Editing Liouville's theorem (complex analysis) (section)

===If ''f'' is less than or equal to a scalar times its input, then it is linear===
Suppose that <math>f</math> is entire and <math>|f(z)|\leq M|z|</math>, for <math>M>0</math>. We can apply Cauchy's integral formula; we have that

:<math>|f'(z)|=\frac{1}{2\pi}\left|\oint_{C_r}\frac{f(\zeta)}{(\zeta-z)^2}d\zeta\right|\leq \frac{1}{2\pi} \oint_{C_r} \frac{|f(\zeta)|}{\left|(\zeta-z)^2\right|} |d \zeta|\leq \frac{1}{2\pi} \oint_{C_r} \frac{M |\zeta|}{\left| (\zeta-z)^2\right|} \left|d\zeta\right|=\frac{MI}{2\pi}</math>

where <math>I</math> is the value of the remaining integral. This shows that <math>f'</math> is bounded and entire, so it must be constant, by Liouville's theorem. Integrating then shows that <math>f</math> is [[Affine transformation|affine]] and then, by referring back to the original inequality, we have that the constant term is zero.