The set of uniqueness for $H^2$ is defined to be a set $E\subseteq \mathbb{D}$ such that if $f\in H^2$ and $f|_E =0$ then $f\equiv 0$. Let $$ k_\lambda(z) =\frac{1}{1-z\ \overline{\lambda}}$$ be the reproducing kernel for $H^2$. I want to prove a statement that is claimed in a text and the statement is : If $\{\lambda_i:i\in \mathbb{N}\}$ be a sequence in $\mathbb{D}$ and $\lambda_i\rightarrow 0$ as $n\rightarrow 0$, then $\{\lambda_i:i\in \mathbb{N}\}$ is a set of uniqueness for $H^2$ and $$ \bigvee \{k_{\lambda_i}:i\in \mathbb{N}\}=H^2$$ Please help.
Uniqueness set for the Hardy space $H^2$
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complex-analysis
functional-analysis
analysis
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0What's the giant V thing? – 2017-01-02
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0It represents the closed linear span of the set $\{k_{\lambda_i}:i\in \mathbb{N}\}$. – 2017-01-08
2 Answers
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The closed linear span $M$ of the $k_{\lambda_j}$ must be all of $H^2$ if the $\lambda_j$ are distinct and converge to $0$. This is because, if $f \perp M$, then $$ 0= \langle f,k_{\lambda_j}\rangle = f(\lambda_j), $$ which, by the identity theorem, implies that $f\equiv 0$. Hence $M^{\perp}=\{0\}$, which proves that the closure $M^c=M^{\perp\perp}$ of $M$ is all of $H^2$.
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Any holomorphic function in $\mathbb D$ that vanishes on such a sequence is $\equiv 0.$ That's not a statement about $H^2,$ really, it's just the identity principle for holomorphic functions.
I take it you want to show the linear span of $\{k_{\lambda_n}\}$ is dense in $H^2?$ That's the same as saying that if $\langle f, k_{\lambda_n}\rangle =0$ for all $n,$ then $f=0.$ Here I'll give you a hint: Write
$$\frac{1}{2\pi}\int_0^{2\pi}f(e^{it})\overline { k_{\lambda_n}(e^{it})}\,dt$$
as a contour integral over $\partial D.$