An inner product which does not produce a Hilbert norm. I have a simple question:
Consider 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 $\langle x,y \rangle = \sum_{k=1}^{+\infty} \frac{x_k y_k}{\sqrt{k}}$. Prove that this space is Euclidean, but not Hilbert.
$\textbf{Idea for to show it:}$
For being Euclidean it is clear from the assumption $\sum_{k=1}^{+ \infty} x_{k}^{2} < + \infty$.
For to not be Hilbert, we have to show that the norm induced by the defined inner product not satisfies in parallelogram equality
$$\|x + y\|^2 + \|x-y\|^2 = 2(\|x\|^2 + \|y\|^2)$$
For example, if we suppose $x=(0, 0, \ldots , 1 , 0 , \ldots , 0)$, for 1 in the $m$th component and $y=(0, 0, \cdots , 1 , 0 , \ldots , 0)$ , for $1$ in the $n$th component for $m <n$, then we have
$$\|x\| = \sqrt{\langle x,x \rangle}= \sqrt{\frac{1}{\sqrt{m}}}$$ and $$\|y\| = \sqrt{\langle y,y \rangle}= \sqrt{\frac{1}{\sqrt{n}}}$$ and $$\|x+y\| = \sqrt{\langle x+y,x+y \rangle}= \sqrt{\frac{1}{\sqrt{n}}} + \sqrt{\frac{1}{\sqrt{m}}}$$ and $$\|x-y\| = \sqrt{\langle x-y,x-y \rangle}= \sqrt{\frac{1}{\sqrt{m}}} - \sqrt{\frac{1}{\sqrt{n}}}.$$
So we will have $\left(\sqrt{\frac{1}{\sqrt{n}}} + \sqrt{\frac{1}{\sqrt{m}}}\right)^2 + \left(\sqrt{\frac{1}{\sqrt{m}}} - \sqrt{\frac{1}{\sqrt{n}}}\right)^2 = 2\left(\sqrt{\frac{1}{\sqrt{m}}}\right)^2 + 2\left(\sqrt{\frac{1}{\sqrt{n}}}\right)^2 $ which will give us $2\left(\frac{1}{n} + \frac{1}{m}\right) = 2\left(\frac{1}{n} + \frac{1}{m}\right)$.
So this example was not our counter example and we need to find an another example which shows the parallelogram equality not satisfies.
Can you please give me a counter example?
Thanks!
 A: An inner product will always induce a norm that satisfies the parallelogram law: you always have 
$$
\|x+y\|^2+\|x-y\|^2=\langle x+y,x+y\rangle+\langle x-y,x-y\rangle=\langle x,x\rangle+\langle y,y\rangle=\|x\|^2+\|y\|^2.
$$
The point of the question is that the norm that your product induces makes the space not complete. So the example you need to find is of a sequence (of sequences) in your space that is Cauchy but doesn't converge. 
To find the counterexample, we need a sequence that converges to something that is not in $\ell^2$. We can achieve that because of the square roots in the denominator. So let us think of an $x$ that is not in $\ell^2$ but such that 
$$
\|x\|^2=\sum_{k=1}^\infty \frac{|x_k|^2}{\sqrt k}<\infty. 
$$
For instance
$$
x=(k^{-3/4})_k
$$
Now consider the sequence $\{x_n\}$ in your space where $x_n$ is the truncation of $x$ to the first $n$ coordinates and zeroes elsewhere:
$$
x_n=\sum_{k=1}^n k^{-3/4} e_k. 
$$
Then, for $m>n$, 
$$
\|x_n-x_m\|^2=\sum_{k=n+1}^m\frac{k^{-3/4}}{\sqrt k}=\sum_{k=n+1}^m\frac1{k^2}.
$$
As this is the tail of a convergent sequence, we conclude that $\{x_n\}$ is Cauchy. But there is no $x$ in your space with $x=\lim x_n$. 
