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Let $A:L^2(0,\pi) \to L^2(0,\pi)$ be defined by $(Af)(x)=\displaystyle\int_{0}^\pi \sin(x-y)f(y)dy$. Find the point spectrum and spectrum of $A$.

I am not sure how to go about this. I thought to start with, I should find the point spectrum (i.e. the set of eigenvalues). So not knowing any techniques for integral operators, I would just set $\displaystyle\int_{0}^\pi \sin(x-y)f(y)dy=\lambda f(x)$ and try to solve for $f$. Unfortunately I have no idea how to, or even if it is possible, to solve for $f$ here. I am wondering if there is a trick specific to this integrand or a more general technique regarding integral operators that I should know about. I have a result that tells me $A$ is compact but it's not clear how this could be used. Any hints would be appreciated.

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  • $\begingroup$ Note that $Af(x) = (\sin\star f)(x)$, where $\star$ denotes convolution. $\endgroup$ – Math1000 Dec 24 '18 at 22:04
  • $\begingroup$ See here: math.stackexchange.com/questions/1117098/… $\endgroup$ – Math1000 Dec 24 '18 at 22:05
  • $\begingroup$ We have never actually seen or used Fourier transforms so far, so it is strange that we would be set such an exercise. I suppose there is no more elementary way of finding the spectrum, or by using the spectral theory for compact operators? $\endgroup$ – AlephNull Dec 24 '18 at 22:09
  • $\begingroup$ $A$ is a compact operator, so it is not invertible. Hence $0$ is in the spectrum. For a compact operator non-zero points in the spectrum are all eigen values. Hence it is enough to find eigen values which has been done by Jacky Chong. $\endgroup$ – Kavi Rama Murthy Dec 24 '18 at 23:24
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    $\begingroup$ @Math1000 If you expand $\sin (x-y)$ as $\sin\,x \cos ,y-\cos\, x \sin\, y$ you will see that the range of $A$ is contained in the span of sin and cos. Any finite rank operator is compact. $\endgroup$ – Kavi Rama Murthy Dec 25 '18 at 0:17
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Observe \begin{align} Tf(x)=\int^\pi_0 \sin(x-y) f(y)\ dy = \left(\int^\pi_0 \cos(y)f(y)\ dy\right) \sin(x)-\left(\int^\pi_0 \sin(y)f(y)\ dy\right)\cos(x) \end{align} which means $\mathcal{R}(T) = \operatorname{span}\{\cos(x), \sin(x)\}$ and $\dim\mathcal{R}(T)=2$. Next, consider $f(x) = A\cos(x)+ B\sin(x)$, then we see that \begin{align} Tf(x) = \frac{\pi A}{2}\sin(x)-\frac{\pi B}{2}\cos(x) = \lambda (A\cos(x)+ B\sin(x)) \end{align} iff $\lambda A = -\frac{\pi B}{2}$ and $\lambda B= \frac{\pi A}{2}$. Solving for $A$ and $B$ yields \begin{align} \lambda^2 +\frac{\pi^2}{4} =0 \ \ \implies \ \ \lambda = \pm \frac{\pi}{2}i \end{align} which means $f_1(x) = \cos(x)-i\sin(x)$ and $f_2(x) = \cos(x)+i\sin(x)$ are both eigenfunctions of $T$.

Remark: Note that if $L^2(0, \pi)$ is viewed as a real vector space, then $T$ has no real eigenvalue other than $0$.

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  • $\begingroup$ Great answer, might be worth mentioning that the first equality is due to $\sin(x-y)=\sin(x)\cos(y) - \sin(y)\cos(x)$ though. At first I thought you invoked integration by parts which was confusing. $\endgroup$ – Math1000 Dec 24 '18 at 23:58
  • $\begingroup$ Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum. $\endgroup$ – AlephNull Dec 25 '18 at 10:38
  • $\begingroup$ @AlephNull Note the operator is compact. $\endgroup$ – Jacky Chong Dec 25 '18 at 19:07
  • $\begingroup$ Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately. $\endgroup$ – AlephNull Dec 25 '18 at 19:15

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