I came across following problem in a prelim question paper.
Let $(X,\mu)$ be a measure space such that $\mu(X)<\infty.$ Let $K(x,y)$ be in $L^2(X\times X)$. Define an operator $K:L^2(X)\to L^2(X)$ by $f\mapsto \int K(.,y)f(y)d\mu(y)$. Consider a sequence $f_n$ in the unit ball of $L^2(X)$, which converges weakly to $f$. Show that $||Kf_n-Kf||_2\to 0$.
One way to prove it would be to argue that $K$ is a compact operator and invoking the standard result from functional analysis that $K$ is compact iff it maps weakly convergent sequences to strongly convergent sequences. The part that $K$ is compact is also pretty standard. We can choose a basis for $L^2(X)$ and then construct a basis for $L^2(X\times X)$ and write $K$ in terms of this basis and argue that $K$ is strong limit of finite rank operators.
Alternatively, we can observe that $Kf_n$ converges to $Kf$ almost everywhere (This follows from weak convergence of $f_n$ to $f$). Therefore, $Kf_n^2$ converges to $Kf^2$ almost everywhere. But, also $|Kf_n|^2(x)\le ||K(x,.)||_2^2$. Applying DCT we can obtain the convergence of norms. Combining this with the fact that $Kf_n$ converges to $Kf$ weakly, we can argue that convergence happens in strong sense.
Any alternate idea is of course a welcome. I am worried because as far as I can see, I am not using the finiteness assumption on measure space and neither does it seem to reduce any work. So, I wanted a clarification if there is a flaw in my argument(s) above or it is tha case that we really don’t need any finiteness assumption on the measure.