TITLE: proof: Hilbert Schmidt operator is compact
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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.