# Showing that spectrum of operator is not compact

1. Dec 14, 2009

### Leitmotif

1. The problem statement, all variables and given/known data

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

Show that $$\sigma(T)$$ is not compact.

2. Relevant equations

None.

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: Dec 14, 2009
2. Dec 14, 2009

### Leitmotif

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. Dec 14, 2009

### Dick

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. Dec 14, 2009

### Leitmotif

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 $$f''(x)=-A^2sin(Ax)$$, so for any integer A, $$-A^2$$ is an eigenvalue, which is still an unbounded set.

5. Dec 14, 2009

### Dick

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. Dec 14, 2009

### Leitmotif

Point taken. Thanks for the input :)