Show that f is constant.

Question

Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an analytic function and let $a \in \mathbb{R}$. Show that if $\Re f(z) \geq a$ for all $z \in \mathbb{C}$, then $f$ is constant.

Now I know that I need to use the Liouville's theorem, to get the partials in respect of $(x,y)$: $u_x=v_y$ and $u_y=-v_x$.

The problem I'm stuck with is what $f$ should be. Will I use $f(z)= u(x,y)+iv(x,y)$?

Answer 1

Let $g(z)=e^{-f(z)}$. Then for each $z\in\mathbb{C}$ we have $|g(z)|=e^{-Re(f(z))}\leq e^{-a}$. So $g$ is entire and bounded, hence constant. From here we also get that $h(z)=e^{f(z)}$ is constant. This means there is a constant $c\in\mathbb{C}$ such that $e^{f(z)}=e^c$ for all $z$. It means for each $z\in\mathbb{C}$ we have $f(z)\in\{c+2\pi in:n\in\mathbb{Z}\}$. Since this set has no accumulation points and $f$ is holomorphic we conclude $f$ is constant.

Answer 2

Hint: For every $z\in \Bbb C$, we have$$ |e^{-f(z)}|=e^{-\Re(f(z))}\le e^{-a}. $$ If $e^{-f(z)}$ is constant, then $\frac{d}{dz}\left(e^{-f(z)}\right) =\cdots $.

$\frac{d}{dz}\left(e^{-f(z)}\right)=-e^{-f(z)}f'(z)=0$, i.e. $f'(z)=0$ for every $z\in \Bbb C$. Hence $f$ must be constant.

Answer 3

It is an immediate consequence of Liouville's theorem that if $f\colon\mathbb{C}\longrightarrow\mathbb{C}$ is analytic and non-constant, then $f(\mathbb{C})$ is dense. Therefore, unless $f$ is constant, you cannot have$$(\forall z\in\mathbb{C}):\operatorname{Re}\bigl(f(z)\bigr)\geqslant a$$for some real number $a$.