Polynomials having all their zeros on the unit circle

Question

Let $P(z)=\sum_{k=0}^na_kz^k$ be a polynomial of degree $n$ having all its zeros on the unit circle. Let $M=\max_{0\leq k\leq n}\lvert a_k\rvert$. The polynomial $P(z)=z^n+1$ has $\max_{\lvert z\rvert=1}\lvert P(z)\rvert= 2$. Intuitively, it appears that $\max_{\lvert z\rvert=1}\lvert p(z)\rvert\geq 2M$. Kindly share your opinion on the validity of this bound $2M$. Can it be further sharpened?

Answer 1

The result follows with $M=|a_n|$ actually as below:

Note that if $P$ has all the roots $w_j$ on the unit circle it follows that $w_j=1/\bar w_j$ so there is $c \ne 0$ st $$P(z)=cz^n\overline {P(1/\bar z)}=c\sum_{k=0}^n \bar a_kz^{n-k} $$ hence $a_k=c \bar a_{n-k}$ and in particular since $a_n \ne 0$ we have $a_0 \ne 0, |c|=1$ so $|a_0|=|a_n|$

But $a_0+a_nz^n=\frac{1}{n}\sum_{\omega^n=1}P(wz)$, so $|a_0+a_nz^n| \le \max_{|\zeta|=1}|P(\zeta)|$ for all $|z|=1$ hence choosing $|z|=1$ st $|a_0+a_nz^n|=|a_0|+|a_n|=2|a_n|$ the result follows

Note also that we only used that $P$ is self inversive ($w_j$ root iff $1/\bar w_j$ root and no zero roots) and actually only that $|a_0|=|a_n|$ which is implied by self-inversivity.

Answer 2

Answer became outdated when the question was edited. Originally $2M$ was stated as an upper not lower bound, due to a typo. Watch your signs!

Yup, as suggested by Fedor in the comments the premise is false. For example, $P(z)=z^4+z^3+z^2+z+1$ has all zeroes on the boundary of the unit disk and all coefficients unity, but $|P(1)|>2$.

If $z$ is inside or on the unti disk, then all polynomial functions (regardless of their zeroes) satisfy $|P(z)|\le\sum|a_n|$ from the Triangle Inequality.