I'm just starting with unbounded operators and I am trying to find sufficient (perhaps also necessary) conditions for $$H(t) = \sum_{i = 1}^n f_i(t) H_i, \qquad \mathrm{dom}\,H(t) = \mathcal{D}$$ to be self-adjoint, where $\mathcal{D}$ a dense subset of the Hilbert space $\mathcal{H}$, $H_i$ are (possibly unbounded) linear operators on $\mathcal{H}$ with $\mathrm{dom}\, H_i \supset \mathcal{D}$, and $f_i:\mathbb{R} \to \mathbb{R}$ 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 $\mathcal{H}$, and if $\mathrm{Ran}\, T = \mathcal{H}$, then $T$ is self-adjoint and $T^{-1}$ is bounded.

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

  1. $H(t)$ is symmetric: for any $\psi,\varphi \in \mathcal{D}$ we have $$\langle \psi, H(t)\varphi \rangle = \sum_{i=0}^n f_i(t) \langle \psi, H_i\varphi \rangle = \sum_{i=0}^n f_i(t) \langle H_i\psi, \varphi \rangle = \langle H(t) \psi, \varphi \rangle, $$ where we have used that every $H_i$ are symmetric in $\mathcal{D}$.
  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 $\tilde{H}(t) = H(t) + M$. This implies $\mathrm{Ran}\,\tilde{H}(t) = \mathcal{H}$, hence it is self-adjoint by the theorem. Then $H(t) = \tilde{H}(t) - M$ 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: $\tilde{H}(t)$ 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 $\mathrm{Ran}\, \tilde{H}(t) \neq \mathcal{H}$. Still, any help to find when $H(t)$ is self-adjoint would be appreciated.

  • $\begingroup$ If $H$ is densely-defined and selfadjoint, but not defined on all of $\mathcal{H}$, then $H-H$ is not selfadjoint. $\endgroup$ – DisintegratingByParts Aug 9 '18 at 19:24

Let $H_1=-\frac{d^2}{dx^2}$ on the domain $\mathcal{D}_1$ 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 $H_2 = -\frac{d^2}{dx^2}$ on the domain $\mathcal{D}_2$ 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 $\mathcal{D}=\mathcal{D}_1\cap\mathcal{D}_2$. It is even true that $H_1+H_2$ is closed on this domain, but it is not selfadjoint.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.