Continuity of the inner-product in a Hilbert space with respect to unordered sums

Question

Let $\{x_\alpha\}_{\alpha\in A}$ be an indexed set of elements in a Hilbert space. We say that the unordered sum $\sum_{\alpha\in A}x_\alpha$ converges to $x$ if for every $\epsilon>0$ there is some finite subset $J_0$ of indices such that $$\|x-\sum_{\alpha\in J}x_\alpha\|<\epsilon$$ for all finite sets $J$ of indices containing $J_0$.

On page 136 of John Hunter's Applied Analysis, or Theorem 11 in the same author's note say that

if the unordered sums $\sum_{\alpha\in A}x_\alpha,\sum_{\beta\in B}y_\beta$ converge in a Hilbert space, then the unordered sum (of inner products) $$\sum_{(\alpha,\beta)\in A\times B}\langle x_\alpha,y_\beta\rangle$$ converges and $$\sum_{(\alpha,\beta)\in A\times B}\langle x_\alpha,y_\beta\rangle= \langle \sum_{\alpha\in A}x_\alpha, \sum_{\beta\in B}y_\beta\rangle$$

---

The book and the note do not provide a proof but say that this follows from continuity of the inner product. However, the continuity of inner product only gives that $$\sum_{\alpha\in A}(\sum_{\beta\in B}\langle x_\alpha,y_\beta\rangle),\ \sum_{\beta\in B}(\sum_{\alpha \in A}\langle x_\alpha,y_\beta\rangle)$$ converge and did not give any information on the unordered sum $$\sum_{(\alpha,\beta)\in A\times B}$$ which involves finite subsets of $A\times B$ that are not necessarily of the form $E\times F\subset A\times B$; and this makes directly proving the claim unclear:

Choose $E_0,F_0$ such that $\|x-\sum_{\alpha\in E}x_\alpha\|<\epsilon$ and $\|y-\sum_{\beta\in F}y_\beta\|< \epsilon$ whenever $E\supset E_0$ and $F\supset F_0$. If $U$ contains $E_0\times F_0$, then there is $E\times F\supset E_0\times F_0$ and \begin{align*} \|\sum_{(\alpha,\beta)\in U}\langle x_\alpha,y_\beta\rangle - \langle \sum_{\alpha\in A}x_\alpha, \sum_{\beta\in B}y_\beta\rangle\| &\leq \|\sum_{(\alpha,\beta)\in U}\langle x_\alpha,y_\beta\rangle - \sum_{(\alpha,\beta)\in E\times F}\langle x_\alpha,y_\beta\rangle\|\\ &+\|\langle \sum_{\alpha\in E}x_\alpha,\sum_{\beta\in F}y_\beta\rangle - \langle \sum_{\alpha\in A}x_\alpha, \sum_{\beta\in B}y_\beta\rangle\| \end{align*} The second term is easy to upper bound, but I don't see a way to bound the first term.

---

I realize the solution provided is incorrect. I used the Cauchy-Schwartz inequality incorrectly. In the accepted answer it wrote $$\sum_{\alpha,\beta}|\langle x_\alpha,y_\beta\rangle|\leq (\sum_\alpha \|x_\alpha\|^2)^{1/2}(\sum_{\beta}\|y_\beta\|^2)^{1/2}<\infty$$ However, it should be $$\sum_{\alpha,\beta}|\langle x_\alpha,y_\beta\rangle|\leq (\sum_{\alpha,{\color{red}\beta}} \|x_\alpha\|^2)^{1/2}(\sum_{{\color{red}\alpha},\beta}\|y_\beta\|^2)^{1/2}=\infty$$ which made the argument invalid.

Thus I would like to re-open the question.

Answer 1

If $\sum_{\alpha\in A}x_\alpha$ converges, then we have $\sum_{\alpha\in A}\|x_\alpha\|^2<\infty$ (See Theorem 3.1 in the note). Thus $\sum_{\alpha,\beta}|\langle x_\alpha,y_\beta\rangle|\leq (\sum_\alpha \|x_\alpha\|^2)^{1/2}(\sum_{\beta}\|y_\beta\|^2)^{1/2}<\infty$ by Cauchy–Schwarz inequality. For complex numbers, we can use the Riemann rearrangement theorem to conclude that $\sum_{\alpha,\beta}\langle x_\alpha,y_\beta\rangle$ converges.