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I had asked this question in the math stackexchange about a year ago. I did not get any response, so I am asking it here.

Given distinct points $x_1,\dots,x_m \in [0,1]^n$ and real numbers $y_1,\dots,y_m$ with $m > 1$, let $\mathbb{P} = \{p \text{ is a real polynomial on } \mathbb{R}^n :p(x_1) = y_1, \dots, p(x_m) = y_m\}.$

Let $\|p\| = \sup_{x \in [0,1]^n} |p(x)|.$

What is $M = \inf\{\|p\| \colon p \in \mathbb{P}\}$?

Clearly we must have $M \geq \max \{ |y_1| , \dots, |y_m| \}$. Does the equality hold?

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    $\begingroup$ Yes, by the Stone Weierstrass theorem. Choose a continuous function with max absolute value on the cube attained on the given points. You can approximate uniformly by polynomials which are in $\mathbb P$ $\endgroup$
    – Pietro Majer
    Commented Nov 1, 2021 at 14:12
  • $\begingroup$ @PietroMajer The sequence of polynomials that converge uniformly to such a function need not obey these constraints in general. Is there a way to overcome that? $\endgroup$
    – Arin Chaudhuri
    Commented Nov 1, 2021 at 16:05
  • $\begingroup$ In fact I think it is not difficult to make a polynomial $p\in \mathbb P $ with $\|p\|_{\infty,I^n}= \max_{j}|f(x_j)|$. A linear combination of $\big(1-\frac{\|x-x_j\|_2^2}{n}\big)^N$, ($j=1,\dots,m$) should work. $\endgroup$
    – Pietro Majer
    Commented Nov 1, 2021 at 18:18

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Let $ F:=\{x_1,..x_m\}\subset I^n$ and $f\in C(I^n)$ such that $\|f\|_{\infty,I^n}=\|f\|_{\infty,F}= \max_{x\in F}|f(x)|$. Note that there exists a finite dimensional linear space of polynomials $V$ such that the restriction map $V\ni p\mapsto p_{|F}\in{ \mathbb R}^F$ is surjective (i.e. any interpolation problem with nodes in $F$ has solution $p\in V$). All norms in finite dimensional spaces are equivalent, so you have $\|p\|_{\infty,I^n}\le C \|p\|_{\infty,F} $ for all $p\in V$. This allows you to correct an approximating sequence $P_k\to f$ given by the Weierstrass theorem by means of $p_k\in V $ so that $P_k-p_k\in\mathbb P$, and $\|p_k\|_{\infty,I^n}=o(1)$.

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