prove no zeroes for entire function

Question

Suppose $f$ is entire and $|f(z)f(-z)|\le1$ for all $z\in C$. Prove that $f$ has no zeroes.

What I did is: since $f$ is entire, so $g(z)=f(z)f(-z)$ is also entire. And we have $|g(z)|\le1$ which is bounded, so from Liouville's THM, we say that $g(z)=constant$. So $f(z)f(-z)=c$, so if $c\ne0$, there is no $z$ s.t. $z$ is zeros of $f$. But what if $c=0?$ I'm puzzled here.

Answer 1

You are right, the constant function $f(z)=0$ is a counterexample for this problem.

Now, if $f$ is not the trivial function, it is easy to show that $f(z)$ entire + $f(z)f(-z)=0$ implies that $f(z)=0$ for all $z$. (hint Identity theorem).

Answer 2

Hint: I guess one assumes $f$ is not identically zero to begin with. Consider a convergent sequence $z_n \to z_0$ (with distinct points). Then, if $c=0$, we conclude that either $f(z_n) = 0$ or $f(-z_n) = 0$ for infinitely many indices $n$. Why does that show that $f \equiv 0$?

Answer 3

Suppose $f\neq 0.$ Then $g(z)=f(z)f(-z)$ is also an entire function. By the Liouville's theorem $g$ must be a constant function. Therefore if $f(z)$ has a zero, then $f(-z)$ has a pole which contradicts the entireness of $f.$