Showing that spectrum of operator is not compact

  • Thread starter Leitmotif
  • Start date
1. The problem statement, all variables and given/known data

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.

2. Relevant equations


3. 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:
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.


Science Advisor
Homework Helper
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?
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.


Science Advisor
Homework Helper
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.
Point taken. Thanks for the input :)

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