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First, let us fix some Notation:

Let $n\in\mathbb{N}$ and $x_i=\cos(\tfrac{(i+1/2)\pi}{(n+1)})$, $i=0,\dots,n$, be the Chebyshev points. Let \begin{align}L_i(x)={\displaystyle\prod_{\substack{0\leq j\leq n\\i\neq j}}}\frac{x-x_j}{x_i-x_j}, \end{align} $i=0,\dots,n$, be the Lagrange polynomials. Now let $f\in C^{\infty}([-1,1],\mathbb{R})$ with $\sup_{x\in[-1,1]}|f(x)|\leq 1$. We can now consider the interpolating polynomial given by \begin{align} P_n(x)=\sum_{i=0}^n f(x_i)L_i(x). \end{align} Now to the question:

By definition of the $L_i$ we know that $P_n$ is a polynomial of degree $n$, i.e. there exists coefficients $c_{n,j}$, $j=0,\dots,n$, such that \begin{align} P_n(x)=\sum_{j=0}^n c_{n,j}x^j. \end{align}
We are looking for an estimate on the size of the coefficients $c_{n,j}$, specifically we would like something like \begin{align} \max_{j=0,\dots,n}|c_{n,j}|\leq \pi(n) \end{align} for some polynomial $\pi$. (Or alternatively an argument of why this is not possible).

As we are not all that familiar with this topic, we have been trying find such a result in the literature. While there is huge amount of work on Lagrange interpolation and Chebyshev polynomials, this kind of estimate does not seem to be of interest for the usual applications. I would, however, find it surprising if no one had ever considered this question. If someone could point us to an answer, we would be very grateful.

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  • $\begingroup$ Crossposted at mathoverflow.net/questions/299735/… $\endgroup$ – Dirk May 9 '18 at 6:23
  • $\begingroup$ The notes inis.jinr.ru/sl/M_Mathematics/MRef_References/… give an explicit formula for the coefficients of the interpolating polynomial in the Chebyshev polynomials and an estimate of the the largest coefficients of the Chebyshev polynomials should lead to a simple upper bound (and my guess is that this estimate may be crude and exponential…). $\endgroup$ – Dirk May 9 '18 at 6:34

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