1
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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 $T_{n}$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.