1

Let $f_n$ be a sequence of functions in $L^2_\text{loc}(\mathbb{R})$ which converge to a function $f\in L^2_\text{loc}(\mathbb{R})$ in the topology of $L^2_\text{loc}(\mathbb{R})$, i.e., $f_n\to f$ in $L^2(K)$ for all compact subsets $K\subset\mathbb{R}$.

A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=\left(\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T|f(x)|^2dy\right)^{1/2}$. Clearly, all Besicovitch almost periodic functions are locally square integrable.

Assuming that $f_n,f$ are Besicovitch almost periodic, can we also conclude that $f_n\to f$ in the seminorm of almost periodic functions, i.e, $|f_n-f|_2\to 0$?

Bernard
  • 175,478
Thangachelli Debopritama
  • 433

1 Answers1

2

No, convergence in $L_{loc}^2(\mathbb{R})$ does not imply convergence in the seminorm. Consider $$ f_n(x) = \sin\left( \frac{2\pi x}{n} \right) $$ Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_{loc}^2(\mathbb{R})$ to the zero function (which of course is Besicovitch almost periodic itself). However, we do not have convergence in the seminorm as (now we use that $f_n$ is $n$-periodic) $$ \vert f_n - 0 \vert_2^2 = \frac{1}{2n} \int_{-n}^{n} \vert f_n(x) \vert^2 dx = \frac{1}{2} \int_{-1}^{1} \vert \sin(2\pi y) \vert^2 dy =1. $$

Severin Schraven
  • 19,179
  • $\sin(x \frac{n+1}{n})$ is another example. To repair it we can try replacing $|f_n-f|_2$ by things like $| (f_n-f) \ast \phi_m|_2$ where $\phi_m(x) = m e^{-\pi m^2 x^2}$ – reuns Mar 10 '19 at 20:19
  • @reuns I would guess that fixing a common frequency module which has finite rank would allow us to conclude as well (however, I am not completely sure). – Severin Schraven Mar 10 '19 at 20:31
  • I meant replacing $ \frac{1}{2m} \int_{-m}^m |f_n(x)-f(x)|^2dx$ by $\int_{-\infty}^\infty (f_n-f)\ast\phi_m(x) e^{-\pi x^2/m^2} dx$. The convolution by $\phi_m$ is to repair your example, the multiplication by $e^{-\pi x^2/m^2}$ is to repair mine. – reuns Mar 10 '19 at 20:45
  • @reuns I meant that if we assumed in addition that all the function in the sequence are quasiperiodic with the same frequency module, then the result might be true. – Severin Schraven Mar 10 '19 at 21:08