That for a polynomial of degree $n$, $\frac{M(r)}{r^n}$ is a non-decreasing function of $r$.

Question

Where the $M(r)=\operatorname{Max}_{\mid z\mid=r}f(z)$, where $f(z)=p_n(x)$, 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 $\displaystyle \frac{M(r)}{r^{n}}\leq\frac{M(R)}{R^n}$ remotely assembles the Hadamard three circle theorem.

Hi, please do not edit a question to invalidate existing answers. I've rolled back to the version answered by mrf below. For the interested reader: the OP asked a follow-up http://math.stackexchange.com/questions/909322 — Willie Wong, Aug 26 '14 at 11:02

score 1 · Accepted Answer · answered Aug 25 '14 at 06:50

1

This is not true. Take for example $f(z) = 1+z^2$. Then $M(0) = 1$ and $M(r) \approx r^2$ for $r$ large. Hence $M(r)/r^2$ blows up at $r=0$, and tends to $1$ as $r\to\infty$.

In fact, for this example, the function $M(r)/r^2$ is actually decreasing:

enter image description here

answered Aug 25 '14 at 06:50

mrf

43,639

No wonder I can't prove it, it isn't even true! – Zhipu 'Wilson' Zhao Aug 25 '14 at 08:16
Then when is it an increasing function? It seems like I remembered the inequality in a totally opposite way – Zhipu 'Wilson' Zhao Aug 25 '14 at 08:25

That for a polynomial of degree $n$, $\frac{M(r)}{r^n}$ is a non-decreasing function of $r$.

1 Answers1

Linked