2
$\begingroup$

Let $f:[-1,1]\rightarrow \mathbb{R}$ be a continuous function. Consider the following greedy algorithm for interpolation:

Set $r_0 = f$.

for $k = 0,1,\ldots,$

Find the location of the global maximum of $|r_k|$, say $x_k\in[-1,1]$.

Let $p_k$ be the polynomial interpolant of $f$ at $x_0,\ldots,x_k$.

Set $r_{k+1} = f - p_k$.

end

I am interested to know if $\|f - p_k\|_\infty\rightarrow0$ as $k\rightarrow \infty$. If $\|f - p_k\|_\infty\not\rightarrow0$ for all continuous $f$, does it when we assume $f$ is analytic on $[-1,1]$ and analytic continuable to some neighbourhood of $[-1,1]$?

$\endgroup$
5
  • $\begingroup$ if $f$ is just continuous, having it be zero on some sub-interval should provide a counter example. $\endgroup$
    – meh
    Commented Nov 20, 2013 at 14:39
  • $\begingroup$ Aginensky, I don't understand how this counter example is constructed. Can you say more? $\endgroup$
    – alext87
    Commented Nov 20, 2013 at 16:56
  • $\begingroup$ Sure, let $f=0$ for $-1\leq x \leq 0$ and let $f(x) = x^2$ for$ 0<x\leq 1$. That is a simple version of my proposed counter example. $\endgroup$
    – meh
    Commented Dec 5, 2013 at 14:13
  • $\begingroup$ @meh Why is it a counterexample? Why can't the error converge to 0 in this case? $\endgroup$
    – Andrei Kh
    Commented Feb 14, 2020 at 13:48
  • $\begingroup$ I believe @meh's example is not a counterexample. $\endgroup$
    – alext87
    Commented Mar 3, 2020 at 17:24

0

You must log in to answer this question.