If $T$ is normal, is ($T^2$ is compact $\Rightarrow T$ compact) true?

Question

I am studying Spectral Theory and Functional Analysis. I know that in a Hilbert space H, if $T$ is self-adjoint then we have: $$T^2 \text{ compact }\Rightarrow T \text{ compact}$$ I want to know if this is true if $T$ is normal.
To show that I need to show that $\|Tx\|^2\leq\|T^2x\|\|x\|$, for all $x\in H$. I find out that showing that $\|Tx\|=\|T^*x\|$ for all $x\in H$ would help. However, I can't seem to find a way to manipulate that. Any hints?

Answer 1

Let $\{Tx_n\}$ be a sequence with $\|x_n\|\leq1$ for all $n$. Since $T^2$ is compact, the sequence $\{T^2x_n\}$ admits a convergent subsequence $\{T^2x_{n_k}\}$. Now $$ \|Tx_{n_k}-Tx_{n_j}\|^2=\|T(x_{n_k}-x_{n_j})\|^2\leq \|T^2(x_{n_k}-x_{n_j})\|\,\|x_{n_k}-x_{n_j}\|\leq2\|T^2x_{n_k}-T^2x_{n_j}\|, $$ which shows that $\{Tx_{n_k}\}$ is Cauchy. Thus $T$ is compact.

Answer 2

Using $\|Tx\|=\|T^\ast x\|$ for all $x$ and Cauchy-Schwarz we have $$ \|Tx\|^2 = \langle Tx,Tx \rangle = \langle T^\ast Tx,x \rangle \le \|T^\ast (Tx)\| \|x\| = \|T(Tx)\| \|x\| = \|T^2 x\| \|x\|. $$