# A function is of bounded variation if and only if the errors of its best approximation by trigonometric polynomials satisfy $\sum\frac{e_n}n<\infty$?

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 (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?

## 1 Answer

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.

• Fair enough. Do you know about the other implication? Sep 24, 2022 at 10:05
• False, too. See the edit. Sep 24, 2022 at 14:04