An inner product which may produce a Hilbert norm in some situations. As it has discussed in linked-to result, the space of sequences $x = (x_1, x_2, \ldots$, $x_i,\ldots) \in \mathbb{R}$, for all $i \in \mathbb{N},$ such that $\sum_{k=1}^{+ \infty} x_k^2 < + \infty,$ with the product $\langle x,y \rangle = \sum_{k=1}^{+\infty} \frac{x_k y_k}{\sqrt{k}}$ is not Hilbert space.
Now the question is that can we extend this result as follows and disscuse the different cases:
Suppose we are given a space of sequences $x=(x_1, \cdots )$ such that  $||x||^2=\sum_{k=1}^{+\infty} \omega_k |x_k|^2 < \infty$ for $\omega = (\omega_1, \cdots )$ for $\omega_i \in \mathbb{R}$ for all $i$. Let's call this space $l_2(\omega)$. In this case our inner product is given by $\langle x,y \rangle = \sum_{k=1}^{+\infty} \omega_k x_k \bar{y_k} $ for $x,y \in l_2(\omega)$. And as we see in our example linked-to result, $\omega = (1, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}} , \cdots )$, What can we say about this general space? Is it Hilbert? 
Thanks! 
$\textbf{A quick discussion around:}$
As @Martin Argerami , mentioned out in a comment, if $\omega_k\to 0$, then we can basically repeat what I did above. If $\omega\to\infty$ we get a convergence that is harder than that of $\ell^2$, so it will likely be a Hilbert space (but now not every $x\in\ell^2$ will be in your space). If $0<\delta<|\omega_k|<M$
for all $k$, you get exactly $\ell^2$, (equivalent norms). Other cases (different accumulation points, say) will generate other phenomena.
Can someone give me that in more details, please?
Thanks!
 A: The answer is yes, $\ell^2(\omega)$ is a Hilbert space. I'll discuss the complex case.
As noted in the comments by @daw, a necessary (and sufficient) condition for $$\langle x, y \rangle_\omega = \sum_{n=1}^\infty \omega_nx_n\overline{y_n}$$ 
to be an inner product on $\ell^2(\omega)$ is that the sequence $\omega =
 (\omega_n)_{n=1}^\infty$ must be real and strictly positive: for every $n \in \mathbb{N}$ consider $e_n$, the $n$-th canonical vector.
$$0 < \langle e_n, e_n\rangle_\omega = \omega_n$$
Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $\ell^2(\omega)$.
We'll just imitate the standard proof that $\ell^2$ is complete. For every $m, n, j \in\mathbb{N}$ we have:
$$\left|x^{(m)}_j - x^{(n)}_j\right| = \frac{1}{\omega_j} \sqrt{\sum_{k=1}^\infty \omega_k \left|x^{(m)}_k - x^{(n)}_k\right|^2} = \frac{1}{\omega_j} \left\|x_m - x_n\right\|_{2, \omega} \xrightarrow{m, n \to \infty} 0$$
so the coordinate sequences $\big(x^{(n)}_j\big)_{n=1}^\infty$ are Cauchy in $\mathbb{C}$ for each $j \in \mathbb{N}$. Since $\mathbb{C}$ is complete, they converge. Set $x^{(n)}_j \xrightarrow{n \to \infty} x^{(0)}_j \in \mathbb{C}$.
Define $x_0 = \big(x^{(0)}_1, x^{(0)}_2, x^{(0)}_3, \ldots\big)$. We wish to show $x_0 \in \ell^2(\omega)$ and $x_n \xrightarrow{n\to\infty} x_0$ with respect to $\|\cdot\|_{2, \omega}$.
Since $(x_n)_{n=1}^\infty$ is Cauchy, it is also bounded so there exists $M > 0$ such that for all $N, n \in \mathbb{N}$ we have:
$$\sum_{k=1}^N \omega_k \left|x^{(n)}_k\right|^2 \le \left\|x_n\right\|_{2, \omega}^2 \le M$$
Letting $n \to\infty$ gives:
$$\sum_{k=1}^N \omega_k \left|x^{(0)}_k\right|^2 \le M \text{ for all } N \in \mathbb{N}$$
Now letting $N \to \infty$ gives:
$$\sum_{k=1}^\infty \omega_k \left|x^{(0)}_k\right|^2 \le M$$
so $x_0 \in \ell^2(\omega)$.
Now pick $\varepsilon > 0$ and let $n_0 \in \mathbb{N}$ be such that for all $m, n, N \in \mathbb{N}$ with $m \ge n$ holds:
$$\sum_{k=1}^N \omega_k\left|x^{(m)}_k - x^{(n)}_k\right|^2 \le \sum_{k=1}^\infty \omega_k\left|x^{(m)}_k - x^{(n)}_k\right|^2 = \|x_m - x_n\|_{2, \omega}^2 < \left(\frac{\varepsilon}{2}\right)^2$$
Letting $m \to \infty$ gives
$$\sum_{k=1}^N \omega_k\left|x^{(0)}_k - x^{(n)}_k\right|^2 \le \left(\frac{\varepsilon}{2}\right)^2$$
Now letting $N \to \infty$ gives:
$$\sum_{k=1}^\infty \omega_k\left|x^{(0)}_k - x^{(n)}_k\right|^2 \le \left(\frac{\varepsilon}{2}\right)^2$$
And taking the square root:
$$\|x_0 - x_n\|_{2, \omega} = \sqrt{\sum_{k=1}^\infty \omega_k\left|x^{(0)}_k - x^{(n)}_k\right|^2} \le \frac{\varepsilon}{2} < \varepsilon$$
We conclude $x_n \xrightarrow{n\to\infty} x_0$ so $\ell^2(\omega)$ indeed is a Hilbert space. 
