# Convergence in the topology of $L^2_\text{loc}$ implies convergence in $B^2$?

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, $$|f_n-f|_2\to 0$$?

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) $$\vert f_n - 0 \vert_2^2 = \frac{1}{2n} \int_{-n}^{n} \vert f_n(x) \vert^2 dx = \frac{1}{2} \int_{-1}^{1} \vert \sin(2\pi y) \vert^2 dy =1.$$
• $\sin(x \frac{n+1}{n})$ 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 $\phi_m(x) = m e^{-\pi m^2 x^2}$ Mar 10, 2019 at 20:19
• 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. Mar 10, 2019 at 20:45