If an entire function $f(z)$ has no zeros and satisfies $|f(z)|\le C_1e^{C_2|z|}$ then prove that $f(z)=e^{az+b}$.

I am trying to apply Liouville's theorem on $f$.

Since $f$ is entire and it has no zero so it can be expressed as $f(z)=e^{g(z)}$ , where $g$ is entire function.

Then from the given relation , $|e^{g(z)}|\le C_1e^{C_2|z|}=e^{\ln C_1+C_2|z|}$. I want to show that $|g(z)|\le A+B|z|$ for some constants $A$ and $B$. But I can't do it. How I can proceed further from this step ?


1 Answer 1


You have obtained $\text{Re}\big(g(z)\big)\leq a+b|z|$ for some $a,b\in\mathbb{R}$. If $b\leq 0$, then the real part of $g$ is bounded from above by $a$, whence $g$ is constant (due to Liouville's Theorem with the entire function $z\mapsto \frac{1}{1+a-g(z)}$), implying that $f$ is also constant. From now on, we assume that $b>0$.

Denote by $B_\epsilon(w)$ the closed disc of radius $\epsilon>0$ centered at $w\in\mathbb{C}$. Let $R$ and $r$ be positive real numbers such that $r<R$. By the Borel-Caratheodory Theorem, we have that $$\big|g(z)\big|\leq \frac{2r}{R-r}\,\sup_{z\in B_R(0)}\,\text{Re}\big(g(z)\big)+\frac{R+r}{R-r}\,\big|g(0)\big|\leq \frac{2r(a+bR)}{R-r}+\frac{R+r}{R-r}\,\big|g(0)\big|$$ for every $z\in B_r(0)$. In particular, if $R:=2r$, we obtain $$\big|g(z)\big|\leq (a+2br)+3\big|g(0)\big|\,.$$ Suppose that $g(z)=\sum_{k=0}^\infty\,t_kz^k$ for all $z\in\mathbb{C}$. Then, $$t_k=\frac{1}{2\pi\text{i}}\,\oint_{\partial B_r(0)}\,\frac{g(z)}{z^{k+1}}\,\text{d}z\,.$$ That is, for all $k\in\mathbb{N}_0$ and $r>0$, we have $$\left|t_k\right|\leq\frac{1}{r^k}\,\sup_{z\in B_r(0)}\,\big|g(z)\big| \leq \frac{1}{r^k}\Big(a+2br+3\big|g(0)\big|\Big)\,.$$ If $k>1$, then taking $r\to\infty$, we conclude that $t_k=0$. That is, $g(z)$ is linear.

More generally, if $f:\mathbb{C}\to\mathbb{C}$ is a nonvanishing entire function such that $\left|f(z)\right|\leq C\,\exp\Big(p\big(|z|\big)\Big)$ for some $p(z)\in\mathbb{R}[z]$, then $f(x)=\exp\big(g(z)\big)$ for some $g(z)\in\mathbb{C}[z]$ such that $\deg(g)\leq\deg(p)$.


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