A compact operator cannot be unbounded. Justify your answer.

Question

A compact operator cannot be unbounded. Justify your answer.

My Attempt:-

Let $H$ be a Hilbert space. Suppose $C:H\to H$ be a compact operator on $H.$ Let $A$ be any bounded set. Then, $\overline{C(A)}$ is compact. Suppose on contrary assume $C(H)$ is unbounded. Hence, for any real number $M$, we can find $x\in H$ such that $\langle Cx,Cx \rangle>M.$ I am trying to find contradiction. Could you help me?

Answer 1

Let $B$ be the unit ball of $H.$ Since $C$ is a compact operator, $\overline{C(B)}$ is compact hence bounded. A fortiori, the subset $C(B)$ is bounded, which proves that $C$ is a bounded operator.

A non-zero operator is never bounded as a map, since a non-zero linear subspace is always unbounded.

Answer 2

Bounded operators are bounded on bounded sets, not on the whole space.

If $C$ is not bounded, then there is a sequence $(x_n)$ in the unit ball such that $\|Cx_n|| \to \infty$. By definition of a compact operator there is a subsequence for which $Cx_{n_k}$ converges in the norm. But convergent sequnces are bounded, so $\|Cx_{n_k}\|$ is bounded. But $\|Cx_{n_k}\| \to \infty$, so we have a contradcition.