# Cauchy estimates

We used a trick a few posts ago that I wanted to expound upon. Let ${f\in{\mathcal H}(U)}$, and ${z_0\in U}$. If we have a circle ${\beta_r}$ of radius ${r}$ centered at ${z_0}$, since it’s compact, there is some maximum value of ${|f(z)|}$ on ${\beta_r}$. Call this ${M_r}$. We then argued that as ${r\rightarrow 0}$, ${M_r\rightarrow 0}$ because ${f}$ is continuous. I want to generalize this result slightly:

Theorem 1 (Cauchy Estimates) In the setup as above,

$\displaystyle \left|f^{(n)}(z_0)\right| \le \frac{n!}{r^n}\cdot M_r$

You can see that, for ${n=0}$, the result is exactly what we already used.

Proof:Recall from the generalized Cauchy integral formula, that

$\displaystyle f^{(n)}(z_0)=\displaystyle\frac{n!}{2\pi i}\int_{\beta_r}\frac{f(w)}{(w-z_0)^{n+1}}dw$

Then we have the estimates:

$\displaystyle \begin{array}{rcl} |f^{(n)}(z_0)| &=& \left|\displaystyle\frac{n!}{2\pi i}\int_{\beta_r}\frac{f(z)}{(z-z_0)^{n+1}}dz\right|\\ &\le& \displaystyle\frac{n!}{2\pi}\int_{\beta_r}\left|\frac{f(z)}{(z-z_0)^{n+1}}\right|\cdot\left|dz\right|\\ &=& \displaystyle\frac{n!}{2\pi}\int_{\beta_r}\frac{\left|f(z)\right|}{r^{n+1}}\cdot\left|dz\right|\\ &\le& \displaystyle\frac{n!}{2\pi}\cdot \frac{M_r}{r^{n+1}}\cdot 2\pi r\\ &=& \displaystyle\frac{n!}{r^n}\cdot M_r \end{array}$

$\Box$