Let $\mu$ be a positive finite measure on the unit circle $\mathbb{T}$. Then the set $span\{z^n:n\in \mathbb{Z}\}$ is dense in $L^2(\mu)$. Let $H^2(\mu)$ be the closure ( in $L^2(\mu)$ ) of the set $span\{z^n:n\geq 0\}$. My question is:
If $H^2(\mu)$=$L^2(\mu)$ holds, can we claim that $\mu$ has no absolutely continuous part with respect to the Lebesgue measure?
No. Recall that the support of $\mu$ is defined as the smallest closed set whose complement has zero measure. Then:
Proof: Apply Runge's theorem to $z^{-1}$ on the support of $\mu.$ $\Box$
So, for example, the uniform distribution $\mu$ on the semicircle $\mathrm{Re}(z)\leq 0$ is non-trivial and absolutely continuous but satisfies $H^2(\mu)=L^2(\mu).$
It's also possible to get full support. By Runge's theorem we can pick a sequence of polynomials $p_n$ such that $|p_n(e^{i\theta})-e^{-i\theta}|\leq 2^{-n}$ for $\theta\in[\tfrac1n,2\pi-\tfrac1n].$ The function $e(z)=\max_n|p_n(z)-z^{-1}|$ is then finite and continuous on $\mathbb T\setminus\{1\}.$ Set $e(1)=+\infty.$ By the $L^2$ dominated convergence theorem, for any measure $\mu$ such that $\int_{\mathbb T} e(z)^2 d\mu<\infty$ we get $p_n\to z^{-1}$ in $L^2(\mu).$ For example we can take $d\mu=e(z)^{-2}d\lambda$ where $\lambda$ is Lebesgue measure.