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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.

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  • $\begingroup$ 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 $n\to\infty$? $\endgroup$ – Pietro Majer Nov 1 '18 at 9:11
  • $\begingroup$ @PietroMajer: I corrected the typo, thanks ... and no condition on nodes ... I want it to happen for every node .. $\endgroup$ – user521337 Nov 2 '18 at 1:24
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Yes, it follows from the following result of S. Bernstein (Quelques remarques sur l'interpolation, Math. Ann. 79 (1918), 1-12):

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 $M=\max_{[-1,1]}f$.

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