# Show that a compact operator is bounded

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.

• can you write your definition correctly. You can not have iff in the definition
– user537667
May 31, 2018 at 10:14
• Of course you can have iff in a definition. What is that for nonsense?
– user370967
May 31, 2018 at 10:31
• What do you mean by "What is that for nonsense?"
– user537667
May 31, 2018 at 10:34
• Every definition works in 2 directions. It's nonsense that you can't have an iff in a definition.
– user370967
May 31, 2018 at 10:37
• That is not intended to be a sarcasm. It is just the culture in your place. We do not use iff. We say $A$ iff $B$ if $A$ and $B$ both have a meaning already.. if we are giving meaning for some word, which means there is no meaning for that word already, we use if... We call a map $T:V\rightarrow W$ a linear operator if (** something happens).... We do not write we call a map $T:V\rightarrow W$ linear if and only if (** something happens).. It is just a culture..
– user537667
May 31, 2018 at 10:45

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 $$\|Tv\|\leq m\|v\|,\ \ \ v\in V.$$

• Why is the image compact? I don't work in a Banach space!
– user370967
Jun 1, 2018 at 7:58
• Take the closure, then. Jun 1, 2018 at 16:30

If $T$ is unbounded, there is a sequence $(x_n)_{n\in\mathbb N}$ of elements of the unit ball in $V$ such that$$(\forall n\in\mathbb N):\bigl\|T(x_n)\bigr\|>n.$$Therefore, the sequence $\bigl(\|T(x_n)\|\bigr)_{n\in\mathbb N}$ is not a Cauchy sequence.

• But maybe it has a cauchy subsequence?
– user370967
May 31, 2018 at 10:00
• I've edited my answer. I hope that everything is clear now. May 31, 2018 at 10:03
• Still I don't see how this implies that sequence has no Cauchy subsequence. Sorry.
– user370967
May 31, 2018 at 10:06
• Because the sequence has no bounded subsequence and therefore no Cauchy subsequence. May 31, 2018 at 10:20