Consider the set of all trigonometrical polynomials of the form $P(x)=\sum_{j=1}^n a_n e^{ix\cdot\xi_n}$, where $\xi_n\in\mathbb{R}^d$. A function is said to be almost periodic in the sense of Bohr if it is a uniform limit of trigonometrical polynomials. Define mean value of a function $f\in L^1_{loc}(\mathbb{R}^d)$ to be the number $$\mathcal{M}(f)=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T f(x)\,dx.$$ Then, a function $u$ is said to be almost periodic in the sense of Besicovitch if there is a sequence of trigonometrical polynomials $P_n$ such that $\mathcal{M}(|u-P_n|^2)\to 0$ as $n\to\infty$. In most texts, it is mentioned that the quantity $(\mathcal{M}(|u|^2))^{1/2}$ is a semi-norm on the set of all Besicovitch almost periodic functions, however I have not found any text which would provide an example of a non-zero Besicovitch almost periodic function $u$ with $\mathcal{M}(|u|^2)=0$.

Is anyone aware of such an example? I am not able to construct such an example and it is certainly not true for trigonometrical polynomials. Of course, I mean examples which are not zero on a set of positive measure.

  • 1
    $\begingroup$ $$f(x)=\left\{ \begin{array}{lc} 1 \mbox{ if } 2^n < x <2^{n}+1 \\ 0 &\mbox{otherwise} \end{array}\right.$$ and $P_n=0$. $\endgroup$
    – N. S.
    Jun 13, 2020 at 18:55

1 Answer 1


Take any compactly suported $L^2$ function $f$ and set $P_n=0$, then $$ \lim_{n\rightarrow \infty} \mathcal{M}(\vert f - P_n \vert^2) = \mathcal{M}(\vert f \vert^2) =0 $$ thus, $f$ is Besicovitch almost periodic such that its seminorm vanishes.

  • $\begingroup$ Thank you for the answer but that is still the zero function upto measure zero. $\endgroup$ Sep 27, 2018 at 18:34
  • $\begingroup$ I am sorry, it was not clear to me that you wanted to exclude this case. My feeling is that we cannot do better than this, but I'll think about it. $\endgroup$ Sep 27, 2018 at 22:10
  • $\begingroup$ @TanujDipshikha I was wrong, there are a lot of such functions, see my edit. $\endgroup$ Sep 28, 2018 at 6:57
  • $\begingroup$ Hi, you can remove all the other answers and keep the Final answer since that is the simplest example. Thanks again. $\endgroup$ Oct 3, 2018 at 1:53
  • $\begingroup$ @TanujDipshikha I edited as you suggested. $\endgroup$ Oct 3, 2018 at 7:38

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