# 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!

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 $x=\lim x_n$.
• Thanks! Yes now I see that what I have shown is for when we are given a norm and have been asked for to show that our space is an Hilbert space according to the given norm. I think this space is $l_2=\{ x=(x_1, x_2, \cdots ) \mid \sum_{i=1}^{+\infty} x_{i}^{2} < \infty \}.$ And we know that in $l_2$ we have an another inner product $\langle x,y \rangle = \sum_{i=1}^{\infty} \bar{x_i} y_i$. But is there such Cauchy sequence? – Nikita Dec 5 '17 at 18:43
• Sorry, I just thought about something. 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 $x,y \in l_2(\omega)$. As we see in our example above, $\omega = (1, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}} , \cdots )$. What can we say about this general space? Is it Hilbert? Thanks! – Nikita Dec 6 '17 at 0:36
• On a quick thought, if $\omega_k\to0$, then we can basically repeat what I did above. If $\omega\to\infty$, you 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. – Martin Argerami Dec 6 '17 at 2:06