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

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

  • 1
    $\begingroup$ What is "the spectral measure of $T$ with respect to $x$ and $y$"? $\endgroup$
    – Nik Weaver
    Oct 2, 2020 at 11:47
  • $\begingroup$ @NikWeaver, thanks for the comment. I have added the definition to the question. $\endgroup$ Oct 2, 2020 at 11:56

1 Answer 1


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.


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