Analyticity: Uniform Limit

Question

Problem

Consider a uniformly bounded sequence over the real line: $$f_n:\mathbb{R}\to\mathbb{C}:\quad|f_n(x)|\leq L$$

Suppose they have analytic continuations with common domain: $$F_n:\Omega\to\mathbb{C}:\quad F_n\restriction_\mathbb{R}=f_n$$

Does their uniform limit have an analytic continuation, too? $$F:\Omega\to\mathbb{C}:\quad F\restriction_\mathbb{R}=f\quad(f_n\stackrel{\infty}{\to}f)$$ (By uniform boundedness this seems very likely; but really?)

Application

An almost modular state is modular: $$A\in\mathcal{A}^\omega:\quad\omega(\sigma^t[A]B)=\omega(B\sigma^{t+i\beta}[A])\quad(B\in\mathcal{A})$$ (Supposed that entire elements are dense.)

Answer 1

Meanwhile I found an applied counterexample...

Consider a Hamiltonian dynamics: $$H:\mathcal{D}\to\mathcal{H}:\quad U(t)\varphi:=e^{-itH}\varphi$$

Regard the entire elements: $$\mathcal{C}^\omega:=\{\varphi:e^{-itH}\varphi=\sum\frac{1}{k!}(it)^kH^k\varphi\}$$

Then even the entire elements are dense: $$\overline{\mathcal{C}^\omega}=\mathcal{H}:\quad U(t)\varphi_n\to U(t)\varphi\in\mathcal{H}$$

But still it has elements that are not analytic at all: $$\mathcal{D}\subsetneq\mathcal{H}:\quad U(z)\varphi_n\nrightarrow U(z)\varphi\notin\mathcal{H}$$

(The weak version boils down to numerical functions.)

Answer 2

No, this is in general not true. Let $D = (-1,1)^2$, let $\Omega=\mathbb{R} \times (-1,1)$, both viewed as subsets of the complex plane, and let $\phi:D \to \Omega$ be the conformal map with $\phi(0)=0$, $\phi'(0)>0$, so that $\phi((-1,1)) = \mathbb{R}$. Using this conformal map, we can translate your problem into a similar problem on $D$, as follows: If we have analytic functions $f_n:D \to \mathbb{C}$, with $f_n \to f$ on $(-1,1)$ uniformly convergent to a bounded limit $f$, does $f$ necessarily have an analytic continuation to $D$? However, by Runge's theorem any function $f$ which is analytic in some neighborhood of $[-1,1]$ is a uniform limit of (complex analytic) polynomials on $[-1,1]$. Now just pick $f$ to have a singularity somewhere in $D$, and let $(f_n)$ be the sequence of polynomials given by Runge's theorem, and you will get a contradiction to your statement.