"Let $\rho$ be a space of complex polynomial and define $<f,g>={1\over 2\pi}\int_{0}^{2\pi}f(e^{it})\overline{g(e^{it})}dt$ for $f,g:\rho\to \Bbb{C}$.
Let $\phi$ be a linear functional on $\rho$ defined by $\phi(f)=f({1\over 2})$. Prove $\phi $ is bounded and find its Riesz representation in the completion of $\rho$ to a Hilbert space."
I am stuck in the middle of the process.
Attempt: $f=\sum_{n=1}^{\infty}{a_z}z^n$. Hence, since $<e^{int},e^{imt}>=0$ for $n\ne m$, I get $<f,f>=\sum^{\infty}_{n=1}{|a_n|^2}$. $||f({1\over 2})||^2\le \sum^{\infty}_{n=1}{|a_n|^2}\sum^{\infty}_{n=1}({1\over 2})^n={1\over 2}\sum^{\infty}_{n=1}{|a_n|^2}$
Therefore ${||f({1\over 2})||\over ||f||}\le \sqrt 2$.
Now I have to find a polynomial $g$ such that $<f,g>=f({1\over 2})$ for all $g$ and it seems complicated. What I can do is simply starting to guess because the calculations are confusing. I tried taking $g(z)=\sum_{n}^{\infty}({1\over 2})^nz^n$ because then I get $<f,g>=\sum ({1\over 2})^na_n=f({1\over 2})$ but now I am not sure because this isn't a Hilbert space, and although I know I was asked to refer to the completion, I can't really be sure if it means I can simply do what I did and finish. That is a big insecurity and I could use a reference.
You are correct.
We can note that our space is isometric to the subspace of $l^2(\mathbb{N})$ comprised of sequences with finitely many non-zero entries, by the isometry $\sum_{n=0}^N a_n z^n\mapsto (a_0, a_1, \ldots, a_N, 0, 0, \ldots)$.
(Note that, indeed, $\left\langle \sum_{n=0}^N a_n z^n, \sum_{n=0}^M b_n z^n\right\rangle = \sum_{n=0}^{\min\{N,M\}}a_n\overline{b_n}$.)
This is readily seen to be a dense subspace, so its completion is $l^2(\mathbb{N})$, and as you noted $$\phi\left(\sum_{n=0}^N a_n z^n\right) = \sum_{n=0}^N a_n \left(\frac{1}{2}\right)^n = \left\langle (a_0, a_1, \ldots, a_N, 0, 0, \ldots),\ (1, \frac{1}{2}, \frac{1}{2^2}, \ldots)\right\rangle$$