TITLE: Continuous functions whose Riemann sums are zero
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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?