The Besicovitch class of $B^p$ almost-periodic functions is defined as the closure of the set of trigonometric polynomials (of the form $t \mapsto \sum\_{n=1}^N a\_n e^{i \lambda\_n t}$ with $\lambda\_1, \dots, \lambda\_n \in \mathbb R$) under the semi-norm $$||f||\_{B^p} := \left(\limsup\_{X \to +\infty} \frac{1}{X} \int\_0^X |f(t)|^p \,\mathrm{d}t\right)^{1/p}.$$


Let $F \in B^1$ be such that there exists $(a\_n)\_{n \in \mathbb N} \in \ell^2(\mathbb N)$ and $(\lambda\_n)\_{n \in \mathbb N} \in \mathbb R^{\mathbb N}$ be such that $$\lim\_{N \to +\infty} ||F-P\_N||\_{B^1} = 0,$$ where $$P\_N := t \mapsto \sum\_{n=1}^N a\_n e^{i \lambda\_n t}.$$ Does $F \in B^2$ follow from the fact that $(a\_n)\_{n \in \mathbb N} \in \ell^2(\mathbb N)$ ?


Litterature on the subject is a bit hard to find since there are so many different classes of almost-periodic functions, of which $B^1$ is one of the largest. Also $|| \cdot ||\_{B^p}$ is only a semi-norm so this might complicate things.



On $B^1$ and $B^2$ almost-periodic functions
**The answer depends on what exactly you mean by the question.** The subtle thing about $B^p$ is that it represents not functions by classes, and the classes depend on $p$.


This is a problem since if you replace $F$ by another representative $F'$ in its class, you could have $F \in B^2$ but $F' \notin B^2$.


The core of the issue, as you guessed, comes from the fact that the Besicovitch seminorm is not a norm.




---


Here is what you can prove:


**Claim** Under the given conditions, there exists a representative $G$ for the class of $F$ (i.e. $\| F-G \|\_{B^1}=0$) such that $G \in B^2$.


**Proof** Since $(a\_n) \in L^2$ you get that $P\_n$ is a Cauchy sequence in $(B^2, \| \, \|\_{B^2})$. Since this space is complete, there exists an element $G \in B^2$ such that
$$
\lim\_N \|P\_N -G \|\_{B^2}=0
$$


Now, the Cauchy--Schwarz inequality gives
$$
\left(\frac{1}{X} \int\_0^X |f(t)| \,\mathrm{d}t\right) \leq \left(\frac{1}{X} \int\_0^X |f(t)|^2 \,\mathrm{d}t\right)^{1/2} \left(\frac{1}{X} \int\_0^X 1^2\,\mathrm{d}t\right)^{1/2}.
$$
which gives
$$
\| \, \|\_{B^1} \leq \| \, \|\_{B^2} \,.
$$


*Side note here: This implies that if $\| G-G'\|\_{B^2}=0$ then $\|G-G'\|\_{B^1}=0$ but the converse is not true, which is the core of the issue.*


From here we get that
$$
\lim\_N \|P\_N -G \|\_{B^1}=0
$$


Since you are given that
$$
\lim\_N \|P\_N -F \|\_{B^1}=0
$$


you get
$$\| F-G \|\_{B\_1}=0$$
as claimed
\qed


**Example where $F \notin B^2$**


It is easy to come up with an example of a function $F$ such that $\|F\|\_{B^1}=0$ but $\|F\|\_{B^2}=\infty$ (for example $F=\sum\_{n \in \mathbb N} b\_n 1\_{[2^n, 2^n+1]}$ will work for the right $b\_n$).


This function trivially satisfies the conditions of your statement with $a\_n=0 \forall n$, but $\| F \|\_{B^2}=\infty$ implies that $F \notin B^2$.


As in the above claim, $F$ does have some $G \in B^2$ in its class, namely $G=0$.


