Vtyjkyuk

I'm just starting with unbounded operators and I am trying to find sufficient (perhaps also necessary) conditions for
$$H(t) = sum_i = 1^n f_i(t) H_i, qquad mathrmdom,H(t) = mathcalD$$
to be self-adjoint, where $mathcalD$ a dense subset of the Hilbert space $mathcalH$, $H_i$ are (possibly unbounded) linear operators on $mathcalH$ with $mathrmdom, H_i supset mathcalD$, and $f_i:mathbbR to mathbbR$ real functions as regular as needed.
I found the following theorem on Rudin's Functional Analysis:

Theorem 13.11. If $T$ is a densely defined symmetric operator on a Hilbert space $mathcalH$, and if $mathrmRan, T = mathcalH$, then $T$ is self-adjoint and $T^-1$ is bounded.

It seems to me that basing on this theorem, if the restrictions to $mathcalD$, $H_i|_mathcalD$, are symmetric and semibounded from below then $H(t)$ is self-adjoint:

1. $H(t)$ is symmetric: for any $psi,varphi in mathcalD$ we have
   $$langle psi, H(t)varphi rangle = sum_i=0^n f_i(t) langle psi, H_ivarphi rangle = sum_i=0^n f_i(t) langle H_ipsi, varphi rangle = langle H(t) psi, varphi rangle, $$
   where we have used that every $H_i$ are symmetric in $mathcalD$.


2. Since all $H_i$ are bounded from below, for every $t$ there exists an $M > 0$ such that $H(t) + M$ is positive definite and thus 0 is in the resolvent set of $tildeH(t) = H(t) + M$. This implies $mathrmRan,tildeH(t) = mathcalH$, hence it is self-adjoint by the theorem. Then $H(t) = tildeH(t) - M$ is self-adjoint for it is the difference between a self-adjoint operator and a bounded one.

That is as far as I can get, but it looks odd that no condition need to be imposed to the functions $f_i$, so if this reasoning were right that means any linear combination of operators both symmetric and semibounded from below is in fact self-adjoint, which sounds too good to be right. I have probably missed something on point 2, but can't figure out what.

Any help finding errors or hints on how to find conditions for $H(t)$ to be self-adjoint would be nice.

EDIT: Ok, I found an error: $tildeH(t)$ being positive definite just means 0 is not in the discrete spectrum, but if i'm not wrong it could still be in the continuous or residual spectrum, which would mean $mathrmRan, tildeH(t) neq mathcalH$. Still, any help to find when $H(t)$ is self-adjoint would be appreciated.

Let $H_1=-fracd^2dx^2$ on the domain $mathcalD_1$ consisting of all twice absolutely continuous functions $f$ on $[0,1]$ with $f''in L^2[0,1]$ and $f(0)=f(1)=0$. Let $H_2 = -fracd^2dx^2$ on the domain $mathcalD_2$ consisting of all twice absolutely continuous functions $f$ on $[0,1]$ with $f''in L^2[0,1]$ and $f'(0)=f'(1)=0$. Both of these operators are densely-defined, non-negative selfadjoint linear operators. And $H_1+H_2$ is symmetric positive-definite and densely-defined on $mathcalD=mathcalD_1capmathcalD_2$. It is even true that $H_1+H_2$ is closed on this domain, but it is not selfadjoint.