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Let $A:L^2([0,1])\to L^2([0,1])$ given by $$ Af(t)=\int_0^1K(s,t)f(s)ds, $$ where $K$ is a mensurable square integrable operator, i.e $\int_0^1\int_0^1|K(s,t)|^2\,dsdt<\infty$.

$A$ is acompact operator: It is known that this is compact operator. In fact, we consider first the case where $K$ is a contínuous Kernel, and observe that $A$ maps $L^2([0,1])$ into $C^0([0,1])$, and the we use the Arzelá Ascoli Theorem in order to conclude tha $A$ is compact. In the general let $K$ a mensurable square integrable Kernel, consider a sequênce $(K_n)$ of contínuos kernels such that $K_n\to K$ in $L^2([0,1]^2)$ norm, this implies that $A_n\to A$ in the operator norm, where $$ A_nf(t)=\int_0^1K_n(s,t)f(s)ds, $$

and then we use que the space of the compact operators is closed.

My question ($A$ is a simetric operator?): My goal is to show that the operator $ A $ is symmetric , in order to show its worth the spectral theorem for $ A $. Please would you criticize my "proof": using the Fubini theorem we have for all $f,g\in L^2([0,1])$ that

\begin{eqnarray} \left<Af,g\right>&=&\int_{0}^{1}Af(t)g(t)dt=\int_{0}^{1}(\int_{0}^{1}K(s,t)f(s)ds)g(t)dt \\ & = & \int_0^1(\int_0^1K(s,t)g(t)dt)f(s)ds \\ & = & \left<f,Ag\right> \end{eqnarray} therefore $A=A^{*}$. My doubt in this demonstration consists in the fact that I assert that $A(s)=\int K(s,t)g(t)dt$, i.e I changed the variable of integration in the definition of A, is that correct??

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    $\begingroup$ It's not correct. $A$ is only symmetric if $K$ has a specific form. Note that $$Ag \colon s \mapsto \int_0^1 K(t,s)g(t)\,dt$$ by the definition of $A$. So which property of $K$ is needed for $A$ to be symmetric? $\endgroup$ – Daniel Fischer Nov 13 '14 at 19:27
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    $\begingroup$ Just by analogy with matrices, I would expect that you need $K(s,t)=K(t,s)$. $\endgroup$ – copper.hat Nov 13 '14 at 19:28
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No; note the definition of $A$ is $\int_0^1 K(s,t)f(s)\,ds$, and what you have written in the next to last line is $\int_0^1 K(s,t)g(t)\,dt$ (as noted by Daniel Fischer above), which is, in fact, just the definition of $A^*g$.

For a counterexample let $K(t,s) = t$ and $f \equiv 1$, $g = t$. Then

$$Af = \int_0^1 t \,ds = t$$

and so

$$\langle Af,g\rangle = \langle t, t\rangle = \int_0^1 t^2 \,dt = \frac13.$$

On the other hand,

$$Ag = \int_0^1 ts \,ds = \frac12 t,$$

so

$$\langle f,Ag\rangle = \langle 1, \frac12 t\rangle = \int_0^1 \frac12 t \,dt = \frac14.$$

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  • $\begingroup$ I believe that mentioning the comments of Daniel Fischer in you answer leave its richest. Thank you. $\endgroup$ – O Empalador de Cabras Nov 13 '14 at 19:54

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