Suppose $H$ is a self-adjoint operator on a Hilbert space having a simple isolated least eigenvalue $0$ with gap $1$ ( $H\Omega = 0$, $\Vert \Omega\Vert = 1$ ), $P$ is a non-negative symmetric operator, $\Omega \in \mathscr{D}(P)$ and for $\lambda>0$, $H + \lambda P$ is essentially self-adjoint. ($P$ is not assumed to be $H$-bounded.)

Then for small $\lambda$, $\overline{H+ \lambda P }$ (the self-adjoint extension of $H + \lambda P$) has a simple isolated least eigenvalue .

In the finite dimensional case it is easy to show that there cannot be orthonormal vectors $\alpha, \beta$ such that $(\alpha, H \alpha)$ and $(\beta, H \beta)$ are both less than $\frac{1}{2}$. The same holds for $H + \lambda P$. But for small $\lambda$, the spectrum of $H + \lambda P$ extends close to $0$, so there must be a simple isolated eigenvalue below $\frac{1}{2}$.

In the case of unbounded operators we can prove this similarly using the self-adjoint properties and the condition $\Omega \in \mathscr{D}(P)$.

**Proof:**
Let $\alpha' \in \mathscr{D}(\overline{H+ \lambda P })$
be a vector based below $A > 0$ on the spectrum of $\overline{(H+ \lambda P) }$.

Since $\overline{H+ \lambda P }$is the closure of $H+ \lambda P$, there are vectors $\alpha \in \mathscr{D}(H+ \lambda P)$ such that $\alpha$ and $(H+\lambda P)\alpha$ are arbitrarily close to $\alpha'$ and $ \overline{(H+ \lambda P) }\alpha'$ respectively. W.l.o.g. (and using $\Omega \in \mathscr{D}(P)$) we can write $\alpha = \sqrt{1-a^2}\Omega + a \alpha^\perp $ where $ (\alpha^\perp, \Omega) = 0$, $0\le a\le 1$and $\Vert \alpha^\perp \Vert = 1$ . Define $\beta', \beta, \beta^\perp, b$ and $B$ similarly, with $\alpha'$ and $\beta'$ based on disjoint intervals (so they are orthonormal) and $B > A > 0$. Then we have:

\begin{array}{drcl} A\ge(\alpha ',\overline{(H+ \lambda P) }\alpha') & \approx & (\alpha,(H+ \lambda P)\alpha)\ge(\alpha,H\alpha)\ge a^2 \\ B\ge(\beta ',\overline{(H+ \lambda P) }\beta') & \approx & (\beta,(H+ \lambda P)\beta)\ge(\beta,H\beta)\ge b^2 \\ 0=(\alpha', \beta') & \approx & (\alpha, \beta) = \sqrt{1-a^2}\sqrt{1-b^2} +ab(\alpha^\perp, \beta^\perp) \end{array} Since $\vert (\alpha^\perp, \beta^\perp) \vert \le 1$, the third relation cannot hold if $a$ and $b$ are $< 1/\sqrt 2$. Thus if $A$ and $B$ are $<\frac{1}{2}$ there cannot be two orthonormal vectors based below $B$ . Thus, if the spectrum of $\overline{H+ \lambda P}$ extends below $A$, there must be only a simple eigenvalue below $A$ (with $(A,B) \subset$ the gap).

Note that the above does not depend on $\lambda$. Let $\psi\in\mathscr{D}(H+\lambda P)$ with $(\psi,\Omega)=0$. Then (again using $\Omega \in \mathscr{D}(P)$)

$$\lim_{\lambda\to0}\frac{(\Omega + \lambda \psi, (H + \lambda P)(\Omega + \lambda \psi))}{||\Omega + \lambda \psi||^2} =0 $$ so for small $\lambda$ the spectrum of $\overline{H+ \lambda P}$ extends arbitrarily close to $0$. $\blacksquare$

Absent a reference, can you suggest a less awkward way to prove this?