I came across the following in Example 3: of these notes. Let $A$ represent the unit circle. If $K(x, y) \in L^2(A \times A)$, then the integral operator
$$Tf = \int_A K(x,y) f(y) dy,$$
is compact. The author says that this can be proven using
- finite rank condtion
- weak convergence definition
I have next to zero experience with showing that operators are compact so I was wondering does anyone know how these proofs go?
Treat $A$ as the interval $[-\pi,\pi]$, where you have the orthonormal basis $\{ \frac{1}{\sqrt{2\pi}}e^{inx}\}_{n=-\infty}^{\infty}$. That gives you an orthonormal basis $\{ \frac{1}{2\pi}e^{inx}e^{imy}\}_{n,m}$ of $L^2(A\times A)$. So, given $\epsilon > 0$. Then, for $K_N=\sum_{n=-N}^{N}\sum_{m=-N}^{N}k_{n,m}e^{inx}e^{imy}$, you can choose $N$ large enough that $\|K-K_N\|_{L^2(A\times A)}$ is as small as you want. Then that gives you a way to approximate your integral operator by the finite rank operator $$ \int_{A} K_N(x,y)f(y)dy=\sum_{n=-N}^{N}\sum_{m=-N}^{N}\langle f,e^{-imy}\rangle e^{inx}. $$ All you have to do is show that the operator norm on $L^2(A)$ of an integral operator is bounded by a constant times the $L^2(A\times A)$ norm of the kernel, which can be done directly. You can show that $\|T\|_{\mathcal{L}(L^2(A))}\le \|K\|_{L^2(A\times A)}$.
To prove Using $\|T\|_{\mathcal{L}(L^2(A))}\le \|K\|_{L^2(A\times A)}$, first apply the Cauchy-Schwarz $$ \left|\int_{A}M(x,y)f(y)dy\right|^2 \le \int_{A}|M(x,y)|^2dy\int_{A}|f(y)|^2dy $$ Then integrating in $x$ to obtain $$ \left\|\int_{A}M(x,y)f(y)dy\right\|_{L^2(A)}^2 \le \int_{A\times A}|M(x,y)|^2dxdy\int_{A}|f(y)|^2dy $$ Taking square roots, $$ \|Tf\|_{L^2(A)}\le \|M\|_{L^2(A\times A)}\|f\|_{L^2(A)}, $$ which is all you need to be able to closely approximate the operator $T$ in the operator norm by a finite rank operator.