Let $n\ge 1$ be an integer, $\mathcal P_n$ be the vector space of all polynomial functions over $[a,b]$, of degree at most $n$.

My question is : Is it true that

$$\inf_{x_0,x_1,...,x_n\in[a,b], x_0<x_1<...<x_n} \sup_{x\in [a,b]} \prod_{i=0}^n |(x-x_i)|=\inf_{P\in \mathcal P_n } \sup_{x\in [a,b]} |x^{n+1}-P(x)|.$$

I can easily see that $$\inf_{x_0,x_1,...,x_n\in[a,b],x_0<x_1<...<x_n} \sup_{x\in [a,b]} \prod_{i=0}^n |(x-x_i)| \ge \inf_{P\in \mathcal P_n } \sup_{x\in [a,b]} |x^{n+1}-P(x)|,$$ but I don't know about the other reverse inequality.

  • 2
    $\begingroup$ Take $[a,b]=[1,2]$, and $n=2$. Then in the right hand side you can always take infimum over polynomials $P(x)=\lambda x^{2}$ with $\lambda \to \infty$. In this case, the left-hand side stays bounded whereas the right-hand side goes to $-\infty$. $\endgroup$ – Paata Ivanishvili Oct 1 '18 at 15:10
  • $\begingroup$ There is a modulus missing on the right, I guess... $\endgroup$ – user111 Oct 1 '18 at 16:08
  • $\begingroup$ @user111: yeah there's a modulus sign .. thanks ... fixed it now $\endgroup$ – user521337 Oct 1 '18 at 22:47
  • $\begingroup$ @PaataIvanishvili: there were some modulus sign missing ... please look at the edited question $\endgroup$ – user521337 Oct 1 '18 at 22:51
  • $\begingroup$ Still does not work. Take $n=1$, and $[a,b]=[0,1]$. Then $\inf_{x_{1} \in [0,1]} \sup_{x\in [0,1]}|x-x_{1}| = \inf_{x_{1}\in [0,1]}\max\{|x_{1}|, |1-x_{1}|\}=1/2$. For the right hand side choose $P(x)=x$, then the right hand side is less than $\sup_{x\in [0,1]}|x^{2}-x|=1/4$. $\endgroup$ – Paata Ivanishvili Oct 2 '18 at 0:57

When $[a,b]=[-1,1]$, the $\inf$ on the right-hand side is attained (uniquely) by the monic Chebyshev polynomial $T_{n+1}$. It is well known that its roots belong to $[-1,1]$ and are simple.

For a general interval $[a,b]$, using the linear map $$f(x)=\left(\frac{2}{b-a}\right)x-\left(\frac{b+a}{b-a}\right)$$ that sends $[a,b]$ onto $[-1,1]$, it is clear that the $\inf$ on the right-hand side is attained by $$ \left(\frac{b-a}{2}\right)^{n+1} T_{n+1}(f(x)),$$ which, of course, has all its (simple) roots in $[a,b]$.

Actually, a more general result holds true. Replace $[a,b]$ with any compact subset $K$ of the complex plane $\mathbb{C}$. Set, for a polynomial $P$, $$\|P\|_{K}=\sup_{z\in K}|P(x)|,$$ and let $T_{n}$ be a polynomial that achieves the minimum of $\|P\|_{K}$ among all monic polynomials of degree $n$.

Claim: All the zeros of $T_{n}$ belong to the convex hull of $K$.

Indeed, assume that $z_{0}$ is a root of $T_{n}$ that does not belong to the convex hull of $K$. Then, $K$ lies in a cone with vertex at $z_{0}$, of opening $<\pi$. Choose a $z_{1}$ on the bisector $L$ of that cone, sufficiently close to $z_{0}$ so that $K$ lies in the half-plane, delimited by the perpendicular to $L$ at $z_{1}$ (and not containing $z_{0}$). Since $$|z-z_{1}|<|z-z_{0}|,\quad z\in K,$$ one gets $\|\tilde T_{n}\|_{K}<\|T_{n}\|_{K}$ where $\tilde T_{n}(z)=T_{n}(z)(z-z_{1})/(z-z_{0})$, which contradicts the assumption on $T_{n}$.


