Show that if a function is bounded by a another specified function, then it is identically zero

Question

The question I am trying to answer is as follows:

Show that if $f$ is an entire function satisfying $|f(z)| \leq C \sqrt{|z|}|\cos(z)|$ for some constant $C$, then $f$ is identically zero. (Hint: compare $f$ to the function $z\cos(z)$.)

My intuition would be to make use of the following Corollary from Louiville’s Theorem:

Corollary: Let $f$ and $g$ be two entire functions. Suppose $|f(z)| \leq M|g(z)|$, for all $z \in \mathbb{C}$, where $M$ is a fixed positive constant. Then $f=cg$

$\bigg($Proof: Clearly if $g(z) = 0$, then $f(z) = 0$ as well. Assume that $z \in \mathbb{C}$ is such that $g(z) \neq 0$. Then we have:

$$\frac{|f(z)|}{|g(z)|}=\bigg|\frac{f(z)}{g(z)}\bigg|\leq M$$

and hence by Louiville’s Theorem, $\frac{f(z)}{g(z)}$ is a constant (since $\frac{f}{g}$ is entire). Suppose $\frac{f}{g} = c \leq M$. Then $f(z) = cg(z)$. $\bigg)$

However, the problems I'm having here is being able to apply the above to my question are as follow:

1.) If we define $|g(z)|=\sqrt{|z|}|\cos(z)|$, then $|g(z)|=0 \ \forall z =\pi(k+\frac{1}{2}), k \in \mathbb{Z}$. This means that $g$ is neither $0$ or not $0$ everywhere. How would one account for this?

2.) Even if we have accounted for the above, we will have shown $f$ is constant, not necessarily identically zero?

Answer 1

Hint: Copy the proof of the Liouville's theorem replacing the constant with $\sqrt{|z|}|\cos z|$.

Answer: Theorem 10.22 in Rudin‘s Real and Complex Analysis states that

If $$f(z)=\sum_{n=0}^{\infty}c_n(z-a)^n \quad (z\in D(a;R))$$ and if $0

Back to your question, suppose $f(z)=\sum_{n=0}^{\infty}c_nz^n$, we have $$\sum_{n=0}^{\infty}|c_n|^2r^{2n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(a+re^{i\theta})|^2d\theta \leq\frac{1}{2\pi}\int_{-\pi}^{\pi}(\sqrt{|re^{i\theta}|})^2d\theta \leq\frac{1}{2\pi}\int_{-\pi}^{\pi}rd\theta=r, \forall r>0.$$ Hence $c_n=0 \forall n>0$, i.e. $f$ is constant. But $|f(0)|\leq 0$, so $f$ is $0$.