2
$\begingroup$

I am doing some functional analysis and am struggling with the following exercise: Let $T$ be an integral operator defined as followed: $$T(f)(s)=\int_0^1(s-t)^2f(t)dt$$ I am asked to it’s spectrum. I did/think the following:

I think this operator is compact thus it suffices to find all the eigenvalues. I.e. all the $\lambda\in\mathbb R$ satisfying: $\lambda f=T(f)$. After I tried some functions for $f$, I noticed that $T$ sends any integrable function to a polynomial of at most degree 2. So the eigenvalues are polynomials of at most degree 2.

Note: this is just my intuition I have not yet been able to make it rigorous. My questions are:

  • Am I on the right track?
  • How can I make this rigorous for a random integrable function $f$ instead of examples?
  • How do I find the eigenvalues given the eigenvectors are polynomials of at most degree 2?

Thank u :)

$\endgroup$
4
  • 1
    $\begingroup$ If $f(t)=1$, $T(f)$ is a cubic polynomial in $s$... $\endgroup$ Dec 24, 2021 at 22:05
  • $\begingroup$ What exactly is the domain of the operator? $\endgroup$ Dec 24, 2021 at 22:27
  • $\begingroup$ The exercise did not state the domain I think $C[0, 1]$, but I also believe $L^2 [0, 1]$ also works. $\endgroup$
    – user974406
    Dec 24, 2021 at 22:45
  • $\begingroup$ Does this answer your question? Spectrum of a self-adjoint integral operator $\endgroup$
    – Jean Marie
    Dec 24, 2021 at 23:08

1 Answer 1

2
$\begingroup$

In fact, $T$ sends integrable functions to polynomials of at most degree 2. To see that this is the case, note that the integral can be expanded to $$ Tf = \left(\int_0^1 f(t)\,dt\right)s^2 - 2\left(\int_0^1 tf(t)\,dt\right)s + \left(\int_0^1 t^2f(t)\,dt\right). $$ Other than that, your argument is correct. To prove that this kind of operator is generally compact, you could use one of the arguments presented here. In this case, however, it suffices to note that the image of $T$ is finite-dimensional.

From there, we can determine the eigenvectors of $T$ as follows. Consider the restriction of $T$ to the space $\Bbb P_2$ of degree-2 polynomials. Relative to the basis $\mathcal B = (1,t,t^2)$, we find that the matrix of the restriction $T|_{\Bbb P_2}$ is given by $$ [T]_{\mathcal B} = \pmatrix{\frac 13 & \frac 14 & \frac 15\\ -1& -\frac 23 & -\frac 12 \\ 1& \frac 12 & \frac 13}. $$

The eigenvalues of this operator are the non-zero elements of the spectrum of $T$.

You should find that these are equal to $-1/6$ and $(5 \pm 3\sqrt{5})/60$.

$\endgroup$
2
  • $\begingroup$ If we restrict our map does that not change the spectrum? Thanks for answering! :) $\endgroup$
    – user974406
    Dec 24, 2021 at 23:02
  • 1
    $\begingroup$ @Main It does not. In general, restricting a map to an invariant subspace yields a subset of the spectrum. You might find it to be a helpful exercise to convince yourself that each eigenvector of this matrix corresponds to a degree-2 polynomial that works as an eigenfunction of $T$. $\endgroup$ Dec 24, 2021 at 23:05

You must log in to answer this question.