# Detecting isolated eigenvalues from local spectral measures

Please note: This question has been edited after it became clear from Christian Remling's answer that the original formulation was far from what I really meant to ask.

Let $$T\ne 0$$ be a self-adjoint operator on a Hilbert space $$H$$, with spectrum $$\sigma(T)$$. For any $$x∈H$$, denote by $$μ_x$$ the spectral measure of $$T$$ at $$x$$, that is the unique Borel measure on $$\sigma(T)$$ such that

$$⟨x,f(T)x⟩ = \int_{\sigma(T)} f(λ) d \mu_{x}(λ) \quad \forall f \in \mathcal{C}(\sigma(T),\mathbb{C}).$$

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

Suppose that there exists a closed subspace $$H_0$$ of $$H$$ and a positive integer $$k$$ such that $$T|_{T^i H_0}$$ is injective for all $$i=0,\dots,k-1$$, and $$H=H_0 \overset{\perp}{\oplus} TH_0 \overset{\perp}{\oplus} \dots \overset{\perp}{\oplus} T^{k-1} H_0,\qquad T^{k}H_0=H_0.$$

If $$k=1$$, then $$T$$ is invertible, hence $$0\in \sigma(T)$$ if and only if the inverse $$T^{-1}$$ is unbounded. Therefore, either $$0\not\in \sigma(T)$$ or $$\lambda=0$$ is an accumulation point of $$\sigma(T)$$.

Suppose now that $$k\ge 2$$, and suppose that there is an orthonormal basis $$X_0$$ of $$H_0$$ such that $$\mu_x(\{0\})>0$$ for all $$x\in X_0$$. Then, one has $$0\in \sigma(T)$$. Under some additional assumptions I can actually show that $$\lambda=0$$ must be an isolated eigenvalue. However, I was actually wondering whether that might always be the case.

Addendum: This is how I would like reason. Let $$X:=X_0\cup TX_0 \cup \dots \cup T^{k-1}X_0.$$

Let $$\mathfrak{M}$$ be the set of discrete positive measures $$\nu$$ on $$X$$ with the following property: $$\forall A\subseteq X,\; \nu(A)=0\implies \nu(TA)=0.$$

Then, for all non-zero $$\nu\in\mathfrak{M}$$, one has $$\nu(X_0)>0$$. Hence $$\int_{X} \mu_{x}(\{0\})d\nu(x) \ge \int_{X_0} \mu_{x}(\{0\})d\nu(x)>0.$$

It seems to me that this could be enough to conclude that $$\lambda=0$$ must be an isolated eigenvalue. However, I can only show this with extra assumptions.

Addendum: It occurred to me that the case $$k\ge 3$$ is not that interesting because one gets $$0=⟨x,T^2 x⟩ = \int_{\sigma(T)} \lambda^2 d \mu_{x}(λ) \implies \mu_{x}=\delta_0.$$ The case $$k=2$$ (which is actually also the one that motivated this question) is thus the only really interesting case.

## 1 Answer

Updated answer: For $$k=2$$ the conditions are contradictory.

We have the decomposition $$H=H_0\oplus H_1$$. You now impose the following conditions: (1) $$TH_0=H_1$$; (2) $$T$$ injective on $$H_0$$ and $$H_1$$; (3) $$T^2 H_0=H_0$$; (4) $$Tu=0$$ for some $$u\notin H_1$$.

Since $$N(T^2)=N(T)$$ for the self-adjoint operator $$T$$, we can rephrase the last condition as: (4') $$T^2u=0$$ for some $$u\notin H_1$$.

Let's write $$T= \begin{pmatrix} 0 & B \\ B^* & C \end{pmatrix} ,$$ and here the components refer to this decomposition of $$H$$; the zero in the $$(1,1)$$ is a consequence of (1) above.

In terms of $$B,C$$, your conditions become: (1') $$B^*$$ surjective (onto $$H_1$$); (2') $$B^*$$ injective (on $$H_0$$) and also if $$By=Cy=0$$, then $$y=0$$; (3') $$BC=0$$, $$BB^*$$ surjective; (4'') $$BB^*x=$$ for some $$x\in H_0$$, $$x\not= 0$$.

Since $$B^*$$ is injective, (4'') shows that $$N(B)\not= 0$$, but since $$N(B)=R(B^*)^{\perp}$$, this contradicts (1').

Comment: The part below answers the original version of the question.

It does follow, but in a rather trivial, disappointing way. Let $$S$$ be the operator $$T^k$$, restricted to its reducing subspace $$H_0$$. Since $$N(T^k)=N(T)$$ for a self-adjoint operator, we have $$N(S)= N(T)\cap H_0$$. By assumption, if $$x\in H_0$$, $$x\not= 0$$, then $$x\notin N(T)^{\perp}$$, so $$x\notin N(S)^{\perp}$$. Thus $$S=0$$ and hence $$T=0$$ as well on $$H_0$$. But then $$H=H_0$$, and your operator was the zero operator all along.

• Mmm, that's something weird going on. This question is motivated by many explicit examples that come up from my research in which clearly $T\ne 0$ and $H\ne H_0$. Commented Oct 30, 2020 at 7:50
• @MaurizioMoreschi: Now it's a different situation, of course. There's a huge difference between "for all $x\in H_0$" (the question you asked originally) and "there is an ONB of $H_0$ for which..." (the edited version). Commented Oct 30, 2020 at 12:44
• If I may make a general comment, please try to avoid this kind of moving target question, where you change it to a different question after the first version has been answered. Commented Oct 30, 2020 at 12:46
• You are perfectly right, sorry. I didn't realize at all that there were something wrong with the formulation until reading your answer. The current formulation is what I really meant. Commented Oct 30, 2020 at 12:52
• @MaurizioMoreschi: Yes, no problem, thanks for clarifying. I was just trying to make a general comment (especially since this is a widespread and somewhat unfortunate phenomenon on MO). Commented Oct 30, 2020 at 14:48