Easy Proof Adjoint(Compact)=Compact

Question

I am looking for an easy proof that the adjoint of a compact operator on a Hilbert space is again compact. This makes the big characterization theorem for compact operators (i.e. compact iff image of unit ball is relatively compact iff image of unit ball is compact iff norm limit of finite rank operators) much easier to prove, provided that you have already developed spectral theory for C*-algebras.

By the way, I'm using the definition that an operator $T\colon H \to H$ is compact if and only if given any [bounded] sequence of vectors $(x_n)$, the image sequence $(Tx_n)$ has a convergent subsequence.

edited for bounded

Answer 1

What you're asking about is called Schauder's theorem.

An operator $T: X \to Y$ is compact if and only if $T^{\ast}: Y^{\ast} \to X^{\ast}$ is compact.

I'm using the following definition of compactness: An operator $T: X \to Y$ between Banach spaces is compact if and only if every sequence $(x_{n}) \subset B_{X}$ in the unit ball of $X$ has a subsequence $(x_{n_{j}})$ such that $(Tx_{n_j})$ converges. This implies that $K = \overline{T(B_{X})} \subset Y$ is compact, as it is sequentially compact and metric. Now let $(\phi_{n}) \subset B_{Y^{\ast}}$ be any sequence and we want to show that $(T^{\ast}\phi_{n})$ has a convergent subsequence. Observe that the sequence $f_{n} = \phi_{n}|_{K}$ in $C(K)$ is bounded and equicontinuous, so by the theorem of Arzelà-Ascoli, the sequence $(f_{n})$ has a convergent subsequence $(f_{n_{j}})$ in $C(K)$. Now observe $$\|T^{\ast}\phi_{n_i} - T^{\ast}\phi_{n_{j}}\| = \sup_{x \in B_{X}} \|\phi_{n_i}(Tx) - \phi_{n_j}(Tx)\| = \sup_{k \in K} |f_{n_i}(k) - f_{n_j}(k)|$$
where the last equality follows from the fact that $T(B_{X})$ is dense in $K$. But this means that $(T^{\ast}\phi_{n_j})$ is a Cauchy sequence in $X^{\ast}$, hence it converges.

I leave the other implication as well as the translation to the Hilbert adjoint to you as an exercise.

Answer 2

Since $H$ is a Hilbert space, I will give another proof using this property.

Assume $T$ is compact and let $(x_n)_{n\in\mathbb{N}}$ be a bounded sequence. As it is bounded, we can construct a weakly convergent subsequence, call it $(x_n)_n$ again in abuse of notation, such that $x_n\to x$ weakly as $n\to\infty$. Now, we want to show that $T^\ast x_n \to T^\ast x$ strongly. Note that $$\begin{aligned}\|T^\ast (x_n - x) \|^2 &= \langle T^\ast (x_n - x), T^\ast (x_n - x)\rangle=\langle x_n - x , T T^\ast (x_n - x)\rangle \\ &\leq \|x_n-x\| \|T T^\ast (x_n - x)\|\leq C \|T T^\ast (x_n - x)\|,\end{aligned}$$ where we used that a weak convergent sequence is bounded. It suffices to show that $T T^\ast (x_n - x)\to 0 $ strongly as $n\to\infty$. For this, note that $T T^* (x_n - x ) \to 0 $ weakly as $T T^*$ is continuous (in the strong and thus also in the weak topology). Since $T$ is a compact operator and limits are unique we also now that $T T^\ast (x_n - x)\to 0 $ as $n\to\infty$, which concludes the proof.

Answer 3

Here is an alternative proof, provided that you know that an operator is compact iff it is the operator-limit of a sequence of finite-rank operators.

Let $T: H \to H$ be a compact operator. Then $T= \lim_n T_n$ where the limit is w.r.t. the operatornorm and $T_n$ is a finite rank operator. Using that the $*$-involution is continuous, we get

$$T^*= \lim_n T_n^*$$

where $T_n^*$ is also a finite rank operator for all $n$. Thus $T^*$ is the limit of finite-rank operators and it follows that $T^*$ is compact as well.