Claim: Let C(S) be a sup-normed Banach space of continuous complex functions. If $S$ is compact set in $\mathbb{C}$ and if $A$ consists of all $f\in C(S)$ that are holomorphic in the interior of $S$,then every component of the interior of $S$ is A-antisymmetric (which means f is constant in every set or point of S).

I thought of proving this claim using the Liouville theorem:

Theorem: Let $f(z)$ be a holomorphic function in $\mathbb{C}$ such that $f(z)\leqslant M\:\forall z\in\mathbb{C}$. Then $f(z)=constant$.

However I cannot apply the theorem since it would require for the claim to state the function is holomorphic in $\mathbb{C}$ while the function is only holomorphic at the interior of $S$.


Is there a more generalized form of the Liouville theorem that would prove the claim? If not, how should I prove this claim?

Thanks in advance!

  • $\begingroup$ I think you missed something important in the definition of $A$-antisymmetry: According to Rudin's Functional Analysis (Definition 5.6) a subset $E$ of $S$ is called $A$-antisymmetric if every $f\in A$ which is real on $E$ is constant on $E$. $\endgroup$ – Jochen Jun 11 at 11:56
  • $\begingroup$ @Jochen Thanks for the comment. However I am not seeing how that definition relates to the imaginary part of the function. I thought there was a need of a proof for the "claim" in Rudin's Functional Analysis. Could you explain your point please? $\endgroup$ – Pedro Gomes Jun 11 at 12:00
  • $\begingroup$ The only real-valued holomorphic functions on an open set are constant. This follows, e.g., from the Cauchy-Riemann equations. $\endgroup$ – Jochen Jun 11 at 12:25

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