# Find point spectrum and spectrum of integral operator

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.

• Note that $Af(x) = (\sin\star f)(x)$, where $\star$ denotes convolution. – Math1000 Dec 24 '18 at 22:04
• – Math1000 Dec 24 '18 at 22:05
• 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? – AlephNull Dec 24 '18 at 22:09
• $A$ 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. – Kavi Rama Murthy Dec 24 '18 at 23:24
• @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. – Kavi Rama Murthy Dec 25 '18 at 0:17

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 $$0$$.
• 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. – Math1000 Dec 24 '18 at 23:58
• 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. – AlephNull Dec 25 '18 at 19:15