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?](https://mathoverflow.net/questions/313706/does-every-non-negative-continuous-function-have-a-non-negative-interpolating-po) 


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



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
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.


