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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 $M^2$?

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    $\begingroup$ I would not call $M^p$ functions (such as your $f$) "almost periodic" at all. $\endgroup$
    – Gerald Edgar
    Commented Mar 2, 2014 at 13:13
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    $\begingroup$ Of course. The $B^p$ functions have all the nice properties, like plenty of $epsilon$-translation numbers, which make them almost periodic. The $M^p$ functions almost just seem like a nice place to house the $B^p$. $\endgroup$
    – Greg Zitelli
    Commented Mar 2, 2014 at 15:14
  • $\begingroup$ Simple-minded question: why does the limit in your defining formula exist? or do you have to build that into your definition of $M^p$? $\endgroup$
    – Yemon Choi
    Commented Mar 2, 2014 at 19:06
  • $\begingroup$ And if the limit exists for $f$ and $g$, does it follow that it also exists for $f+g$? Greg, you say it is a Banach space, but perhaps you should explain. $\endgroup$
    – Gerald Edgar
    Commented Mar 2, 2014 at 19:19
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    $\begingroup$ The definition is careless. Although relying solely on a supremum might get me into trouble, I think everything will work properly with a limsup (Besicovitch uses these upper means in his own investigation of the almost periodics). Then we look at locally $p$-integrable functions finite under this seminorm (a real seminorm this time), which will be a Banach space after modding out and completing (if necessary). $\endgroup$
    – Greg Zitelli
    Commented Mar 2, 2014 at 19:59

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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:

$f(x)$
f

$g(x)$
g

$f(x)+g(x)$
f+g

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): $$ \limsup_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2} =\frac{2}{\sqrt{3}}, $$

$$ \liminf_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2} =\frac{\sqrt{2}}{\sqrt{3}} $$

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With the "limsup" definition, as suggested by Jean Van Schaftingen, we contradict the parallelogram law, since $$ \|f\|_{M^2} = \|g\|_{M^2} = \frac{1}{\sqrt{2}},\qquad \|f+g\|_{M^2} = \|f-g\|_{M^2} = \frac{2}{\sqrt{3}} $$

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    $\begingroup$ In the definition above, $M^p$ is defined by the limsup being finite, so I think that $M^p$ is indeed a vector space. However your example might show that the semi-norm induced by this limsup on $M^2$ is not Euclidean as it would not satisfy the parallelogram identity. $\endgroup$
    – Jean Van Schaftingen
    Commented Mar 3, 2014 at 9:01
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    $\begingroup$ When questions are changed, answers may not longer apply... editing to say I answered the original question. $\endgroup$
    – Gerald Edgar
    Commented Mar 3, 2014 at 14:44

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