Let $H^2(\omega)$ be the closure of the set of analytic on the unit circle $T$ polynomials, with respect to the scalar product $$ \langle f, g\rangle_\omega = \oint_T f \bar{g} \omega \,\mathrm{d}z, $$ where $\omega$ is a non-negative absolutely continuous non-constant weight on $T$.
Is it the case that $H^2(\omega)\subset H^2$ ? I feel that it's not, but can't construct a counterexample.
On
Let $K$ be the kernel of the measure $\omega$: $$K = \{g\in L^2: ||g||_\omega=0\}. $$ When there are points on $T$ where $\omega =0$, the set $K$ is non-empty. If $g\in K$, then $\bar{g} \in K$ as well, so there exists $h\in K$ such that $h\not\in H^2$.
Take any $f\in H^2$, then a sequence $f_k$ that converges to $f$ also converges to $f+h$ almost everywhere with respect to $\omega$-measure. Therefore $f+h\in H^2(\omega)$, while not being in $H^2$.
(Thanks to discussion with reuns, I clarified a misstatement in my original question)
If $f = \sum_n c_n e^{2i\pi n x}$ is periodic and $L^2[0,1]$ then $|c_n|$ is bounded so that $F_k(z)= \sum_n c_n z^n e^{-\pi n^2/k^2}$ is analytic on $\Bbb{C}^*$.
Moreover $$\lim_{k \to \infty}\|F_k(e^{2i\pi x})- f\|_{L^2[0,1]}^2 =\lim_{k \to \infty} \sum_n |c_n|^2 (1-e^{-\pi n^2/k^2})^2 = 0$$
$H^2$ is the subspace with $c_n = 0$ for $n < 0$, the entire functions are dense in it.
If I took this $e^{-\pi n^2/k^2}$ regulator it is because $$F_k(e^{2i\pi x}) = \int_{-\infty}^\infty k e^{-\pi k^2 y^2} f(x-y)dy$$ so $F_k(e^{2i\pi x})$ is also an approximation of $f$ in every norm.