Continuous functions whose Riemann sums are zero

Question

here is my question : Let's denote $F$ the following set $$ F=\left\{ f \in \mathcal{C}^0([0,1],\mathbb{C}) / \forall n \in \mathbb{N}^*, \sum_{k=0}^{n-1}f \left( \dfrac{k}{n} \right) =0 \right\}. $$ $F$ contains the line spanned by $f_0 : x \in [0,1] \mapsto \sin (2 \pi x)$ (but not $\cos$ because of $n=1$), but is it bigger ? I hope not, but I may be wrong. Thanks in advance.

Answer 1

In fact one can prove that $$S^\prime=\left\{ f \in \mathcal{C}^0([0,1],\mathbb{C}) \mid \forall x \in [0,1/2), f(1/2+x) = -f(1/2-x) \text{ and } f(0)=0 \right\}$$

is included in $F$.

The proof is a consequence of the fact that for $f \in \mathcal{C}^0([0,1],\mathbb{C})$ satisfying for all $x \in [0,1/2)$, $f(1/2+x) = -f(1/2-x)$, you have $$\sum_{k=0}^{n-1}f \left( \dfrac{k}{n} \right) =f(0).$$

In particular the function $$g(x) = \begin{cases} x & 0 \le x < 1/4 \\ -x + 1/2 & 1/4 \le x < 3/4 \\ x-1 & 3/4 \le x \le 1 \end{cases}$$ belongs to $S^\prime$. Therefore $S^\prime$ is much bigger than $\operatorname{span}\{\sin(2 \pi t)\}$.

And one can prove that it exists non trivial even (at $1/2$) continuous functions belonging to $F$. See the answer to Are those two subsets of $\mathcal{C}^0([0,1],\mathbb{R})$ equal?