Show that there exists an entire function $h$ such that $\lim_{n\to\infty}{h(nz)}=0$ for all $z\ne0$

Question

Show that there exists an entire function $h$ such that $\lim_{n\to\infty}{h(nz)}=0$ for all $z\ne0$. The following construction is in Walter Rudin's Real and Complex Analysis Chapter 16, Exercise 11.

For $\alpha>0$, define $\Gamma_\alpha(t)=\left\{\begin{matrix}-t-\pi i,t\le-\alpha\\\alpha+\frac{\pi i t}{\alpha},|t|<\alpha\\ t+\pi i,t\ge\alpha\\ \end{matrix}\right.$. Let $f_\alpha (z)=\frac{1}{2\pi i}\int_{\Gamma_\alpha}{\frac{\exp(e^w)}{w-z}dw}$, for all $z\in\Omega_\alpha$, the component of $\mathbb C-\Gamma_\alpha^*$ containing the origin. ($\Gamma_\alpha^*$ is the graph of $\Gamma_\alpha$.)

By Cauchy's formula, I have proved that there exists an entire function $f$ whose restriction to $\Omega_\alpha$ is $f_\alpha$.

I have trouble proving that $f(r)\to\infty$ when $r$ tends to infinity along the real axis. Since once this is proved, then $g=f\exp(-f)$ will have the desired property.

I try to estimate the integral directly but cannot deal with the integral on $|t|<\alpha$. Can anyone help me?

Answer 1

Hint: Use Cauchy's formula again to move the contour past $z=r$. For more details see below.

By Cauchy's formula $f(r) = \exp(e^r) + \frac{1}{2\pi i}\int_{\Gamma_1} \frac{\exp(e^w)}{w-r}\,dw$. For $r>2$ we can bound the latter integral just by $C\int_{\Gamma_1} |\exp(e^w)|\,dw<\infty$.