2
$\begingroup$

On a Hilbert space $\cal H$, consider an essentially self-adjoint operator $A\colon Dom(A)\to {\cal H}$, and a vector $\psi\in\bigcap_{n=1}^\infty Dom(A^n)$. Without further assumptions, can we say that the associated spectral measure $\mu_\psi$ is the unique Borel measure satisfying $$\langle \psi,A^n\psi\rangle = \int\lambda^n\mu_{\psi}(d\lambda),$$ for all $n\ge 0$ ?

$\endgroup$
3
$\begingroup$

The answer is no. There are random variables with all moments finite, whose distributions are not determined by their moments (e.g. the log-normal distribution).

Let $X$ be such a random variable and let $\mu$ be its distribution. Then the operator of multiplication by $x$ on the space $L^{2}(\mathbb{R},\mu)$ has $\psi:= 1$ (the constant function) in the domain of all of its powers, but the spectral measure in this case is $\mu$ and it is not determined by moments.

$\endgroup$

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.