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Define $$ \ell^2 = \{(z_n)\in \mathbb{C}^{\mathbb{N}}: \sum_{j=1}^{\infty}|z_j|^2<+\infty\}.$$ One can show that $\ell^2$ is a $\mathbb{C}$-vector space and, moreover, that $\ell^2$ is an inner product space for $$ \langle(z_n),(u_n)\rangle=\sum_{j=1}^{\infty}z_j\overline{u_j}.$$ It's not too challenging to show that this map is indeed an inner product, but I'm also trying to show that it is well-defined; i.e. that $$ |\langle(z_n),(u_n)\rangle|<+\infty,\quad \forall(z_n),(u_n)\in \ell^2.$$ I want to show something like this $$|\langle(z_n),(u_n)\rangle|^2 = \left| \sum_j z_j\overline{u_j}\right|^2\le \dots\le \left(\sum_j |z_j|^2\right)\left( \sum_j|u_j|^2\right) < +\infty.$$ I can't use Cauchy-Schwarz' inequality since I have yet to show that $\ell^2$ is an inner product space.

Any hints?

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$\sum\limits_{n=1}^{N}|u_nv_n| \leq \sqrt {\sum\limits_{n=1}^{N}|u_n|^{2}} \sqrt {\sum\limits_{n=1}^{N}|u_n|^{2}} $ for all $N$ (by Cauchy-Schwarz inequality in $\mathbb C^{N}$). Since the right side is bounded it follows that the series $\sum u_nv_n$ is absolutely convergent.

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  • $\begingroup$ So essentially you're using a fact in a finite inner product space and take $N\to\infty$ to link the result to $\ell^2$? $\endgroup$
    – MyWorld
    Oct 2, 2020 at 7:43
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    $\begingroup$ Yes, that is the only reasonable way to prove this result. @Zachary $\endgroup$ Oct 2, 2020 at 7:44
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    $\begingroup$ I'm having trouble with the first line: CS in $\mathbb{C}^N$ would yield $$ |\sum_{n=1}^N u_n\overline{v_n}|\le \sqrt{()}\sqrt{()},$$ with the RHS as you've written down. I tried to use the triangle inequality: $| |u_1v_1|-|u_2v_2|-\dots-|u_nv_n| | \le |\sum_{n=1}^N u_n\overline{v_n}|$, but this isn't what I'm looking for. $\endgroup$
    – MyWorld
    Oct 2, 2020 at 8:21
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    $\begingroup$ @Zachary You need not apply C-S to $(u_n)$ and $(v_n)$. You can apply it to $(|u_n|)$ and $(|v_n|)$. $\endgroup$ Oct 2, 2020 at 8:26

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