Support of the complex measures associated to the spectral measure Sorry for the long winded but trivial question. I am new to the topic and unsure. 
Let $u$ be a bounded self adjoint linear operator on a Hilberspace H and let sp(u) denote its spectrum. By the spectral measure theorem (for example the one for normal operators) there is a unique projection-valued measure E on the sp(u), such that $\int id_{sp(u)} dE = u$. For every pair of vectors $\xi, \nu$ in H we define a complex measure via $E_{x, y}(S) :=\langle E(S)\xi, \nu\rangle$.
What can we say about the support of these associated measures, for example are they compact? I would say yes they are, as, to my understanding, we have $supp E_{x, y} \subseteq supp E$ and supports of measures are closed per definition.
Is that right? If so, can we say more about which subset of the spectrum they are?
Thank you very much in advance. 
 A: If $A$ is a bounded selfadjoint operator on a Hilbert space $\mathcal{H}$, then the support of the spectral measure $E$ is bounded, because it is the spectrum of $A$. If $x\in\mathcal{H}$, then $\mu_{x,x}(S)=\langle E(S)x,x\rangle$ defines a real measure with support that is contained in the support of $E$, which means that $\mu_{x,x}$ is supported on subset of the spectrum of $A$. Finally, if $x,y\in\mathcal{H}$, then the polarization identity gives
$$
          \mu_{x,y}(S)=\frac{1}{4}\sum_{n=0}^{\infty}\mu_{x+i^ny,x+i^ny}(S),
$$
which is also supported in the spectrum of $A$.
If $x\in\mathcal{H}$, let $M_x$ be the closure of the linear space of orbits $\{x,Ax,A^2x,\cdots\}$. The restriction of $A$ to $M_x$ is selfadjoint, and the support $S_x$ of $\mu_{x,x}$ is the spectrum of the restriction of $A$ to $M_x$. You can see this because $A : M_x\rightarrow M_x$ is represented by multiplication by $\lambda$ on $L^2(S_x,\mu_{x,x})$. Indeed, $x\mapsto 1$, $Ax\mapsto \lambda$, $\cdots$, $A^nx\mapsto \lambda^n$, etc. under this correspondence. Each $m\in M_x$ is mapped to a unique $\hat{m}(\lambda)\in L^2(\sigma(A|_{M_x}),\mu_x)$, and $Am$ is mapped to $\lambda \hat{m}(\lambda)$. That is, $\widehat{Am}=\lambda\hat{m}(\lambda)$, which is an abstract sort of Fourier transform, where the operator is transformed to multiplication by a variable. The spectrum of $A|_{M_x}$ is the support of $\mu_x$, which will always be a closed subset of $\sigma(A)$.
