Compact operator on $L^{2}$

Question

Let $K(t,s)$ be a real-valued function of two real variables, and let $T: L^2(a,b) \to L^2(a,b)$ be defined by $(Tf)(t) = \int_{a}^{b} K(t,s) f(s) ds$ where $$K(t,s) = \sum_{j=1}^{n} \phi_{j}(t) \psi_{j}(s)$$

I want to prove that $T$ is a compact operator. If $K$ was a continuous function in both variables, one can simply apply Arzela-Ascoli theorem, however I am not sure how to proceed in this case.

Thanks.

Answer 1

Okay:

$\|Tf\|_2^2=\int_a^b|\sum_{j=1}^n\int_a^b\phi_j(t)\psi_j(s)f(s)ds|^2dt\le\int_a^b|\sum_{j=1}^n|\phi_j(t)|\int_a^b|\psi_j(s)f(s)|ds|^2dt$

$\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\int_a^b|\sum_{j=1}^n \phi_j(t)|^2dt\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\|\sum_{j=1}^n \phi_j(t)\|_2^2$

Thus: $\|Tf\|\le\sup_j\|\psi_j\|_2\|\sum_{j=1}^n \phi_j(t)\|_2\|f\|_2$

I.e. $T$ is bounded (and thus continous) and also it is clearly linear now:

$T(f)=\int_a^bK(t,s)f(s)ds=\int_a^b\sum_{j=1}^n\phi_j(t)\psi_j(s)f(s)ds=\sum_{j=1}^n\phi_j(t)\int_a^b\psi_j(s)f(s)ds$

$=\sum_{j=1}^na_j\phi_j(t)$, i.e $Ran(T)=span\{\phi_j\}$ which is $n$ dimensional and finite, so $T$ is a bounded linear finite rank operator and thus is compact.

I should probably mention that I have used holders inequality, i.e. If $f,g\in L^2$, then $fg\in L^1$ and $\|fg\|_1\le\|f\|_2\|g\|_2$.

And if $\phi_1,...,\phi_n\in L^2$, then $\sum_{j=1}^n\phi_j\in L^2$.

Answer 2

Since $$ \int_{a}^{b}dt\int_{a}^{b}ds|K(t,s)|^{2}<\infty $$ $K(t,s)$ is a Hilbert-Schmidt operator and such operators are compact.