Spectrum of operator from L^2 to L^2

[veraguth]

Homework Statement

Find spectrum and eigenvalues of operator from L^2(-1,1) to L^2(-1,1)

T(f)(t) = ∫(t+s)^2f(s)ds

The integral is taken over [-1,1]

2. The attempt at a solution

I have already proven that this operator is self-adjoint and compact. However, I have now idea how to find spectrum. What is have tried is to apply definition (unable to solve integral equation) and the representation in Hilbert space in a form of series.

I am an absolute beginner in functional analysis.

Thank you in advance for any help

 

[micromass]

So an eigenvector f would satisfy

\lambda f(t)=\int_{-1}^1 (t+s)^2f(s)ds

for all t. Expanding the right side gives us

\lambda f(t)=t^2\int_{-1}^1 f(s)ds+2t\int_{-1}^1 sf(s)ds+\int_{-1}^1 s^2f(s)ds

So we see from this that f must be a quadratic polynomial. This eases the computations a whole lot.

 

[veraguth]

Thank you very much.

I managed to find the eigenvalues (4/3, 2/3 +- √5) and eigenvectors for those values.

However, what is still a kind of 'don't-know-what-to-do' problem is how to find eigenvectors for 0, which is also in the spectrum of this operator.

A hint on what to do now would be wonderful.

And again - thanks for quick help.

 

[micromass]

Thank you very much.

I managed to find the eigenvalues (4/3, 2/3 +- √5) and eigenvectors for those values.

However, what is still a kind of 'don't-know-what-to-do' problem is how to find eigenvectors for 0, which is also in the spectrum of this operator.

A hint on what to do now would be wonderful.

And again - thanks for quick help.

That 0 is in the spectrum doesn't mean that it's an eigenvalue. For a compact, self-adjoint operator it is true that every nonzero element of the spectrum is an eigenvalue, but 0 might not be an eigenvalue (or it might be)

That 0 is in the spectrum is easily seen by showing that T is not invertible (a compact operator on a Hilbert space is never invertible).
If 0 were an eigenvalue, then there should exist an f in L² such that

\int_{-1}^1(t+s)^2f(s)ds=0

for all t. Can you find conditions on f for this to be true?

 

[veraguth]

Thank you for the response.

Ok, I have proven that this operator is compact. Therefore, it's spectrum consist of set of eigenvalues and 0.

Maybe I messed up the names a bit. I would like to find the kernel of the operator described above. I thought that as the 0 belongs to the spectrum, it those are also named 'eigenvectors'

A help on this part of task would be also something great for me.

 

[micromass]

0 can be in the spectrum and still the kernel could be trivial.

To find the kernel, you'll have to solve

\int_{-1}^1 (t+s)^2f(s)ds=0

 

[veraguth]

0 can be in the spectrum and still the kernel could be trivial.

To find the kernel, you'll have to solve

\int_{-1}^1 (t+s)^2f(s)ds=0

that's exactly what I know. Just don't know how to do it.

 

[micromass]

that's exactly what I know. Just don't know how to do it.

Make an expansion such as in post 2 to get conditions on f.