If $f(z)$ is entire function such that $|f(z)f'(z)| \leq 1$ then $f(z)$ is constant

Question

If $f(z)$ is entire function such that $|f(z)f'(z)| \leq 1$ then $f(z)$ is constant.

Choose $g(z) = \frac {(f(z))^2}{2}$, $g'(z) = f(z)f'(z) \implies |g'(z)| \leq 1$(given condition)

So $g'(z)$ is analytic and bounded. By lioville theorem we have $g'(z) = A$ where A is a constant

$f(z)f'(z) = A$

$f(z) df(z) = A dz$

Integrating we get

$f(z) = \sqrt{Az+b}$ where A,b are constants

But the question asks $f(z)$ to be proved as constant. But i got different. Pls have a look at where i made mistake. It looks correct to me

Answer 1

You proved correctly that $g(z)=Az+B$ for some numbers $A$ and $B$. In other words,$$f^2(z)=2Az+2B.$$Therefore, $\lim_{z\to\infty}\left|\frac{f^2(z)}{z^2}\right|=0$. But$$\lim_{z\to\infty}\left|\frac{f^2(z)}{z^2}\right|=0\iff\lim_{z\to\infty}\left|\frac{f(z)}z\right|^2=0\iff\lim_{z\to\infty}\left|\frac{f(z)}z\right|=0.$$So, $f$ is a polynomial function and its degree is $1$, at most. But, since its square doesn't have degree $2$, $f$ is, in fact, a constant polynomial.

Answer 2

If $2g(z)=f^2(z)$, then every zero of $g$ is a zero of $f$.

Since $g'(z)=f(z)f'(z)$, every zero of $f$ is a zero of $g'$.

Therefore, every zero of $g$ is a double zero, at least.

Now $g(z)=Az+B$ has a simple zero if $A\ne0$.

Therefore, $A=0$ and $f$ is constant.