Cauchy's estimate

Bounds the derivatives of an analytic function in terms of a bound on the function itself.

If $f$ is analytic on a domain containing the closed disk $|z-z_0|\le r$, and $|f(z)|\le M$ on the circle $|z-z_0|=r$, then for every nonnegative integer $n$,

$|f^{(n)}(z_0)|\le \frac{n!M}{r^n}.$

Proof

Apply the ML estimate to the generalized Cauchy integral formula with $C$ = circle of radius $r$ around $z_0$. The contour has length $2\pi r$, and the integrand magnitude is bounded by $M/r^{n+1}$. So

$|f^{(n)}(z_0)|=\left|\frac{n!}{2\pi i}{\oint }_C\frac{f(z)}{(z-z_0{)}^{n+1}}dz\right|\le \frac{n!}{2\pi }\cdot \frac{M}{r^{n+1}}\cdot 2\pi r=\frac{n!M}{r^n}.■$

What it says

A function that doesn’t grow too fast can’t have wildly large derivatives. Larger $r$ (bigger circle where $|f|$ is bounded by $M$) means a smaller bound on derivatives. Smaller $r$ means tighter neighborhood for the bound, but a larger numerical bound on each derivative.

Consequences

Liouville’s theorem is one line from here. If $f$ is entire and $|f|\le M$ everywhere, then for any $z_0$ and any $r>0$, the estimate with $n=1$ gives $|f^{\prime }(z_0)|\le M/r$. Letting $r\to \infty$: $f^{\prime }(z_0)=0$. Since $z_0$ was arbitrary, $f^{\prime }\equiv 0$, so $f$ is constant.

Taylor series convergence radius. Cauchy’s estimate is used in the proof that the Taylor series of an analytic function around $z_0$ converges on the largest open disk around $z_0$ where $f$ is analytic, with radius exactly the distance to the nearest singularity.