# Spectrum of a self-adjoint operator and spectral measures

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

$$\overline{\bigcup_{j\in J} \text{Supp}(\mu_{e_j e_j})} = \sigma(T)?$$

• What is "the spectral measure of $T$ with respect to $x$ and $y$"? Oct 2, 2020 at 11:47
• @NikWeaver, thanks for the comment. I have added the definition to the question. Oct 2, 2020 at 11:56

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.