Liouville’s Theorem

If $$f$$ is entire and bounded, $$f$$ is constant.

• Since $$f$$ is bounded, $$f(z) \leq M$$ uniformly on $${\mathbf{C}}$$.
• Apply Cauchy’s estimate for the 1st derivative: \begin{align*} {\left\lvert {f'(z)} \right\rvert} \leq { 1! {\left\lVert {f} \right\rVert}_{C_R} \over R } \leq {M \over R}\overset{R\to\infty}\longrightarrow 0 ,\end{align*} so $$f'(z) = 0$$ for all $$z$$.

Suppose $$f$$ is entire and bounded. Under an affine change of variables in the domain and range, $$f(0) = 0$$ and $${\left\lvert {f(z)} \right\rvert} \leq 1$$, the claim is that $$f\equiv 0$$. The function $$g(z) \coloneqq f(Rz)$$ satisfies the Schwarz lemma, so $${\left\lvert {f(Rz)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {f(w)} \right\rvert} \leq {\left\lvert {w} \right\rvert}/R\overset{R\to\infty}\longrightarrow 0$$.

Exercises

Show that an entire doubly periodic function is constant.

Show that if $$f, g$$ are entire with $${\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {g(z)} \right\rvert}$$, then $$f(z) = cg(z)$$ for some constant $$c$$.

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 $$c = h = f/g \implies f=cg$$.

Show that if $${\left\lvert {f(z)/z^n} \right\rvert}$$ is bounded for $${\left\lvert {z} \right\rvert}\geq R$$, then $$f$$ is a polynomial of degree at most $$n$$. What happens if this bound holds on all of $${\mathbf{C}}$$?

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 $$f(z) = cz^n$$.

Find all entire functions $$f$$ satisfying \begin{align*} {\left\lvert {f(z)} \right\rvert} \geq {\left\lvert {z} \right\rvert} + 1 &&\forall z\in {\mathbf{C}} .\end{align*}

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.

Let $$f$$ be entire and suppose that for $${\left\lvert {z} \right\rvert} \geq M$$, \begin{align*} {\left\lvert {f} \right\rvert} \leq A + B{\left\lvert {z} \right\rvert}^k \end{align*} for some constants $$A, B$$ and $$k$$. Show that $$f$$ is a polynomial of degree at most $$k$$.

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 $$n>k$$.

Suppose $${\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {g(z)} \right\rvert}$$ for all $$z$$. What conclusion can you draw?

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 $$c$$.

Suppose $$f$$ is entire and for every $$z$$, \begin{align*} {\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {\sin(z)} \right\rvert} .\end{align*} Characterize all possibilities for $$f$$.

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 $${\left\lvert {c} \right\rvert} \leq 1$$.

Let $$f$$ be entire with $$\operatorname{im}(f) \cap{\mathbb{D}}_r(z_0)$$ empty. Show $$f$$ must be constant without using the Casorati-Weierstrass or Picard theorems.

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.

Show that if $$f$$ is entire and $${\left\lvert {f(z)} \right\rvert} > 1$$ for all $$z$$, then $$f$$ is 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 $$f$$.

Let $$f$$ be an entire function. Assume that for some $$k \in \mathbb{N}$$, and sufficiently large $$|z|$$, we have that $$|f(z)| \leq A+B|z|^{k}$$. Prove that $$f$$ is a polynomal of degree at most $$k$$.

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 $$k$$.

Show that if $$f:{\mathbf{C}}\to {\mathbf{C}}$$ is nonconstant and entire then $$f({\mathbf{C}})$$ is dense in $${\mathbf{C}}$$.

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.

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