Sachchidanand Prasad

07-02-2023

Problem: Let $f:\mathbb{C} \to \mathbb{C}$ be an entire function. Let us write \[ f(z) = u(x,y) + \iota v(x,y),~z=x+\iota y. \]

  1. Show that if $u$ is positive valued for all $x,y\in \mathbb{R} $, that is, $u(x,y)>0$, then $f$ is constant.
  2. Show that $\left\vert u(x,y) \right\vert \lt \left\vert v(x,y) \right\vert $ for all $x,y\in \mathbb{R} $, then $f$ is constant.
  3. Can we still draw the same conclusion if less than is replaced by greater than in both the parts?

Solution: We will use the Liouville's theorem. Let us recall the same.

Let $f:\mathbb{C} \to \mathbb{C}$ be a entire function. Suppose that $f$ is bounded, that is, there exists a real number $M\gt 0$ such that $|f(z)|\lt M$ for all $z$. Then $f$ is a constant function.

1. We take a function $g(z) = e^z$. Note that \[ \left\vert e^{-f(z)} \right\vert = \left\vert e^{-u} e^{-\iota v} \right\vert = \left\vert e^{-u} \right\vert \lt e^0 = 1. \] Thus, $e^{-f(z)}$ is a constant function. Since $f(z)$ is an entire function and composition of two entire functions is an entire function, so $e^{-f(z)}$ is an entire function. Thus using the Liouville's theorem, we conclude that $e^{-f(z)}$ is constant and hence, $f(z)$ is constant.

2. Choose the function $g(z) = e^{z^2}$. We note that \begin{align*} \left\vert e^{f(z)^2} \right\vert & = \left\vert e^{u^{2+2\iota u v - v^2} } \right\vert \\ & = \left\vert e^{u^2 - v^2}e^{2\iota uv} \right\vert \\ & = \left\vert e^{u^2 -v^2} \right\vert \\ & \lt e^0 \quad \boxed{u(x,y)^2 \lt v(x,y)^2}\\ & = 1. \end{align*} Thus, $e^{f(z)^2}$ is a constant function. Since $f(z)$ is an entire function and composition of two entire functions is an entire function, so $e^{f(z)^2}$ is an entire function. Thus using the Liouville's theorem, we conclude that $e^{f(z)^2}$ is constant which implies $f(z)^2$ is constant and hence, $f(z)$ is a constant function.

2. Choose the function $g(z) = e^{z^2}$. We note that \begin{align*} \left\vert e^{f(z)^2} \right\vert & = \left\vert e^{u^{2+2\iota u v - v^2} } \right\vert \\ & = \left\vert e^{u^2 - v^2}e^{2\iota uv} \right\vert \\ & = \left\vert e^{u^2 -v^2} \right\vert \\ & \lt e^0 \quad \boxed{u(x,y)^2 \lt v(x,y)^2}\\ & = 1. \end{align*} Thus, $e^{f(z)^2}$ is a constant function. Since $f(z)$ is an entire function, so is $f(z)^2$ and composition of two entire functions is an entire function, therefore, $e^{f(z)^2}$ is an entire function. Thus using the Liouville's theorem, we conclude that $e^{f(z)^2}$ is constant which implies $f(z)^2$ is constant and hence, $f(z)$ is a constant function.

3. If we replace by greater than sign then also both conclusion will remain same. For the first, we can take $g(z)=e^{f(z)}$ and for the other one, we take $e^{-f(z)^2}$.