Polynomial coefficients are bounded by 1 Suppose $P(z)$ is a polynomial satisfying $|P(z)|\leq 1$ for all $|z|=1$, prove that all its coefficients are bounded by $1$. 
So far, I've been thinking something like this: 
proof: Suppose $g(z)=\frac{P(z)}{z}$. We observe that 
$$
|g(z)|=\left|\frac{f(z)}{z}\right|\leq 1 \quad \forall z\in \partial D_{1}(0). 
$$
First, we see $g$ has a singularity at $z=0$, however it is removable since it is bounded there. Thus, $g$ is analytic in $D_1(0)$. Second (I'm not sure if I can apply the MMP), since $g$ attains its maximum there, by MMP it must be a constat. 
Any hint or observation would be very welcome.
Thanks 
 A: Note that if $Q$ is an arbitrary polynomial which is not the constant zero, and $M$ the maximum of $Q$ on the unit circle which is strictly positive because $Q$ is not the zero polynomial, then $P=\frac{Q}{M}$ satisfies the hypothesis so there are tons of such polynomials. 
$P(z)=\frac{1+z}{2}$ shows that it is simply not true that $P(0)$ must be zero as incorrectly proved in the OP post.
To prove that the coefficients are bounded by $1$ under the hypothesis here, the simple identity:
If $P(z)=a_0+a_1z+..a_nz^n$, then $2\pi ia_k=\int_{|z|=1}P(z)z^{-k-1}dz$ and the triangle inequality for integrals suffices. 
(much more it is true but less trivially to prove, for example $|a_0|+|a_n| \le 1$ if $P$ has degree $n$ and satisfies $|P(z)| \le 1$ if $|z|=1$ which obviously implies that if $|a_n|=1$ then $P(z)=a_nz^n$ by a simple induction; similar inequalities $|a_j|+|a_k| \le 1$ are true for other pairs of coefficients $a_j, a_k, j\ne k$ but under some arithmetic conditions regarding $|j-k|$)
A: One can use Parseval's identity:

Let $f(z) = \sum_{n=0}^\infty a_n z^n$ be holomorphic in $B(0,R)$. Then
  $$
 \sum_{n=0}^\infty |a_n|^2 r^{2n} = \frac{1}{2 \pi }\int_0^{2\pi}|f(re^{it})|^2dt.
$$ 
  for $0 \le r < R$.

For an elementary  proof see for example Proof of Parseval's identity.
It follows that if $f$ is holomorphic in a neighborhood of the closed unit disk and  satisfies $|f(z)| \le 1$ for $|z| = 1$ then
$$
\sum_{n=0}^\infty |a_n|^2  \le 1
$$
which in particular implies that $|a_n| \le 1$ for all $n$.
