# Show $f$ is a complex polynomial of degree at most 2

Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function and

$$\displaystyle\min\{\left|f'(z)\right|,\left|f(z)\right|\}\leq \left|z\right|+2$$ for all $z\in\mathbb{C}$.

How to see that $f$ is a polynomial in $z$ of degree at most 2?

I can only see it when $\left|f(z)\right|\leq \left|z\right|+2$ by using Cauchy estimate. How to handle $f'(z)$ part together?

• Divide by $|z|$ and use Liouville's theorem?! Commented Mar 31, 2014 at 21:38

By adding a constant, you can assume that $f(0) = 0$ . For an arbitrary point $z \in \mathbb{C}$ consider the minimal part of the line segment $L_z$ from $z$ to $0$ which connects $z$ to a point $z_0$ with $|f(z_0)| \le |z_0| + 2$. (If this is already satisfied for $z_0 = z$, the line segment will just be the point $z = z_0$.) For any $w \in L_z$ by assumption $|f'(w)| \le |w| + 2$. Integrating $f'$ along $L_z$ and using the fundamental theorem for line integrals, you get by standard estimates that $|f(z)|$ grows at most quadratically in $|z|$. Then proceed using Cauchy's estimates.