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Math Help - Inequality

  1. #1
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    Inequality

    Lemma. Let  f(z) be analytic inside and on the boundary of the disk  |z-z_0| < \rho_0 and satisfy  |f(z)| \leq M_0 for  |z-z_0| \leq \rho_0 . If  |f(z_0)| = M_0 then  |f(z)| = M_0 for all  z in the disk.

    Proof. Consider any circle  |z-z_0| = \rho \leq \rho_0 . By the Cauchy Integral Formula we know that  f(z_0) = \frac{1}{2 \pi i} \oint_{|\xi-z_0| = \rho} \frac{f(\xi) \ d \xi}{\xi - z_0} . For each  \xi on this circle with  |f(\xi)| < M_0 or  |f(\xi)| = M_0 . Assume that there is a  \xi_1 on this circle for which  |f(\xi_1)| < M_0 . Then since  |f(\xi)| is continuous there must be a whole arc  C_1 of the circle on which  |f(\xi)| < M_0 for  \xi on  C_1 . Also  |f(\xi)| \leq M_0 on the rest of the circle. Write  f(z_0) = \frac{1}{2 \pi i} \left[ \int_{C_1} \frac{f(\xi) \ d \xi}{\xi- z_0}+ \int_{\text{rest}} \frac{f(\xi)}{\xi-z_0} \ d \xi \right] .


    From here how do we establish that  |f(z_0)| < M_0 , a contradiction? From the M-L Inequality, we know that  |f(z_0)| < \frac{1}{2 \pi i} \left(2 \pi \rho M_0+ 2 \pi \rho M_0 \right) = \frac{2 \rho M_0}{i}. How do we get this to  M_0 ?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Lemma. Let  f(z) be analytic inside and on the boundary of the disk  |z-z_0| < \rho_0 and satisfy  |f(z)| \leq M_0 for  |z-z_0| \leq \rho_0 . If  |f(z_0)| = M_0 then  |f(z)| = M_0 for all  z in the disk.
    This is similar to the maximum-modulos theorem. By Cauchy's theorem we know that:
    f(z_0) = \frac{1}{2\pi i} \oint_{|z-z_0|=r} \frac{f(z)}{z-z_0}dz \text{ where }0<r\leq \rho
    Therefore, (by definition of contour integral)
    f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) d\theta
    Thus,
    M_0 = |f(z_0)| \leq \frac{1}{2\pi } \int_0^{2\pi} |f(z_0 + re^{i\theta}) |d\theta  \leq \max\{ |f(z_0+re^{i\theta})| : \theta\in [0,2\pi]\}

    We have shown that on maximum value of |f| on the disk |z-z_0| = r is at least M_0. However, |f|\leq M_0 and so the maximum value of |f| must be equal to M_0. Therefore, it follows that,
    \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + re^{i\theta})| d\theta = M_0
    The only way that integral can equal to M_0 is it at each z on |z-z_0|=r we have |f|=M_0.
    Thus, |f| = M_0 for |z-z_0| = r.
    But r was arbitrary and by shrinking and expanding r we get the whole disk |z-z_0|\leq \rho with |f| = M_0.
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