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

Question

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

Answer 1

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.