Let $\mathcal{H}$ a Hilbert space, $A$ a bounded self-adjoint operator on the space, $W \subseteq \mathcal{H}$ a closed vector subspace that it $A$-invariant, i.e. $AW \subseteq W$. Then there is a result (which we will assume here), that $\sigma(A|_W) \subseteq \sigma(A)$, where $\sigma(B)$ denotes the spectrum of $B$, and that $f(A|_W) = f(A)|_W$, where we are using the respective functional calculi associated to $A|_W$ and $A$ resp.

I have the following conjecture: For each Borel $E \subseteq \sigma(A|_W)$, the spectral subspace $V_E$ corresponding to $E$ (which can be defined as the image of the projection $1_E(A)$) is nontrivial iff it has a nontrivial intersection with $W$.

I don't know if this is true, but seems to be used as a step of a proof I'm trying to work through (which I asked about here, with little luck given the complexity; though if I get a proof of this proposition I can figure out the linked problem). If you want, you can assume that $\mathcal{H}$ is separable. Note that in the case that $W=0$, the spectrum is empty and the statement holds trivially.

Note: I first asked about if the subspace if nontrivial iff it has a nontrivial intersection with $W$. I then changed it to "nontrivial iff not orthogonal to $W$". I now realize that these are the same, as if $x \in V_E$ and $(x,w) \neq 0$ for some $w \in W$, then $1_E(A)w \neq 0$ and $1_E(A)|_Ww = 1_E(A|_W)w$, so $1_E(A)w \in V_E \cap W$.

Edit: I have accepted Martin Argerami's answer, though look at the comments as things were clarified there.


1 Answer 1


Because $A$ is selfadjoint, the subspace $W$ is not only invariant, but reducing. That is, you can write $$ A=\begin{bmatrix}A_0&0\\0&B\end{bmatrix}, $$ where $A_0=A|_W$. From this you can see that $$ 1_E(A)=\begin{bmatrix}1_E(A_0)&0\\0&1_E(B)\end{bmatrix}. $$ So, if $V_E$ is non-trivial, this means that $E$ isn't. Then $1_E(A_0)\ne0$. And so $W\cap V_E$ contains the range of $1_E(A_0)$.

  • $\begingroup$ I'm not familiar with reducing subspaces of operators, but I will think about it/look in to it. Can you explain how this isn't a counterexample though: Consider $L^2([-1,1],m) \oplus L^2([-1,1],m)$ with $m$ Lebesgue measure. Let $A = X \oplus I$ where $X,I$ are bounded self adjoint on $L^2([-1,1],m)$ with $Xf(x) := xf(x)$ and $If = f$. Then let $W := L^2([-1,1],m) \times \{0\}$. Then consider the spectral subspace associated with $\{1\}$. This should contain precisely the functions $(0,f)$ I believe. But this has a trivial intersection with $W$. $\endgroup$ Commented Apr 16, 2020 at 18:37
  • $\begingroup$ The set $E=\{1\}$ has measure zero. So $1_E=0$. "Non trivial" for a set $E$, in this context, means "positive measure". $\endgroup$ Commented Apr 16, 2020 at 18:42
  • $\begingroup$ Just because the set has measure zero doesn't mean the spectral subspace is the 0 subspace. Even though $E$ is a null set, in this case $V_E$ is distinctly non-trivial, right? $\endgroup$ Commented Apr 16, 2020 at 18:44
  • $\begingroup$ Yes, you are right. If you allow that (and it does make sense, in your example), then your conjecture is false. $\endgroup$ Commented Apr 16, 2020 at 19:02
  • $\begingroup$ Ok, thank you. Does this not then break the proof I talk about in the linked question in the way I describe in the comments? (Though I suppose any discussion of the question should take place over there) $\endgroup$ Commented Apr 16, 2020 at 19:05

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