Show $K$ is a compact operator

Question

Consider the Hilbert space $L^{2}((-1,1))$ with standard inner product, and let the integral operator $K:L^{2}((-1,1)) \to L^{2}((-1,1))$ be defined by \begin{equation*} (Kf)(x) := \int_{-1}^{1}\frac{1}{1+\left(\frac{xy}{2}\right)^{2}}f(y)dy,\qquad f \in L^{2}((-1,1)). \end{equation*} Show that $K$ is a compact operator, for instance by showing how $K$ can be approximated by a sequence of finite rank operators.

My approach: Define the kernel $\kappa(x,y) = \frac{1}{1+\left(\frac{xy}{2}\right)^{2}}$. Construct a finite rank approximation $\kappa_{n}(x,y) = \sum_{i=1}^{n}\kappa_{i}(y)e_{i}(x)$ and $(K_{n}f)(x) = \int_{-1}^{1}\kappa_{n}(x,y)f(y)dy$. Then we show that this converges in the $\Vert \cdot \Vert_{L^{2}}$ norm. As the limit of finite rank operators is compact we can conclude that $\kappa(x,y)$ is compact. However, I don't know how to create such a finite rank approximation

Answer 1

Define $g_n(t) :=\sum_{j=0}^n(-1)^j t^{2j} $. Then for any $t\in (-1,1)$, $\lim_{n\to +\infty}g_n(t)=1/\left(1+t^2\right)$. Define $$(K_n f)(x)=\int_{-1}^1g_n\left(xy / 2\right)f\left(y\right)\mathrm dy.$$ Observe that $$\left|g_n\left(\frac{xy}2\right)-\frac 1{1+\left(\frac{xy}2\right)^2} \right| \leq\sum_{j=n+1}^{+\infty} \left|\frac{xy}2\right|^{2j}\leq \sum_{j=n+1}^{+\infty} 2^{-2j} $$ and $K_n$ is a finite rank operator to get the wanted conclusion.