# Cauchy’s Inequality

For $$z_0 \in D_R(z_0) \subset \Omega$$, setting $$M \coloneqq\sup_{z\in \gamma}{\left\lvert {f(z)} \right\rvert}$$ so $${\left\lvert {f(z)} \right\rvert}\leq M$$ on $$\gamma$$ \begin{align*} {\left\lvert { f^{(n)} (z_0) } \right\rvert} \leq \frac{n !}{2 \pi} \int_{0}^{2 \pi} \frac{ M } {R^{n+1}} R \,d\theta = \frac{M n ! }{R^n} .\end{align*}

The $$n$$th Taylor coefficient of an analytic function is at most $$\sup_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f} \right\rvert}/R^n$$. \begin{align*} {\left\lvert {c_k} \right\rvert} \ll {{\left\lVert {f} \right\rVert}_\infty \over R^k} .\end{align*}

• Given $$z_0\in \Omega$$, pick the largest disc $$D_R(z_0) \subset \Omega$$ and let $$C = {{\partial}}D_R$$.
• Then apply the integral formula.

\begin{align*} \left|f^{(n)}(z_0)\right| &= {\left\lvert { \frac{n !}{2 \pi i} \int_{C} \frac{f(\zeta) }{(\zeta-z_0)^{n+1}} \,d\zeta} \right\rvert} \\ &=\left|\frac{n !}{2 \pi i} \int_{0}^{2 \pi} \frac{f\left(z_0 + r e^{i \theta}\right) r i e^{i \theta} }{\left(r e^{i \theta}\right)^{n+1}} \,d\theta\right| \\ &\leq \frac{n !}{2 \pi} \int_{0}^{2 \pi}\left|\frac{f\left( z_0 +r e^{i \theta}\right) r i e^{i \theta}}{\left(r e^{i \theta}\right)^{n+1}}\right| \,d\theta\\ &=\frac{n !}{2 \pi} \int_{0}^{2 \pi} \frac{\left|f\left(z_0 +r e^{i \theta}\right)\right|}{r^{n}} \,d\theta\\ &\leq \frac{n !}{2 \pi} \int_{0}^{2 \pi} \frac{M}{r^{n}} \,d\theta\\ &=\frac{M n !}{r^{n}} .\end{align*}

## Exercises

Show that if $$f$$ is entire and $${\left\lvert {f(z)} \right\rvert} \in { \mathsf{O}}({\left\lvert {z} \right\rvert}^p)$$ for $${\left\lvert {z} \right\rvert}$$ sufficiently large, then $$f$$ is a polynomial of degree at most $${\left\lfloor p \right\rfloor}$$.

The basic idea: \begin{align*} {\left\lvert {c_k} \right\rvert} &\leq {k!\over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f(\xi) \over (\xi - 0)^{k+1}} \right\rvert}\,d\xi\\ &\leq {k! \over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R}{ {\left\lvert {\xi} \right\rvert}^{p} \over {\left\lvert {\xi} \right\rvert}^{k+1} }\,d\xi\\ &= {k! \over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R} {1\over {\left\lvert {R} \right\rvert}^{k+1-p}} \,d\xi\\ &= {k! \over 2\pi} {1\over {\left\lvert {R} \right\rvert}^{k+1-p}} \cdot 2\pi R \\ &= { \mathsf{O}}(1/R^{k-p}) ,\end{align*} which converges to $$0$$ as $$R\to \infty$$ provided that $$k-p>0$$, so $$k>p$$. So any coefficient $$c_k$$ for $$k\geq {\left\lfloor p \right\rfloor}$$ vanishes.

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