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, $|f_n-f|_2\to 0$?
Best Answer
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) $$ \vert f_n - 0 \vert_2^2 = \frac{1}{2n} \int_{-n}^{n} \vert f_n(x) \vert^2 dx = \frac{1}{2} \int_{-1}^{1} \vert \sin(2\pi y) \vert^2 dy =1. $$