Trying to understand part of Liouville's Theorem regarding integral inequality

Question

Im trying to understand Liouville's Theorem. I know its a very short proof, but I still got stuck on one part. Would anyone mind trying to explain?

The part I don't understand is the later part of this.

$f'(a)=$ [Cauchy integral formula] = $\big |\frac{1}{2\pi i} \int_\gamma \frac{f(z)}{(z-a)^2}dz\big | \leq \frac{1}{2 \pi} max_{|z-a|=r} \frac{|f(z)|}{r^2} 2 \pi r$

I can see that it has something to do with the integral inequality, but cant see how

$\big |\int_\gamma \frac{f(z)}{(z-a)^2}dz\big | \leq max_{|z-a|=r} \frac{|f(z)|}{r^2} 2 \pi r$

can can be seen as $|\int_{\gamma}f(z)dz| \leq \int_\gamma |f(z) dz|$

Answer 1

Let $I(a)$ be the magnitude of the integral as given by

$$I(a)=\left|\int_\gamma \frac{f(z)}{(z-a)^2}\,dz\right|$$

where $\gamma$ is the contour $|z-a|=r$. Then we have the estimates

$$\begin{align} I(a)&= \left|\int_\gamma \frac{f(z)}{(z-a)^2}\,dz\right| \\\\ &\le \int_\gamma \left|\frac{f(z)}{(z-a)^2}\right|\,|dz|\\\\ &\le \max_{|z-a|=r}\left(f(z)\right) \int_0^{2\pi} \frac{1}{r^2}\,r\,d\phi\\\\ &=\frac{2\pi}r\,\max_{|z-a|=r}\left(f(z)\right) \end{align}$$

where we used the parametrization $z=a+re^{i\phi}$ for $z$ on $\gamma$ (i.e., on $\gamma$, $dz=ire^{i\phi}$, $|dz|=r$, and $|z-a|=r$). In addition, we used the assumption of the theorem that $|f|$ is analytic and bounded (i.e., $|f(z)|\le M$ for all $z$).

And we are done!

Answer 2

Assuming that $\gamma(t)=a+re^{it}$, then you use the fact that$$\left|\int_\gamma\varphi(z)\,\mathrm dz\right|\leqslant\max_{|z-a|=r}|\varphi(z)|\times\operatorname{length}(\gamma).$$In this case, $\operatorname{length}(\gamma)=2\pi r$, and$$\max_{|z-a|=r}\left|\frac{f(z)}{(z-a)^2}\right|=\frac{\max_{|z-a|=r}|f(z)|}{r^2}.$$