# Show that inner product on $\ell^2$ is well-defined

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?

$$\sum\limits_{n=1}^{N}|u_nv_n| \leq \sqrt {\sum\limits_{n=1}^{N}|u_n|^{2}} \sqrt {\sum\limits_{n=1}^{N}|u_n|^{2}}$$ 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.
• So essentially you're using a fact in a finite inner product space and take $N\to\infty$ to link the result to $\ell^2$? Oct 2, 2020 at 7:43
• 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. Oct 2, 2020 at 8:21
• @Zachary You need not apply C-S to $(u_n)$ and $(v_n)$. You can apply it to $(|u_n|)$ and $(|v_n|)$. Oct 2, 2020 at 8:26