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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.

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user148502
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    Your "It's known" characterization is not correct. This characterizes bounded operators, not compact ones. For instance, the identity operator has this property but is not compact. – Nate Eldredge Aug 23 '14 at 17:36
  • You can prove your result using the fact that in a Hilbert space an operator is compact if and only if it is the norm limit of finite rank operators. – David Mitra Aug 23 '14 at 17:54
  • Arzela-Ascoli led to the notion of a compact operator. The result dates back to the late 19th century, before a compact operator was even defined. Why would you want to avoid Arzela-Ascoli? – Disintegrating By Parts Aug 24 '14 at 03:38

2 Answers2

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Here is an answer using the idea that Cauchy sequences are convergent in complete spaces.

MathIsHard
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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.

Martin Argerami
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