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

  • $\begingroup$ You mean that $(x_n)$ is a bounded sequence. $\endgroup$ May 26, 2011 at 5:06
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    $\begingroup$ This seems to me a rather strange way of proving that result. In my opinion the spectral theorem requires quite a bit more machinery than what you call big characterization theorem (proved e.g. as Theorem 3.3.3 in Pedersen's Analysis now, where you can see that this is essentially just the definition of compactness). $\endgroup$
    – t.b.
    May 26, 2011 at 5:27
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    $\begingroup$ I hope the proof below is easy enough for your taste. What I was trying to say is that the spectral theorem is a much stronger and much more difficult result than the characterization theorem for compact operators, so it should be the case that there is an easy proof of the characterization theorem when the spectral theorem is assumed known. $\endgroup$
    – t.b.
    May 26, 2011 at 6:28

3 Answers 3


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.

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    $\begingroup$ Thanks. I realize that this is a weird order in which to prove things. I'd seen the theorem on compact operators before and now I'm taking a class on C*-algebras. We developed all the spectral theorem stuff and started on K(H) afterward (or B_0(H) for the Pedersenite). I recognize that it's not a good way to prove that theorem on compacts, it's just sort of a nifty thing that falls out of the C* stuff. $\endgroup$
    – Daliath
    May 27, 2011 at 3:51
  • $\begingroup$ Oh, I see! Then I take back what I said and encourage you: There are many things that you can prove with $C^{\ast}$-algebras and related techniques, which renders the things rather algebraic and a bit magical. It's only the very beginning and it's amazing how an identity like $\|a^\ast a\| = \|a\|^2$ can be so powerful (even if it's often not really needed). $\endgroup$
    – t.b.
    May 27, 2011 at 4:48
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    $\begingroup$ @Daliath: One of the classic applications of the spectral theorem is Wiener's $1/f$-theorem. Look at any of the references on the Wikipedia page to see how easy the result is when using the spectral theorem, then look at the original paper by Wiener (also in the reference list) to see how difficult that theorem was for Wiener to establish. It's simply an amazing illustration of the power of the spectral theorem and $C^{\ast}$-algebras. $\endgroup$
    – t.b.
    May 27, 2011 at 5:29
  • $\begingroup$ Do we need to consider the double dual to show the reverse implications? $\endgroup$
    – Xiang Yu
    Dec 29, 2015 at 16:58
  • $\begingroup$ @XiangYu: yes. See also arxiv.org/abs/1010.1298 $\endgroup$ Apr 13, 2017 at 13:42

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.


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.

  • $\begingroup$ Why $*$-involution is continuous? $\endgroup$ May 2 at 5:39
  • $\begingroup$ @onepotatotwopotato Since $\|x^*\| = \|x\|$. It is even isometric. $\endgroup$
    – J. De Ro
    May 2 at 16:46

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