# Find $\|x\|$ when $\langle x,e_k \rangle=1/2^k$

I am trying to learn about Hilbert spaces. In this I have a small problem in my textbook that says:

Find $$\|x\|$$ when $$\langle x,e_k \rangle=1/2^k$$ where $$(e_k)_{k \in \mathbb{N}}$$ is an orthonormal basis.

In my book I have a statement that says: $$\|x\|^2 = \sum_{n=1}^\infty |\langle x,e_n\rangle|^2$$

And so I get: $$\|x\|^2 = \sum_{n=1}^\infty (\frac{1}{2^k})^2= \sum_{n=1}^\infty (\frac{1}{2})^{2k}$$ $$= \sum_{n=1}^\infty (\frac{1}{2^2})^{k}= \frac{1}{1-1/4}-(1/4)^0=4/3-1=1/3$$

However I am in doubt if this is correct because I don't know why we take $$\|x\|$$ squared. My textbook doesn't explain why and I am worried that this is actually an error in the textbook and the final result of $$1/3$$ is actually wrong

Any input would be appreciated

• There's a square because it's an analogue of the Pythagorean theorem. There's generally squares attached to norms in Hilbert space theory because $\langle x, x\rangle = \lVert x\rVert^2$ is a useful identity. – minimalrho Oct 15 at 18:20
• It's just like in ordinary geometry, where the length of a vector $(a,b,c)$ is $\sqrt{a^2+b^2+c^2}$. – Lord Shark the Unknown Oct 15 at 18:35
• You got $\|x\|^2 = 1/3$ and this is correct. Now take the square root to get $\|x\|$. – amsmath Oct 15 at 18:36

Suppose $$H$$ be Hilbert space and $$E$$ be orthonormal set in $$H$$. Then the following are equivalent:
(i) $$E$$ is an orthonormal basis.
(ii) $$\textbf{(Fourier expansion)}$$ For every $$x \in H, x=\sum_{u \in E} \langle x,u \rangle u.$$
(iii) $$\textbf{(Parseval's Formula)}$$ For every $$x \in H, \| x \|^{2}=\sum_{u \in E} |\langle x,u \rangle |^{2}.$$
Proof can be found in any standard functional analysis book. In your case $$E=(e_{k})_{k \in \mathbb{N}}~$$ is orthonormal basis, so result follows.