Entire Function with Bounded Real Part is Constant
Theorem
Let $f : \C \to \C$ be an entire function.
Let the real part of $f$ be bounded.
That is, there exists a positive real number $M$ such that:
- $\cmod {\map \Re {\map f z} } < M$
for all $z \in \C$, where $\map \Re {\map f z}$ denotes the real part of $\map f z$.
Then $f$ is constant.
Proof
Let $g : \C \to \C$ be a complex function with:
- $\ds \map g z = e^{\map f z}$
By Derivative of Complex Composite Function, $g$ is entire with derivative:
- $\ds \map {g'} z = \map {f'} z e^{\map f z}$
We have:
\(\ds \cmod {\map g z}\) | \(=\) | \(\ds e^{\map \Re {\map f z} }\) | Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part | |||||||||||
\(\ds \) | \(\le\) | \(\ds e^{\cmod {\map \Re {\map f z} } }\) | Exponential is Strictly Increasing | |||||||||||
\(\ds \) | \(<\) | \(\ds e^M\) | Exponential is Strictly Increasing |
So $g$ is a bounded entire function.
By Liouville's Theorem, $g$ is therefore a constant function.
We therefore have, by Derivative of Constant: Complex:
- $\map {g'} z = 0$
for all $z \in \C$.
That is:
- $\map {f'} z e^{\map f z} = 0$
Since the exponential function is non-zero, we must have:
- $\map {f'} z = 0$
for all $z \in \C$.
From Zero Derivative implies Constant Complex Function, we then have that $f$ is constant on $\C$.
$\blacksquare$