Is it zero?

If $f$ is an even continuous function defined on $[−1, 1]$ and all its midpoint Riemann sums are zero ( i.e $\displaystyle \sum_{k=1}^{n} f \left ( \frac{2 k-1}{n} - 1 \right ) \cdot \frac{2}{n} = 0$ for every $n \in \mathbb{N}$ ), then is $f \equiv 0$ ?

Solution

Let $a_n: \mathbb{N} \cup \{ 0 \} \rightarrow \mathbb{R}$ be an absolutely summable sequence, and use it to define a function $f:[-1, 1] \rightarrow \mathbb{R}$ as

$f(x) = \sum_{n=0}^{\infty} a_n \cos n \pi x$

It is evident that $f$ is even. By the Weierstrass M-Test $\sum \limits_{n=0}^{\infty} a_n \cos n \pi x$ converges uniformly on $[-1, 1]$ to $f$ . Hence, $f$ is continuous, and using
the fact that $\int_{-1}^{1} \cos n \pi x \, \mathrm{d}x = 0$ we have that

$\begin{align*} \int_{-1}^{1} f(x) \, \mathrm{d}x &= \int_{-1}^{1} \sum_{n=0}^{\infty} a_n \cos n \pi x \, \mathrm{d}x \\ &= \sum_{n=0}^{\infty} \int_{-1}^{1} a_n \cos n \pi x \, \mathrm{d}x \\ &= 2 a_0 \end{align*}$

Next, we make two observations:

If $\displaystyle \sum_{k=1}^{N} f \left ( \frac{2k-1}{N} - 1 \right ) \cdot \frac{2}{N} = 0$ for every $N \in \mathbb{N}$ then by the definition of the Riemann integral we have that
$\begin{align*} 2a_0 &= \int_{-1}^{1} f(x) \, \mathrm{d}x \\ &= \lim_{N \rightarrow +\infty} \sum_{k=1}^{N} f \left ( \frac{2k-1}{N} - 1 \right ) \cdot \frac{2}{N} \\ &= 0 \end{align*}$

which yields $a_0=0$ .
Using the identity
$\sum_{k=1}^{N} \cos \left ( n \pi \left ( \frac{2k-1}{N} - 1 \right ) \right ) = \left\{\begin{matrix} (-1)^{n(N+1)/N } N & , & N \mid n \\ 0 & , & N \nmid n \end{matrix}\right.$

we obtain the following string of bi-implications for every $N \in \mathbb{N}$

$\begin{aligned} \sum_{k=1}^{N} f \left ( \frac{2k-1}{N} - 1 \right ) \cdot \frac{2}{N} =0 &\Leftrightarrow \sum_{k=1}^{N} \left ( \sum_{n=0}^{\infty} a_n \cos \left ( n \pi \left ( \frac{2k-1}{N} - 1 \right ) \right ) \frac{2}{N} \right ) =0 \\ &\Leftrightarrow \sum_{n=0}^{\infty} \left ( \sum_{k=1}^{N} a_n \cos \left ( n \pi \left ( \frac{2k-1}{N} - 1 \right ) \right ) \frac{2}{N} \right ) =0 \\ &\Leftrightarrow \sum_{n=0}^{\infty} a_n \sum_{k=1}^{N} \cos \left ( n \pi \left ( \frac{2k-1}{N} - 1 \right ) \right ) \frac{2}{N}=0 \\ &\Leftrightarrow \sum_{\substack{n \in \mathbb{N} \cup \{0 \} \\ N \mid n}} a_n (-1)^{n(N+1)/N} N \cdot \frac{2}{N} = 0 \\ &\Leftrightarrow \sum_{\substack{n \in \mathbb{N} \cup \{0 \} \\ N \mid n}} a_n (-1)^{n(N+1)/N} = 0 \end{aligned}$

In order for all of the mid-point Riemann sums of $f$ to be zero, it is thus
necessary and sufficient that

(i) $a_0=0$ and
(ii) $\displaystyle \sum_{\substack{n \in \mathbb{N} \cup \{0 \} \\ N \mid n}} a_n (-1)^{n(N+1)/N} = 0$

For this reason we choose the sequence $a_n$ as follows

$a_n = \left\{\begin{matrix} 0 & , & \text{if there exists no} \; j \in \mathbb{N} \cup \{0 \} \; \text{such that} \; n= 2^j \\\\ -1 & , & n=1\\\\ \dfrac{1}{n} & , & \text{if there exists a} \; j \in \mathbb{N} \; \text{such that} \; n= 2^j \end{matrix}\right.$

Finally, $f$ is continuous , even , non zero and its Riemann mid-points are zero. To see that $f$ is non zero we simply calculate $f(1)$ ;

$f(1) = -\cos \pi + \sum_{j=1}^{\infty} \frac{\cos 2^j \pi}{2^j} = 1 + \sum_{j=1}^{\infty} \frac{1}{2^j} = 2$