I am wondering whether the following fact, for which I know well the proof when $H$ is a Schroedinger operator (see Lieb-Loss, Analysis, Chapter 11), is also true in the general setting used below, and if so, how to prove it:
Let $H$ be a self-adjoint and bounded below (possibly unbounded, though) operator on a Hilbert space $\mathcal{H}$, with dense domain $\mathcal{D}(H)$. As customary, $H[\cdot,\cdot]$ denotes the associated form on the form domain $\mathcal{D}[H]$ (which is therefore densely defined, closed, bounded below with the same bound as $H$). Assume that the minimisation problem
$$ \inf \{ H[f,f]\,:\, f\in\mathcal{D}[H], \|f\|=1 \} $$
is actually solved by a minimiser $f_0\in\mathcal{D}[H]$ (that is. the inf is a min), and denote by $E_0$ the corresponding value of the inf. Claim: $f_0\in\mathcal{D}(H)$ and $H f_0 = E_0 f_0$.
Any ideas? It is not just min-max, because nothing is said on whether $H$ admits eigenfunctions.
Let $E(\cdot)$ be the projection-valued measure associated with $H$. Then $$D[H]=\{f\mid\int|\lambda|d\langle E(\lambda)f,f\rangle<\infty\}\ \ \ \ \mbox{and}\ \ \ \ H[f,f]=\int\lambda d\langle E(\lambda)f,f\rangle.$$
It is easy to show that $M=\inf_{||f||=1,f\in D[H]} H[f,f]$ if and only if $E$ is supported on $[M,+\infty).$ In particular, the support of every measure $\mu_f=\langle E(\lambda)f,f\rangle,\ f\in D[H]$ is contained in $[M,+\infty).$
If $f_0\in D[H],||f_0||=1$ then $\mu_{f_0}(\{M\})=1$ and $$H[f_0,f_0]=\int_{\mathbb R}\lambda d\mu_{f_0}=\int_{[M,+\infty)}\lambda d\mu_{f_0}\geq M\cdot \mu_{f_0}([M,+\infty))=M.$$ The equality is possible iff $$\mu_{f_0}([M,+\infty))=\mu_{f_0}(\{M\})=1.$$
Hence $f_0$ is a minimizer iff $\mu_{f_0}(\{M\})=\langle E(\{M\})f_0,f_0\rangle=\langle f_0,f_0\rangle=1,$ i.e. $E(\{M\})f_0=f_0.$ By properties of the spectral measure we have $f_0\in D(H)$ and $Hf_0=Mf_0.$