Take any interval $[a,b]$. We want to show that
$$ \inf_{a\leq x_{0}<\ldots <x_{n} \leq b} \sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}| = \inf_{P \in \mathcal{P}_{n}}\sup_{x \in [a,b]} |x^{n+1}-P(x)| $$ Notice that $|x^{n+1}-P(x)| = \prod_{j=0}^{n}|x-z_{j}|$ for some $z_{j} \in \mathbb{C}$. Moreover by varying the polynomial $P$ we can make $z_{j}$ to be an abitrary complex numbers (if $\mathcal{P}_{n}$ are real polynomials then whenever $z_{j}$ is complex for some $j$ there will be its conjugate counterpart as well so we will have terms of the form $|x-z_{j}|^{2}$)

Next since $|x-z_{j}|\geq |x-\Re z_{j}|$ it is always better to choose $z_{j}$ to be real numbers. Thus we only need to verify that $$ \inf_{a\leq x_{0}<\ldots <x_{n} \leq b} \sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}| = \inf_{x_{0},x_{1}, \ldots, x_{n}}\sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}|. $$ Becuase of the symmetry in the right hand side without loss of generality we can assume that $x_{0}\leq x_{1}\leq \ldots \leq x_{n}$. It is clear that the right hand side is attainable for some $x,x_{0},\ldots, x_{n}$. Next, consider the function $f(x)= \prod_{j=0}^{n}|x-x_{j}|$ on $[a,b]$, and fix points $x_{0}\leq \ldots \leq x_{n}$. Let $x_{\ell}$ be the smallest point such that $x_{\ell}>b$. By moving $x_{\ell}$ towards the point $b$ the value of the function $f(x)$ decreasis no matter where $x$ is located in $[a,b]$. Similarlry if $x_{q}$ is the largest point such that $x_{q}<a$ then by moving $x_{q}$ towards $a$ the value of $f(x)$ decreasis. This means that $$ \inf_{x_{0},x_{1}, \ldots, x_{n}}\sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}| = \inf_{a\leq x_{0}\leq x_{1}\leq \ldots\leq x_{n}\leq b}\sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}| $$ and this proves the claim because it really does not matter whether we take infimum in the righ hand side over $x_{0}<x_{1}...$ or $x_{0}\leq x_{1}...$ by continuity. For example if you have a contiuous function $g(x_{1}, x_{2})\geq 0$ then one can easily see that $\inf_{a\leq x_{0}<x_{1}\leq b}g(x_{1},x_{2}) =\inf_{a\leq x_{0}\leq x_{1}\leq b}g(x_{1},x_{2})$ and you can iterate this equality for the rest of the variables.

  • $\begingroup$ there are typo s ... in your very first equality of the statement you want to prove, you've two $\inf$ s and then so on at each and every step ... $\endgroup$ – user521337 Oct 2 '18 at 2:12
  • $\begingroup$ you say "beacuse of the symmetry of RHS, we can assume $x_0<x_1<...<x_n$ ..." is that really true or do you mean $x_0\le x_1\le...\le x_n$ ? $\endgroup$ – user521337 Oct 2 '18 at 2:17
  • $\begingroup$ It does not matter we have $<$ or $\leq$ because of the $\inf$. $\endgroup$ – Paata Ivanishvili Oct 2 '18 at 2:28
  • $\begingroup$ I have updated the answer. It should not play a role that the infimum is taken with respect to domain $x_{0}<x_{1}$ or $x_{0}\leq x_{1}$ because everything is continuous and if it happens that in the second case the infimum is attained (notice that by compactness it is always attained) for some $x^{*}_{0}$ and $x^{*}_{1}$ such that $x^{*}_{0}=x^{*}_{1}$ then by taking the infimum over the domain $x_{0}<x_{1}$ you can take sequence $x_{0}^{k}, x_{1}^{k}$ such that $x_{0}^{k}<x_{1}^{k}$ and $\lim_{k \to \infty} x_{j}^{k} \to x_{j}^{*}$ for each $j=0,1$ then by continuouty you get the result. $\endgroup$ – Paata Ivanishvili Oct 2 '18 at 13:52

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