Suppose $f$ is entire and there exist constants $a, b$ such that $|f(z)| \leq a|z| + b$ for all $z \in \mathbb{C}$. Prove that $f$ is a linear polynomial (that is, of degree less than or equal to $1$).
Here's what I tried.
Proof. Since $f(z)$ is entire, it assumes the form of its Taylor expansion about zero for all $z$ in $\mathbb{C}$. That is,
$$f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$$
for all $z$. Let $R > 0$ be given. The Cauchy Estimate gives
$$|f^{(n)}(0)| \le \frac{n!}{R^n}(a|z|+b)$$
for $|z| = R$. So letting $R\to\infty$, $f^{(n)}(0) = 0$ for $n\ge 2$. So f is a linear polynomial.
Best Answer
Or think about estimates on the Taylor coefficients you get from the Cauchy Integral Formula.