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.

  • 1
    $\begingroup$ The equivalent form $$\log M(r)-\log M(R)\geq n(\log r - \log R) $$ recalls en.wikipedia.org/wiki/Jensen%27s_formula ,too. $\endgroup$ – Jack D'Aurizio Aug 26 '14 at 2:38
  • $\begingroup$ I assume you mean $\max_{|z| = r} |f(z)|$ and not $\max_{|z| = r} f(z)$? $\endgroup$ – Willie Wong Aug 26 '14 at 11:24
  • $\begingroup$ @williewong yes. It was w typo $\endgroup$ – Wilson of Gordon Aug 26 '14 at 13:06

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.

  • $\begingroup$ You are right. Thanks. $\endgroup$ – Wilson of Gordon Aug 26 '14 at 13:12
  • $\begingroup$ I still don't know how to prove the desired result... please tell me... $\endgroup$ – Aolong Li Sep 14 '17 at 11:13

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