Suppose that $ f $ is entire and that for each $ z $, either $ |f(z)| \leq 1 $ or $ |f^\prime (z) |\leq 1 $. Prove that $ f $ is a linear polynomial.

My question is in the title. I'm a little lost in how to solve this problem. There is a hint associated with the problem that states the following:

Use a line integral to show that $ |f(z)| \leq A + |z| $ where A = $\max\{1, |f(0)|\}$

Thanks for the help.

We may as well suppose $|f(z)| > 1$

Take the line $[0,z]$ and let $d = \sup \{ w : w\in (0,z), |f(w)| \leq 1\}$ and $d = 0$ if this set is empty. Then $|f'(z)| \leq 1$ on $[d,z]$.

Then $$ |z| \geq |z - d| \geq \int_d^z |f'(w)| dw \geq \left|\int_d^z f'(w) dw \right| = |f(z) - f(d)| $$

Now $|f(d)| = \max \{ |f(0)|, 1 \}$. Because of the inequality $|a - b| \geq ||a| - |b|| \geq |a| - |b|$, we therefore have $|z| + |f(d)| \geq |f(z)|$

Excuse the notation, once I restrict to the line $[0,z]$, I borrow a lot of the language from real numbers over.