If $A$ is self-adjoint, then is $\langle \psi |A \psi \rangle \in \sigma(A)$

Question

If $A$ is self-adjoint (not necessarily bounded), then is $\langle \psi |A \psi \rangle \in \sigma(A)$ where $\psi$ is a normalized vector in the domain of $A$? This seems to be true, but I can't think of a relatively easy way to prove it.

Answer 1

If $A=A^*$ has unit eigenvectors $\psi_1$,$\psi_2$ with corresponding eigenvalues $\lambda_1\ne\lambda_2$, then $\psi_1\perp\psi_2$ and the following is a unit vector $$ \psi(t) = t\psi_1+\sqrt{1-t^2}\psi_2, \;\;\; 0 \le t \le 1. $$ Furthermore, $$ \langle A\psi(dt),\psi(t)\rangle=\langle \lambda_1t\psi_1+\lambda_2\sqrt{1-t^2}\psi_2,t\psi_1+\sqrt{1-t^2}\psi_2\rangle \\ = \lambda_1t^2+\lambda_2(1-t^2), \;\; 0 \le t \le 1. $$ It follows that $\langle A\psi,\psi\rangle$ includes every value in $[\lambda_1,\lambda_2]$, even when there are no elements of the spectrum between $\lambda_1$ and $\lambda_2$.