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This is supposed to be an upper bound counterpart for the Cauchy inequality. Let $f$ be entire and $M(R) = sup_{|z|=R} |f(z)|$ and $A(R) = sup _{n≥0} |a_n|R^n$ prove that $2A(2R)$ ≥ $M(R)$

I used Cauchy formula on circle $|z| =2r$ : $M(R) =$ $1/2{\pi}i\int f(z)/(z-z_0) dz $ for some $z_0$ on $|z| = R$

Then I tried to compare it with $2A(2R)$ = $2 sup _{n≥0} [ 1/2{\pi}i\int f(z) e^{-ni\phi} d\phi]$

Everything looks quite close, but is not really there. Can someone help me?

Khoa
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1 Answers1

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For all $z$, $|z| = R$ implies $|f(z)| \le \sum\limits_{n = 0}^\infty |a_n|R^n \le A(2R)\sum\limits_{n = 0}^\infty \frac{1}{2^n} = 2A(2R)$. Hence $M(R) \le 2A(2R)$.

kobe
  • 41,901