I am interested in proving that a bounded linear operator $T$ over a separable Hilbert space $H$ is compact if and only if $T^*$ (its adjoint operator) is compact.
I know that there exists other alternative proofs of this fact, but I want to know if the way which I am following can works.
My attempt:
I have proved before that $T$ is compact iff $\{P_nT\}\to T$. Where, if $\{e_n:n\in \mathbb{N}\}$ is an orthonormal basis of $H$, $P_n(x)=\sum_{k=1}^n(x|e_k)e_k$.
So, the proof of the result would be complete if I prove that $\{P_nT^*\}\to T^*$. However, this is the part that I don't see.
Could someone help me?
Thanks.
Best Answer
Well, if $P_nT\to T$ in norm, then as taking adjoint is isometric, we get $T^*P_n\to T^*$ in norm, so $T^*$ is a norm-limit of finite-rank operators and hence compact. The converse also follows applying the above argument to $T^*$.