Prove that $S$is a closed subspace of $H^2$ invariant under multiplication by $z$. Find the inner function $F$ such that $S=FH^2$ Let ${\alpha_n}$ be a sequence of points in the open unit disc such that $\sum(1-|\alpha_n|)<\infty$. Let $S$ be the set of all functions $f$ in $H^2$ spaces such that $f(\alpha_n)=f'(\alpha_n)=0$ for each $n$. Prove that $S$ is a closed subspace of $H^2$ invariant under multiplication by $z$. Find the inner function $F$ such that $S=FH^2$.
 A: I think the others gave you unnecessarily a hard time: your question is very clear for anyone knowing what Hardy spaces are (and that's one of your tags).
Proof that $S$ is invariant under $M_z$ (multiplication by $z$): take a function $f \in H^2$ such that $f(\alpha_i)=0$ and $f'(\alpha_i)=0$ for some $i \in N$ (natural numbers). Define $g(z)=zf(z)$ ($g=M_zf$). Evaluate $g$ and $g'$ at $z=\alpha_i$ and you will see that the result is again 0. So if you start with a function in $S$, the result is again in $S$.
To proof that $S$ is closed in an elementary way, take a sequence of functions $\{f_n\}$ such that $f_n \in S \forall n$ and suppose that $f_n \rightarrow f$ (in Hardy norm). Since Hardy norm convergence is stronger than uniform convergence over compacta, that implies $f_n$ converge pointwise to $f$ and $f'_n$ converge pointwise to $f'$ on compact subsets of the disc. Hence $f$ is in $S$.
To find the inner function $F$ that makes $S= FH^2$, note that that function must have a zero of order two at each of the points $\alpha_i$, and that no other condition is required. So the square of the Blaschke product with zeros at those points is exactly your function $F$. I let you do the details to prove this, but anyway, the fact that the $\alpha_i$'s satisfy the Blaschke condition guarantees that $F$ is a valid Blaschke product.
