It is known that every Lipschitz function $f \colon [-1,1] \to \mathbb R$ can be expressed as a series in the Chebyshev polynomials $$f = \sum_{n = 0}^\infty a_n T_n $$ which is absolutely convergent under the $\lVert \, \cdot \, \rVert_\infty$ norm (see Approximation Theory and Approximation Practice by Lloyd Trefethen, 2018, p.19-20). I'm wondering whether the series converges under the Lipschitz norm, i.e. $$ \Big\lVert \sum_{n = N}^\infty a_n T_n \Big\rVert_\text{Lip} \xrightarrow{N \to \infty} 0 $$ (and if so, is convergence unconditional? absolute?) where the Lipschitz norm is defined by $$\lVert f \rVert_\text{Lip} := |f(-1)| + \sup_{x \neq y}\bigg|\frac{f(y)-f(x)}{y-x} \bigg|$$ I'm interested in this since $(\text{Lip}[-1,1], \lVert \, \cdot \, \rVert_\text{Lip})$ is a Banach space, while $(\text{Lip}[-1,1], \lVert \, \cdot \, \rVert_\infty)$ is not. Thank you.
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No, the space of Lipschitz functions on an infinite metric space is non-separable so it can't have a Schauder basis.
answered Apr 20, 2022 at 16:24
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$\begingroup$ Thanks for your answer, I wasn't aware of that. Do you know where I could find that result? Would it make a difference if I only consider bounded Lipschitz functions? $\endgroup$ Commented Apr 20, 2022 at 16:29
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$\begingroup$ It's explained in here math.stackexchange.com/a/2651998/18015 $\endgroup$ Commented Apr 20, 2022 at 17:48
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1$\begingroup$ @EmilioFerrucci Nik Weaver's book books.google.com/books?id=IBZeDwAAQBAJ is one of the first references that comes to my mind when I look for a result about Lipschitz algebras. $\endgroup$ Commented Apr 22, 2022 at 17:24
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