Why the interpolation polynomials for $f=|x|,x\in[-1,1]$ will oscillate near the endpoint of $[-1,1]$ as $n$ increases?

I know that one explanation of the Runge's phenomenon is that the interpolation error is not bounded which make use of the derivates of $f$.Wiki Runge's phenomenon

However, when $f=|x|$, it is not in differentiable at $x=0$. So, how to understand why the interpolation polynomials oscillate near $x=1$ and $x=-1$?

  • $\begingroup$ One way of thinking about it is to consider arbitrarily localized mollifications of $f(x)=|x|$, where the localization is so small that it is invisible to the nodes that you select. In other words, if you select nodes uniformly spaced $h$ apart, you change $f$ only on $[-h/2,h/2]$ (and not at $0$) so that it is smooth. The interpolant that you get for this function is exactly the same as the interpolant you get for $f$ itself, and in this case you see the bad error bound again. Of course, this is not a proof, because you can do this even for a function that was nice in the first place... $\endgroup$ – Ian Sep 20 '18 at 15:34
  • $\begingroup$ @lan Thanks for your idea. But if smooth the $f(x)=|x|$ near $x=0$, I guess the high derivative is still bounded by one and then the error bound could converges to 0 if takes $n$ to $\infty$. Am I wrong at some point? $\endgroup$ – whereamI Sep 20 '18 at 16:41
  • $\begingroup$ Oh no, because of non-analyticity of $|x|$ you will have very large high order derivatives near $x=0$ in the mollification. $\endgroup$ – Ian Sep 20 '18 at 17:03
  • $\begingroup$ @Ian would mind explaining a little bit more about why the high order derivatives shall be large? Sine the answer here math.stackexchange.com/questions/2093842/… says that the limit of first derivative of mollified at x=0 exists and converges to 0. $\endgroup$ – whereamI Sep 20 '18 at 17:30
  • $\begingroup$ I didn't say first. I said high order. The high order derivatives are what come into play when talking about error of high order interpolants. And the intuitive reason is easy, the function is close to $|x|$ which is a non-analytic function. $\endgroup$ – Ian Sep 20 '18 at 19:58

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