Definition: A linear operator $T: V \to W$ is compact if and only if the image of the unit ball in $V$ is precompact (= every sequence
has a cauchy subsequence $\iff $ totally bounded).Prove: Let $T: V \to W$ be a compact linear operator. Show that $T$ is
bounded.
My attempt:
Suppose $T$ is not bounded. Then,
$$\forall M > 0: \exists v_M \in V: \Vert T v_M \Vert > M \Vert v_M \Vert$$
I then tried to construct a sequence without cauchy sequence in $\{Tv \mid \Vert v \Vert \leq 1\}$
So, let $p > q$. Then, $$\left\Vert T \frac{v_p}{\Vert v_p \Vert} – T \frac{v_q}{\Vert v_q \Vert}\right\Vert \geq \left|\frac{\Vert Tv_p \Vert}{\Vert v_p \Vert} – \frac{\Vert Tv_q \Vert}{\Vert v_q \Vert}\right|$$
but was unable to conlude something because off the minus sign.
Any ideas?
EDIT: This is not a duplicate, as other posts use other definitions of compact operators.
Best Answer
It follows directly from the definitions. Since $T$ is compact, the image $T(V_1)\subset W$ of the closed unit ball of $V$ is precompact. Consider the cover $W\subset \bigcup_n W_n$, where $W_n$ is the ball of radius $n$. As $\overline{T(V_1)}$ is compact, the cover has a finite subcover, which means that there exists $m$ with $\overline{T(V_1)}\subset W_m$. In particular, $\|Tv\|\leq m$ for all $v\in V_1$, which leads to $$ \|Tv\|\leq m\|v\|,\ \ \ v\in V.$$