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

Question

Show that the operator $(Af)(t)=\int_{0}^{1}\min\{s,t\}f(s)ds$ is compact in $L_2[0,1]$. (not $L_2[0,1]^2$).

Our definition of compact operator is: operator $K$ is compact if for bounded sequence $(x_n)_{n=1}^{\infty}\in X$, $\{Kx_n\}$ has a Cauchy subsequence.

Also, it is concluded that $K$ is compact if the image of the unit ball under $K$ is pre-compact[also called relatively compact], meaning every sequence in $K(B_1(0))$ has a cauchy sub-sequence.

With the general $\int_{0}^{1}K(t,s)f(s)ds$ operator with continuous $K(t,s)\in C_1[0,1]^2$ it is much easier using a Theorem by Arzela about uniformly bounded and equicontinuous functions.

But here I am really not sure what to do. Does it matter whether $\min\{s,t\}$ is continuous?

Should the general approach finding a subsequence that is Cauchy? Suppose we have a uniformly bounded sequence $(f_n)$, i.e. $\forall n \sqrt{\int_{0}^{1}|f_n|^2dt}<M$ for some $M$, if we assume by contradiction that ${Af_n}$ is not precompact, meaning that out of a countable collection of differences $||Af_m-Af_n||$, there is no subsequence of differences which converges to\is identically $0$, hence there exist $\epsilon$ and $\{Af_{n_k}\}$ for which $|Af_{n_m}-Af_{n_N}|>\epsilon$ for all $n_N,n_m$.

So if $0<\epsilon<\int_{0}^{1}{|\int_{0}^{1}\min\{s,t\}f_{n_m}(s)ds-\int_{0}^{1}\min\{s,t\}f_{n_N}(s)ds|}^2 dt\le \int_{0}^{1}|\int_{0}^{1}\min\{s,t\}(f_{n_m}(s)-f_{n_N}(s))ds|^2 dt\le \int_{0}^{1}\int_{0}^{1}|\min\{s,t\}(f_{n_m}(s)-f_{n_N}(s))|^2ds dt\le \int_{0}^{1}t^2dt\int_{0}^{1}|(f_{n_m}(s)-f_{n_N}(s))|^2ds$. Here I got stuck: how can I show the last expression is smaller than $\epsilon$? Is my approach correct?

What confuses me about the finite rank property is:

a. How is it applied in $L_2[0,1]$? A function $K(t,s)$ continuous in both variables is approximable by $K_n(t,s)$ which is the partial sum $\sum^{n}_{1}\sum^{n}_{1}a_{ij}\phi_i\phi_j$ with $\{{\phi_i}{\phi_i}\}$ being orthonormal basis in $L_2([0,1]^2)$. But here we deal with $L_2[0,1]$.

b. Why is it the case that finite rank operator are compact? In my lecturer's notes it's said that a finite rank operator is compact because it maps the unit ball to a bounded subspace of finite dimension. But for $||f||\le 1$, $\int_{0}^{1}|\int_{0}^{1}\min\{s,t\}f(s)ds|dt\le \int_{0}^{1}|\int_{0}^{1}f(s)ds|dt$. What kind of subspace does it define and why is it of finite dimension?

Answer 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.