# Find maximum of absolute value of polynomial in the unit circle

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?

• A penny for your thoughts? jk no penny. – RKD Jul 13 '17 at 19:50
• For real polynomials, you may find tye answer in the paper of Erdös projecteuclid.org/download/pdf_1/euclid.bams/1183511456, For complex polynomials the best constant may still be unknown. – Felix Klein Jul 13 '17 at 21:10
• @FelixKlein: the optimal constant should not be too far from $(1+\sqrt{2})^n$, but to find the maximal polynomials looks like an extremely tough problem. – Jack D'Aurizio Jul 13 '17 at 22:13

$f(x)=\frac{1}{x^2+1}$ 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 $T_{n+1}(x)$.

• I think you're answering a slightly different question to the one asked. The question is if there exists a $C_n$ that is a uniform bound for all such polynomials of a given degree $n$. You seem to discuss the behaviour of these constants as a function of $n$, and show that if these exist then they must diverge at least exponentially. – bjorne Jul 13 '17 at 20:41
• @bjorne: you are correct, indeed. I am just showing that if a bound of the wanted type exists, it is quite weak since the involved constants grow exponentially as functions of the maximum degree. – Jack D'Aurizio Jul 13 '17 at 21:17
• @bjorne: in the last addendum I also proved that OP's claim (with a probably horribly sub-optimal constant) follows by Lagrange's interpolation and known results about Chebyshev polynomials. – Jack D'Aurizio Jul 13 '17 at 21:48