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Spectrum of operator from L^2 to L^2

  • Thread starter veraguth
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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
 

Answers and Replies

  • #2
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So an eigenvector f would satisfy

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

for all t. Expanding the right side gives us

[tex]\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[/tex]

So we see from this that f must be a quadratic polynomial. This eases the computations a whole lot.
 
  • #3
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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.
 
  • #4
22,097
3,278
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

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

for all t. Can you find conditions on f for this to be true?
 
  • #5
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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.
 
  • #6
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0 can be in the spectrum and still the kernel could be trivial.

To find the kernel, you'll have to solve

[tex]\int_{-1}^1 (t+s)^2f(s)ds=0[/tex]
 
  • #7
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0 can be in the spectrum and still the kernel could be trivial.

To find the kernel, you'll have to solve

[tex]\int_{-1}^1 (t+s)^2f(s)ds=0[/tex]
that's exactly what I know. Just don't know how to do it.
 
  • #8
22,097
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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.
 

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