text: # 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