11
$\begingroup$

Let $p(z)$ be a monic polynomial of degree $n$.

Prove that $\max\limits_{|z|=1}|p(z)|\geq 1$ and that equlity holds if and only if $p(z)=z^n$.

I first observed that $p^{(n)}(z)=n!$. Therefore by Cauchy's formula

$$n!=p^{(n)}(0)=\frac{n!}{2\pi i}\int\limits_{|z|=1}\frac{p(z)}{(z-0)^{n+1}}dz\implies 1\leq\frac{1}{2\pi}\int\limits_{|z|=1}|p(z)|dz\leq\max\limits_{|z|=1}|p(z)|$$

Now, clearly if $p(z)=z^n$ then $\max\limits_{|z|=1}|p(z)|= 1$, but what about the converse?

$\endgroup$
0

3 Answers 3

14
$\begingroup$

Let $p(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_0$.

Then, $\forall z\neq 0:p(\frac{1}{z})=z^{-n}+a_{n-1}z^{-n+1}+\cdots+a_0$.

Let $q(z)=z^n\cdot p(\frac{1}{z})=1+a_{n-1}z+\cdots+a_0z^n$.

Observe that $$\max\limits_{|z|=1}|q(z)|=\max\limits_{|z|=1}|z^n\cdot p(1/z)|=\max\limits_{|z|=1}|p(1/z)|=\max\limits_{|z|=1}|p(z)|$$

The last step is due to $\{1/z\mid z\in\mathbb{C}:|z|=1\}=\{z\mid z\in\mathbb{C}:|z|=1\}$.

Now, $q(0)=1$, so $|q(0)|=1$ and, by the maximum principle, $\max\limits_{|z|=1}|q(z)|\geq1$.

Finally, $\max\limits_{|z|=1}|p(z)|\geq1$.

Now, if $p(z)=z^n$ then clearly $\max\limits_{|z|=1}|p(z)|=1$.

Suppose that $\max\limits_{|z|=1}|p(z)|=1$, then also $\max\limits_{|z|=1}|q(z)|=1$.

Again, by the maximum principle, we have that $q\equiv 1$.

Therefore, $\forall z\neq 0:p(1/z)=1/z^n$, so $\forall z\neq 0:p(z)=z^n$.

Finally from continuity $p(0)=0$ and we get that $\forall z\in\mathbb{C}:p(z)=z^n$.

$\endgroup$
2
  • $\begingroup$ Little question why at the last part you can conclude that from MMP $q\equiv 1$? $\endgroup$
    – convxy
    May 26, 2020 at 16:26
  • $\begingroup$ @ronkurman By the maximum principle, as the polynomial attains its maximum value on the boundary, inside the ball (i.e. $|q(0)|=1$). $\endgroup$ Jun 12, 2020 at 11:09
3
$\begingroup$

You've shown that the mean value of $|p(z)|$ on the unit circle is at least $1$, so to show that the maximum is greater than $1$, you just have to do is show that $|p(z)|$ is not constant.

Suppose for the sake of contradiction that $|p(z)|$ is constant on the unit circle. Then $iz p'(z)$, the directional derivative of $p(z)$ in a direction tangent to the unit circle at $z$, must be an imaginary multiple of $p(z)$ everywhere on the unit circle; that is, the (necessarily meromorphic!) function $q(z) = \frac{p(z)}{z p'(z)}$ must be real wherever $|z| = 1$. By a corollary of the maximum modulus principle, $q(z)$ is equal everywhere to some constant $c$, and we can write $p(z) = cz p'(z)$. By comparing leading coefficients of $p(z)$ and $zp'(z)$ we get that $c$ has to equal $1/n$, but in that case, none of the nonzero trailing coefficients of $p(z)$ and $czp'(z)$ match up, so $p(z) = \frac{1}{n} z p'(z)$ can only hold if $p(z) = z^n$.

$\endgroup$
1
  • 1
    $\begingroup$ I have posted a somewhat nicer argument. $\endgroup$ May 13, 2018 at 15:31
0
$\begingroup$

Consider $p(x)=xq(x)+k$ where $q$ is a degree $n-1$ polynomial. Then

$$|p(x)|^2 = |x g(x)|^2 + 2 \Re(xg\bar{k}) + |k|^2 = |x g(x)|^2 + 2 |xg(x)| |k| \cos \arg(xg(x) \bar{k}) + |k|^2.$$

If we can show that $x g(x) \bar{k}$ has a argument in $[-\pi/2,\pi/2]$, then we can write $$1 = |p(x)|^2 \geq 1 + 0 + |k|^2$$ as $|xg(x)|^2 \geq 1$, and this is a contradiction if $k \neq 0$.

I'm only going to sketch the rest of a potential argument. I am not sure if this works. First, I would make an induction argument that for any $g(x)$ monic the range of $\arg(g(x))$ is an interval greater than $\pi$, and since $\arg x$ is increasing on $|x|=1$ going clockwise, $\arg(g(x)xk) = \arg(g(x)) + \arg(x) + \arg(k)$ must also have a range of length greater than $\pi$. Which means there is a value reached in the interval $[-\pi/2, \pi/2]$.

$\endgroup$
1
  • 1
    $\begingroup$ I have posted a somewhat nicer argument. $\endgroup$ May 13, 2018 at 15:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .