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Please note: This is a reformulation of a previous question of mine. The old question has been already answered to, so I prefer asking a new one. However, it looked like the old formulation did not reflect correctly what I really wanted to ask.

Let $T$ be a self-adjoint operator on a Hilbert space $H$, with spectrum $σ(T)$. For any $x∈H$, denote by $μ_x$ the spectral measure of $T$ at $x$, that is the unique Borel measure on $σ(T)$ such that

$$ ⟨x,f(T)x⟩=∫_{σ(T)}f(λ)dμ_x(λ) \quad ∀f∈\mathcal{C}(σ(T),C).$$

Then, one can prove that $$ \overline{\bigcup_{x∈X}Supp(μ_x)}=σ(T) $$ for any orthonormal basis X of H.

Suppose that there exists an orthogonal decomposition $H=H_0 \oplus H_1$, according to which $T$ decomposes as $$ T = \begin{pmatrix} 0 & B \\ B^* & 0\end{pmatrix}.$$

Moreover, suppose that there exists an orthonormal basis $X_0$ of $H_0$ such that $μ_x(\{0\})>0$ for all $x∈X_0$.

Is this enough to conclude that $\lambda=0$ must be an isolated eigenvalue of $T$? I can prove this with some additional assumptions, but I suspect it might always be the case.

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    $\begingroup$ It's still not working; if anything, it's becoming more hopeless. Since $N(T)=N(T^2)$, your assumption simply says that $N(BB^*)\not= 0$, and of course this won't imply that $0$ is isolated in the spectrum. (For example, take $H_0=H_1$, and $B$ self-adjoint, with $0$ an eigenvalue and also $0\in\sigma_{ess}(B)$.) $\endgroup$
    – Christian Remling
    Commented Oct 30, 2020 at 23:51
  • $\begingroup$ @ChristianRemling Isn't my assumption saying that for any $x\in X_0$ we have $v\in\ker(T)$ with $\langle v,x \rangle \ne 0$? The formula I know for $\mu_x(\{0\})=\max_{v\in \ker(T), \|v\|=1} \langle v,x \rangle$. $\endgroup$
    – Maurizio Moreschi
    Commented Oct 31, 2020 at 8:30
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    $\begingroup$ Yes, you assume that there is an ONB such that all its elements are non-orthogonal to $N(T)$, but that isn't really asking for anything because if $v$ is an arbitrary non-zero vector from a Hilbert space, you can always find an ONB such that $\langle x, v\rangle\ \not=0$ for all its elements. $\endgroup$
    – Christian Remling
    Commented Oct 31, 2020 at 14:09
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    $\begingroup$ Also, you forgot to take absolute values and the square, the correct formula is $\mu_x(\{ 0\} )= \|Px\|^2=\max_{v\in N(T), \|v\|=1} |\langle v,x\rangle |^2$, with $P$ being the projection onto $N(T)$ (since $E(\{ 0\} )=P$). $\endgroup$
    – Christian Remling
    Commented Oct 31, 2020 at 14:20
  • $\begingroup$ @ChristianRemling I think that I now see where I was too naive in trying to abstract a spectral theoretic statement from the specific graph theoretic setting I was working in. Maybe it could still be possible, but I'll have to be much more careful about the properties on T that I need to require. Thank you for your answer and comments, they have been of great help. What do you think I should do with this question? Do you want to post your comment as an answer, or should I delete the question at once? $\endgroup$
    – Maurizio Moreschi
    Commented Nov 2, 2020 at 10:52

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