Cauchy inequality

The Cauchy inequality states that for a complex-valued function $f(z)$, if $|f(z)|\le M$ on the circle $|z-z_0|=R$, then:

$$\left|f^n(z_0)\right|\le \frac{n!}{R^n}M$$

for any $n\in \mathbb{Z}$. This is easily extendible to Liouville’s theorem.

Proof

From Cauchy’s integral formula for derivatives:

$$f^n(z_0)=\frac{n!}{2\pi i}{\oint }_{|z-z_0|}\frac{f(z)}{(z-z_0{)}^{n+1}}dz$$

We define $g(z)$:

$$g(z)=\frac{f(z)}{(z-z_0{)}^{n+1}}$$

And want to bound, for $R=|z-z_0|$:

$$|g(z)|=\frac{|f(z)|}{|z-z_0{|}^{n+1}}\le \frac{M}{R^{n+1}}$$

Then by the ML inequality, where $L=2\pi R$ is the length of $C$:

$$|f^n(z_0)|=\frac{n!}{2\pi }\left|\oint g(z)dz\right|\le \frac{n!}{2\pi }\frac{M}{R^{n+1}}2\pi R=\frac{n!}{R^n}M$$ $$|f^n(z_0)|\le \frac{n!}{R^n}M$$

As given above.

See also

• Not to be confused with the Cauchy-Schwarz inequality