Given any sequence of interpolating nodes, can we find a continuous function $f$ whose interpolating polynomials doesn't converge to $f$ point-wise

Question

Let $[a,b]$ be an interval in real line . Given any function $f:[a,b]\to \mathbb R$ and set $A \subseteq [a,b]$ of size $n+1$, there exists a unique polynomial $p_{f,A,n}(x)$ of degree $n$ such that $f(a)=p_{f,A,n}(a),\forall a\in A$. Such a polynomial is called the interpolating polynomial of $f$ with nodes $A$. Due to this, given $A\subseteq [a,b]$ with $n+1=|A|$, we can consider a linear map $X_{A,n} : C[a,b]\to C[a,b]$ such that $X_{A,n}(f)=p_{A,f,n}$ . Then $X_{A,n}$ is a continuous map and also a projection i.e. $X^2_{A,n}=X_{A,n}$. It can be shown that there is a constant $c$ such that $||X_{A,n}||_{\infty} \ge c \ln (n)$ for any set $A\subseteq [a,b]$ of size $n+1$, where $||.||_{\infty}$ is the $L^{\infty}$ operator norm induced from the $L^{\infty}$ norm on $C[a,b]$. Since $(C[a,b],||.||_{\infty})$ is a Banach space, it follows from Uniform boundedness principle, that given any sequence of subsets $\{A_n\}_{n=1}^\infty$ of $[a,b]$ with each $A_n$ having size $n+1$, $\exists f\in C[a,b]$ such that $||X_{A_n,n}(f) - f||_{\infty} $ does not converge to $0$.

This can be rephrased as: Given any sequence $\{A_n\}_{n=1}^\infty$ of subsets of $[a,b]$, with each $A_n$ of size $n+1$, there exists a continuous function $f$ on $[a,b]$ such that the sequence of interpolating polynomials $p_{f,A_n,n}(x)$ doesn't converge to $f$ uniformly on $C[a,b]$.

My question is the following: Given any sequence $\{A_n\}_{n=1}^\infty$ of subsets of $[a,b]$, with each $A_n$ of size $n+1$, does there exist a continuous function $f$ on $[a,b]$ such that the sequence of interpolating polynomials $p_{f,A_n,n}(x)$ doesn't converge to $f$ point-wise at atleast one point ?

Answer 1

The result that you mention in the first part of your question is a classical result by Faber

G. Faber, Uber die interpolatorsche Darstellung stetiger Funktionen, Jahresber. der deutschen Math. Verein. 23 (1914), 190-210.

Of course, this result does not exclude pointwise convergence. This question was negatively answered by S. Bernstein in

S. Bernstein, Quelques remarques sur l'interpolation, Math. Ann. 79 (1918), 1-12; doi: 10.1007/BF01457173.

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 $$ { \limsup _ { n \rightarrow \infty } } \left| L _ { n } \left( f, X , x _ { 0 } \right) \right| = \infty. $$ The next natural question is the possibility of divergence on a set of positive measure. Such a result was obtained by Marcinkiewicz and Grunwald (independently) for the particular scheme $T$ of Chebyshev nodes :

There exists a function continuous $f$ such that $$ { \limsup _ { n \rightarrow \infty } } \left| L _ { n } \left( f, T , x \right) \right| = \infty \quad \text { for all } \quad x \in [ - 1,1 ]. $$ There is an explicit construction of such a function in the book by I.P. Natanson, Constructive Function Theory, Vol. 3, pp. 35-46.

P. Erdos made the conjecture that this negative result holds for an arbitrary scheme $X$ of points in $[-1,1]$. This was proved in

P. Erdos and P. Vertesi, On the almost everywhere divergence of Lagrange interpolatory polynomials for arbitrary system of nodes, Acta Math. Acad. Sci. Hungar. 36 (1980), 71-89 and 38 (1981), 263; doi: 10.1007/BF01897094.

For any scheme of points $X$ in $[-1,1]$, there exists a function $f$ such that $$ { \limsup _ { n \rightarrow \infty } } \left| L _ { n } \left( f, T , x \right) \right| = \infty \quad \text {almost everywhere in} \quad x \in [ - 1,1 ], $$ and the divergence set is of second category. Note that divergence everywhere is not possible just by considering a newtonian scheme i.e. a scheme that repeats points ($A_{n}\subset A_{n+1}$, $n\geq1$, in your notation).

A very nice book about classical interpolation of functions is

Szabados, J., Vértesi, P., Interpolation of functions. World Scientific Publishing Co., Teaneck, NJ, 1990.

Answer 2

The answer is certainly yes. But it may not be that pessimistic due to the following fact: for any continuous function, there always exists a configuration of interpolation points such that the Lagrangian interpolation converges.