Let $H$ be a Hilbert space and let $T:H\rightarrow H$ be a continuous self-adjoint operator.
The spectral theorem asserts that any $u\in H$ admits a finite positive measure $\mu_u$ on $\sigma(T)$ so that $$\left<T^nu,u\right> = \int_{\sigma(T)} x^n d\mu_u(x)$$ for all $n\in\mathbb{Z}$.
The Lebesgue decomposition theorem asserts that every measure $\mu$ decomposes as $\mu_{ac}+\mu_{s}$, where $\mu_{ac}$ is absolutely continuous with respect to Lebesgue and $\mu_s$ is singular to Lebesgue.
According to this Wikipedia page, the subset $H_{ac}$ of all vectors whose spectral measure is absolutely continuous with respect to Lebesgue is a subspace.
My questions are:
Why is it true? More concretely, I do not see why it is closed to summation. If $\mu_u,\mu_v$ are absolutely continuous with respect to Lebesgue, why is $\mu_{u+v} = \mu_u + \mu_v + 2Re(\mu_{u,v})$ also absolutely continuous with respect to Lebesgue?
If it will not follow from the answer, I would also like to know whether it is really necessary to assume that $T$ is self-adjoint. For instance, will it still be true if $T$ is unitary?
Thank you!
You have, since $1_E(T)$ is a projection,
$$ |\mu_{u,v}(E)| =|\langle 1_E(T)u,v\rangle|\leq\|1_E(T)u\|\,\|v\|=\langle 1_E(T)u,u\rangle^{1/2}\,\|v\| =\mu_u(E)^{1/2}\,\|v\|. $$
As for unitaries, the Spectral Theorem works for normal operators, so everything should work.