Find all entire functions that satisfy following inequality

Question

Find all entire functions that satisfy following inequality:

$$ |f(z)| \leq |z| e^{\Re(z)} $$ for all $ z \in \Bbb C $

Answer 1

With

$z = x + iy, \tag 1$

since $\vert e^{iy} \vert = 1$ we have

$e^{\Re(z)} = e^x = \vert e^x \vert = \vert e^x \vert \vert e^{iy} \vert = \vert e^{x + iy} \vert = \vert e^z \vert; \tag 2$

thus

$\vert f(z) \vert \le \vert z \vert e^{\Re(z)} = \vert z \vert \vert e^z \vert = \vert z e^z \vert, \tag 3$

which implies

$f(0) = 0; \tag 4$

it follows then that both $z^{-1}f(z)$ and $e^{-z}z^{-1}f(z)$ are also entire and that (3) yields

$\vert z^{-1} e^{-z} f(z) \vert \le 1, \forall z \in \Bbb C; \tag 5$

therefore, via Liouville's theorem (a bounded entire function is constant),

$z^{-1} e^{-z} f(z) = c \in \Bbb C, \tag 6$

where in fact

$\vert c \vert \le 1, \tag 7$

whence

$f(z) = cze^z, c \in \Bbb C, \; \vert c \vert \le 1. \tag 8$

.

Answer 2

Observe that $|e^z|=e^{x}$.

Therefore, you can rearrange this to get $\frac{|f(z)|}{|e^z|}$, and if $f$ is entire, so is $\frac{f(z)}{e^z}$.

This means that $\frac{|f(z)|}{|e^z|} \leq |z|$ for all z. I believe you must've seen in class that if g is an entire function s.t. $|g| \leq |z|$, then we can find constants a,b s.t. $g(z)=(az+b)$

EDIT: Actually, you may not need b, ie g(z)=az. If g is entire, then you can find a Taylor series about the origin with infinite radius of convergence, ie $g(z)=a_0+a_1z+...$.

We have $|g(0)|=|a_0| \leq |0|$, so $a_0=0$. Similarly, show that $a_i =0$ for i greater than 1.(use cauchy's inequalities)

Answer 3

$|\frac{f(z)}{e^z}|\leq|z|\implies f(z)=e^z(az+b)$(by Liouville theorem) $\implies f(z)=e^z(az)$ as $f(0)=0$.