4
$\begingroup$

Let's consider a self-adjoint operator $A$ (not necessarily bounded) on a Hilbert space which is bounded from below, with domain $D$ and whose resolvent is compact. Then, the spectrum consists solely of isolated eigenvalues which are given (in increasing order) by the min-max principle:

\begin{equation} \lambda_k = \min_{\substack{V \subset D\\ \dim V = k}} \max_{\substack{x \in V \\ x \neq 0}} \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle}, \ k \in \mathbb{N}. \end{equation}

The proof I know shows $\lambda_k \geq \min \max \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle}$ and $\lambda_k \leq \min \max \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle}$ by using a orthonormal basis of eigenvectors.

But how can we really write "min" and "max" instead of "inf" and "sup", i.e. why is the minimum and maximum really attained? Does anybody have a proof or a source for this assertion?

Edit: I asked the question for the minimum separately since I want the possibility to start a bounty there and accept the answer here at the same time. See here: Why is the Minimum in the Min-Max Principle for Self-Adjoint Operators attained? .

$\endgroup$
2
  • $\begingroup$ Do you mean that $A$ is bounded below, or that some resolvent operator is? $\endgroup$ Commented Jul 9, 2016 at 1:36
  • $\begingroup$ The operator $A$ is bounded from below. I edited the question. Thank you. $\endgroup$
    – NiU
    Commented Jul 9, 2016 at 9:05

2 Answers 2

2
$\begingroup$

For the first part: If $V$ has dimension $k$ then we have $$\max_{\substack{x \in V \\ x \neq 0}} \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle} = \max_{\substack{x \in V \\ x \neq 0}} \langle \, \frac{x}{\Vert x \Vert} , A \frac{x}{\Vert x \Vert} \rangle = \max_{\substack{x \in V \\ \Vert x \Vert = 1}} \langle \,x , Ax \rangle.$$ This is a maximum because the unit sphere is compact in $V$ since $V$ is of the finite dimension $k$. I have no idea how you can get an argument for the $\min$ though.

$\endgroup$
2
  • $\begingroup$ Perfect, thank you. This works, even if $A$ is not bounded on its whole domain since we restrict to a finite-dimensional setting where $A$ will be bounded (therefore continuous). $\endgroup$
    – NiU
    Commented Jul 9, 2016 at 17:58
  • $\begingroup$ Yes, it's a pretty standard trick. The part with the $\min$ doesn't seem to be that obvious. $\endgroup$
    – Yaddle
    Commented Jul 9, 2016 at 18:37
0
$\begingroup$

The minimum is attained because you can pick $V:=\mathrm{ker}(\lambda_k-A)$, which has dimension $k$. Then for all $x\in V$, with $\|x\|=1$ you have $\langle x,Ax\rangle=\lambda_k$

$\endgroup$
2

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .