# For every table of interpolating nodes, there is a positive continuous function whose interpolating polynomials are not positive infinitely often

Fix an interval $$[a,b]$$. Is it true that for every table of interpolating nodes $$\{x_{0,n},x_{1,n}...,x_{n,n}\}_{n=1}^{\infty}$$, there exists a continuous function $$f:[a,b]\to (0,\infty)$$ such that the sequence of interpolating polynomials $$p_n(x)$$ of $$f$$ at those points satisfy $$(-\infty,0]\cap p_n([a,b])\ne \phi$$ for infinitely many $$n$$ ?

NOTE: If this is true, then it would imply that for every table of interpolating nodes $$\{x_{0,n},x_{1,n}...,x_{n,n}\}_{n=1}^{\infty}$$, there is a continuous function $$f:[a,b]\to (0,\infty)$$ such that the sequence of interpolating polynomials $$p_n(x)$$ of $$f$$ does not converge uniformly to $$f$$ on $$[a,b]$$. i.e. a positive answer to my question implies Faber's theorem.

• Did you mean $({\bf -\infty},0]\cap p_n([a,b])\neq\emptyset$ ? Do you assume something on the points $A_n:=\{x_{i,n}\}_{ 0\le i\le n}$ or can they be any sequence of subsets $A_n$ of $[a,b]$ of size $n+1$? For instance can we exclude $\operatorname{diam}(A_n)\to 0$ as $n\to\infty$? – Pietro Majer Nov 1 '18 at 9:11
• @PietroMajer: I corrected the typo, thanks ... and no condition on nodes ... I want it to happen for every node .. – user521337 Nov 2 '18 at 1:24

For an arbitrary scheme $$X=\cup_{n} A_{n}$$ of points in $$[-1,1]$$, there exist a continuous function $$f$$ and a point $$x_{0}$$ in $$[-1,1]$$ such that $${ \limsup _ { n \rightarrow \infty } } \left| L _ { n } \left( f, X , x _ { 0 } \right) \right| = \infty.$$
If there is a subsequence $$n_{k}$$ with $${ \lim_ { n_{k} \rightarrow \infty } } ~L _ { n_{k} } \left( f, X , x _ { 0 } \right) = -\infty,$$ consider the positive function $$g(x)=f(x)-m+1>0$$, where $$m=\min_{[-1,1]}f$$.
If there is a subsequence $$n_{k}$$ with $${ \lim_ { n_{k} \rightarrow \infty } } ~L _ { n_{k} } \left( f, X , x _ { 0 } \right) = \infty,$$ consider the positive function $$g(x)=-f(x)+M+1>0$$, where $$M=\max_{[-1,1]}f$$.