# Property of sequence of eigenvalues of an operator.

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

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.
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 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
The text passage discusses the properties of eigenvalues of a positive (self-adjoint) operator and whether it is possible for the eigenvalues to neither tend to infinity nor be bounded. It also addresses the implications of such a scenario on the boundedness of the operator and the possibility of reordering eigenvectors.
no_question

the passage does not contain a direct mathematical question that requires a specific answer. instead, it is a discussion or explanation of a mathematical concept. the comments at the end are clarifications or further thoughts on the topic, but they do not form a structured question with an answer provided in the text.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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system
# Property of sequence of eigenvalues of an operator.

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.

-
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

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.
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 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
user
The text passage discusses the properties of eigenvalues of a positive (self-adjoint) operator and whether it is possible for the eigenvalues to neither tend to infinity nor be bounded. It also addresses the implications of such a scenario on the boundedness of the operator and the possibility of reordering eigenvectors.

TYPE: NO_QUESTION

The passage does not contain a direct mathematical question that requires a specific answer. Instead, it is a discussion or explanation of a mathematical concept. The comments at the end are clarifications or further thoughts on the topic, but they do not form a structured question with an answer provided in the text.