# proof: Hilbert Schmidt operator is compact

Consider the Hilbert Schmidt operator $$K: L^2(\Omega) \rightarrow L^2(\Omega)$$, $$\Omega \subset \subset \mathbb R^N$$, with $$k \in L^2(\Omega \times \Omega)$$ and $$f \in L^2(\Omega)$$,

$$(Kf)(x) := \int_\Omega k(x,y)f(y)\, dy.$$

I want to show that the Hilbert Schmidt operator $$K$$ is a compact operator. Therefore I'm using this characterization.

Let $$X$$, $$Y$$ be normed linear spaces and $$X$$ reflexive. A continuous linear operator $$T: X \rightarrow Y$$ that maps weakly convergent sequences onto strongly convergent sequences is compact.

(We already know that $$K$$ is well-defined as is proven here.)

My question here is, isn't it obvious that $$K$$ is compact?

• We know that $$K$$ is linear and bounded, hence continuous.
• Every continuous map takes weakly convergent sequences to weakly convergent sequences.
• The norm itself is also continuous.
• Weak convergence together with convergence of the norms implies convergence.

Thus $$K$$ is compact. Am I missing something here? Or better: What am I missing here?

$$\,$$

I'm also adding the proof from the textbook for completeness:

Proof. Let $$(f_n)_{n \in \mathbb N} \subset L^2(\Omega)$$ a weakly convergent sequence, then $$(f_n)_{n \in \mathbb N}$$ is bounded. That is, $$\exists C > 0$$ such that $$||f_n||_{L^2(\Omega)} \leq C$$, $$\forall n \in \mathbb N$$. By Fubini's theorem we have for almost every $$x\in \Omega$$ that $$|| k(x,\cdot) ||_{L^2(\Omega)} = \int_\Omega |k(x,y)|^2 \, dy < \infty .$$

Thus for almost every $$x \in \Omega$$ we have

\begin{align} \lim_{n \rightarrow \infty} (Kf_n)(x) & = \int_\Omega k(x,y)f_n(y) \, dy = \lim_{n \rightarrow \infty} \langle k(x,\cdot), f_n \rangle_{L^2(\Omega)} \\ & = \langle k(x,\cdot), f \rangle_{L^2(\Omega)} = \int_\Omega k(x,y)f(y) \, dy = (Kf)(x) \end{align}

By Cauchy-Schwarz's inequality we have $$(Kf_n)(x) \leq ||f_n||_{L^2(\Omega)} \int_\Omega |k(x,y)|^2 \, dy \leq C \, \int_\Omega |k(x,y)|^2 \, dy$$

Hence by Lebesgue's dominant convergence theorem we have convergence of the norms $$\lim_{n \rightarrow \infty} \int_\Omega |(Kf_n)(x)| \, dx = \int_\Omega |(Kf)(x)| \, dx ,$$ that is $$|| Kf_n ||_{L^2(\Omega)} \rightarrow || Kf ||_{L^2(\Omega)}\, \, (n\rightarrow \infty)$$. Since weak convergence together with (strong or normal) convergence of the norms implies (strong) convergence, $$K$$ is compact.

• How would your argument not apply to a general bounded operator? Not all bounded operators are compact. Jan 22, 2017 at 18:15
• agreed. but where is the error in the argument? Jan 22, 2017 at 19:57
• it would be good to add a citation for the textbook where you found the proof you wrote Jun 26, 2018 at 22:28

The norm is continuous as a map $\|\cdot\|: (X,\|\cdot\|_X)\to \mathbb R$ but not when defined on $X$ with its weak topology. This is where your general argumentation fails. The proof from your textbook is fine, however one can in general show that every Hilbert-Schmidt operator is already compact: One can represent the finite rank operators in a Hilbert space as a tensor product. We have several natural norms on this space, whose completions lead to several classes of operators (nuclear operators, Hilbert-Schmidt operators and compact operators) and those norms dominate each other in such a way that we have the inclusions nuclear operator is Hilbert-Schmidt operator is compact operator.
• what exactly do you mean by "... but not when defined on $X$ with its weak topology"? the weak topology is not metrizable in general, hence exists no correponding norm or am I mistaken? Jan 23, 2017 at 23:56
• In general not, but the weak topology is, as the name suggests, a topology. Consider $l^2(\mathbb N)$. Then $\delta_n$ converges weakly to zero, but $\|\delta_n\|=1$, hence the norm is not continuous wrt the weak topology. Jan 24, 2017 at 7:03