Liouville generalisation

Question

Suppose $f: \mathbb{C} \rightarrow \mathbb{C}$ is holomorphic on $\mathbb{C}$ and $$\lim_{z \rightarrow \infty} \frac{|f(z)|}{|z|^2} = 0$$ Prove that $f'' = 0 $.

I tried to argue that $f(z) \leq \epsilon |z|^2$ for some $0< |z| < \delta$ and then use a similar proof here (section beginning: if $f$ is less than or equal to a scalar...).

But apparently this doesn't work, because our bound is not uniform and depends on $z$.

However I am not sure how to proceed now, so any hints would be good.

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Update: I have also tried to consider the integral formula for $$f''(z) = \frac{1}{\pi i} \int_{|w-z| = R} \frac{f(w)}{(w-z)^3} dw$$

and then if I take absolute values, I don't seem to get a $|z|^2$ inside where I can use the result. The denominator gives an $R^3$, which doesn't seem too helpful either..

Answer 1

Hint: Since we are given that $f$ is entire, use the Maclaurin series of $f$ to write $f(z) = a+bz + z^2 g(z)$ for some entire $g$. Then Liouville's theorem.

Answer 2

Two ways:

1. $f(z)-f(0)-f'(0)z$ is an analytic function with a double zero at $z=0$, so $g(z)=(f(z)-f(0)-f'(0)z)/z^2$ is analytic. It is bounded on $z<R$ as it is analytic, and bounded on $z>R$ since $f(z)/z^2$ is (the other two terms obviously also $\to 0$). Use Liouville to conclude.

2. The Cauchy Integral formula gives us the bound $$ \lvert f''(w) \rvert = \frac{1}{2\pi} \left\lvert \int_{\lvert w-z \rvert = r} \frac{f(z)}{(z-w)^3} \, dz \right\rvert \leq \sup_{\lvert z-w \rvert = r} \frac{ \lvert f(z) \rvert}{\lvert z-w \rvert^2}. $$ But for $\lvert z \rvert > 2\lvert w \rvert$, $$ \lvert z-w \rvert^2 \geq \lvert z \rvert^2 - \lvert w \rvert^2 = \lvert z \rvert^2 \left(1 - \frac{\lvert w \rvert^2}{\lvert z \rvert^2}\right) \geq \frac{3}{4}\lvert z \rvert^2, $$ so $$ \lvert f''(w) \rvert \leq \sup_{\lvert z-w \rvert = r} \frac{ \lvert f(z) \rvert}{\lvert z-w \rvert^2} \leq \frac{4}{3} \sup_{\lvert z-w \rvert = r} \frac{ \lvert f(z) \rvert}{\lvert z \rvert^2} \to 0. $$ as $r \to \infty$.