Cauchy's Derivative Inequalities

Cauchy's Derivative Inequalities

Recall from the Cauchy's Integral Formula for Derivatives page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ is analytic on $A$, and $\gamma$ is a simple closed piecewise smooth and positively oriented curve in $A$ such that the inside of $\gamma$ is in $A$ then for $k = 0, 1, 2, ...$ we have that:

(1)
\begin{align} \quad f^{(k)}(z_0) = \frac{k!}{2\pi i} \int_{\gamma} \frac{f(z)}{(z - z_0)^{k+1}} \: dz \end{align}

We will now use this result to prove some useful inequalities.

 Theorem 1 (Cauchy's Derivative Inequalities): Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ be analytic. Let $z_0 \in A$ and let $r > 0$ be such that $D(z_0, r) \subseteq A$, i.e., the open disk $D(z_0, r)$ is contained in $A$. If there exists an $M > 0$ such that $\mid f(z) \mid \leq M$ on the circle $C_{z_0, r}$ defined by $\mid z - z_0 \mid = r$ then for each $k = 0, 1, 2, ...$ we have that $\displaystyle{\mid f^{(k)}(z_0) \mid \leq \frac{k! M}{r^k}}$.
• Proof: Let $M > 0$ be such that $\mid f(z) \mid \leq M$ on $C_{z_0, r}$.
• The circle $C_{z_0, r}$ given by $\mid z - z_0 \mid = r$ is a closed piecewise smooth and positively oriented curve in $A$ such that the inside of this circle, $D(z_0, r)$ is contained in $A$. So by Cauchy's integral formula for derivatives we have that for each $k = 0, 1, 2, ...$ that:
(2)
\begin{align} \quad \mid f^{(k)}(z_0) \mid &= \biggr \lvert \frac{k!}{2\pi i} \int_{C_{z_0, r}} \frac{f(z)}{(z - z_0)^{k+1}} \: dz \biggr \rvert \\ & \leq \frac{k!}{2\pi} \int_{C_{z_0, r}} \frac{\mid f(z) \mid}{\mid z - z_0 \mid^{k+1}} \: \mid dz \mid \\ & \leq \frac{k!}{2\pi} \int_{C_{z_0, r}} \frac{M}{r^{k+1}} \: \mid dz \mid \\ & \leq \frac{k!}{2\pi} \cdot \frac{M}{r^{k+1}} \int_{C_{z_0, r}} \: \mid dz \mid \\ \end{align}
• Now $z = z_0 + re^{it}$ for $t \in [0, 2\pi]$ so $dz = ire^{it}$ and $\mid dz \mid = r \: dt$. Thus:
(3)
\begin{align} \quad \mid f^{(k)}(z_0) \mid & \leq \frac{k!}{2\pi} \cdot \frac{M}{r^{k+1}} \int_0^{2\pi} r \: dt \\ & \leq \frac{k!}{2\pi} \frac{M}{r^k} \cdot 2\pi \\ & \leq \frac{k! M}{r^k} \quad \blacksquare \end{align}