Are those two subsets of $\mathcal{C}^0([0,1],\mathbb{R})$ equal? This question is a subsequent question of this one.
Consider the two subsets $S,F \subseteq \mathcal{C}^0([0,1],\mathbb{R})$:
$$F=\left\{ f \in \mathcal{C}^0([0,1],\mathbb{R}) / \forall n \in \mathbb{N}^*, \sum_{k=0}^{n-1}f \left( \dfrac{k}{n} \right) =0 \right\},$$
and
$$S=\left\{ f \in \mathcal{C}^0([0,1],\mathbb{R}) / \forall x \in [0,1/2), f(1/2+x) = -f(1/2-x) \text{ and } f(0)=0 \right\},$$
One can prove that $S \subseteq F$.
Question: do we also have $F \subseteq S$? And what if we replace $\mathbb R$ with $\mathbb C$?
 A: Besicovitch proved that $F\neq S$ using a fractal construction. For references see the introduction to P. Codecà, and M. Nair. "A note on a result of Bateman and Chowla." Acta Arithmetica 93.2 (2000): 139-148.
Remark. If $f\in F$ is nice enough (e.g. smooth) as a function on the unit circle then it must be in $S.$ To show this, it suffices to consider the even part $(f(x)+f(1-x))/2$ so $f(x)=f(1−x).$ The Fourier coefficients are then real-valued and satisfy $\hat f(p)=\hat f(−p).$ We get $\hat f(0)=0$ since the Riemann sum of $f$ must be zero. The condition that $\sum_{k=0}^{n-1}f(k/n)=0$ gives $\sum_{k=0}^{n-1}\sum_p \hat f(p)e^{2\pi pk/n}=0,$ but $\sum_{k=0}^{n-1}e^{2\pi pk/n}$ is $n$ when $p$ is a multiple of $n$ and is zero otherwise. So $\sum_p \hat f(pn)=0$ for all $n\in\mathbb N^*.$ By Möbius inversion (on the reverse divisibility poset) we have $\hat f(p)=\sum_{n,m\geq 1}\mu(n)\hat f(pnm)$ for all $p\geq 1,$ which implies that $\hat f(p)=0$ for each $p.$ For this argument to be rigorous we needed $f$ to be nice enough for the Fourier series to converge and to justify the Möbius inversion, e.g. $\sum_p|\hat f(p)|\sigma(p)<\infty$ where $\sigma$ is the divisor function.
