If $f$ is entire and $|f(z)| \le A + B |z| ^{1/7}$, then $f$ is constant

Question

Let $f$ be an entire function. Suppose there exist constants $A$, $B$ such that $|f(z)| \le A + B|z|^{1/7}$ for every $z \in \Bbb C$. Prove that $f$ is constant.

Answer 1

If $|z| > R > 0$, then the inequality given by the problem implies that

$$ \tag{*} \left| \frac {f(z)} z \right| \le \frac A {|z|} + \frac B {|z| ^{\frac 6 7}} ,$$

so

$$\lim _{z \to \infty} \frac {f(z)} z = 0 .$$

If $f(z) = \sum \limits _{n \ge 0} a_n z^n$ and $g(z) = \dfrac {f(z) - a_0} z$, the above implies that

$$\lim _{z \to \infty} | g(z) | = \lim _{z \to \infty} \left| \frac {f(z) - a_0} z \right| \le \lim _{z \to \infty} \left| \frac {f(z)} z \right| + \lim _{z \to \infty} \left| \frac {a_0} z \right| = 0 ,$$

which means that for sufficiently large $R>0$, $g$ is bounded on $|z| > R$.

On the other hand, the set $|z| \le R$ is compact and since $g$ is continuous (because it is holomorphic), by Weierstrass's extreme value theorem $g$ must be bounded on $|z| \le R$.

We conclude that $g$ is bounded on the whole $\Bbb C$, and since $g$ is entire, Liouville's theorem implies that $g$ must be constant - so it must be equal to $a_1$.

This means that $f(z) = a_0 + a_1 z$, so plugging this back into (*) we get that

$$\left| \frac {a_0} z + a_1 \right| \le \frac A {|z|} + \frac B {|z| ^{\frac 6 7}} ,$$

whence, letting $z \to \infty$, we get $a_1 = 0$, so that $f = a_0$, a constant.

Answer 2

Hint: prove that the function $$ g(z)=\frac{f(z)-f(0)}{z} $$ is entire and apply Louiville's theorem to it.