Let $T$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$, with spectrum $\sigma(T)$. For any $x,y\in \mathcal{H}$, denote by $\mu\_{xy}$ the spectral measure of $T$ with respect to $x$ and $y$, that is the unique Borel measure on $\sigma(T)$ such that


$$ \langle x,f(T)y\rangle = \int\_{\sigma(T)} f(\lambda)d \mu\_{xy}(\lambda) \quad \forall f\in \mathcal{C}(\sigma(T),\mathbb{C}).$$


Then, one can prove that
$$ \overline{\bigcup\_{x,y\in \mathcal{H}} \text{Supp}(\mu\_{xy})} = \sigma(T). $$


Let now $\{e\_j\}\_{j\in J}$ be an orthonormal basis of $\mathcal{H}$. Then, it is easy to see that $\mu\_{e\_i e\_i}\ge 0$ for all $i$ and that $\mu\_{e\_i e\_j}(\sigma(T))$ equals 1 if $i=j$ and 0 otherwise.


Since $\{e\_j\}\_{j\in J}$ is an orthonormal basis for $\mathcal{H}$, I wanted to try and prove that it is possible to recover $\sigma(T)$ from the supports of the measures $\{\mu\_{e\_i e\_i}\}\_i$. This would be straightforward if the measures $\mu\_{e\_i e\_j}$ were all to be positive, however I don't see why this would be the case. My question is thus: is it true that


$$ \overline{\bigcup\_{j\in J} \text{Supp}(\mu\_{e\_j e\_j})} = \sigma(T)? $$



Spectrum of a self-adjoint operator and spectral measures
Yes, it's true.


I prefer to work with positive measures, so I only deal with $x=y$ (the $\mu\_{x,y}$ have to reason to be positive otherwise). This is not problematic, as the spectrum $\sigma(T)$ is also the closure of $\cup\_x \mathrm{Supp}(\mu\_{x,x})$. So we have to show that the support of $\mu\_{x,x}$ is contained in the closure of $\cup\_i \mathrm{Supp}(\mu\_{e\_i,e\_i})$.


If $x = \sum\_i x\_i e\_i$ belongs to the linear span of the $e\_i$'s, you can write $\mu\_{x,x} = \sum\_i x\_i \overline{x\_j} \mu\_{e\_i,e\_j}$. So if for $\varepsilon = (\varepsilon\_i)\_i \in \{-1,1\}^J$ we denote $x(\varepsilon)= \sum\_i \varepsilon\_i x\_i e\_i$, then the average of $\mu\_{x(\varepsilon),x(\varepsilon)}$ over $\{-1,1\}^J$ is $\sum\_i |x\_i|^2 \mu\_{e\_i,e\_i}$. This shows that the support of $\mu\_{x,x}$ is contained in the support of $\sum\_i |x\_i|^2 \mu\_{e\_i,e\_i}$, that is in the union $\cup\_i \mathrm{Supp}(\mu\_{e\_i,e\_i})$. By approximating any $x$ by a sequence of vectors in the linear span of the $e\_i$'s, you get the result.


