This is a part of some proof of the courants minimax principle, where I can't get my head arround.
What is supposed to be shown is actually
$\inf \sigma_{\mathrm{ess}}(A)\geq \inf\limits_{U\subset D(A), \mathrm{dim}(U)=n}\left( \sup\limits_{x\in U, \|x\|=1}\langle Ax,x \rangle\right)$, when $A$ has atmost $n-1$ eigenvalues (multiplicity included) below $\inf \sigma_{\mathrm{ess}}(A)$ ($n\geq 1$).
The excact conditions I mention now:
Fix $n\geq 1$. Let $A:D(A)\subset H \rightarrow H$ be a selfadjoint linear operator, $H$ a (seperable) Hilbert space. Assume further that $A$ is bounded from below i.e. $\langle Ax,x\rangle\geq c\langle x,x\rangle$ for all $x\in H$, where $c\in\mathbb{R}$ is fixed. Let $\Omega\mapsto P_\Omega$ be the spectral measure of $A$. Let $s:=\inf \sigma_{\mathrm{ess}}(A)$ and assume that $A$ has atmost $n-1$ eigenvalues below $s$, where we count the eigenvalues according to their multiplicity i.e. if $\lambda$ would be an eigenvalue, with multiplicity $k$, we would count lambda $k$-times. Now let $U\subset P_{(-\infty,s+\varepsilon]}H$ be some $n$-dimensional subspace, where $\varepsilon>0$ is arbitrary.
Note that we can pick such a subspace because $\mathrm{dim}\left( P_{(-\infty,s+\varepsilon]}H\right)\geq n$.
Now the one step which implies the assertion at the very beginning, but where I can't get my head arround is:
Why does $\sup\limits_{x\in U, \|x\|=1}\langle Ax,x \rangle\leq \varepsilon+s$ hold?
Note: $\sigma_{\mathrm{ess}}(A)=\sigma(A)\setminus \sigma_{\mathrm{disc}}(A)$, where $\sigma_{\mathrm{disc}}(A)$ contains all the isolated eigenvalues of $A$ with finite multiplicity.
After thinking a while, I finally got it.
The spectral theorem
$$\langle A x, x\rangle= \int_{\sigma(A)}t E_x(dt)$$
where $E_x(\Omega):=\langle x, P_\Omega x\rangle$ implies the assertion.
Because with that, we get for $y:=P_{(-\infty,s+\varepsilon]}x$ normed, the estimate
$$\langle A y,y \rangle=\langle A|_{P_{(-\infty,s+\varepsilon]}H}y, y\rangle\leq \sup\sigma( A|_{P_{(-\infty,s+\varepsilon]}H})\leq s+\varepsilon $$
Since $ A|_{P_{(-\infty,s+\varepsilon]}H}$ is selfadjoint, $\sigma( A|_{P_{(-\infty,s+\varepsilon]H}})\subseteq \sigma(A)\cap (-\infty,s+\varepsilon]$ and $E_y(\sigma(A))=1$.
Note that the spectral measure of $A|_{P_{(-\infty,s+\varepsilon]}H}$ is given by $P_{\Omega\cap (-\infty, \varepsilon +s]}$ and $P_{\Omega\cap (-\infty, \varepsilon +s]}\leq P_{\Omega}$