TITLE: Besicovitch Almost Periodic Functions a subspace of what?
QUESTION [5 upvotes]: The common example of a nonseparable Hilbert space comes from the collection of Besicovitch almost periodic function spaces. Starting with $L^p_{\text{loc}}(\mathbb{R})$ we look at those elements finite under the seminorm
$$
\|f\|_{M^p} = \limsup_{R\to\infty} \left(\frac1{2R}\int_{-R}^R \left|f(x)\right|^pdx\right)^{1/p}
$$
Modding out (and completing?) by the zero elements nets us a Banach space for $1 \le p <\infty$, and a Hilbert space with the natural inner product for $p=2$. We'll call this space $M^p$.
We can see that $M^2$ is an example of a nonseparable Hilbert space because the collection $e^{i\xi x}$ is orthonormal for all $\xi \in \mathbb{R}$. We can look at the subspace $B^p\subseteq M^p$ of elements spanned by these functions, called the Besicovitch almost periodic functions.
We can see that $B^2\neq M^2$ since there are functions like 
$$
f(x) = \left\{\begin{align}1 \ \ \ \ \  x\ge0 \\ -1 \ \ \ \ \ x < 0\end{align}\right.
$$
which is orthogonal to all $e^{i\xi x}$, and $\|f\|_2 = 1$.
Question: I can't seem to find any discussion of $M^p$ independent from $B^p$. Is there a standard name for $M^p$? Is there a convenient description of an orthonormal basis for $M^2$?

REPLY [2 votes]: OK.  Perhaps a reason $M^p$ is not often studied is: it is not even a  vector space.
(using the original defintion with lim not limsup.)  
Define functions $f$ and $g$ as follows:
$f(x)=0$ if $x<1$,
$f(x)=1$ if $x \ge 1$ and $\{x\}< 1/2$; here, $\{x\} = x-\lfloor x\rfloor$ is the fractional part
$f(x)=-1$ if $x \ge 1$ and $\{x\} \ge  1/2$.  
$g(x)=0$ if $x<1$,
$g(x)=f(x)$ if $x \ge 1$ and $\lfloor \log_2(x)\rfloor$ is even,
$g(x)=-f(x)$ if $x \ge 1$ and $\lfloor \log_2(x)\rfloor$ is odd.  
Some graphs:
$f(x)$
 
$g(x)$
 
$f(x)+g(x)$
 
But note:
$$
\lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)|^2\right)^{1/2} =
\lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |g(x)|^2\right)^{1/2} = \frac{1}{\sqrt{2}}
$$
both exist, while
$$
\lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2}
$$
does not exist.  In fact (do some computations):
$$
\limsup_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2}
=\frac{2}{\sqrt{3}},
$$  
$$
\liminf_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2}
=\frac{\sqrt{2}}{\sqrt{3}}
$$
added
With the "limsup" definition, as suggested by Jean Van Schaftingen, we contradict the parallelogram law, since
$$
\|f\|_{M^2} = \|g\|_{M^2} = \frac{1}{\sqrt{2}},\qquad
\|f+g\|_{M^2} = \|f-g\|_{M^2} = \frac{2}{\sqrt{3}}
$$