Liouville theorem

Every bounded entire function is constant.

Let $f,g$ be entire functions, with $g$ not identically zero. If $|f(z)|\le C|g(z)|$ everywhere, then $f/g$ is bounded and holomorphic away from the isolated zeros of $g$. The removable singularity extends it across each zero, giving a bounded entire function. Liouville's theorem makes it constant, so $f=cg$ with $|c|\le C$. If $g$ is identically zero the inequality forces $f=0$. For example, domination by $|\cos z|$ forces a constant multiple of $\cos z$.

Suppose $D\subseteq \mathbb{C}$ is open and admits an injective holomorphic map $\Phi :D\to \mathbb{C}$ whose image is bounded. Every entire function $f:\mathbb{C}\to D$ is constant: the function composition $\Phi \circ f$ is a bounded entire function, hence constant by Liouville theorem, and injectivity of $\Phi$ then forces $f$ to be constant. Surjectivity of $\Phi$ onto a disk is unnecessary. For example $z\mapsto z/(1-z)$ is a conformal map from $\mathbb{C}\setminus [0,1]$ onto $(\mathbb{C}\setminus [0,\infty ))\setminus \{-1\}$. A square-root branch with argument in $(0,2\pi )$ followed by the Cayley transform between the half-plane and disk maps this to the punctured unit disk. It therefore supplies a bounded injective holomorphic coordinate on the interval complement, giving the same rigidity even though that domain is not simply connected.

If an entire function $g$ satisfies $\mathrm{Re}(cg)\ge b$ for constants $c\ne 0$ and real $b$, then $e^{-cg}$ is entire and has modulus at most $e^{-b}$. The Liouville theorem makes it constant, and differentiating gives $g^{\prime }=0$ since the exponential never vanishes. The conclusion also applies after rotating and translating any containing half-plane; separate bounds on the real and imaginary parts are unnecessary.

If an entire function $f=u+iv$ has $uv$ bounded above, then $f$ is constant. Indeed $F=\exp (-i{f}^2)$ is entire and $|F|=\exp (2uv)$ is bounded, so the Liouville theorem makes $F$ constant. Differentiating gives $f{f}^{\prime }=0$. The derivative of $f^2$ is therefore zero, and continuity on the connected complex plane makes $f$ itself constant. No separate boundedness of $u$ or $v$ is required.

The image of every nonconstant entire function is dense in $\mathbb{C}$. If a disc about $w$ were omitted, then $1/(f-w)$ would be bounded and entire, so Liouville's theorem would make $f$ constant.

If an entire function satisfies $|h(z)|\le |\mathrm{Re}z{|}^{-1/2}$ away from the imaginary axis, then $h=0$. Cauchy's formula on $|z|=R$ bounds every Taylor coefficient by a constant timeswhich tends to zero as $R\to \infty$.

An entire function bounded by a polynomial in the radius is itself a polynomial, by Cauchy estimates.