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

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One Response to Cauchy estimates

  1. Pingback: Liouville’s theorem « Andy Soffer

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