For which $n$, can we find a sequence of $n+1$ distinct points s.t. the interpolating polynomial of every +ve continuous function is itself +ve

Question

Fix an interval $[a,b]$. For which integers $n>1$, does there exist $n+1$ distinct points $\{x_0,x_1,...,x_n\}$ in $[a,b]$ such that for every continuous function $f:[a,b] \to (0,\infty)$, the unique interpolating polynomial $p_n(x)$ of $f$ at the nodes $\{x_0,x_1,...,x_n\}$ satisfy $p_n(x)\ge 0,\forall x\in [a,b]$ ?

compare with Does every positive continuous function have a non-negative interpolating polynomial of every degree?

In the present question, we do not want to let the nodes vary with the function.

Answer 1

Let $n \ge 2$. Given any points $x_0 < x_1 < \dots < x_n$, there is a quadratic function positive at all those points, but negative somewhere in $[x_0,x_n]$.

Indeed, let $c \in [x_0,x_n]$ be any point other than those $n+1$ points. There is a quadratic $\phi(x) = -1+m(x-c)^2$ that is positive at those points. Simply take $m$ large enough.

The function $f$ to be approximated has values $f(x_j) = \phi(x_j)$, linear between, and constant on $(-\infty,x_0]$ and on $[x_n,\infty)$. So $f$ is continuous on $(-\infty,\infty)$ and $f(x) > 0$ for all $x \in (-\infty,\infty)$.

The interpolating polynomial $p_n$ with $p_n(x_j) = f(x_j)$ for $j=0,1,\dots, n$ is actually $\phi$ itself, so $p_n(c) = -1$.

For $n=1$, take $x_0,x_1$ the two endpoints. Your interpolating polynomial is degree $1$, the graph is a straight line, so if it is positive at the endpoints, then it is also positive at all points between.