# Uniform convergence of analytic functions on Real subspace yields convergence on open neighbourhood of real subspace

I'm guessing following Theorem would be right but I can't conclude my proof nore find any reference proving it.

Let $V$ be a real subspace of a complex Banach space $E$, such that $E$ is the complexification of $V$, and $U$ an open neighborhood. Let $f_n:U\to\mathbb{C}$ be analytic such that the sum $\sum_{n\geq0} |f_n|$ converges locally uniformly on $V$ then there is an open neighborhood $U'$ of $V$ in $U$ such that $\sum_{n\geq0}f_n$ converges to an analytic function on $U'$.

My thoughts are that, since $f_n$ are analytic and I have uniform convergence on the real subspace the sum is a real analytic function on the real subspace and hence can be extended to an open neighborhood $U'$ of $V$. Then I try to show that this function is, maybe on a smaller $U'$, the limit of the sum. The way I try to show it is

• I take a point in $V$, since the sum is real analytic it has a Taylor expansion with positive convergence radius around that point.
• Using the absolute convergence I show that the Taylor coefficients of the sum are the sum of the Taylor coefficients of the $f_n$'s.

Now I would need to show that by this, the analytic continuation on the ball where the Taylor series converges is actually the sum but I can only show that if the sum converges then it is represented by this Taylor series.

Can one finish this proof or is there a counter example explaining my difficulty to show this?

I can only think of a counterexample if you allow the domain to shrink. In that case you can take an essential singularity and let it move closer and closer to the real axis. This way the uniform limit will not be real-analytic again.

For $\epsilon> 0$ set

$$g_\epsilon (x) = e^{-1/(x^2+ \epsilon)}$$

and

$$f(x) = \begin{cases} e^{-1/x^2},& x\neq 0,\\ 0,& x=0. \end{cases}$$

Recall that the function

$$h(x) = \begin{cases} e^{-1/x},& x> 0,\\ 0,& x\leq 0. \end{cases}$$

is smooth and has bounded derivative. Therefore, $g_\epsilon \rightarrow f$ in the supremum norm for $\epsilon \rightarrow 0$ as

$$\Vert f - g_\epsilon \Vert_{\sup} \leq \Vert h' \Vert_{\sup} \cdot \epsilon.$$

Set now $\epsilon_j = (1/2)^j$

$$f_0 = g_1, \qquad f_n = g_{\epsilon_n} -g_{\epsilon_{n-1}} \text{ for } n\geq 1.$$

Then, we have for $N\geq 1$

$$\left\Vert \sum_{n=0}^N f_n - f \right\Vert_{\sup} = \Vert g_{\epsilon_N} - f \Vert_{\sup} \rightarrow 0, \qquad N \rightarrow \infty.$$

We even have that

$$\sum_{n\geq 0} \Vert f_n \Vert_{\sup}$$

converges as $\Vert f_n \Vert_{\sup} = \Vert g_{\epsilon_n} -g_{\epsilon_{n-1}} \Vert_{\sup} \leq (1/2)^{n-1} \Vert h' \Vert_{\sup}$.

Thus, $f$ is the uniform limit of real-analytic functions, but it is not real-analytic at the origin.

Added: There is the following counterexample to the real question. Take $f$ as above. By the Stone-Weierstrass approximation theorem exists for every $n\in \mathbb{N}$ a polynomial $p_n$ such that

$$\sup_{x\in [-n, n]} \vert f(x) - p_n(x) \vert < \frac{1}{2^n}.$$

Let now $K\subset \mathbb{R}$ a compact set. Choose $N\in \mathbb{N}$ such that $K\subseteq [-N,N]$. Then we have for $n\geq N$

$$\sup_{x\in K} \vert f(x) - p_n(x) \vert \leq \sup_{x\in [-n,n]} \vert f(x) - p_n(x) \vert < \frac{1}{2^n} \leq \frac{1}{2^N}.$$

Thus, $p_n$ converges locally uniformly to $f$. Set now $f_0=p_0$ and for $n\geq 1$ set $f_n = p_n - p_{n-1}$. Then we have

$$\sum_{j=0}^n f_j = p_n$$

which converges locally uniformly to $f$. Furthermore, for every compact set $K\subseteq \mathbb{R}$ choose $N\in \mathbb{N}$ such that $K\subseteq [-N, N]$. For every $n\geq N$ we have

$$\sup_{x\in K} \vert f_n(x) \vert \leq \sup_{x\in K} \vert p_n(x) - f(x) \vert +\sup_{x\in K} \vert p_{n-1}(x) - f(x) \vert < \frac{1}{2^{N-1}} .$$

Hence, $\sum_{n\geq 0} \vert f_n \vert$ converges locally uniformly.