2
$\begingroup$

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 $n\geq N_0(\epsilon)$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – reuns Aug 20 '19 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.