let $p$ be a polynomial of degree $n$ and let $t(r)=\max\{|p(z)|:|z|=r\}$. Show that $t(r)\geq t(s)$ is $0<r<s$. Equality holds iff $p(z)=z^n$.
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Do you mean $t(r) \geq t(s)$ if $0 < s < r$ instead? – Antonio Vargas Jul 03 '13 at 17:40
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show that t(r)/r^n \geq t(s)/s^n, – rajesh Jul 03 '13 at 17:46
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This doesn't seem right. For large $|z|$, $p(z)$ is huge (and $|p(z)|$ increases rapidly), so I don't see how you would have $t(r)=t(s)$. – Alexander Jul 03 '13 at 18:31
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there was a typing error. correct question is show that $t(r)/r^n \geq t(s)/s^n$ – rajesh Jul 03 '13 at 18:36
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This seems like it may be a duplicate of Domination of complex-value polynomial by highest power – bryanj Jul 04 '13 at 00:57
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Let $p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_0$. Note that $t(r) = \max \limits _{|z| = r} \displaystyle \Bigg| \frac{p(z)}{z^n} \Bigg |$.
Claim: For every $r$, $t(r) \ge |a_n|$.
Proof: Use the Cauchy integral formula to write $$ |a_n| \le \frac{1}{2 \pi} \Bigg|\int \limits_{|z| = r} \frac{f(\zeta)}{\zeta ^{n+1}}\ \text{ }d\zeta \Bigg| \le \frac{1}{2 \pi} 2 \pi r \max \limits _{|z| = r} \Bigg| \frac{p(z)}{z^{n+1}} \Bigg | = \max \limits _{|z| = r} \Bigg| \frac{p(z)}{z^{n}} \Bigg | = t(r) $$
Now that we have this the rest of the proof is easy. Now let $r < s < R$. The Maximum Modulus Theorem says that on the annulus $r \le |z| \le R$, the maximum modulus of $\displaystyle \frac{p(z)}{z^n}$ is attained on $|z| = r$ or $|z| = R$, for all $R > s$.
If $t(s) \gt t(r)$, then by the Claim we'd have $t(s) \gt |a_n|$. This is a contradiction because from $R$ large enough $t(s) > t(R)$, and so modulus of $\displaystyle \frac{p(z)}{z^n}$ does not take its maximum on the boundary of the annulus.
We've shown that $t(r) \ge t(s)$ when $r < s$. This same argument also showed that if $t(r) = |a_n|$ then $t(s') = t(r) = a_n$ whenever $s' > $r (we already know it can't be less than $|a_n|$, and it can't be more than $|a_n|$ by the reasoning above). So, if $t(r) = |a_n|$ then $t(s') = |a_n|$ whenever $s' \ge r$. Then $\displaystyle \frac{p(z)}{z^n}$ is constant on $|z| \ge r$ (since modulus reaches its maximum on interior points of some annulus).
Finally, suppose $t(r) \gt |a_n|$ and $t(r) = t(s)$ for $r \lt s$. Then for some $R > s > r$, $t(R) < t(s)$, and so the maximum modulus of $\displaystyle \frac{p(z)}{z^n}$ on the annulus $r |\le |z| \le R$ occurs on $|z| = r$. But in this case as well there is in interior point on $|z| = s$ where the maximum modulus is attained. This shows that $\displaystyle \frac{p(z)}{z^n}$ is in fact constant - which is in fact a contradiction, since then we couldn't have $t(R) < t(r)$.
So, whenever $t(s) = t(r)$ we wind up with $\displaystyle \frac{p(z)}{z^n} = K$ a constant. Taking the limit as $z \to \infty$ shows that $K = a_n$.
bryanj
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