# Several questions on the mini-max theorem for self-adjoint operators

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: Questions:

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.

$$\text{2a.}$$ (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.