I'm studying the book Uniform Algebras from Gamelin and I have some doubts about the proof of Theorem 1.1.

Theorem: Let $K$ be a compact subset of $\mathbb C$. If $f \in C(K)$ extends to be continuously differentiable in a neighborhood of $K$, and if $\partial f / \partial \overline z = 0$ on $K$, then $f \in R(K)$. Where $R(K)$ is the algebra of all functions in $C(K)$ which can be approximated uniformly on $K$ by rational functions with poles off $K$.

Proof: We can extend $f$ to be continuously differentiable on the complex plane, and to have compact support.

By Cauchy-Green's formula, $$ f(z) = \frac{-1}{\pi} \int \int \frac{\partial f}{\partial \overline \zeta} \frac{1}{\zeta - z}\, dx\, dy $$

where the integral is extended over $\mathbb C$. Let $\mu$ be a measure on $K$ which is orthogonal to $R(K)$.

By Fubini's Formula:

$$ \int f d\mu = \frac{-1}{\pi} \int \int \frac{\partial f}{\partial \overline \zeta}\, \left [ \int_K \frac{1}{\zeta - z}\, d\mu(\zeta) \right ]\, dx\, dy $$

When $z \in \mathbb C \setminus K$, the inner integer is zero. When $z \in K, \, \partial f / \partial \overline z = 0$. Then, $\int fd\mu = 0$.

Hence every continuous linear functional on $C(K)$ which is orthogonal to $R(K)$ is also orthogonal to $f$.

  1. How can I proof that such an element $g$ in the dual $C(K)'$ is orthogonal to $f \in C(K)$?

  2. And with this, how can I use the Hahn-Banach theorem to conclude that $f \in R(K)$?


  • 1
    $\begingroup$ I assume "measure orthogonal to family of functions" means that all these functions have zero integral under the measure? $\endgroup$ – Hagen von Eitzen Jan 30 '17 at 19:00
  • 1
    $\begingroup$ Ad 1: By assumption $f$ is cont. diffable on some open $U\supset K$. Find $r>0$ such that $|x-y|>r$ for all $x\in K$ and $y\notin U$. Find smooth $g\colon \Bbb C\to [0,1]$ with $g(x)=1$ for $x\in K$, $g(x)=0$ if $d(x,K)\ge r$. Then replace $f$ with $f\cdot g$ $\endgroup$ – Hagen von Eitzen Jan 30 '17 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.