I'm reading the book Spectral Theory of Bounded Linear Operators by Carlos Kubrusly. The author made a claim which was not immediateley clear to me so I tried to prove it myself.
If $\{E_{\gamma}\}_{ \gamma \in \Gamma } $ is an orthogonal family of orthogonal projections (i.e. $\mathcal{R}(E_{\alpha}) \perp \mathcal{R}(E_{\beta})$ for $\alpha \neq \beta$) and $$\sum\limits_{\gamma \in \Gamma}E_{\gamma}x=x\; \text{for every}\, x \in \mathcal{H} $$
then $\{E_{\gamma}\}_{ \gamma \in \Gamma }$ is called a resolution of the identity on $\mathcal{H}$.
The author claims, without proof, that the sum $\sum\limits_{\gamma \in \Gamma}E_{\gamma}x$ has only a countable number of nonzero vectors. Which is what I want to prove with the help of the following theorem
Let $S\subset \mathcal{H}$ be an orthonormal system of a Hilbert space. Then the set $$S_{x}:= \{ y \in S: \langle x,y \rangle \neq 0 \} $$
is at most countable.
Proof
Since $E_{\gamma}$ is an orthogonal projection $E_{y}$ is self-adjoint ($E_{\gamma}^{\ast}=E_{\gamma}$) and idempotent ($E_{\gamma}E_{\gamma}=E_{\gamma}$), and hence $E^{\ast}_{\gamma}E_{\gamma}=E_{\gamma}E_{\gamma}=E_{\gamma}$.
From $E_{\gamma}x=0 \Leftrightarrow ||E_{\gamma}x||^{2} = 0$ we conclude $$E_{\gamma}x \neq 0 \Leftrightarrow 0\neq||E_{\gamma}x||^{2} =\langle E_{\gamma}x,E_{\gamma}x \rangle=\langle x,E_{\gamma}^{\ast}E_{\gamma}x \rangle =\langle x, E_{\gamma}x\rangle.$$
Therefore it suffices to show that for a given $x \in \mathcal{H}$ $\langle x, E_{\gamma}x\rangle\neq 0$ for at most countable many $\gamma \in \Gamma$.
Let $S:=\left\{ \frac{E_{\gamma}x}{||E_{\gamma}x||} : \gamma \in \Gamma \right\}$. $S$ is an orthonormal subset of $\mathcal{H}$ since $\mathcal{R}(E_{\alpha}) \perp \mathcal{R}(E_{\beta})$ for $\alpha \neq \beta$. With the above theorem it follows that the set $\{y \in S : \langle x, y \rangle=||E_{\gamma}x||^{-1}\langle x, E_{\gamma}x \rangle \neq 0 \}$ is at most countable, which concludes the proof.
Is this proof correct? Are there other/easier proofs the see why the claim is true?
Best Answer
Your argument is fine.
A more canonical way to prove this would be to use the same idea that shows that a convergent series of nonnegative numbers can only have countably many nonzero terms.
Let $$G=\{\gamma\in\Gamma:\ \|E_\gamma x\|>0\},\qquad\qquad G_n=\{\gamma\in\Gamma:\ \|E_\gamma x\|>\frac1n\}.$$ Then $$ G=\bigcup_{n\in\mathbb N}G_n. $$ If $G$ is uncountable, then at least one of the $G_n$ is infinite (actually uncountable, but infinite is enough for the argument). Say, $G_{n_0}$ is infinite. Now, for any $k\in\mathbb N$, there exists $F_k\subset G_{n_0}$ with $|F|=k$. Then $$ \Big\|\sum_{\gamma\in F_k}E_\gamma x\Big\|^2=\sum_{\gamma\in F_k}\|E_\gamma x\|^2\geq\frac k{n_0^2}. $$ This prevents the series from being convergent: given any finite set $F\subset\Gamma$ and $k\in\mathbb N$, we have $$ \Big\|\sum_{\gamma\in F\cup F_{kn_0^2}}E_\gamma x\Big\|^2\geq k. $$ This shows that the net of partial sums cannt be convergent. The contradiction implies that $G$ is countable.