Let $\mathcal P\_n$ be the set of trigonometric polynomials of degree less than or equal to $n$ and let $\lVert\cdot\rVert\_\infty$ be the supremum norm. The error of the best approximation of $f$ of degree $n$ is defined as


$$e\_n(f)=\inf\_{p\in\mathcal P\_n}\lVert f-p\rVert\_\infty.$$


A theorem of Bernstein says that if


$$\sum\_{n=1}^\infty \frac{e\_n(f)}n n^\alpha<\infty$$


for $\alpha>0$ then $f$ is $\alpha$-Hölder. I conjecture that this extends to the $\alpha=0$ case with the statement that $f$ is continuous and of bounded variation instead, and also that the reciprocal holds.


There's a result of Newman and Rivlin in [Approximation of monomials by lower degree polynomials](https://doi.org/10.1007/bf01835995) (the lemma starting on the bottom of the second page, p. 452) that you can use to prove that the function


$$F\_k(x)=\sum\_{h=0}^\infty f\_k(h)\cos(hx)$$


where


$$f\_k(x)=-\frac{d}{dx}\frac 1{\log^k(x)}$$


and the exponents are composition, satisfies


$$e\_n\gg\frac{1}{\log(n)\log^2(n)\dotsm \log^{k-1}(n)(\log^k(n))^2}.$$


And it *really* looks like $\sum\frac{e\_n}n$ is on the verge of diverging but doesn't quite get there.


Is the conjecture true?



A function is of bounded variation if and only if the errors of its best approximation by trigonometric polynomials satisfy $\sum\frac{e_n}n<\infty$?
This is not true. If $f(x)=x \sin( \frac 1x)$ near the origin, then $f$ is $1/2$ Holder continuos and $e\_n(f) \approx \frac{1}{\sqrt n}$ by Jackson theorem, so that the series $\sum\_n \frac{e\_n(f)}{n}$ converges. But is not of bounded variation, since $f' \not \in L^1$.


EDIT


Also the converse does not hold. Take $f(x)=-\frac{1}{\log x}$ for $x$ close to 0, which is continuous and increasing. If $P\_n$ realizes the best approximation then
$$
|f(0)-f(\delta)| \leq |f(0)-P\_n(0)|+|P\_n(0)-P\_n(\delta)|+|P\_n(\delta)-f(\delta)| \leq 2e\_n(f)+Cn \delta,$$
where $C \geq \|P\_n\|\_\infty$ for every $n$, using Bernstein inequality $\|P'\_n\|\_\infty \leq n\|P\_n\|\_\infty$. Taking $\delta=n^{-2}$ we get
$$\frac{1}{2\log n} \leq 2e\_n(f)+\frac Cn$$ and then $e\_n(f) \geq \frac{1}{8\log n}$ for large n and the series $\sum\_n \frac{e\_n(f)}{n}$ diverges.


