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!

  • $\begingroup$ Kinda, sorta, not quite related: math.stackexchange.com/q/271199/7003 $\endgroup$ – cardinal Jan 6 '13 at 22:17
  • $\begingroup$ @cardinal: Though not related, the method of solution in the accepted answer might help! $\endgroup$ – Haskell Curry Jan 6 '13 at 22:26
  • $\begingroup$ Some thoughts: One immediately obtains from the frst few $n\in\{1,2,3,4,6\}$ that $f(-1)=0$, $f(0)=0$, $f(-\frac13)=-f(\frac 13)$, $f(-\frac12)=f(\frac12)$, $f(-\frac23)=f(\frac23)$. You want to show that $g\colon[0,1]\to\mathbb R$, $x\mapsto f(x)+f(-x)$ is zero. Note that the $n$th sum for $g$ is the $2n$th sum for $f$ (because $f(-1)=0$). So we already have $g(0)=g(\frac13)=g(\frac12)=g(\frac23)=0$. However, from $n=5$ on it seems a bit more complicated. $\endgroup$ – Hagen von Eitzen Jan 6 '13 at 22:49
  • $\begingroup$ @Hagen: Yes. This leads to a dead end. $\endgroup$ – Haskell Curry Jan 6 '13 at 22:55
  • $\begingroup$ @everyone: It suffices to consider even continuous functions only. Every continuous function on $ [-1,1] $ can be written as the sum of an even continuous function and an odd continuous function: $ f = f_{\text{even}} + f_{\text{odd}} $. As I have mentioned in the wording of the problem, we know that $ \mathcal{R}^{f_{\text{odd}}}_{n} = 0 $ for all $ n \in \mathbb{N} $. Therefore, $ \mathcal{R}^{f}_{n} = 0 \iff \mathcal{R}^{f_{\text{even}}}_{n} = 0 $ for all $ n \in \mathbb{N} $. $\endgroup$ – Haskell Curry Jan 10 '13 at 23:48

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.

  • $\begingroup$ I need to check the details first. I kindly request your patience. :) $\endgroup$ – Haskell Curry Jan 7 '13 at 9:30
  • $\begingroup$ The solution looks good! $\endgroup$ – Haskell Curry Jan 10 '13 at 22:39
  • $\begingroup$ In the end, a Fourier expansion did help! $\endgroup$ – Haskell Curry Jan 10 '13 at 22:48

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