Suppose that $\{e_k\}_{k \in \Gamma}$ is an orthonormal basis of a Hilbert space $V$ and $f,g \in V$.
We know that $f = \sum_{k \in \Gamma} \langle f, e_k \rangle e_k$, where the summation is an unordered sum.
How can we prove that $\langle f, g\rangle = \sum_{k \in \Gamma}\langle f, e_k \rangle\langle e_k, g \rangle$?
My try:
Given $\epsilon>0$, let's us chose the superset of $\Omega_1, \Omega_2$ (both finite subsets of $\Gamma$, by def. of unordered sum), calling it $\Omega$, such that we have for all $\Omega'_1 \supset \Omega_1$ and $\Omega'_2 \supset \Omega_2$:
$$|\langle\sum_{k \in \Gamma} \langle f, e_k \rangle e_{k}, g \rangle – \langle\sum_{k \in \Omega'_1} \langle f, e_k \rangle e_k, g \rangle|
< \epsilon /2 $$
and $$|\sum_{k \in \Omega'_2}\langle f, e_k \rangle \langle e_k, g \rangle-\sum_{k \in \Gamma}\langle f, e_k \rangle\langle e_k, g \rangle|< \epsilon /2$$
Then for all $\Omega' \supset \Omega$,
\begin{align*}
|\langle f, g\rangle – \sum_{k \in \Gamma}\langle f, e_k \rangle\langle e_k, g \rangle| &\leq |\langle\sum_{k \in \Gamma} \langle f, e_k \rangle e_k, g \rangle – \langle\sum_{k \in \Omega'} \langle f, e_k \rangle e_k, g \rangle| \\
& + |\langle\sum_{k \in \Omega'} \langle f, e_k \rangle e_k, g \rangle-\sum_{k \in \Gamma}\langle f, e_k \rangle\langle e_k, g \rangle| \\
&< \epsilon
\end{align*}
The problem is that the book suggests using the Cauchy-Schwarz inequality, and I don't see the need to use it in this setting… Have I done something wrong?
Best Answer
What is typically meant by the convergence of such a series is as follows. Let $X$ be a normed space and let $\mathcal{F}$ denote the set consisting of finite subsets of a set $\Gamma$. The series $(\sum_{\alpha\in F} x_{\alpha} )_{F\in\mathcal{F}}$ of terms indexed by $\Gamma$ converges to $x$ in $X$ if for all $\varepsilon > 0$ there is some $F_{0}\in \mathcal{F}$ such that whenever $F\in\mathcal{F}$ and $F_{0} \subseteq F$, $\| \sum_{\alpha \in F}x_{\alpha} - x \| < \varepsilon$. Another way to see this is that $(\mathcal{F}, \subseteq )$ forms a directed set and the convergence defined above is equivalent to the net $(\sum_{\alpha\in F} x_{\alpha} )_{F\in\mathcal{F}}$ converging to $x$ in $X$. All of this is discussed on pages 5-6 of the notes you sent through in the comments.
Note that in your proposed solution, $\sum_{k\in\Gamma}\langle f, e_{k}\rangle$ is a scalar if it exists while $g$ is in the Hilbert space $V$, so it does not make sense to talk about $\langle \sum_{k\in\Gamma}\langle f, e_{k}\rangle, g\rangle$.
Let's instead start off with an observation. If $F$ is a finite subset of $\Gamma$, by the linearity of the first entry of the inner product,
$$\sum_{k\in F}\langle f, e_{k}\rangle \langle e_{k}, g\rangle = \left\langle \sum_{k\in F} \langle f, e_{k}\rangle e_{k}, g\right\rangle .$$
As a result, for each finite subset $F$ of $\Gamma $,
$$ \sum_{k\in F}\langle f, e_{k}\rangle \langle e_{k}, g\rangle - \langle f, g\rangle = \left\langle \sum_{k\in F} \langle f, e_{k}\rangle e_{k}, g\right\rangle - \langle f,g\rangle = \left\langle \sum_{k\in F} \langle f, e_{k}\rangle e_{k} - f, g\right\rangle . $$
Note that the left-most expression is the one that is supposed to be estimated as the set $F$ becomes larger under the notion of set inclusion and that the right-most expression can be estimated by the Cauchy-Schwarz inequality. Let $\varepsilon > 0$. Let $\mathcal{F}$ denote the finite subsets of $\Gamma$. As the series $(\sum_{k\in F}\langle f, e_{k}\rangle e_{k})_{F\in\mathcal{F}}$ converges to $f$ in $V$, there is a finite subset $F_{0}$ of $\Gamma$ such that if $F$ is a finite subset of $\Gamma$ containing $F_{0}$, $\|\sum_{k\in F}\langle f, e_{k}\rangle e_{k} - f\| < \varepsilon$. It follows that if $F$ is a finite subset of $\Gamma$ containing $F_{0}$, by the Cauchy-Schwarz inequality,
\begin{align*} \left|\sum_{k\in F}\langle f, e_{k}\rangle \langle e_{k}, g\rangle - \langle f, g\rangle \right| &= \left| \left\langle \sum_{k\in F} \langle f, e_{k}\rangle e_{k} - f, g\right\rangle \right| \\ &\leq \left\|\sum_{k\in F}\langle f, e_{k}\rangle e_{k} - f \right\| \|g\| \\ &\leq \varepsilon \|g\|. \end{align*}
By the definition of the convergence of unordered sums, it follows that $(\sum_{k\in\Gamma}\langle f, e_{k}\rangle \langle e_{k}, g\rangle )_{F\in\mathcal{F}}$ converges to $\langle f, g\rangle$.