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Let $n$ be a given positive integer and let $f(x) = \sum_{k=0}^{n} c_kT_k(x)$, where $c_i \in \mathbb{R}$, $0 \leq i \leq n$. If

$|f(x)| \leq 1$, for $|x| \leq 1$, is it possible to get the maximum of $|c_i|$?

For polynomials, bounds were obtained by Markov in 1892 as explained in these previous questions polynomial1 and polynomial2, which show that there is a relation with Chebyshev polynomials. However, it is curretly unclear to me how to translate these Theorems to Chebyshev expansions.

I have done some research and found some interesting Theorems in a paper from Majidian which also states that $c_i$ decays to zero for $n \rightarrow \infty$. I was hoping that this would lead to some close bounds on $c_i$ as well.

Majidian H. On the decay rate of Chebyshev coefficients. Appl Numer Math 2017;113:44–53. https://doi.org/10.1016/j.apnum.2016.11.004.

However, I'm not a mathematician and have problems interpreting the conditions of all these Theorems. A simple explanation in layman's terms or written out example for let's say $c_6$ would help me a lot! Thanks.

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  • $\begingroup$ One has $\int_{0}^{2 \pi } T_k( \cos \theta) \cos ( \ell \theta ) d\theta $ is $\pi$ if $\ell$ is $\pm k$ and $0$ otherwise. So $\int_{0}^{2\pi} f(\cos \theta) \cos(\ell \theta) d\theta=\pi c_\ell$ and thus $|c_\ell| \leq \frac{1}{ \pi} \int_{0}^{ 2\pi} | \cos(\ell \theta) | d \theta = \frac{4}{\pi}$. So $c_\ell$ can never be more than $\frac{4}{\pi}$ for any $\ell$. Is this the kind of bound you are looking for? $\endgroup$
    – Will Sawin
    Commented Apr 16, 2021 at 18:41
  • $\begingroup$ Yes, I think this is the kind of bound I need. Thank you! Do you happen to have a source that I could refer to? $\endgroup$
    – NickVO
    Commented Apr 26, 2021 at 7:36

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