Find all entire functions with $|f(z)|\geq e^{|z|}$ for all $z \in \mathbb{C}$.

Question

Find all entire functions with $|f(z)|\geq e^{|z|}$ for all $z \in \mathbb{C}$.

I don't think there is any such entire function, and here is my thought: since $\Re(z) \leq |z|$, we know $|e^z|\leq e^{|z|}\leq |f(z)|$ for all $z$. Consider $g(z)=\frac{f(z)}{e^z}$. Since $e^z$ is never $0$, $g(z)$ is an entire function. But $|g(z)| \geq 1$ for all $z\in \mathbb{C}$, so there is a contradiction.

I was wondering is there any hole in the preceding argument?

Answer 1

From the inequality it follows that $f$ is never zero.

Therefore $1/f$ is entire and $|1/f|\leq e^{-|z|}\leq1$. Therefore $f$ ought to be constant.

But $e^{|z|}\rightarrow\infty$ as $z\rightarrow \infty$.

...

You probably don't want that, but in a sense $f(z)=\infty\in S^2$ is constant, therefore analytic everywhere, and $|f(z)|\geq e^{|z|}$.

Answer 2

It's a bit more complicated than necessary. $|f(z)| \ge e^{|z|} \ge 1$, so why not just use $f$ instead of $g$?

And, by the way, what are you contradicting?

Answer 3

It looks fine to me. Assuming $|f(z)|\geq e^{|z|}$ we get that $g(z)\stackrel{\text{def}}{=}f(z) e^{-z}$ is an entire function whose modulus is always $\geq 1$, but that contradicts Picard's little Theorem or just Casorati-Weierstrass Theorem, stating that if $f$ is a non-constant entire function then $f(\mathbb{C})$ is dense in $\mathbb{C}$.