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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 $S=FH^2$.

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  • $\begingroup$ What is $\,H^2\,$ ?? $\endgroup$ – DonAntonio Apr 1 '13 at 18:13
  • $\begingroup$ Hilbert space $H^2$ $\endgroup$ – Matema Tika Apr 1 '13 at 18:17
  • $\begingroup$ What Hilbert space?! $\endgroup$ – DonAntonio Apr 1 '13 at 18:18
  • $\begingroup$ And what is an "inner function"?? $\endgroup$ – DonAntonio Apr 1 '13 at 18:18
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    $\begingroup$ I've already explained in the past: I demand that any student in high-school level or above be at least able to explain what is he asking. If you can't then you first must understand the basic elements in your question and then pose the question... $\endgroup$ – DonAntonio Apr 1 '13 at 18:24
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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.

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