$(Af)(t)=\int_{0}^{1}\min\{s,t\}f(s)ds$ is compact in $L_2[0,1]$

Solution 1:

Set $K(s,t)=\min\{s,t\}:[0,1]^2\to[0,1]$. Compactness of your operator doesn't have to do with continuity of $K$ (note that $K$ is indeed continuous), but with the fact that $\int_{[0,1]^2}K^2<\infty$. Let's review the general case.

Exercise 1: Let $H$ be a Hilbert space with ONB $\{e_n\}$ and let $T\in B(H)$ be an operator satisfying $\sum_{n}\|Te_n\|^2<\infty$. Show that $T$ is compact. Such operators are called Hilbert Schmidt operators. (hint: approximate $T$ by the natural sequence of finite rank operators that one thinks of)

Now let $(X,\mu)$ be a measure space and $K\in L^2(X\times X,\mu\otimes\mu)$. Define $A:L^2(X)\to L^2(X)$ by $Af(x)=\int_XK(x,y)f(y)d\mu(y)$. Then, $A$ is Hilbert-Schmidt. To see this, let $\{e_n\}$ be an ONB of $L^2(X)$.

Exercise 2: Show that, if $\{e_n\}$ is an ONB of $L^2(X,\mu)$, then the functions $\{e_{n,m}\}_{n,m}$ defined by $e_{n,m}(x,y):=e_n(x)e_m(y)$ form an ONB of $L^2(X\times X)$.

Now using Fubini's theorem we have for any $n,m$: $$\langle K,e_{n,m}\rangle=\int_{X\times X}K(x,y)e_n(x)e_m(y)d\mu\otimes\mu(x,y)=\int_X\int_XK(x,y)e_n(x)e_m(y)d\mu(x)d\mu(y)=\int_XAe_m(x)e_n(x)d\mu(x)=\langle Ae_m,e_n\rangle$$ thus $$\sum_m\|Ae_m\|^2=\sum_m\sum_n|\langle Ae_m,e_n\rangle|^2=\sum_m\sum_n|\langle K,e_{n,m}\rangle|^2=\sum_{m,n}|\langle K,e_{n,m}\rangle|^2=\|K\|^2<\infty.$$

Comments:

(1) We made use of Parseval's identity for Hilbert spaces in the above equations, namely if $(e_n)$ is an ONB for the Hilbert space $H$, then $\|h\|^2=\sum_n|\langle h,e_n\rangle|^2$ for all $h\in H$.

(2)For ease, one can assume separability to deal with countable bases. The non-separable case is completely analogous. One just has to be a little more careful when justifying summations in the last line of equalities, which nevertheless are still true if $n,m$ are indices running over an arbitrary index set.