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For a polynomial $p(z)=a_0+\ldots+a_nz^n$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$.

I get $|p(z)|\geq |a_n|R^n$, which makes that factor of $1/2$ useless. So I must have done something wrong.

$$|\frac{\partial^n{f}}{\partial z^n}(P)|\leq \frac{\sup_{z\in \overline{D}(P,r)} |f(z)|k!}{r^k}$$. This is the Cauchy estimate.

$$|\frac{\partial^n{p}}{\partial z^n}(z)| = |a_n n!|$$ $$|\frac{\partial^n{p}}{\partial z^n}(0)|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$|a_n n!|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$R^n |a_n|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|$$

There exist a $z_0$, such that $|z_0|>0$ and $$\sup_{\zeta\in \overline{D}(0,|z_0|)} |p(\zeta)| = |p(z_0)|$$.

This is saying there exist a disk centered at the origin, such that $|p|$ is maximum on some point on the boundary.

Let $R=|z_0|$, we have $R^n |a_n|\leq |p(z_0)|$

Edit: I know what I got wrong. $z_0$ is just one point on the circle $|z| = R$. One can chose another point $z_1$ on the circle, such that $p(z_1)<p(z_0)$.

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    $\begingroup$ The estimate you describe can't hold in general. For example, if $p(z) = z^n - 1$, then $|p(R)| < |R|^n$ for all $R > 1$. In any case, the problem is much easier than you're making it: it's just an application of the triangle inequality. $\endgroup$ May 17, 2011 at 6:23
  • $\begingroup$ A good counterexample, thanks. The Cauchy estimate in the beginning is clearly applyable since $p$ is entire. then which step was wrong? $\endgroup$
    – Chao Xu
    May 17, 2011 at 6:36
  • $\begingroup$ There are several things that are fishy here. For one thing: Where do you use that $|z|$ is sufficiently large? Also, the last displayed equality is certainly wrong, as the left hand side only depends on $|z|$, while the right hand side depends on $z$ (the right hand side is not constant in general). Plug $z = R$ and $z = -R$ into Qiaochu's $p(z) = z^n - 1$ with $n$ odd, for example. Also, I don't follow the conclusion "then $|p(0)|$ would be the largest value". $\endgroup$
    – t.b.
    May 17, 2011 at 7:03
  • $\begingroup$ Now I see what I got wrong. I did some editing, hope this clear it up. I was trying to say there exist a disk centered at the origin, such that $|p|$ is maximum at some point on the boundary. $\endgroup$
    – Chao Xu
    May 17, 2011 at 7:26

2 Answers 2

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Let me prove a slightly stronger property than the one you ask about since this can be done in one go:

Let $p(z) = a_{n} z^n + a_{n-1} z^{n-1} + \cdots + a_{0}$ be a complex polynomial with $a_{n} \neq 0$. There exists $R \gt 0$ such that $\frac{1}{2}|a_{n}|\,|z|^n \leq |p(z)| \leq \frac{3}{2} |a_{n}|\,|z|^n$ for all $z$ with $|z| \geq R$.

This is clearly true if $n = 0$, so let's assume $n \geq 1$. Put $q(z) = |a_{n-1}|\,|z|^{n-1} + \cdots + |a_0|$. By the triangle inequality we have the estimates $$|a_{n}|\,|z|^n - q(z) \leq |p(z)| \leq |a_{n}|\,|z|^n + q(z) \qquad \text{for all } z \in \mathbb{C}.$$ For $k \leq n-1$ we have $|z|^{k} \leq |z|^{n-1}$ if $|z| \geq 1$, so $$q(z) \leq (|a_{n-1}| + \cdots + |a_{0}|) |z|^{n-1} \qquad \text{for all } |z| \geq 1.$$ Putting $$R = \max{\left\{1, \frac{2}{|a_{n}|}(|a_{n-1}| + \cdots + |a_{0}|)\right\}}$$ we get the desired $R$ since for $|z| \geq R$ we then have $q(z) \leq q(z) \cdot \frac{|z|}{R} \leq \frac{|a_{n}|}{2} |z|^{n}$.

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    $\begingroup$ Of course, since $R^n \leq |z|^n$ for $|z| \geq R$ (as $R \geq 1$)we get the desired inequality $\frac{1}{2}|a_{n}|\,R^n \leq |p(z)|$ for all $|z| \geq R$. $\endgroup$
    – t.b.
    May 17, 2011 at 8:33
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    $\begingroup$ This looks much neater than mine. :) $\endgroup$
    – Chao Xu
    May 17, 2011 at 8:35
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    $\begingroup$ @Chao: Maybe, but you still figured it out yourself! Perhaps I should have waited a bit longer... The point of this estimate is of course that eventually every polynomial grows like its leading term, i.e., we have estimates from below and above. $\endgroup$
    – t.b.
    May 17, 2011 at 8:38
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Thanks for Qiaochu Yuan's hint. The solution:

Let $p(z) = q(z) + a_nz^n$.

Lemma: Let $c\neq 0$. $|cz^n| > |q(z)|$ for all $z\in \{z : |z|\geq r_c\}$, where $r_c$ is sufficiently large.

Proof: If $|z|=r$, then $|cz^n| = |c|r^n$. $$|q(z)| = |\sum_{i=0}^{n-1} a_iz^i| \leq \sum_{i=0}^{n-1}|a_iz^i| = \sum_{i=0}^{n-1}|a_i|r^i \leq \sum_{i=0}^{n-1} \max(|a_i|) r^i =\max_{i=0}^{n-1}(|a_i|) \sum_{i=0}^{n-1} r^i$$ when $r > 1$, $$\max_{i=0}^{n-1}(|a_i|) \sum_{i=0}^{n-1} r^i \leq \max_{i=0}^{n-1}(|a_i|) (n-1) r^{n-1}$$

If $r > \max(\frac{\max_{i=0}^{n-1}(|a_i|)(n-1)}{|c|},1)$ $$\max_{i=0}^{n-1}(|a_i|) (n-1) r^{n-1} \leq |c|r^n$$

$|p(z)| = |a_n z^n+q(z)|\geq |a_n z^n| - |q(z)| \geq |a_n z^n| - |c z^n| = (|a_n|-|c|)|z^n|$

Let $c = |\frac{a_n}{2}|$. For all $z$ such that $|z|=R>r_{a_n/2}$, this shows $|p(z)| \geq |a_n|R^n/2$.

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    $\begingroup$ +1: This now looks quite okay. See my answer for a slightly simpler way of putting it (and proving a stronger statement). But the idea is essentially the same. $\endgroup$
    – t.b.
    May 17, 2011 at 8:30

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