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I was wondering if anything was known about the following:

Let $\mathbb{D}^2=\lbrace x^2+y^2< 1 \rbrace \subset \mathbb{R}^2$ be the open unit disk. Consider now the Green's functions $G(z; p)$ of this disk. I.e. here $p\in \mathbb{D}^2$ and $G(z;p)$ is smooth and harmonic in $\bar{\mathbb{D}}^2\backslash \lbrace p \rbrace$, vanishes on the boundary and has the property that $H(z;p) =G(z; p)-\log |z-p|$ is smooth.

Now consider the set of functions: \begin{equation} S=\lbrace f\in C^\infty(\partial \mathbb{D}^2): f= \sum_{i=1}^n \lambda_i \partial_{\nu} G(z; p_i), \lambda_i \in \mathbb{R}, p_i \in \mathbb{D} \rbrace \end{equation} Here $\partial_{\nu} G(z; p_i)$ is the normal derivative on $\partial \mathbb{D}^2$.

My question is what can be said about the set $S$? In particular, is there any hope that it is dense in $L^2(\partial \mathbb{D}^2)$?

Playing around with things all I got was a big mess so any references would be appreciated.

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1 Answer 1

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Yes indeed, $S$ is dense in $L^2(\partial \mathbb{D})$.

This is because any $g\in L^2(\partial \mathbb{D})$ has an $L^2$ harmonic extension $h$ to $\mathbb{D}$, and $$h(p)=\frac{1}{2\pi}\int_{\partial \mathbb{D}} g(z) \partial_{\nu} G(z; p) dz\;\;\;(*)$$ by Green's formula. Hence if $g$ is orthogonal to $S$, $h(p)=0$ for all $p$, and $g=0$.

Implicit are the non obvious results that there is a continuous trace from $L^2$ harmonic function in $\mathbb{D}$ to $L^2$ functions on the boundary and that the trace of $h$ given by $(*)$ is indeed $g$. This should be found in many books treating harmonic analysis or one variable complex analysis, like the references here, or Rudin's Real and complex analysis.

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I thought it was something obvious like that. –  Rbega Sep 4 '10 at 18:45

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