Uniform limit of continuous analytic functions is analytic

Question

Let $f_n:\mathbb C \to \mathbb C$ be functions that are continuous on the closed unit disk and analytic on the open unit disk. Suppose that $f_n$ converges uniformly to some $f$ on the the closed unit disk. Prove that $f$ is analytic on the open unit disk.

I'm looking for a way to prove this without resorting to complex analysis theorems.

Here are my thoughts so far : pick $z_0$ in the open unit disk and $\epsilon >0$. There's some $N$ such that $n\geq N\implies ||f_n-f||<\epsilon$. For each $n\geq N$, there's some $\delta_n>0$ and $(a_{n,k})_k$ such that $\forall z\in B(z_0,\delta_n), f_n(z) = \sum_{k=0}^\infty a_{n,k}(z-z_0)^k$

Since $||f_n-f_m||$ is Cauchy, I'm wondering if it's possible to build a convergent subsequence of $(a_{n,k})_n$ for each $k$. If that is true, I expect the limits to be the coefficients of $f$ as a power series.

Another issue is that it could be that $\delta_n \to 0$.

I haven't used yet that the $f_n$ are continuous on the boundary of the circle.

Any help is appreciated (preferably without complex analysis).

Answer 1

So far, there is no difference in your approach and what one might try for functions analytic on $[0,1].$ But by the Weierstrass approximation theorem, the result fails there.

Answer 2

Let $C$ be the positively oriented unit circle. By the Cauchy integral formula, for $z$ in the open disk $$ f_n(z) = \dfrac{1}{2\pi i} \oint_C \dfrac{f_n(\zeta)\; d\zeta}{\zeta-z}$$ so that taking the limit as $n \to \infty$, $$ f(z) = \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta)\; d\zeta}{\zeta-z}$$ Now the difference quotient

$$ \eqalign{\dfrac{f(z) - f(w)}{z-w} &= \dfrac{1}{2\pi i} \oint_C \frac{f(\zeta)}{z-w} \left( \frac{1}{\zeta - z} - \frac{1}{\zeta - w}\right) \; d\zeta\cr & = \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta) \; d\zeta}{(\zeta-z)(\zeta-w)}\cr & \to \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta) \; d\zeta}{(\zeta-w)^2} \ \text{as}\ z \to w}$$ which says $f$ is complex differentiable at $w$ with derivative $$f'(w) = \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta) \; d\zeta}{(\zeta-w)^2}$$

Answer 3

If we allow complex analysis theorems:

Let $\Delta$ be any triangle in $D(0,1)$. We know (by Cauchy's theorem) that $\forall n, \int_{\partial \Delta} f_n = 0$. So by uniform convergence, $\int_{\partial \Delta} f = 0$. Moreover, uniform convergence preserves continuity. Therefore, by Morera's theorem, $f$ is holomorphic on $D(0,1)$.