How to use Cauchy-Schwarz inequality to show the scalar product $\langle\xi,\eta\rangle=\sum_{i=1}^{\infty}\omega_i\xi_i\bar\eta_i$ is well-defined This is part of a problem in which I have to prove that the space of sequences defined below is a Hilbert space. I have a doubt with the first part
In the book I have:
Given the
scalar product $\langle\xi,\eta\rangle=\sum_{i=1}^{\infty}\omega_i\xi_i\bar\eta_i$, defined on the space of sequences $\xi=(\xi_1,\xi_2,...)$ of complex numbers $\xi_i$ such that $\sum_{i=1}^{\infty}\omega_i|\eta_i|^2<\infty$ and $\omega_i >0$ for all $i$.
The scalar product is well-defined because, using the Cauchy-Schwarz inequality in $
\mathbb{C}^k$
\begin{align}
\sum_{i=1}^k|\omega_i\xi_i\bar\eta_i|=\sum_{i=1}^k(\sqrt\omega_i|\xi_i|)(\sqrt \omega_i|\eta_i|) &\le \Bigl(\sum_{i=1}^k\omega_i|\xi_i|^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^k\omega_i|\eta_i|^2\Bigr)^{1/2}\\[0.3cm]
& \le \Bigl(\sum_{i=1}^{\infty}\omega_i|\xi_i|^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^{\infty}\omega_i|\eta_i|^2\Bigr)^{1/2}
\end{align}
and thus $$\tag1\sum_{i=1}^{\infty}\omega_i|\xi_i\bar\eta_i|< \infty .$$
How I see it:
I don't think the Cauchy-Schwarz inequality is being applied correctly because the module $|\cdot|$ should be applied to the whole sum not to every term, so I would write:
\begin{align}
\Bigl|\sum_{i=1}^k\omega_i\xi_i\bar\eta_i\Bigr|=\Bigl|\sum_{i=1}^k(\sqrt\omega_i\xi_i)(\sqrt \omega_i\eta_i)\Bigr|
&\le \Bigl(\sum_{i=1}^k\omega_i|\xi_i|^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^k\omega_i|\eta_i|^2\Bigr)^{1/2}\\[0.3cm]
&\le \Bigl(\sum_{i=1}^{\infty}\omega_i|\xi_i|^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^{\infty}\omega_i|\eta_i|^2\Bigr)^{1/2}.
\end{align}
and thus $$\tag2\Bigl|\sum_{i=1}^k\omega_i\xi_i\bar\eta_i\Bigr|< \infty.$$
So my questions are
1) Is the book applying correctly the C-S inequality or is it doing something else?
2) Is my approach correct?
3) How do I conclude that the series  $\langle\xi,\eta\rangle=\sum_{i=1}^{\infty}\omega_i\xi_i\bar\eta_i$ converges from  $(1)$ or $(2)$ or from each one (if they are both correct)? $(1)$ tells me the sequence converges absolutely, but I don't know if the space of sequences I am working with is complete yet, so I can't conclude from that the series converges I guess. And from $(2)$ I don't know what to do either.
 A: No. Bounding sums in absolute value does not show that your series converges. Think
$$
\Big|\sum_{j=1}^k(-1)^j\Big|\leq1.
$$
Your estimate still works, because what you want to show, to prove that the series converges, is that the sequence of partial sums is Cauchy; this is equivalent to having small tails. So if you write your estimate as, for $m<n$,
$$
\Big|\sum_{i=1}^n\omega_i\xi_i\bar\eta_i-\sum_{i=1}^{m}\omega_i\xi_i\bar\eta_i\Big|=\Big|\sum_{i=m+1}^n\omega_i\xi_i\bar\eta_i\Big|\le \Bigl(\sum_{i=m+1}^{n}\omega_i|\xi_i|^2\Bigr)^{1/2}\Bigl(\sum_{i=m+1}^{n}\omega_i|\eta_i|^2\Bigr)^{1/2},
$$
now you can use the fact that both series on the right converge to prove that  your series converges.
The first approach is not wrong, though, and in fact it's the standard one. What you get from
$$
\sum_{i=1}^k\omega_i\,|\xi_i\bar\eta_i|\le \Bigl(\sum_{i=1}^{\infty}\omega_i|\xi_i|^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^{\infty}\omega_i|\eta_i|^2\Bigr)^{1/2}
$$
is, since the right-hand-side does not depend on $k$, that the sequence of partial sums $\sum_{i=1}^k\omega_i\,|\xi_i\bar\eta_i|$ is increasing and  bounded, thus convergent. So the series $\sum_{i=1}^\infty\omega_i\,|\xi_i\bar\eta_i|$ converges, which is precisely the absolute convergence of the series $\sum_{i=1}^\infty\omega_i\,\xi_i\bar\eta_i$ in $\mathbb C$, which I assume you know it's complete. At this stage you are not discussing the completeness of your inner product space, just the existence of these series in $\mathbb C$.
A: There is a measure theory perspective for this which generalizes nicely to other settings: Let $\mu$ be the counting measure defined on the powerset of $\mathbb{N}$. Then $d\nu = \omega\,d\mu$ is a measure on the powerset of $\mathbb{N}$. If $\xi, \eta \in L^2(\nu)$, then
$$(\xi, \eta)_{L^2(\nu)} = \int_{\mathbb{N}}\xi_i\overline{\eta_i}\,d\nu(i) = \int_{\mathbb{N}}\xi_{i}\overline{\eta_i}\omega_i\,d\mu(i) = \sum_{i = 1}^{\infty}\xi_i\overline{\eta_i}\omega_i = \langle\xi, \eta\rangle.$$
Therefore the space you are considering is exactly $L^2(\nu)$. Since $L^2$ is always a Hilbert space, your space is a Hilbert space.
Note that the $\omega_i > 0$ hypothesis was not used above. The hypothesis is there to ensure that if a sequence is $0$ $\nu$-a.e., then all of it's terms are zero, i.e. so that $\xi = 0$ in $L^2(\nu)$ implies $\xi_i = 0$ for all $i$.
