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Problem: Describe the Fredholm operators of the first kind which are self-adjoint.

Attempt: As is standard when showing something is self-adjoint, I want to show that $\forall f,g\in L^2([a,b])$ we have $\langle Kf,g\rangle=\langle f,Kg\rangle$, where $K$ denotes the Fredholm operator of the first kind. We recall it here:

$$(Kf)(s)=\int_a^bk(s,t)f(t)dt$$

where $K:L^2([a,b])\to L^2([a,b])$ and $k(s,t)$ is defined for $s,t\in[a,b]$. Further, for each $f\in L^2([a,b])$ the map $s\to\int_a^bk(s,t)f(t)dt$ is continuous on $[a,b]$.

So $\forall f,g\in L^2([a,b])$ lets consider:

$$\langle Kf,g\rangle=\int_a^b\left(\int_a^bk(s,t)f(t) dt\right)g(s)ds.$$

After consulting a reference, I see that for $\langle f,Kg\rangle$ we should get:

$$\langle f,Kg\rangle=\int_a^bf(t)\left(\int_a^bk(t,s)g(s)ds \right)dt$$

But I'm not sure that I quite get how to bridge the gap between them. With the question in mind, it seems to me that the "description" of the Fredholm operators of the first kind on $[a,b]$ should be when the kernel, $k$, is symmetric (I have also gleaned this from my reference.)

Here are the central apprehensions I have:

  1. What, exactly, allows you to switch the order of integration here?

  2. What is the significance of $k(s,t)$ switching to $k(t,s)$ (that is, of the kernel being symmetric) and what justifies this happening here?

If there are any other key observations that you feel are important in understanding how we move from $\int_a^b\left(\int_a^bk(s,t)f(t) \right)g(s)dt$ to $\int_a^bf(t)\left(\int_a^bk(t,s)g(s) \right)dt$ I'm sure they can only help to bolster my comprehension.

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  • $\begingroup$ You are talking about self-adjoint operators on something which isn't a Hilbert space. Shouldn't $K$ be defined on $L^2(a,b)$? $\endgroup$ – Lorenzo Quarisa Apr 30 '18 at 19:03
  • $\begingroup$ Yes, I believe you are right. The textbook from which this problem is drawn introduced these operators earlier for $C([a,b])$ alone, but you are entirely right. Thanks for pointing this out, I will edit the question. $\endgroup$ – Jeremy Jeffrey James Apr 30 '18 at 19:08
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Let $k\in L^2((a,b)\times(a,b))$. By Cauchy-Schwartz we have $$s\mapsto \int_a^b k(s,t)f(t)dt \in L^2(a,b) $$ Hence $K$ is a well defined bounded linear operator on the Hilbert space $L^2(a,b)$, with $\|K\|=\|k\|$. By Cauchy-Schwartz again $$s\mapsto g(s)\int_a^bk(s,t)f(t)dt\in L^1(a,b) $$ The same goes with the reverse order of integration. Hence by Fubini's theorem we may exchange the order of integration.

The condition of self-adjointness is then equivalent to the simmetricity of $k$, i.e. $k(t,s)=k(s,t)$ for a.e. $s,t\in [a,b]$.

I assumed that your functions are real valued. If they are complex valued, then the steps are identical except the inner product on $L^2(a,b)$ puts a complex conjugate on one of the two elements and the condition becomes $k(t,s)=\overline{k(s,t)}$ for a.e. $s,t$.

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