Control of values of an entire function in a strip around the real line

Question

Consider an entire function $f: \mathbb C \to \mathbb C$ such that $f|_{\mathbb R}(x)\to 0$ as $x \in \mathbb R \to \pm\infty.$ Does that imply that for each $T>0,$ we have $f(x+iy) \to 0$ as $x\to \pm\infty$ uniformly for $y\in [0,T]?$

This looks like the result of controlling the value of a holomorphic function by its value on the boundary of a region. But $\mathbb R$ is not the boundary of a region on the Riemann sphere. Moreover, I could not find many good examples. Examples like $e^{-z^2}$ does satisfy the condition above.

Such uniform convergence is very useful for estimating integrals of the form $$ \int_{\mathbb R} f(x) dx $$ where we can just instead estimate $\int_{\mathbb R} f(x+iy) dx$ for some suitable value of $y.$ This can be very helpful.

Answer 1

Not at all. Consider the function $\sin z^2/z$, which is entire because the singularity at zero is removable. On any horizontal line $z=x+iy$ with $y\neq 0$, we have $$ \frac{\sin z^2}{z}=\frac{e^{-2yx}e^{i(x^2-y^2)}-e^{2yx}e^{-i(x^2-y^2)}}{2i(x+iy)}. $$ and one of the terms in the numerator tends to zero and another to infinity exponentially fast as $x\to\pm \infty$.

Answer 2

The answer is negative. A simple example is $e^{e^{-z^2}}-1$. In general, you can approximate almost anything by an entire function: the precise statement is given by Arakelyan's theorem

Answer 3

You need a growth condition on $f$, namely if it is of exponential type $|f(z)| \le Ce^{a|z|}$ then the result holds; for example, $\frac{\sin z}{z}$ satisfies this; note that the counterexamples below are of (growth) order $2$ or infinity respectively.