Does Schwarz Inequality guarantee that the expectation value for an operator exists and is finite?

Question

For two given square-integrable wave functions $\phi(x)$ and $\psi(x)$, Schwarz Inequality states that $$|\int_a^bdx\phi^*\psi|\le\sqrt{\int_a^bdx\phi^*\phi\int_a^bdx\psi^*\psi}.$$ This guarantees that the inner product between the functions exists and is finite. What about the expectation value? Does the relation guarantee the same thing when I am working instead with the integral form of expectation value for some operator $A$?

I think it does, since we are just taking the inner product between a wave function and that wave function worked on by $A$. But is my thought correct?

Answer 1

Not for an arbitrary observable. For an explicit counter example, see this question.

It is, however, easy to see that this is true when the observable is a bounded operator, i.e. one in which for any $\vert\psi\rangle$ with $\Vert\psi\Vert<\infty$, we have $\Vert A\vert \psi\rangle\Vert <\infty$. This is simply because \begin{align} \vert \langle \psi\vert A \vert \psi\rangle\vert \leq \big\Vert\psi\big\Vert~ \big\Vert A\vert \psi\rangle \big\Vert, \end{align} and $\Vert \psi\Vert < \infty$, $\Vert A\vert \psi\rangle\Vert <\infty$, with the latter being guaranteed since $A$ is bounded.