I am currently reading Hilbert Spaces and confused about a thing. Say, $C=\{e_\alpha : \alpha\in\mathcal{A}\}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $\sum_{\alpha\in\mathcal{A}}e_\alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?
Uncountable sum of vectors in a Hilbert Space
functional-analysishilbert-spaces
Best Answer
The sum $\sum_{\alpha\in A} v_\alpha$ does make sense. It is defined to converge to $L \in H$ if $$\forall \varepsilon > 0 \,\exists F_0 \subseteq A \text{ finite such that }\forall F \subseteq A \text{ finite}, F \supseteq F_0 \text{ we have} \left\|\sum_{\alpha\in F}v_\alpha - L\right\| < \varepsilon$$
However, the sum $\sum_{\alpha\in A} e_\alpha$ of an orthonormal set only converges if $A$ is finite.
Indeed, assume that $\sum_{\alpha\in A} e_\alpha = L$. For $\varepsilon = \frac12$ there exists $F_0 \subseteq A$ finite such that for all $F \subseteq A$ finite, $F \supseteq F_0$ implies $\left\|\sum_{\alpha\in F} e_\alpha -L\right\| < \frac12$.
For any $F_0 \subseteq F \subseteq A$ finite we have $$\sqrt{|F \setminus F_0|}= \left\|\sum_{\alpha\in F\setminus F_0}e_\alpha\right\| = \left\|\sum_{\alpha\in F}e_\alpha - \sum_{\alpha\in F_0}e_\alpha\right\| \le \left\|\sum_{\alpha\in F}e_\alpha - L\right\| +\left\|L- \sum_{\alpha\in F_0}e_\alpha\right\| < 1$$
so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.