Let's $H$ be a Hilbert space and $T:H \rightarrow H$ a compact map. Show that the adjoint operator $T^*$ is also compact.
I've been thinking a lot about this problem and this is what I've done:
It's known that $T$ is compact iff for every $D \subset H$ bounded, $\overline{T(D)}$ is compact.
Now, if $T':H' \rightarrow H': y'\mapsto y'\circ T$ is the transpose operator of $T$, we can write (by definition) as $T^*=\psi^{-1}\circ T' \circ \psi$, where $\psi:H \rightarrow H'$ is the Fréchet-Riesz map (bijective isometry).
Then, given $D \subset H$ bounded: $$\overline{T^*(D)}=\overline{\psi^{-1}\circ T' \circ \psi (D)}=\overline{\psi^{-1}\circ (T'( \psi))(D)}=\overline{\psi^{-1}\circ (\psi \circ T)(D)}=\overline{(\psi^{-1}\circ \psi) \circ T (D)}=\overline{T(D)}$$ that is compact by hypothesis.
But I know that this proof must be necessary incorrect, because it would mean that $T=T^{*}$, what is false in general. I'd appreciate if someone help me to solve this problem. I've seen a similar question in: Easy Proof Adjoint(Compact)=Compact, but I'd like a proof avoiding using Ascoli-Arzelá. Thank you for your help.
2 Answers
Here is an answer using the idea that Cauchy sequences are convergent in complete spaces.
If you look at the Polar Decomposition, you can write $T=U(T^*T)^{1/2}$, and it follows that $U^*$ maps the closed range of $T$ onto the closed range of $T^*$. So if $T(X)$ is compact for a ball $X$, so is $T^*(X)=UT(X)$.
The Polar Decomposition can be used here in a second way, if you use the fact that compact operators form a closed ideal. Indeed, if $T$ is compact then so is $T^*T$, and thus $(T^*T)^{1/2}$. So $T^*=(T^*T)^{1/2}U$ is compact.
