Points with Nonzero Spectral Measure are Countable

Question

This is from Conway's functional analysis textbook, questioin IX.2.10

Show that if $\mathscr{H}$ is separable Hilbert space, there are at most a countable number of points $\left\{z_n\right\}$ in $\sigma(N)$ such that $E\left(z_n\right) \neq 0$, where $E$ is the spectral measure corresponding to $N$.

I'm aware that the spectral measure of a relatively open subset of $\sigma(N)$ is nonzero. Also, that a separable Hilbert space has a countable orthonormal basis. But I'm anaware that how to use separability over here.

Any help is appreciated.

Answer 1

Let $A=\{z\in\sigma(N): E(\{z\}) \ne 0\}$, and $S$ be a countable dense subset of $\mathscr{H}$.

Let $\mathcal{P}(S)$ be the power-set of $S$ and define a mapping $\tau: A\to \mathcal{P}(S)$ as follows $$\tau(z) = \{s\in S:\exists v\, \left(v\in \mathscr{H}\land E(\{z\})(v)=v\land \|v\|=1\land \|v-s\|<\frac{1}{2}\right)\}$$

Claims

1. For each $z\in A$, $\tau(z)\ne\emptyset$.
2. For $z,w\in A$ such that $z\ne w$, $\tau(z)\cap\tau(w)=\emptyset$.

Since $S$ is countable, $S$ can be well-ordered, so there is a function $f:A\to S$ such that $f(z)\in\tau(z)$ for all $z\in A$. (For example, choose the least element in a fixed well-order of $S$.) By the second claim, $f$ is injective, and since $S$ is countable it follows that $A$ is countable.$\Box$

Now to prove the claims. Start with the second claim. Suppose $s\in \tau(z)\cap\tau(w)$. This means there exist $v_z, v_w\in \mathscr{H}$ such that $\|v_z\|=\|v_w\|=1$, and $\|v_z-s\|, \|v_w-s\| < \frac{1}{2}$ and $E(\{z\})(v_z) = v_z$ and $E(\{w\})(v_w) = v_w$. By the triangle inequality $\|v_z-v_w\|<\frac{1}{2}+\frac{1}{2} = 1$. Moreover, since $\{z\}\cap\{w\}=\emptyset$, by the properties of spectral projections $E(\{z\})E(\{w\}) = E(\{w\})E(\{z\}) = E(\{z\}\cap\{w\})=E(\emptyset) = 0$. But then $$\langle v_z, v_w\rangle = \langle E(\{z\})v_z, E(\{w\})v_w\rangle = \langle E(\{w\})^* E(\{z\})v_z, v_w\rangle = \langle E(\{w\}) E(\{z\})v_z, v_w\rangle = 0$$ So $v_z, v_w$ are two orthogonal unit vectors, and then $\|v_z-v_w\|=\sqrt{2}$, contradicting the bound $\|v_z-v_w\|<1$ derived earlier. This establishes the second claim.

For the first claim, for $z\in A$, $E(\{z\})\ne 0$ by assumption, so there is some non-zero vector $v$, which may be normalized such that $E(\{z\})(v) = v$. Since the set $S$ is assumed dense, there is some $s\in S$ such that $\|v-s\| < \frac{1}{2}$. Hence $\tau(z)$ is non-empty.