# Maximum principle - prove that function is increasing/decreasing

$P$ is an $n$-degree polynomial . For $r>0$ let's define: $$M(r,P)=\sup_{|z|=r} |P(z)|$$ Prove that function $r\mapsto M(r,P)$ is increasing and function $r\mapsto\frac{M(r,P)}{r^n}$ is decreasing.

I think I should use the principle of maximum. Have you got any hints?

• Certainly, you can. The first proposition is straightforward, as you suggest. For the second, you might note $\frac{M(r,P)}{r^n} = \sup_{|z|=r}\left|\frac{1}{z^n}P(z)\right|$ and $\frac{1}{z^n}P(z) = Q\left(\frac{1}{z}\right)$, for a different polynomial $Q$. Commented Jun 14, 2017 at 15:43
• @JonathanY. thank you for answer. But how to show that the function is increasing/decreasing? Should I use (sup|P(z)-P(x)|(z-x)>0? Or derivative? There is a big black hole in my head.
– json
Commented Jun 14, 2017 at 15:51
• Perhaps you can edit your question with what you know the maximum principal says, as it relates to $P(z)$? Commented Jun 14, 2017 at 15:53

I will answer the first part of the question, and give a hint for the second.

By the maximum modulus principle, $\left|P(z)\right|$ achieves no maximum in the domain $|z|<r$ (I'm assuming $n>0$, so $P$ isn't constant). However, as a continuous function, it does a achieve a maximum in the compact domain $|z|\le r$. This implies that $\max_{|z|\leq r}\left|P(z)\right| = \max_{|z|=r}\left|P(z)\right| = M(r,P)$, and it isn't achieved anywhere in $|z|<r$.

In particular, for all $|z_0| < r$ we have $$\left|P(z_0)\right| \leq \max_{|z|\leq |z_0|}\left|P(z)\right| < \max_{|z|=r}\left|P(z)\right| = M(r,P)$$ Put differently, for $r<R$ and all $|z|=r$ we have $\left|P(z)\right|<M(R,P)$. A direct consequence is that $M(r,P)<M(R,P)$.

For the second part, as mentioned in comments, we can develop $$\frac{M(r,P)}{r^n} = \sup_{|z|=r}\left|\frac{1}{z^n}P(z)\right| = \sup_{\left|\frac{1}{z}\right|=\frac{1}{r}}\left|Q\left(\frac{1}{z}\right)\right|$$ for a different polynomial $Q(z)$ (which?). We can now use part (a) to conclude (how?).

• Thank you very much! So does it mean that $Q(1/z)=P(1/z)$. Am I wrong? And than from maximum principle and the first part Q(1/z) is increasing so P(z) is decreasing?
– json
Commented Jun 14, 2017 at 16:59
• @json $Q(z)$ isn't quite $P(z)$. Try to denote $P(z) = \sum_{k=0}^n a_k z^k$ and find $\frac{1}{z^n}P(z)$. Commented Jun 14, 2017 at 17:52
• Ok, I see so $\frac{1}{z^n}P(z)=\sum_{k=0}^{n} \frac{a_k z^k}{z^n}$
– json
Commented Jun 14, 2017 at 18:18
• And now I can use the (a) to Q(1/z) and Q(1/z) is a increasing function, am I right?
– json
Commented Jun 14, 2017 at 18:20
• Once you have $Q$, note that $\sup_{|1/z|=1/r}|Q(1/z)| = M(1/r, Q)$. Commented Jun 14, 2017 at 18:21