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Mar 5, 2014 at 17:39 vote accept Greg Zitelli
Mar 4, 2014 at 19:48 comment added Delio Mugnolo also, I would refrain from calling this the common example of a nonseparable Hilbert space.
Mar 3, 2014 at 14:52 comment added Gerald Edgar If it is, in fact, incomplete, I think that would answer your question.
Mar 2, 2014 at 20:05 answer added Gerald Edgar timeline score: 4
Mar 2, 2014 at 20:00 history edited Greg Zitelli CC BY-SA 3.0
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Mar 2, 2014 at 19:59 comment added Greg Zitelli 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).
Mar 2, 2014 at 19:19 comment added Gerald Edgar 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.
Mar 2, 2014 at 19:06 comment added Yemon Choi Simple-minded question: why does the limit in your defining formula exist? or do you have to build that into your definition of $M^p$?
Mar 2, 2014 at 15:14 comment added Greg Zitelli 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 $B^p$.
Mar 2, 2014 at 13:13 comment added Gerald Edgar I would not call $M^p$ functions (such as your $f$) "almost periodic" at all.
Mar 2, 2014 at 3:20 history asked Greg Zitelli CC BY-SA 3.0