calculus - On Vanishing Riemann Sums and Odd Functions

Let $f: [-1,1] \to \mathbb{R}$ be a continuous function. Suppose that the $n$-th midpoint Riemann sum of $f$ vanishes for all $n \in \mathbb{N}$. In other words,
$$\forall n \in \mathbb{N}: \quad \mathcal{R}^{f}_{n} := \sum_{k=1}^{n} f \left( -1 + \frac{2k - 1}{n} \right) \cdot \frac{2}{n} = 0.$$
Question: Is it necessarily true that $f$ is an odd function?

It is easy to verify that if $f$ is an odd continuous function, then $\mathcal{R}^{f}_{n} = 0$ for all $n \in \mathbb{N}$. However, is the converse true?

This is part of an original research problem, so unfortunately, there is no other source except myself. With someone else, I managed to obtain the following partial result.

Theorem If $f$ is a polynomial function and $\mathcal{R}^{f}_{n} = 0$ for all $n \in \mathbb{N}$, then $f$ has only odd powers, which immediately implies that $f$ is an odd function.

The proof relies on properties of Bernoulli polynomials and Vandermonde matrices.

For the general case, I was thinking that Fourier-analytic tools might help, such as Poisson summation. A Fourier-analytic approach seems promising, but it has limitations and might not be able to fully resolve the question.

Would anyone care to offer some insight into the problem? Thanks!

Answer

Take the function $f(x)=\sum_{j\geq1} \alpha_j \cos(\pi j x)$. Then its $n$-th midpoint Riemann sum is
$$\begin{align} 0 = R_n f &= \sum_{j\geq1}\alpha_j \sum_{1\leq k\leq n}\frac{2}{n} \cos\left( \pi j\left( -1 + \frac{2k-1}{n}\right)\right) \\&= \frac{2}{n}\sum_{j\geq1}\alpha_j \sum_{1\leq k\leq n} (-1)^j\cos\left(\pi j(2k-1)/n\right) \\&= \frac{2}{n} \sum_{j\geq1} \alpha_j \frac{\sin \pi j}{\sin (\pi j/n)} \end{align}$$
where (by Mathematica)
$$\sum_{1\leq k\leq n}\cos\frac{\pi j(2k-1)}{n} = \frac{\cos \pi j\sin \pi j}{\sin (\pi j/n)}$$

and when I write $\sin \pi j/\sin(\pi j/n)$ I mean the limit as $j$ approaches its integer value (so no division by zero).

Now, when $n=1$, the condition is
$$0 = \sum_{j\geq1} \alpha_j$$
and when $n>1$, the condition is
$$0 = \sum_{j\geq1} \alpha_j(-1)^{(j/n)}[n\backslash j].$$

The condition $R_n f=0$ is only nontrivial when there are $j$ such that $n\backslash j$ and $\alpha_j\neq0$. So suppose that $\alpha_j\neq0$ only when $j$ is a power of 2, so that the function is
$$f(x) = \sum_{k\geq0} \beta_k \cos(\pi 2^k x).$$
Then the only $n$ that impose any conditions on $\alpha_k$ are the powers of 2.

If $n=2^m$, $m>0$, then the condition is
$$\beta_m - \beta_{m+1}-\beta_{m+2}-\cdots = 0,$$
and for $n=1$ the condition is
$$\sum_{k\geq0} \beta_k = 0.$$

Pick $\beta_0 = -1, \beta_k = 2^{-k}$ ($k\geq1$).
The condition for each $n=2^m$ and also $n=1$ will be satisfied, the function
$$f(x) = -\cos\pi x+\sum_{k\geq1} 2^{-k} \cos(\pi 2^k x)$$
is clearly even and nonzero, and $R_nf=0$ for every $n$.

If the Fourier series is finite, the function must then be zero.