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 1


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.

  • $\begingroup$ 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 $H\ne H_0$. $\endgroup$ Commented Oct 30, 2020 at 7:50
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    $\begingroup$ @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). $\endgroup$ Commented Oct 30, 2020 at 12:44
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    $\begingroup$ 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. $\endgroup$ Commented Oct 30, 2020 at 12:46
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    $\begingroup$ 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. $\endgroup$ Commented Oct 30, 2020 at 12:52
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    $\begingroup$ @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). $\endgroup$ Commented Oct 30, 2020 at 14:48

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