Suppose $f$ is entire such that $f(z+1)=-f(z)$ for all $z$ and $f(0)=0$ and $|f(z)|\leq e^{\pi Im(z)}$ prove $f(z)=c\sin (\pi z)$.
First it is easy to show that $f(n)=0$ for all natural numbers and we need to use Liouville’s $$\frac{|f(z)|}{e^{\pi |Im(z)|}}\leq 1$$
But the denominator is not entire so I want to find a function which entire and have a bigger modulus value I think it should be $\sin(\pi z)$. How to prove that $e^{\pi |Im(z)|}\leq |\sin(\pi z)|$.
Thanks
On
Posting this answer here, just to demonstrate how to solve a slightly harder problem:
An entire function $f$ satisfies the following conditions:
- $f(0)=0$
- $f(z+1)=-f(z)$
- $|f(z)|\le \exp(\pi\cdot|\Im(z)|)$
Find $f$.
The following theorems/identities will be used:
$$\frac{\Gamma(T+a)}{\Gamma(T)}\sim T^a \qquad(T\to\infty)$$ $$\Gamma(1-a)\Gamma(a)=\frac\pi{\sin \pi a}$$ and Hadamard factorization theorem (HFT):
If $f$ is an entire function of finite order $\rho$, and $\{c_n\}$ is the sequence of non-zero zeroes of $f$ repeated according to multiplicity, then $$f(z)=z^m e^{q(z)}\prod_{n=0}^\infty E_p\left(\frac z{c_n}\right)$$ where $q$ is a polynomial of degree less than or equal to $\rho$, and $p=[\rho]$.
$$\rho=\limsup_{r\to\infty}\frac{\ln\ln \lVert f(z)\rVert_{\infty, |z|<r}}{\ln r} \le\lim_{r\to\infty}\frac{\ln\ln e^{\pi r}}{\ln r}=1$$
Thus, $0\le\rho\le1$.
Let the sequence $\{a_n\}$ be the non-zero zeroes of $f$ on $\mathbb C$ repeated according to multiplicity.
Since $f(\mathbb Z)=0$, there is at least one sub-sequence $\{\cdots,-2,-1,1,2,\cdots\}$ in $\{a_n\}$. Let $\{b_n\}$ be the sequence $\{a_n\}-\{\mathbb Z_{\ne0}\}$ (removing one subsequence of non-zero integers).
By HFT, $$f(z)=z^m e^{q(z)}\prod_{n=0}^\infty E_p\left(\frac z{b_n}\right)\cdot\lim_{N\to\infty}\prod^N_{n=-N,n\ne 0}E_p\left(\frac zn\right)$$ where $m\ge 1$ (due to $f(0)=0$).
No matter $p=[\rho]=0$ or $1$, it can be shown that $$f(z)=g_p(z)\lim_{N\to\infty}\prod^N_{n=-N,n\ne 0}\left(1-\frac zn\right)$$ where $g_p$ is entire and has a zero at $z=0$.
$$\begin{align} \lim_{N\to\infty}\prod^N_{n=-N,n\ne 0}\left(1-\frac zn\right) &=\lim_{N\to\infty}\prod^N_{n=-N,n\ne 0}\frac{n-z}{n} \\ &=\lim_{N\to\infty}\frac1{-z}\frac{\Gamma(N-z+1)}{\Gamma(-N-z)}\frac{(-1)^N}{\Gamma^2(N+1)} \\ \end{align} $$
After applying Gamma reflection formula, we obtain $$-\frac1z\lim_{N\to\infty}-\frac{\sin\pi z}\pi\frac{\Gamma(N-z+1)}{\Gamma(N+1)}\frac{\Gamma(N+z+1)}{\Gamma(N+1)} $$
which is asymptotic to $$\sim\frac{\sin\pi z}{\pi z}(N+1)^{-z}(N+1)^z\to \frac{\sin\pi z}{\pi z}$$
Therefore, $$f(z)=g_p(z)\frac{\sin\pi z}{\pi z}:= h(z)\sin\pi z$$
Note that the singularity of $\frac{g_p(z)}{\pi z}$ at $0$ is removable and thus $h(z)$ is entire.
Using condition 3 again, we obtain $$|h(z)|\le\left\vert\frac{\exp(\pi|\Im(z)|)}{\sin\pi z}\right\vert$$
As shown in @J_P's answer, RHS is bounded for sufficiently large $\Im(z)$. Call this region $Z$. Moreover, @J_P showed that $h$ is bounded on $\mathbb C\setminus Z$ due to continuity and periodicity of $f$.
Hence, $h(z)$ is entire and bounded on $\mathbb C$. By Liouville theorem, $h(z)\equiv c$.
As a result, $$f(z)=c\sin\pi z$$
You can't prove $e^{\pi|\Im(z)|}\leq|\sin(\pi z)|$ because it isn't true, pick $z=0$ for example. But this approach can still work, so long as we choose our domains carefully.
Here's one way to do it: let $S$ be the vertical strip defined by $\Re(z)\in[-1/2,1/2]$. Define: $$ g(z)=\frac{f(z)}{\sin(\pi z)} $$ If we can show $g$ is bounded on $S$, then by the functional equation $f(z+1)=-f(z)$, we will have that $g$ is bounded on every vertical strip $\Re(z)\in[-1/2+k,1/2+k]$ by the same constant, so it will be bounded on all of $\mathbb{C}$ and thus constant.
Let's look at $\sin(x+iy)$: $$ \sin(x+iy)=\sin(x)\cos(iy)+\sin(iy)\cos(x)=\sin(x)\cosh(y)+i\sinh(y)\cos(x) $$ Thus for $y$ large enough and some positive $C$: $$ |\sin(x+iy)|^2 =(\sin(x)\cosh(y))^2+(\sinh(y)\cos(x))^2=\\ =\sin^2(x)(\cosh^2(y)-\sinh^2(y))+\sinh^2(y)=\sin^2(x)+\sinh^2(y)>Ce^{-2|y|} $$ So we get that for $\Im(z)$ large enough, $$ |\sin(\pi z)|>C'e^{\pi|\Im(z)|} $$ This means that $g$ is bounded on $S$ if we're far enough away from the origin: $$ \left\vert\frac{f(z)}{\sin(\pi z)}\right\vert<\left\vert\frac{f(z)}{C'e^{\pi\vert\Im(z)\vert}}\right\vert\leq\frac{1}{C'} $$ Call the set where this holds $Z\subset S$. But the closure of $S-Z$ is compact and $g$ is continuous (it has a removable singularity at $0$), so it is also bounded there. Hence $g$ is bounded on $S$ and so $g$ is constant.