Let $f$ be analytic in the unit disk, and $f=u+iv=\sum a_n z^n$ is its Taylor series. If $a_n=p_n+iq_n$, then$$p_n=\dfrac{1}{r^n\pi}\int^{2\pi}_0u(r,\theta)\cos n\theta d\theta\,\,,\,\,q_n=\dfrac{1}{r^n\pi}\int^{2\pi}_0u(r,\theta)\sin n\theta d\theta$$ Prove that if $A(r)=\max_{|z|=r}u$, then $$|a_n|r^n \leq \max\{4A(r),0\}-2u(0)$$ I have no ideas about that.
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I would start with $$p_n \le \dfrac{1}{r^n\pi}\int^{2\pi}_0A(r)\cos n\theta d\theta$$ and similarly for $q_n$ – Paul Sinclair Jan 08 '18 at 20:25
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And $A(r)$ is constant then the integral is zero? – mnmn1993 Jan 09 '18 at 05:58
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First it will be pointed out that $$q_n=\dfrac{1}{r^n\pi}\int^{2\pi}_0u(r,\theta)\sin n\theta d\theta$$ should be $$q_n={\color{red}-}\dfrac{1}{r^n\pi}\int^{2\pi}_0u(r,\theta)\sin n\theta d\theta.$$ Anyway we start with $$ a_nr^n=(p_n+iq_n)r^n=\frac{1}{\pi}\int^{2\pi}_0 u(r,\theta)e^{-i n\theta} d\theta. $$ So we have $$ |a_n|r^n=\frac{1}{\pi}\left|\int^{2\pi}_0 u(r,\theta)e^{-i n\theta} d\theta\right|\le \frac{1}{\pi}\int^{2\pi}_0 |u(r,\theta)| d\theta. $$ On the other hand, since $u(0)=\frac{1}{2\pi} \int_0^{2\pi} u(r,\theta )d\theta $ we have $$ |a_n|r^n+2u(0)\le\frac{1}{\pi}\int^{2\pi}_0 \left(|u(r,\theta)|+u(r,\theta )\right) d\theta.\tag{1}$$ If $A(r)\le 0$, then $|u(r,\theta)|+u(r,\theta )=0$ and so RHS of $(1)=0.$ If $A(r)> 0$, then$$ \text{RHS of }(1)\le \frac{1}{\pi}\int^{2\pi}_0 2A(r) d\theta =4A(r).$$ Threfore we have $$ |a_n|r^n+2u(0)\le \max \{4A(r),0\}. $$
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