# Prove without Liouville's theorem: $f$ is entire, $\forall z \in \mathbb C: |f(z)| \leq |z|$, then $f=a \cdot z$, $a \in \mathbb C, |a| \leq 1$

Prove without Liouville's theorem: $f$ is entire, $\forall z \in \mathbb C: |f(z)| \leq |z|$, then $f=a \cdot z$, $a \in \mathbb C, |a| \leq 1$

What I tried so far: $f$ is entire, so $f(z)= \Sigma _{n=0}^\infty a_nz^n$, and then $|\Sigma _{n=0}^\infty a_nz^n| \leq |z| \Rightarrow |\frac {\Sigma _{n=0}^\infty a_nz^n}{z}| \leq 1 \Rightarrow |\Sigma _{n=0}^\infty a_nz^{n-1}| \leq 1$

How can I continue from here?

• Isn't the statement more or less equivalent to Liouville? Assume the problem statement is true, then here's a proof of Liouville: Let $f$ be entire and bounded, say $|f(z)|<M$ for all $z$. Then apply the problem statement to $\frac zMf(z)$ to conclude $\frac zM f(z)=az$ for some $a$. But then $f(z)=Ma$ is constant. Commented May 30, 2015 at 15:27
• Can you use Cauchy's integral formula? Commented May 30, 2015 at 15:30
• Yes I can. So I just get with integral evaluation that $|f'(z_0)| \leq \frac {|z|}{R}$, R being the radius of a circle around some $z_0$, and then $R \rightarrow \infty$, which means $f'(z_0)=0$, so it is $f$ is constant? Or I can't really say that...? Where does the $z$ come? Or should I do this trick for $g(z)= \frac {f(z)}{z}$ Commented May 30, 2015 at 15:34

Note that $|f(z)| \leq |z|$ implies $|f(0)| \leq |0|$ and thus $f(0) = 0$. If $f(z) = \sum \limits _{n=0} ^\infty a_n z^n$, this means that $a_0 = 0$, so $f(z) = z g(z)$, where $g(z) = \sum \limits _{n=1} ^\infty a_n z^{n-1}$. By hypothesis, you get $|g(z)| \leq 1$. Now, you don't want to use Liouville's theorem, so we turn to Cauchy's inequality: for any $r>0$ and $n>1$ we have $|a_{n+1}| = |g ^{(n)} (0)| \leq {\sup \limits _{|z| = r} |g(z)| \over r^n} = {1 \over r^n}$. Since $r>0$ may grow arbitrarily large, this shows that $a_{n+1}$ may get arbitrarily small when $n>1$, and thus $a_{n+1} = 0$ for $n>1$. So, $f(z) = a_0 + a_1 z$. But we have proved right at the beginning that $a_0 = 0$, so $f(z) = a_1 z$.

• Actually the Liouvilles' proof :) Commented May 30, 2015 at 16:07

Consider $g(z) = z f(\frac{1}{z})$. Then $g$ is bounded hence by Riemann's extension theorem $g$ is holomorphic even at $0$. Then changing $z$ to $\frac{1}{z}$ we get $z g(\frac{1}{z}) = f(z)$. Now taking the Taylor's series at zero we get that $f(z) = az + b$. But $b$ must be zero as follows from the original inequality $|f(z)| \leq |z|$.

We have $f(0) = 0,$ so we can write $f(z)= \sum_{n=1}^{\infty}a_nz^n.$ From the given we have

$$\int_0^{2\pi} |f(re^{it})|^2\,dt \le 2\pi r^2.$$

On the other hand, by orthogonality of the exponentials,

$$\int_0^{2\pi} |f(re^{it})|^2\,dt = \sum_{n=0}^{\infty}2\pi|a_n|^2r^{2n}.$$

This sum grows faster than $r^2$ unless $a_n=0, n>1.$ Thus $f(z) = a_1z$ and we're done.

Set $g(z)=\frac {f(z)}{z}$ for $z \neq 0$ and $g(z)=A, A \in \mathbb C$ for $z=0$. So we get that $g(z)$ is entire. Let there be some $z_0 \in \mathbb C$, and circle with radius $R$ centered in $z_0$, $C_R$. So by Cauchy Integral Formula:
$g'(z_0)= \frac {1}{2 \pi i} \int_{C_R} \frac{g(z)}{(z-z_0)^2}$
$$|g'(z_0)|= |\frac {1}{2 \pi i} \int_{C_R} \frac{g(z)}{(z-z_0)^2}| = |\frac {1}{2 \pi i} \int_{C_R} \frac{ \frac {f(z)}{z}}{(z-z_0)^2}|= |\frac {1}{2 \pi i} \int_{C_R} \frac{f(z)}{z(z-z_0)^2}| \leq \frac {1}{2 \pi } \cdot 2 \pi R \cdot \frac {1}{R^2} \cdot max_{z \in C_R}|\frac {f(z)}{z}|= \frac {1}{R}\cdot 1= \frac {1}{R}$$
So we'll take circle $C_R$ with infinite radius, and we get that $g'(z_0)=0$ for every $z_0$, so $g(z)$ is constant. So $g(z)=a$ for some $a \in \mathbb C$, so $f(z)=a \cdot z$, and $|a| \leq 1$.