# Proving that an operator $T$ on a Hilbert space is compact

Let $H$ be a Hilbert space, $T:H \to H$ be a bounded linear operator and $T^{*}$ be the Hilbert Adjoint operator of $T$. Show that $T$ is compact if and only if $T^{*}T$ is compact.

My attempt:

Suppose first that $T$ is compact. The Hilbert Adjoint operator of $T$ is bounded therefore $T^{*}T$ is compact.

How can i proceed with the converse part ?

Here is an outline of a proof in the direction you seek; you should be able to fill is the details:

Suppose $T^* T$ is compact.

1. Show that $\sqrt{ T^* T}$ is compact.

2. Use the Polar Decomposition to write $T = U \sqrt{T^* T}$ for some partial isometry $U$.

3. Conclude that $T$ is compact.

Suppose that $f = \text{w}-\lim_{n \to \infty} f_n$. So we have $\lim_{n \to \infty} \| T^* T (f_n-f)\|=0$ because $T^*T$ is compact. Also, we know that sequence $\{f_n-f\}_{n=1}^{\infty}$ is bounded, so we have \begin{align*}\lim_{n \to \infty} \|T(f_n-f)\|^2 = \lim_{n \to \infty} \langle T^*T (f_n-f),f_n-f\rangle \leqslant \limsup_{n \to\infty} \|T^*T(f_n-f)\|\|f_n-f\| = 0, \end{align*} that is, $\text{s}-\lim_{n \to \infty} Tf_n = Tf$. So, $T$ is compact.

• @johny No. That is just arbitrary sequence of elements of our Hilbert space which is weakly convergent. I used that bounded operator is compact iff operator maps weakly convergent sequences into strongly convergent sequences. Sep 5 '15 at 8:01
• Thanks,I just had to review the definition of weak convergence. Sep 5 '15 at 8:03