# Domination of complex-value polynomial by highest power.

### The problem:

Let $P(n)$ be a polynomial of degree $n$. Let $$M(r):= \underset{|z|\le r}{\mbox{sup}} \hspace{2mm} \left|P(z)\right|.$$ I desire to establish that $$r\mapsto \frac{M(r)}{r^n}$$ for $r>0$ is non-increasing as a function of $r$.

At least I believe this to be true and necessary to conclude a long homework assignment :P.

### 1st attempt at solution

I think it is reasonable to assume the polynomial has no constant term, for adding a constant term as far as I understand can only change the $M(r)$ term by a constant. So $P(z)$ can be taken to fix the origin. Then considering, say the holomorphic function, $$\frac{P(rz)}{M(r)}$$ the function should take the disk to the disk, so we can apply Schwartz's Lemma, and get the inequality: $$\left|\frac{P(rz)}{M(r)}\right|\le |z|$$ From here I would hope to pick a nice $z$ in the disk to establish something useful... I'm at a loss.

### 2nd attempt at solution

I did start thinking, establishing an inequality between expressions doesn't quite get me that an expression is non-increasing. What would get me there is taking derivative. I think it's clear $M(r)$ is smooth. So we can differentiate the function in question (with respect to $r$) and get that the derivative is:

$$\frac{M'(r)r^n - n r^{n-1}M(r)}{r^{2n}}=\frac{M'(r)r - n M(r)}{r^{n+1}}$$

We want that this derivative is $\le 0$. That is, that

$$M'(r) \le \frac{n M(r)}{r}$$

The RHS sort of makes me think of what $M'(r)$ would look like. $P'(z)$ would of course be almost $\frac{n P(z)}{z}$, except of course for the constant term of $P(z)$ which disappears when differentiating.

-

If $f$ is entire, then by the maximum modulus principle $\max_{|z|=r}|f(z)|$ is non-decreasing in $r$. Take $f(z) = z^n P(\tfrac{1}{z})$. This is an entire function (a polynomial). So $\max_{|z|=r} r^n|P(\tfrac{1}{z})|$ is non-decreasing in $r$. In other words, $\max_{|z|=r}|P(z)|/r^n$ is non-increasing in $r$.

-

Define $g(z) = f(z)/z^n$. Then $g$ is analytic on $\mathbb C \setminus \{0\}$, and $g(z) \rightarrow a_n$, the leading coefficient of $f$, as $z\rightarrow \infty$.

Suppose to the contrary that $M(r_1)/r_1 ^n < M(r_2)/r_2^n$ for some $r_1 < r_2$. Equivalently,

$$\sup\{|g(z)| : |z| = r_1 \} < \sup\{|g(z)| : |z| = r_2 \}$$

By the maximum modulus principle, $g$ attains its maximum modulus on the annulus $A_r = \{r_1 \leq |z| \leq r\}$ on the boundary of $A$ for each $r > r_2$. For each such $r$, we have $|g(w)| > \sup\{|g(z)| : |z| = r_1\}$ for some $w\in A_r$ (specifically, one with $|w| = r_2$) by hypothesis, so $g$ must attain its maximum modulus on $A_r$ somewhere on $\{|z| = r\}$. Taking the limit as $r\rightarrow \infty$, we get $|g(z)| < |a_n|$ on $A_r$ for each $r$, and in particular on $\{|z| = r_1\}$.

But, by the Cauchy integral formula, we have

$$|a_n| =|\frac{1}{2\pi}\int_{|z| = r_1} \frac{f(z)}{z^{n+1}} dz | \leq \frac{1}{r_1^n} M(r_1) = \sup\{|g(z)| : |z| = r_1\} < |a_n| .$$

I feel like the RHS of your last inequality should be multiplied by $2\pi$, which would only give $|a_n|<2\pi|a_n|$, not so useful. – Marcel T. Apr 24 '12 at 23:58
I edited it, there should be a $2\pi$ by the first integral. We are integrating over a curve of length $2\pi r$. Therefore, when we take the absolute value of the integral, we get that it is at most $2\pi r$ times the supremum of the modulus of the integrand. The $2\pi$'s cancel out, so the rest of the equation stays as is. – user15464 Apr 25 '12 at 0:08
I still do not see the strict inequality, why cannot the $|g(z)|$ approach the $|a_n|$? I'm beginning to doubt if what I wanted to prove holds after all. – Marcel T. May 7 '12 at 19:45