bound on entire function If f is an entire function such that for each fixed $z$ either $|f(z)|\leq 1 $ or $|f'(z)|\leq 1$ then prove that $f$ is a linear polynomial.
I guess something have to do with integration, but not getting what to do.
 A: For any $z \in \mathbb{C} \setminus \{0\}$, consider the set $T$ and the number $\alpha$
defined by:
$$T = \big\{ t \in [0,1] : |f(tz)| \le 1 \big\}
\quad\text{ and }\quad
\alpha = \begin{cases} 
0, & T = \emptyset\\
\sup_{t \in T} t, & \text{otherwise}\end{cases}$$
It is clear $|f'(tz)| \le 1$ whenever $t \ge \alpha$. From this, we see
$$|f(z) - f(\alpha z)| = \left| \int_{\alpha}^1 f'(tz) z dt \right|
\le \int_{\alpha}^1 | f'(tz) z | dt \le |z|(1-\alpha) \le |z|$$
If $\alpha \ne 0$, then $|f(\alpha z)| \le 1$. This implies for all $z$, we have
$$|f(z)| \le |f(\alpha z)| + |f(z)- f(\alpha z)| \le \max(|f(0)|, 1) + |z|$$
For any $\zeta \in \mathbb{C}$ and $R \in \mathbb{R}$ such that $R > |\zeta|$.
By Cauchy integral formula for $f''(z)$, we have
$$f''(\zeta) = \frac{2!}{2\pi i}\int_{|z| = R} \frac{f(z)}{(z-\zeta)^3} dz\\
\implies
|f''(\zeta)| 
\le  \frac{1}{\pi}\int_{|z| = R} \frac{|f(z)|}{|z-\zeta|^3} |dz|
\le   \frac{2R(\max(|f(0)|,1) + |R|)}{(R-|\zeta|)^3}
$$
Sending $R \to \infty$, we find $f''(\zeta) = 0$ for any $\zeta$. As a result, $f(z)$ is a linear polynomial.
A: Write your function as a power series.
In the first case, you get that $f$ is bounded. Use Liouville's theorem, what do you get?
If now, the derivative is bounded, what can you say about the derivative of an entire function? Is the derivative also holomorphic? Use Liouville's theorem again. 
A: $f$ is entire, so $f$ is analytic over the entire $\mathbb{C}$. This means that
$$f(z)=\sum^{\infty} a_n(z-z_0)^n$$
Now for starters, if $|f(z)|\leq 1$ over its entire domain, then it is bounded, and so by Liouville's Theorem it is just a constant (i.e. $f(z)=C$, and $C$ is just a complex number).
If, however, $|f'(z)|\leq 1$, then that means the derivative of that power-series is bounded. That means that must be a constant, and so the $a_1$ term (the $(z-z_0)$ coefficient) would be the only term that lives after saying that $f'(z)$ is constant. 
