Density result in Hilbert space Assume that $b\in \mathbb{C}$ such that $0<\vert b \vert <1$. We consider the familly $f_{p}=\{1,b^{p},b^{2p},b^{3p},b^{4p},...,b^{np},...)$. How can one prove that $\operatorname{Span}(f_{p}, \ p\in \mathbb{N}^{*})$ is dense in $l^{2}(\mathbb{C})$? Thanks.
 A: Suppose that $f=\{ f_{0},f_{1},f_{2},\cdots\} \in l^{2}$ is orthogonal to every such $f_{p}$. Let
$$
              F(z)=\sum_{n=0}^{\infty}f_{n}z^{n},\;\;\; |z| < 1.
$$
$F$ converges absolutely in $|z| < 1$ because the Cauchy-Schwarz inequality implies
$$
        \left[\sum_{n=0}^{\infty}|f_{n}||z|^{n}\right]^{2} \le \sum_{n=0}^{\infty}|f_{n}|^{2}\sum_{n=0}^{\infty}|z|^{2n} = \|f\|^{2}\frac{1}{1-|z|^{2}},
$$
which means that $F$ is a holomorphic function on the open unit disk. Because $f$ is orthogonal to every $f_{p}$ as you have defined, then
$$
       F(b^{p})=\sum_{n=0}^{\infty}f_{n}(b^{p})^{n} = (f,f_{p})=0.    
$$
Because $0 < |b| < 1$, and the above holds for all $p=1,2,3,\cdots,$ then $F$ has infinitely many zeros in the open unit disk $D$, with a cluster point at $0$. By the identity theorem for holomorphic functions, $F$ is identically $0$, which forces $f_{n}=0$ for all $n$. That is, $f=0$. So your subspace is dense in $l^{2}$ because there is no non-zero vector $f$ which is orthogonal to everything in your subspace.
