1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ A penny for your thoughts? jk no penny. $\endgroup$ – RKD Jul 13 '17 at 19:50
  • 1
    $\begingroup$ 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. $\endgroup$ – Felix Klein Jul 13 '17 at 21:10
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Jul 13 '17 at 22:13
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – bjorne Jul 13 '17 at 20:41
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Jul 13 '17 at 21:17
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Jul 13 '17 at 21:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.