I am reading the proof of mini-max theorem for bounded self-adjoint operators following "Unbounded Self-adjoint Operators on Hilbert Space" by Konrad Schmüdgen and it seems a totally mess.

Given a bounded below self-adjoint operator $A$ on a Hilbert space $\mathcal{H}$, we define the following sequences:

$$\mu_n(A) := \sup_{\mathcal{D} \in \mathcal{F}_{n-1}}\inf_{x \in \mathcal{D}(A), \|x\| = 1, x\perp \mathcal D}\langle Ax,x\rangle,$$ where $\mathcal{F}_{n-1}$ is the set of all finite dimensional subspaces of $\mathcal{H}$ with dimension at most $n-1$.

If $\sigma_{ess}(A) = \emptyset,$ where $\sigma_{ess}(A)$ denotes the essential spectrum of $A$, then we define $\{\lambda_n(A)\}$ as the sequence of isolated eigenvalues of $A$ with finite multiplicity (counting the multiplicity).

If $\sigma_{ess}(A) \neq \emptyset,$ one defines $$\alpha := \inf \{\lambda : \lambda \in \sigma_{ess}(A)\}$$ and define the same sequence $\lambda_n$ for the isolated eigenvalues with finite multiplicity on the bottom of $\sigma_{ess}(A)$. The rest of the elements of the sequence are $\alpha$. If there are no eigenvalues on the bottom of $\sigma_{ess}(A)$ we make $\lambda_n(A) := \alpha, ~\forall n.$

The mini-max principle states that:

$$\mu_n(A) = \lambda_n(A) = \inf \{\lambda : \dim E_A((-\infty,\lambda))\mathcal H \geq n\},$$ where $E_A$ is the unique spectral measure that represents $A$.

We denote $E_A((-\infty,\lambda))\mathcal{H} := \mathcal E_{\lambda}$ and $d(\lambda) := \dim \mathcal{E}_{\lambda}.$

Now one starts the proof:

It is easy to see that the equality follows for $n=1$ assuming an intermediate lemma that proves that:

$$\mu_n = \inf\{\lambda : d(\lambda) \geq n\}.$$

The idea is proceed by induction, we assume the statement holds for $1,2,3,\ldots,n-1$ and try to prove it holds for $n$. Here I paste the proof:

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1) Why does $\mu_n$ is isolated? This does not need to be true at all. This Proposition the author referes just says that the support of the spectral measure $E_A$ coincides with the spectrum of $A$ and that the points on the spectrum are points where the identity resolution is not continuous.

2) It seems to me that by the induction hypothesis $\mu_n \geq \lambda_n$, otherwise, $\mu_n = \mu_{j}$ for some $j \leq n-1.$ Is this right? (Note that the book does not even mention where he is using the hypothesis induction).

3) Here is the most concerning point: it says that if one assumes that there is $\lambda \in (\lambda_n,\mu_n)$, then $d(\lambda) \geq n.$ Why? It does not seem to be true according to what we know of $\mu_n.$

Thank you.


2 Answers 2


The problem is that the book contains a wrong reference (possible typo). The Proposition that solves the problem is Proposition 8.11, not 5.10.


Here are the answers to your remaining questions:

$\text{2a.}$ (How to conclude $\mu_n=\lambda_n$ in Case I?) The contradiction implies that $\lambda_n\geq\mu_n$, while $\leq$ follows from the fact that $\lambda_n$ is an eigenvalue in $(-\infty,\mu_n]$.

$\text{2b.}$ (On the use of induction.) I believe the induction hypothesis is not used. The proof simply checks the assertion for each $n$.

$\text{3.}$ (Why is $d(\lambda)\geq n$ in Case I?) This is because $\lambda > \lambda_n$, so the interval $(-\infty,\lambda)$ contains the eigenvalues $\lambda_1,\cdots,\lambda_n$. By the way, $\text{dim} E_A (-\infty,\mu_n]\mathcal{H} \geq n$ implies existence of $n$ eigenvalues because we are in Case I.


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