Show that there exists $x \in \mathcal{H}$ such that $\langle x,e_n \rangle = \alpha^n$

Question

In my homework I have the following problem.

Let $(e_n)_{n=1}^\infty$ be an orthonormal basis for a Hilbert space and $\alpha \in \mathbb{R}$. Show that there exists $x \in \mathcal{H}$ such that $\langle x,e_n \rangle = \alpha^n$ for all $n \geq 1$ iif $|\alpha| < 1$ and determine $||x||$ for $\alpha=1/3$

My approach is the following: $$x=\sum_{n=1}^\infty \langle x,e_n \rangle e_n=\sum_{n=1}^\infty \alpha^n \cdot e_n$$

I am thinking to use the geometric series so that $||x||=\frac{1}{1-1/3}-(1/3)^0=\frac{1}{2}$

But it gives me a problem. Because I am ignoring the multiplication of the $e_n$ and not sure if this is allowed in the geometric summation formula even though it is orthogonal

Another formula in the book is $||x||^2 = \sum_{n=1}^\infty |\langle x,e_n \rangle|^2=\sum_{n=1}^\infty ((1/3)^2)^n$ but using this formula I get a strange result with $(1/(1-1/9)-1)^{1/2}=\sqrt{1/8}$

Any hint would be appreciated

Answer 1

$\|x\|^{2}=\sum\limits_{k=1}^{\infty} |\langle x, e_n \rangle|^{2}$. So in this case we get $\sum\limits_{k=1}^{\infty} |\alpha|^{2n}=\frac {|\alpha|^{2}} {1-|\alpha|^{2}}$.

We always have $\langle x , e_n \rangle \to 0$ so $|\alpha| <1$ is necessary. Conversely, if $|\alpha| <1$ then the series $\sum \langle x , e_n \rangle e_n$ converges in the norm and defines an element $x$ of $H$ which satisfies $ \langle x , e_n \rangle =\alpha_n$ for all $n$.

To prove convergence of the series when $|\alpha| <1$ note that $\|\sum\limits_{k=j}^{l} \alpha_k e_k\|^{2}=\sum\limits_{k=j}^{l} |\alpha_k|^{2} \to 0$ as $l >j \to \infty$ (by convergence of the geometric series $\sum |\alpha_k|^{2}$) . Use completeness of the Hilbert space.