sequence uniformly convergent on the boundary of a bounded set in $\mathbb{C}$

Question

Let $(f_n)$ be a sequence of functions which are analytic on a bounded region $A\subset\mathbb{C}$ and continuous on the closure $Cl(A)$. Suppose that the sequence is uniformly convergent on the boundary $Bd(A)$. Prove that $(f_n)$ is uniformly convergent to a function on $A$.

This is an exercise in the book Basic Complex Analysis by J.E. Marsden and M.J. Hoffman, and is not homework. I am just studying by myself. The book shows a hint: use the Maximum Modulus Theorem.

I don't know how to approach the problem because I don't know how to construct the limit function. By hypothesis, there is a function $f$ defined on $Bd(A)$ such that $f_n\to f$ uniformly on $Bd(A)$ but How to extend this function to $A$? Using paths-connectedness of $A$? Do you have a solution for this problem?

Answer 1

I found out that this problem is not that hard.

By hypothesis there exists a function $g:Bd(A)\to\mathbb{C}$ such that for every $\varepsilon>0$ there is $N \in\mathbb{N}$ such that $n\geq N$ implies that $|f_n(z)-g(z)|<\varepsilon$ for all $z\in Bd(A)$. This is equivalent to the following: for every $\varepsilon>0$ there is $N_\varepsilon \in\mathbb{N}$ such that $m,n\geq N_\varepsilon$ implies that $|f_n(z)-f_m(z)|<\varepsilon$ for all $z\in Bd(A)$.

Let $\varepsilon>0 $ be given and consider the corresponding $N_\varepsilon\in\mathbb{N}$. For $m,n\geq N_\varepsilon$ we have that the function $f_n-f_m$ is analytic on $A$ and continuous on $Cl(A)$. Therefore, because of the Maximum Modulus Principle, there is $z_0\in Bd(A)$ such that $|f_n(z)-f_m(z)|\leq|f_n(z_0)-f_m(z_0)|$ for every $z\in A$. Since $|f_n(z_0)-f_m(z_0)|<\varepsilon$ we have shown that for every $\varepsilon>0$ there is $N_\varepsilon \in\mathbb{N}$ such that $m,n\geq N_\varepsilon$ implies that $|f_n(z)-f_m(z)|<\varepsilon$ for all $z\in A$.

This final statement is equivalent to say that the sequence converges uniformly on $A$.