Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $K(s,t)$ be a real-valued function of two real variables, and let $T: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be defined by $(Tf)(s) = \int_\mathbb{R} K(s,t) f(t) dt$.

If $||K||_{L^2({\mathbb{R}^2})} < \infty$, can we say that $T$ is a compact operator?

I think this is true if we are looking at a bounded domain for $K$ and $f$ (by an application of the Arzela-Ascoli theorem), but I am not sure if it is true in general.

share|improve this question
    
Got it. Thanks for the suggestion. –  user1736 Jan 3 '11 at 5:29

1 Answer 1

up vote 7 down vote accepted

Hint:

1) Approximate $K(s,t)$ by functions of the form $\sum_{k=1}^n g_k(s)h_k(t)$.

2) A norm limit of finite-rank operators is compact.

For further reference, your operator is called a Hilbert-Schmidt integral operator. A more general version of this problem (where $\mathbb{R}$ with Lebesgue measure is replaced by an arbitrary $\sigma$-finite measure space) is Problem 173 in Halmos's Hilbert space problem book. The book contains a hint and (in case you give up or solve it but want more information) a solution.

share|improve this answer
    
I'm not sure I have a complete understanding of the solution yet, but it looks like it's all there in front of me to sort out. Thanks for the help and also for the reference! –  user1736 Jan 3 '11 at 5:32
    
@user1736: You're welcome. Are there particular parts of the hints you're stuck on? I can elaborate (for example by making "approximate" precise or saying a few words on how $g_k$ and $h_k$ might be chosen). –  Jonas Meyer Jan 3 '11 at 5:39
    
Well, I'm guessing that by "approximate $K(s,t)$" you mean to use simple functions representable by the $g_k$ and $h_k$ right? This should be possible because $K$ is an $L^2$ function. Then, if we denote $K_n$ the approximation made by a sum of $n$ terms, and $T_n$ the corresponding operator, these $T_n$ should be have finite rank (this might be dumb of me, but can you explain why this part is true?). –  user1736 Jan 3 '11 at 5:49
    
@user1736: Yeah basically. There are many ways to get the $g_k$ and $h_k$, and using simple functions is one of the most straighforward that also generalizes to other measure spaces. (Here you could even use continuous functions if you want, but it won't really make things easier.) You could also use an orthonormal basis of $L^2(\mathbb{R})$ to get a doubly index orthonormal basis of the form $e_i(s)e_j(t)$ for $L^2(\mathbb{R}^2)$, whose finite linear combinations are dense. In any case, it is important that $g_k$ and $h_k$ are in $L^2$. (to be continued) –  Jonas Meyer Jan 3 '11 at 5:55
    
@user1736: The precise sense of approximation I recommend for starters is in the norm of $L^2(\mathbb{R}^2)$. You can show that $\|T-T_n\|\leq \|K-K_n\|_2$, where the norm on the left is the operator norm (which you ultimately want to go to $0$) and the norm on the right is the $L^2$ norm. To see that the operators in question have finite rank, first note that $\int g(s)h(t)f(t)dt=g(s)\langle f,\overline{h}\rangle$, so that the range of the integral operator corresponding to $g(s)h(t)$ is the span of $g$ (unless $h=0$). Taking finite linear combinations of such gives finite dim'l ranges. –  Jonas Meyer Jan 3 '11 at 6:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.