# Let $f(z)$ be entire and $|f(w)-f(z)|\le R|w-z|$ for arbitrary $w, z$ in C and $R>0$. Prove that $f(z)$ is a polynomial of degree less than $2$.

From an old examination paper:

Let $f(z)$ be entire and $|f(w)-f(z)|\leq R|w-z|$ for arbitrary $w, z$ in $\mathbb C$ and $R>0$. Prove that $f(z)$ is a polynomial of degree less than 2.

I have absolutely no idea where to start so any help would be much appreciated!

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If you rewrite the inequality as

$$\frac{|f(w) - f(z)|}{|w - z|} \leq R$$

Then what does this tell you about the derivative of $f$?

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great I didn't even see that thank you! – katherinebarry Nov 3 '12 at 23:51
@katherinebarry I'm glad that it helped you =) – Adrián Barquero Nov 4 '12 at 0:24

Since $\dfrac{f(w) - f(0)}{w}$ is a bounded entire function (set $z = 0$, and continue analytically to $w=0$), by Liouville's theorem it's constant.

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(+1): It's almost an entire function--it has a removable singularity at $w=0$. You should fix that. – Cameron Buie Nov 3 '12 at 18:23
@CameronBuie Yes, of course. – Cocopuffs Nov 3 '12 at 18:43