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Prove the Circumferential Mean Value Theorem: If $f$ is analytic at $z_0$, then $$f(z_0)= \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0+re^{i \theta} ) d \theta$$ for sufficiently small $r>0$.

Question: Can I use Cauchy's Integral Formula $f(z_0)= \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(z)}{z-z_0}$ to prove this? I'm hesitant to apply it because $f$ is analytic at just a point (instead of a domain), but it would make the proof very easy

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  • $\begingroup$ I think one must have $\;f\;$ analytic for a whole circle, or open domain in general, about $\;z_0\;$ . $\endgroup$ – DonAntonio May 9 '16 at 7:13
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    $\begingroup$ $f$ analytic at $z_0$ means that it is analytic on $|z-z_0| < r$ for $r$ small enough. analytic (or holomorphic) at only one point doesn't exist. $\endgroup$ – reuns May 9 '16 at 7:15
  • $\begingroup$ @user1952009 Thank you. I thought that, by definition, a power series with radius of convergence equal to zero could be an example of an analytic function at one single point, but I think that's a clinical case. $\endgroup$ – DonAntonio May 9 '16 at 7:19
  • $\begingroup$ @iser1952009 Awesome, thanks $\endgroup$ – user167857 May 9 '16 at 7:19
  • $\begingroup$ @Joanpemo : you mean a function which is smooth at $0$ but its Taylor series at $0$ has radius of convergence $0$ ? there is an example Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic but you won't say it is analytic (you'll say that it is smooth) $\endgroup$ – reuns May 9 '16 at 7:23
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Put $\;z:=z_0+re^{it}\;,\;\;t\in[0,2\pi]\;$ , so

$$dz=rie^{it}dt=i(z-z_0)dt$$

so

$$\frac1{2\pi}\int_0^{2\pi}f(z_0+re^{it}) dt=\frac1{2\pi i}\oint_{|z-z_0|=r}\frac{f(z)}{z-z_0} dz\stackrel{\text{Cauchy Int. Formula}}=f(z_0)$$

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