Find maximum of absolute value of polynomial in the unit circle

Question

Consider a polynomial $p(x)$ of fixed degree $n$ which satisfies following condition $$ \max_{-1 \le x \le 1} |p(x)| \le 1. \;\;\;\;\;\;(*) $$ What can we say about the maximum of absolute value of this polynomial in the unit circle $$ \max_{|z| \le 1} |p(z)| ? $$ Is it true that $$ \max_{|z| \le 1} |p(z)| \le C_n $$ for any polynomial $p(x)$ of degree $n$ which satisfies (*) and some constant $C_n$?

If so, what is the exact value of $C_n$ and how does the maximal polynomial look like?

Answer 1

$f(x)=\frac{1}{x^2+1}$ is bounded by $1$ on $[-1,1]$ but has a simple pole at $x=\pm i$.
By considering a truncation of its Taylor series at the origin, we get that $$ p_n(x)=\sum_{k=0}^{n}(-1)^k x^{2k} $$ is bounded by $1$ on $[-1,1]$ but $$ \max_{|z|\leq 1}|p_n(z)|=\max_{|z|=1}|p_n(z)| \geq |p_n(i)| = n+1. $$ It follows that $\max_{x\in[-1,1]}|p(x)|$ does not tell us really much about $\max_{|z|=1}|p(z)|$.
For instance, by considering $p_n(z)=T_n(z)$ with $T_n$ being a Chebyshev polynomial of the first kind we also have $\max_{x\in[-1,1]}|p_n(x)|=1$ but $$ \max_{|z|=1}|p_n(z)|\approx \frac{1}{2}(1+\sqrt{2})^n. $$ In particular there is no hope of proving something stronger than

$$ \max_{|z|\leq 1}|p(z)|=\max_{|z|=1}|p(z)|\leq \max_{x\in[-1,1]}|p(x)|\cdot C^{\partial p} $$ for some positive constant $C$.

Markov's and Remez' inequalities seems deeply related.
Indeed, the weaker inequality

$$ \max_{|z|\leq 1}|p(z)|\leq \underbrace{(\partial p+1)(1+\sqrt{2})^{\partial p+1}}_{C_n} \max_{x\in[-1,1]}|p(x)| $$

can be simply proved by applying Lagrange interpolation with respect to the nodes given by the roots of the Chebyshev polynomial $T_{n+1}(x)$.