Exercise :
Let :
$$p(z) = \sum_{k=0}^{n}a_kz^k$$
a polynomial of degree $n$.
If :$$|p(z)|\leq M ,\forall |z|<1$$ show that :
$$|p(z)| \leq M|z|^n, \forall|z|\geq 1$$
Attempt :
Let :
$$q(ζ) := ζ^np\bigg(\frac{1}{ζ} \bigg), |ζ| \leq 1$$
Then :
$$z:=\frac{1}{ζ}$$
which means :
$$q\bigg(\frac{1}{z}\bigg) = \bigg(\frac{1}{z}\bigg)^np(z), |z|\geq1$$
$$\Leftrightarrow$$
$$q\bigg(\frac{1}{z}\bigg) = z^{-n}p(z), |z|\geq1$$
But after here I cannot see how to use the given inequality, that $|p(z)|\leq M ,\forall |z|\leq1$, since we have $|z|\geq 1$.
First of all, I would like to ask if the $|z| \geq 1$ that is being asked is wrong.
Secondly, if it isn't wrong, I would really appreciate any thorough help/solution or tip to continue on with solving it, because it's a part of my collection of exercises for the preparation of the semester exams.
Thanks a lot for your time !
Kind regards,
Charalampos Filippatos, NTUA.
I think you are on to something: $q(z)$ is also a polynomial. It satisfies $|q(z)|\le M$ on the boundary of the unit disc, so also inside it. Thus if $|z|>1$, $|q(1/z)|\le M$ and so $|p(z)|\le M|z|^n$.