Prove that a function is constant

[Tsakanikas Nickos]

If \( \displaystyle f \) is entire and such that \( \displaystyle \lim_{z \to \infty} \frac{f(z)}{z} = 0 \), then show that \( \displaystyle f \) is constant.

[Tolaso J Kos]

Since \( \displaystyle \frac{f(z)}{z} \) has a removable singularity at \( \infty \) and the value is \( 0 \) this implies that \( f(z) \) is holomorphic in \( \infty \). Hence it is a holomorphic mapping from \( \widehat{\mathbb{C}} \to \mathbb{C} \) hence constant (because the image is compact , not open.)

[Tolaso J Kos]

It suffices to prove that \( f'(z)=0 \). Since \( \displaystyle \lim_{z\rightarrow +\infty}\frac{f(z)}{z}=0 \) forall \( \epsilon >0 \) then there exists \( n \in \mathbb{C} \) so large that if \( z \geq n \) then \( \displaystyle \left | \frac{f(z)}{z} \right |<\epsilon \). Let \( C_R \) be a circle about the origin. For an arbitrary \( z \) satisfying \( \left | z \right |\leq \dfrac{R}{2} \) we get:

$$\left | f'(z) \right |\leq \frac{1}{2\pi i}\oint \limits_{C_R}\frac{f(\xi)}{\xi^2}\,d\xi \leq \cdots \leq 4\sup \left \{ \left | \frac{f(\xi)}{\xi} \right |, \;\; \xi \geq R \right \}$$

Sending \( R \rightarrow +\infty \) (meaning that circle dilates , covering all of \( \mathbb{C} \) ) we get that \( f'(z)=0 \) and the result follows.