Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.