On $B^1$ and $B^2$ almost-periodic functions 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.
 A: 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 $G=0$.
