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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);

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

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