Analytic Extension of bounded real analytic functions to complex analytic ones

Question

If a real analytic function $f$ is bounded on real line $\Bbb R$: given a finite open strip containing $\Bbb R$, does there always exist a holomporphic function $h$ which is bounded on the given finite strip, such that $h = f$ on real line?

Answer 1

No. You can't extend$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\dfrac1{1+x^2}\end{array}$$analytically to the strip $\left\{z\in\mathbb C\,\middle|\,\operatorname{Im}z\in(-2,2)\right\}$.

Answer 2

If you meant

Given $f$ analytic and bounded on $\Bbb{R}$ can it be extended to a function analytic and bounded on some horizontal strip $|\Im(z)| < r$

then the answer is no, look at $$f(z)=\sum_{n=1}^\infty \frac{e^{-n}}{z-n-\frac{i}n}$$ having some poles on every horizontal strip $|\Im(z)| < r$.

If you meant

Given $g$ entire and bounded on $\Bbb{R}$ is it bounded on some horizontal strip $|\Im(z)| < r$

then the answer is no, look at $$g(z) = e^{iz^2}, \qquad |g(a-ib)| = e^{2ab},\qquad h(z)=2\cos(z^2),\qquad|2\cos(z^2)|\ge e^{2|ab|}-1$$

Answer 3

For $f(z)=e^{iz^2},$ we have $|f(x+iy)|=e^{-2xy}.$ Thus $f$ is bounded on $\mathbb R,$ but on every horizontal line below the real axis, $f$ blows up exponentially as $x\to \infty.$