# An entire function whose real part is bounded above must be constant.

Greets

This is exercise 15.d chapter 3 of Stein & Shakarchi's "Complex Analysis", they hint: "Use the maximum modulus principle", but I didn't see how to do the exercise with this hint rightaway, instead I knew how to do it with the Casorati-Weiestrass Theorem, here is my answer:

Define $g(z)=f(1/z)$ for $z\neq{0}$,then by the hypothesis we must have that for any $\epsilon>0$ $g(D_{\epsilon>0}(0)-\{0\})$ is not dense in $\mathbb{C}$, then the singularity at $0$ of $g$ is not essential, this implies $f$ must be a polynomial, but if $f$ is a non-constant polynomial, it is easy to see that its real part must be unbounded, so $f$ must be constant.

I would like to know an answer with the the maximum modulus principle.

Thanks

• Nov 19, 2014 at 15:52

As other posters have commented, the standard approach here would be to invoke Liouville's Theorem. One way to do this is to consider the entire function $e^{f(z)}$.

Observe that $|e^{f(z)}| = e^{\Re f(z)}$, which is bounded by our assumption on $\Re f(z)$.

Then $e^{f(z)}$ is an entire bounded function, and hence (by Liouville's Theorem) constant.

From this, we conclude that $f(z)$ is constant as well.

• I was wondering why would $e^{f(z)}$ being constant will imply $f(z)$ is constant... Unlike real exponential, complex exponential is not one one.. It is periodic. So, i was thinking it need not be true.. After some time i realized it is true... Justification is as follows: We have $e^{f(z)}=C$ for all $z\in \mathbb{C}$.. Taking derivative both sides we see that $e^{f(z)}f'(z)=0$ for all $z\in \mathbb{C}$.. As $e^{f(z)}$ is never zero we have $f'(z)=0$ for all $z\in \mathbb{C}$ and so $f(z)$ is constant function..
– user312648
Apr 7, 2016 at 5:56
• @user312648: Thanks! Aug 3, 2018 at 16:30
• The "standard" way of noting that $\exp \circ f$ being constant implies $f$ is so is to note that the preimage of every singleton is discrete. Indeed, if $\exp \circ f$ is constant, then $f$ must take values in some set of the form $c + 2 \pi i \Bbb Z$. Since $f$ is continuous and $\Bbb C$ connected, $f$ must be constant. Oct 31, 2021 at 12:55
• @AryamanMaithani Thanks! Nov 6, 2021 at 19:28

A simpler way is to use Liouville's Theorem: consider $g(z) = 1/(1+b - f(z))$ where $\text{Re}(f(z)) \le b$.

• can you please supply more details as to how this proof would go? Thank you.
– user100463
Mar 25, 2016 at 1:31
• @ALannister Note that $g(z)$ is a bounded entire function, by Liouville's Theorem we have $g(z)$ is constant and so is $f(z)$.
– Bach
Jul 12, 2019 at 4:49

Let $f(z)$ be an entire function with real part bounded,then

$f(z)$ is entire$\implies \phi(z)=e^{f(z)}$ entire

$\implies \vert\phi(z)\vert=\vert e^{f(z)}\vert= e^{Re(f(z))} \le e^M$ ,Where $Re(z) \le M$ for some fixed $M \in \mathbb R$

$\implies \phi(z)$ is constant,by Liouville's theorem

$\implies\phi'(z)=e^{f(z)}f'(z)=0\forall z\in \mathbb C$

$\implies f'(z)=0$ in $\mathbb C$,since $e^{f(z)}\neq 0$ in $\mathbb C$

$\implies f(z)$ is constant

• What is wrong with this answer for which it is downvoted? Oct 21, 2017 at 5:46
• Might be that it's extremely similar to the other answer given here.
– JKEG
May 2, 2019 at 11:23