# Holomorphic function at the interior of a compac set

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 $$S$$.

Questions:

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

• 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 $E$. – Jochen Jun 11 at 11:56
• @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? – Pedro Gomes Jun 11 at 12:00
• The only real-valued holomorphic functions on an open set are constant. This follows, e.g., from the Cauchy-Riemann equations. – Jochen Jun 11 at 12:25