# Convergence in the topology of $B^2$ implies convergence in $L^2_{loc}(\mathbb{R}^d)$

Regarding the question here, I wanted to ask whether the statement is true the other way round.

Let $$f_n$$ be a sequence of functions in the Besicovitch space of almost periodic functions $$B^2(\mathbb{R}^d)$$ converging to a function $$f\in B^2(\mathbb{R}^d)$$ in the norm of $$B^2(\mathbb{R}^d)$$, i.e., $$\mathcal{M}(|f_n-f|^2)\to 0$$ as $$n\to\infty$$.

Recall that the Besicovitch space of almost periodic functions is defined as the closure of trigonometrical polynomials in the seminorm given by $$\displaystyle|g|_{B^2}=(\mathcal{M}(|g|^2))^{1/2}=\left(\limsup_{T\to\infty}\frac{1}{|TY|}\int_{TY}|f(y)|^2\,dy\right)^{1/2}$$, where $$Y=[-1,1]^d$$.

Question: Then, it is clear that the functions $$f_n$$ and $$f$$ belong to $$L^2_{loc}(\mathbb{R}^d)$$. Is it true that $$f_n\to f$$ in $$L^2(K)$$ for all compact subsets $$K\subseteq\mathbb{R}^d$$?

In particular, I had some difficulties with the proof given in Corduneanu's book "Almost Periodic Oscillations and Waves". He argues that since $$\mathcal{M}(|f_n-f|^2)\to 0$$, for all $$\epsilon>0$$, there is $$N(\epsilon)$$ such that for all $$n\geq N(\epsilon)$$, we have $$\displaystyle\lim_{T\to\infty}\frac{1}{|TY|}\int_{TY}|f_n(y)-f(y)|^2\,dy<\epsilon.$$ Then it is claimed that there exists $$T_0(\epsilon)$$ such that for all $$T>T_0(\epsilon)$$, $$\displaystyle\frac{1}{|TY|}\int_{TY}|f_n(y)-f(y)|^2\,dy<\epsilon,$$ or $$\displaystyle\int_{TY}|f_n(y)-f(y)|^2\,dy<\epsilon|TY|.$$ From this he claims that the sequence $$f_n$$ is convergent in $$L^2(TY)$$. However, I feel that the $$T_0(\epsilon)$$ would also depend on $$n$$. How does one choose $$T_0$$ uniformly for all $$n\geq N_0(\epsilon)$$?

• If you define your space as being given by the norm $\|f\|^2 =\sup_T \frac{1}{T}\int_{TY}|f(y)|^2\,dy$ then the claim is obvious. Otherwise it is not. – reuns Aug 20 '19 at 14:20