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Let $X,Y=L^2(0,1)$, $k\in C^0([0,1]^2)$. Define $$ K:X\to Y,\,\,\,\,\,Kf(x):=\int_0^1k(x,y)f(y)dy\,\,\,\,\forall\, f\in L^2(0,1). $$ I have to show that $K$ is compact.

My idea is to prove that $K$ is the limit of finite-rank operators. But I don't know exactly which kind of operators should I have to consider.

avati91
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3 Answers3

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To show that $K$ is compact using elementary arguments, let $\{ f_{n} \}_{n=1}^{\infty}$ be a bounded sequence in $L^{2}[0,1]$ with $\|f_{n}\| \le M$. For every $\epsilon > 0$, there exists $\delta > 0$ such that $|k(x,y)-k(x',y')| < \epsilon$ whenever $|x-x'|+|y-y'| < \delta$. Therefore, $\{ Kf_{n}\}$ is a sequence of continuous functions for which $$ |Kf_{n}(x)-Kf_{n}(x')| \le \int_{0}^{1}|k(x,y)-k(x',y)||f_{n}(y)|\,dy \\ \le \epsilon \int_{0}^{1}|f_{n}(y)|\,dy \le \epsilon\|1\|\|f_{n}\| \le M\epsilon,\;\;\; |x-x'| < \delta. $$ That shows that $\{ Kf_{n} \}$ is an equicontinuous family of continuous functions on $[0,1]$. So, there exists a subsequence $\{ Kf_{n_{k}}\}$ that converges uniformly to a continuous function $g$. Since uniform convergence implies convergence in $L^{2}[0,1]$, it follows that $\{ Kf_{n_{k}}\}$ converges in $L^{2}[0,1]$. Therefore $K$ is compact because the image of a bounded sequence always contains a convergent subsequence.

Disintegrating By Parts
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  • Why "Therefore,$ {Kf_{n}}$ is a sequence of continuous functions for which ......"? – Emptymind Dec 31 '17 at 07:19
  • Why you used that the operator $K$ is continuous ? – Emptymind Dec 31 '17 at 07:23
  • what do you mean by "an equicontinuous family "? – Emptymind Dec 31 '17 at 07:25
  • Look at https://en.wikipedia.org/wiki/Arzelà–Ascoli_theorem in the first section "Statement and first consequences." Equicontinuity is described there. – Disintegrating By Parts Dec 31 '17 at 18:36
  • Is it the operator $K$ inside the integral $\int_0^1$ or the kernel $k(x,y)$? – blux Jun 20 '18 at 13:37
  • @blux : I corrected the typos, though the context made it clear how it was being used. – Disintegrating By Parts Jun 20 '18 at 13:43
  • @Intuition To your first question, you just use the Hölder’s inequality and the fact that $[0,1]$ is of finite measure and $f\in L^2[0,1]$. – Sam Wong Jun 27 '18 at 14:54
  • @DisintegratingByParts Hi, if we reduce the condition posted on $k$ to $k\in L^1[0,1]$, can we modify the proof slightly so that it still works? – Sam Wong Jun 27 '18 at 14:59
  • @KitWong : The operator norm of $K$ is bounded by the $L^1$ norm of $k$. So you can approximate $K$ in the operator norm by a kernel operator with continuous symbol $k$, and that operator is compact. The compact operators is a norm closed set. – Disintegrating By Parts Jun 27 '18 at 15:53
  • @DisintegratingByParts Thanks, though I have not learned the knowledge of kernel operator & continuous symbol. I will review this answer if I learn the knowledge you used. – Sam Wong Jun 27 '18 at 16:13
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Using Stone-Weierstass theorem, the function $k$ can be approximated uniformly in $[0,1]^2$ by linear combinations of functions of the form $(x,y)\mapsto u(x)v(y)$, where $u,v\colon [0,1]\to \mathbb R$ are continuous functions.

For such functions, the kernel operator has a finite rank. It remains to show that uniform convergence of the kernel implies convergence for the operator norm.

Davide Giraudo
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  • Why does Stone-Weierstrass thm implies that fact? Then, I know that I have to write the approximating operators in the form $K_nf(x):=\sum_{i=1}^n\alpha_i\langle f, u_i\rangle v_i$, where $(u_i)_i$ and $(v_i)_i$ are orthonormal basis of $L^2$. Which are the orthonormal basis in your example? – avati91 Nov 01 '14 at 14:43
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    We use the theorem with the algebra generated by the functions of the form $(x,y)\mapsto u(x)v(y)$ (it satisfies the assumptions of the theorem). – Davide Giraudo Nov 01 '14 at 15:33
  • What about how to prove that the approximating operators are of the form of my previous comment? – avati91 Nov 02 '14 at 18:41
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Actually, if $k\in L^2(0,1)^2$, then you may consider $K$ as a limit of a sequence of finite rank operators. Since $L^2(0,1)$ is a separable Hilbert space, there exists a Hilbert basis denoted by $\{e_i\}_{i=1}^{\infty}$. Define $T_n (n\in\mathbb{N}^{+})$ as $$ T_nf:=\sum\limits_{i=1}^{n}\langle f, e_i \rangle e_i,\quad \forall f\in L^{2}(0,1). $$ Then, $K_n:=T_nKT_n$ is a finite rank operator. And $K_n$ is compact for each $n$. You need to show that $\lim\limits_{n\rightarrow \infty}||K_n-K||=0$. Then, you conclude that $K$ is compact, since the space of compact operators mapping $L^2(0,1)$ into $L^2(0,1)$ is closed.

Lin Xuelei
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