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For a positive (self adjoint) operator $A$ with eigenvalues $\lambda_k$, is it possible to have the case when neither $\lambda_k\to \infty$ or $sup_k \lambda_k<\infty$ for example if a subsequence tends to $\infty$ and another subsequence stays bounded? In a paper I am reading (Continuity of $l^2$-valued Ornstein-Uhlenbeck Processes), they only check 2 cases.

If so can one order the eigenvectors so that it is not the case?

Thanks.

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So you mean $\limsup \lambda_k=+\infty$ and $\liminf \lambda_k<\infty$. –  1015 Mar 20 '13 at 15:45
    
Yes. That's right. –  David Mar 20 '13 at 15:47

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The operator will not be bounded, as soon as $\limsup \lambda_k=+\infty$ (equivalently $\sup \lambda_k=+\infty$). Take a subsequence $\lambda_{n_k}$ tending to $+\infty$. For each $k$ take a norm $1$ eigenvector $x_k$ associated with $\lambda_{n_k}$. Then $$ \|A\|\geq \|Ax_k\|=\lambda_{n_k}\longrightarrow +\infty. $$ The fact that $\liminf \lambda_k<\infty$ will not change that. Neither the fact that $A$ is positive, self-adjoint, or whatever.

But you can construct unbounded examples, of course. It suffices to take the diagonal operator $\mbox{diag}(\lambda_k)$ in any orthonormal basis. Then the domain is the susbspace of all vectors $x=(x_k)$ such that $\sum_k \lambda_k^2x_k^2$ converges. This will depend on the sequence, but of course it will always contain the vectors with finitely many nonzero coordinates.

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I am not sure I understand what you are saying. The operator $A$ is allowed to be unbounded. –  David Mar 20 '13 at 15:56
    
@David You did not say that. Then the answer is even easier. See my edit. Just take the diagonal operator. I hope I understood your question... –  1015 Mar 20 '13 at 16:00
    
Oh yes. Then is it possible to choose the eigenvectors so that it is not the case? –  David Mar 20 '13 at 16:07
    
I forgot to ask this in the original question but I have made an edit. –  David Mar 20 '13 at 16:15
    
@David What do you mean, order the eigenvectors? That's like reordering the $\lambda_k$'s. But you still have $\limsup =+\infty$ and $\liminf<\infty$. So permute the elements of the diagonal accordingly. This is still an example. –  1015 Mar 20 '13 at 16:48

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