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Note: My previous question was wrong. This is the opposite of it, and I hope it is more sensible.

Where the $\displaystyle M(r)=\operatorname{Max}_{∣z∣=r} \mid f(z) \mid$ , where $f(z)=p_n (z)$ , a polynomial of degree $n$ .

My first attempt: maybe this is related to the Cauchy's inequality of estimating derivatives. Maybe consider the integral $\displaystyle \int\frac{f(z)}{z^{ n+1}} \mathrm{d}z$ ?

Another attempt: The inequality$\displaystyle \frac{M(r)}{r^ n} \geq\frac{M(R)}{R^n}$ remotely assembles the Hadamard three circle theorem.

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Zhipu 'Wilson' Zhao
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1 Answers1

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I assume that you intend the definition to be $$ M(r) = \max_{|z| = r} |f(z)| $$ as otherwise $f(z)\in\mathbb{C}$ and I am not sure how you are ordering $\mathbb{C}$.

In this case observe that since $f$ is a polynomial of degree $n$, there exists $n+1$ complex constants such that $$ \frac{f(z)}{z^n} = \sum_{j = 0}^{n} c_j z^{-j} $$ Let $w = z^{-1}$ we have $$ \frac{f(z)}{z^n} = \sum_{j = 0}^{n} c_j w^{j} = g(w) $$ In particular $g(w)$ is a polynomial of degree at most $n$ and so is holomorphic.

Apply the maximum principle to $g(w)$ on the disc $\{ |w| \leq R\}$ we have that the maximum of $|g(w)|$ is achieved on the circle $|w| = R$. Now, recall that

$$ \frac{M(r)}{r^n} = \max_{|z| = r} \left| \frac{f(z)}{z^n} \right| = \max_{|w| = r^{-1}} |g(w)| $$

we see immediately the desired result.

Willie Wong
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