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Showing that spectrum of operator is not compact

  • Thread starter Leitmotif
  • Start date
  • #1
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Homework Statement



Let [tex]X=C[0,\pi][/tex].
Define [tex]T:\mathcal{D}(T) \to X[/tex], Tx = x" where
[tex]\mathcal{D}(T) = \{ x \in X | x(0)=x(\pi)=0 \}[/tex].

Show that [tex]\sigma(T)[/tex] is not compact.

Homework Equations



None.

The Attempt at a Solution


Well, functions sin(Ax) and sin(-Ax), for A=0,1,2,... are in the domain, and eigenvalues are all integers, which are not bounded, aka. not compact. But I am not sure how to show this in general case. Any help in the right direction is appreciated.
 
Last edited:

Answers and Replies

  • #2
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Yikes, the preview looked fine, but not sure what happened with that LaTeX code there. The operator is second derivative, Tx=x", and the domain is all functions in X so that x(0)=x(Pi)=0.
 
  • #3
Dick
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What do you mean 'general case'? I think you've shown that the second derivative operator T is unbounded, so it can't be compact. What else is there to show?
 
  • #4
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Well, I guess I came up with an example of a function that's in the domain, and whose set of eigenvalues are not bounded, and eigenvalues are in the spectrum. I just wasn't really sure if that is enough to show that it's true for ALL functions in the domain.

Also, I noticed I made a mistake. If f(x)=sin(Ax), then [tex]f''(x)=-A^2sin(Ax)[/tex], so for any integer A, [tex]-A^2[/tex] is an eigenvalue, which is still an unbounded set.
 
  • #5
Dick
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That what is true for ALL functions in the domain? Compactness or boundedness are properties of the operator T, not of specific functions. Your example shows T is not a bounded operator, I think it's just fine.
 
  • #6
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Point taken. Thanks for the input :)
 

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