Hilbert-Schmidt integral operator is compact.
In what follows, we will prove that every Hilbert-Schmidt integral operator is compact. Formally, let $k\in L_2([0,1]\times[0,1])$ and $T:L_2[0,1]\to L_2[0,1]$ be defined by $$Tf(x):=\int_0^1k(x,y)f(y)dy.$$ Then $T$ is compact operator. Although the proof may seem a bit long at first sight, it is proceeds naturally, utilizing approximation of $k$ with continuous functions. $\textbf{Proof}$. We approximate $k$ with a sequence of continuous functions, then prove that each of the corresponding operators has image contained in $C[0,1]$, then show that each of these operators is compact when $C[0,1]$ is equipped with the $\sup$ norm. This will be done via the Arzela-Ascoli theorem. Finally show that these converge uniformly (in the operator norm) to the original operator, and since they are compact, the original must be as well. The Cauchy-Bunyakovsky-Schwartz inequality is used throughout the proof. Thus we start by choosing a sequence $(k_n)_{n\ge 1}$ of continuous functions con