some sequence $s_n$ in $S$. Then $x_n$ admits a convergent subsequence.
Best Answer
The only problematic part of your proof is the implication $3)\implies 2)$. You should argue as follows
Assume $S$ is bounded. Consider arbitrary $(y_n)_{n\in\mathbb{N}}\subset\overline{u(S)}$, then $(y_n)_{n\in\mathbb{N}}$ is also bounded. For each $n\in\mathbb{N}$ we can find $x_n\in S$ such that $$ \Vert y_n-u(x_n)\Vert\leq 2^{-n}.\tag{*} $$ Since $(y_n)_{n\in\mathbb{N}}$ is bounded, so does $(u(x_n))_{n\in\mathbb{N}}$. Since $3)$ holds, we have convergent subsequence $u(x_{n_k})_{k\in\mathbb{N}}$. From $(*)$ is follows that $(y_{n_k})_{k\in\mathbb{N}}$ converges to the same limit as $u(x_{n_k})_{k\in\mathbb{N}}$. Thus we constructed convergent subsequence of arbitrary sequence in $\overline{u(S)}$. So $u(S)$ is relatively compact.