Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
I thought it was something obvious like that. –  Rbega Sep 4 '10 at 18:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.