Liouville’s Theorem

Write $h \coloneqq f/g$, so $h$ is meromorphic with ${\left\lvert {h} \right\rvert} \leq 1$. Moreover, $h$ can only have singularities where $g (z_k) = 0$, but is bounded by 1 in punctured neighborhoods about any such $z_k$. So any such singularities are removable, and $h$ extends over the singularities by Riemann’s removable singularity theorem to give an entire function. Now $h$ is bounded and entire, thus constant, so $c = h = f/g \implies f=cg$.

Use that $f$ is entire to Laurent expand at $z=0$ to get $f(z) = \sum_{k\geq 0}c_k z^k$ everywhere. Claim: $c_{n+k} = 0$ for all $k\geq n+1$ By the formula for Taylor coefficients, it suffices to show $f^{(n+k)}(0) = 0$ for all $k\geq n+1$. Apply the Cauchy estimate on a curve of radius $R\gg 1$: $$\begin{align*} {\left\lvert { f^{n+k} (0)} \right\rvert} &\leq {(n+k)! \over 2\pi} \int_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f(\xi) \over \xi^{n+k+1}} \right\rvert}\,d\xi\\ &\leq {(n+k)! \over 2\pi} \int_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {M \over \xi^n \xi^{k+1}} \right\rvert}\,d\xi\\ &= {(n+k)! \over 2\pi} \int_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {M \over R ^{k+1}} \right\rvert}\,d\xi\\ &= {(n+k)! \over 2\pi} {M\over R^{k+1}} \cdot 2\pi R \\ &= { \mathsf{O}}(1/R) \to 0 .\end{align*}$$

If this holds on all of ${\mathbf{C}}$, then $h(z) \coloneqq f(z)/z^n$ is constant and thus $f(z) = cz^n$.

The inequality implies $f$ has no zeros, so $g(z) \coloneqq 1/f(z)$ is entire. Moreover it is bounded on ${\mathbf{C}}$, since $$\begin{align*} {\left\lvert {g(z)} \right\rvert} \leq {1\over {\left\lvert {z} \right\rvert} + 1} \leq 1 ,\end{align*}$$ so $g\equiv c$ is constant by Liouville. This means $f\equiv c$ is constant, but $\lim_{z\to \infty}g(z) = 0$ forces $c=\infty$, so there are no such entire functions.

Apply a Cauchy estimate over a contour of radius $R> M$ to obtain $$\begin{align*} {\left\lvert {f^{(n)}(0)} \right\rvert} \leq n!{A+B R^k \over R^n} \ll 1/R^{n-k} ,\end{align*}$$ and if $n>k$ then this goes to zero in $R$ and $c_n = 0$ for all $n>k$.

Write $h(z) \coloneqq f(z)/g(z)$, then ${\left\lvert {f} \right\rvert}\leq 1$ is bounded. Provided the zeros of $g$ do not have a limit point, the singularities of $h$ are isolated and thus removable. By Riemann’s removable singularity theorem, $h$ extends to an entire function. By continuity, ${\left\lvert {h(z)} \right\rvert}\leq 1$ on ${\mathbf{C}}$ and is thus bounded. By Liouville $h$ is constant, making $f = cg$ for some $c$.

Write $g(z) \coloneqq f(z) / \sin(z)$, which is meromorphic with singularities at the zeros of $\sin(z)$ and bounded by 1. By boundedness, these singularities are removable, so $g$ extends to a bounded entire and thus constant function. So $f(z) = c\sin(z)$ where ${\left\lvert {c} \right\rvert} \leq 1$.

Write $g(z) \coloneqq f(z) - z_0$, so ${\left\lvert {g(z)} \right\rvert} \geq r$. Now ${\left\lvert {1/g(z)} \right\rvert} \leq 1/r$ for all $z$, so $1/g$ is bounded. Moreover it is entire since $f(z) \neq z_0$ for any $z$, and so $1/g = c$ is constant. Now unwind to get $f(z) = z_0 + {1\over c}$, which is also constant.

The inequality forces $f\neq 0$ anywhere, so $1/f$ is entire and bounded by 1. By Liouville, $1/f$ is constant, and thus so is $f$.

By induction on $k$, suppose that if $f$ is degree $k-1$ with ${\left\lvert {f} \right\rvert} \leq A + B{\left\lvert {z} \right\rvert}^{k-1}$ outside of a large enough disc. Let $f$ be degree $k$, and consider $$\begin{align*} g(z) \coloneqq \begin{cases} f'(0) & z=0 \\ {f(z) - f(0) \over z-0} & z=0. \end{cases} .\end{align*}$$ Then outside of a large enough disc, $$\begin{align*} {\left\lvert {g} \right\rvert} \sim {A+B{\left\lvert {z} \right\rvert}^k \over {\left\lvert {z} \right\rvert}} \leq_{\sim} B{\left\lvert {z} \right\rvert}^{k-1} ,\end{align*}$$ and moreover by the MMP $g$ is bounded inside such a disc. So $g$ is a polynomial of degree at most $k-1$ by hypothesis, making $f$ degree at most $k$.

Supposing not, then there is some ${\mathbb{D}}_R(w) \cap f({\mathbf{C}})$ empty. Then $g(z) \coloneqq{1\over f(z) - w}$ is bounded in this disc and reflects to an entire bounded function, thus constant. Then if $g$ is constant, $f$ is constant.