Re: Sum of periodic functions with incommensurate periods


David C. Ullrich wrote:
> On Fri, 14 Oct 2005 08:45:31 -0500, David C. Ullrich
> <ullrich@xxxxxxxxxxxxxxxx> wrote:
>
> >On Thu, 13 Oct 2005 09:12:43 EDT, Victor <victor_88@xxxxxxx> wrote:
> >
> >>Greetings!
> >>
> >>I'm desperately looking for a proof (or a disproof) of the following fact.
> >>
> >>Let f(x) and g(x) be two periodic functions R->R with least positive periods T1 and T2 respectively, and T1 and T2 are incommensurate (say, T1=1, T2=sqrt(2) ). Then f(x)+g(x) is not periodic.
> >>
> >>There are absolutely no restrictions on f(x) and g(x), they are not necessarily continious or have any other properties, they are just functions from R to R.
> >>
> >>Is is true? Is yes, then how can it be proven?
> >>
> >>Thank you very much in advance!
> >
> >I mentioned this to the irritatingly smart guy upstairs. A few hours
> >later he told me he'd given a counterexample - I found this
> >sufficiently irritating that I came up with my own counterexample
> >a few hours later.
> >
> >The counterexample is not so hard, surprised nobody's posted an
> >example yet. Our two examples are based on the same idea, although
> >his was more elegant - I'll give my version.
>
> Arriving at the office I found that the guy in question
> (one Anthony Kable, definitely a very dangerous sort of
> guy to have around) had vastly simplified things using
> the same idea. What follows is what he said, somewhat
> simplified:
>
> Say a, b, c are rationally independent reals. Let G be
> the subgroup of R generated by a, b, c.
>
> Define f by f(x) = 0 if x is not in G, and
>
> f(ia + jb + kc) = jb + kc (i, j, k integers).
>
> Then f certainly has period a, and it has no smaller
> period: If p is not in G then f(b+p) = 0 while
> f(b) = b; on the other hand if p is in G and
> f(p) = f(0) then f(p) = 0, which implies p = ia
> for some integer i.
>
> Similarly define g by g(x) = 0 if x is not in G, and
>
> g(ia + jb + kc) = ia - kc (i, j, k integers).
>
> Then g has smallest period b. And f + g = ia + jb,
> so f + g has smallest period c.
>
> >I don't know whether f and g can be measurable
>
> The above is measurable.
>
> >(They can't be tempered distributions...)
>
> I was aware that this wasn't quite right when I wrote
> it. For example the above functions are certainly
> tempered distributions, being 0 almost everywhere.
> What f and g cannot be is tempered distributions,
> neither of which is constant _as_ a distribution;
> considered as distributions both the functions above
> are identically 0.
>
> We don't know whether f and g can be _bounded_ measurable
> _functions_.
>

Can't you just choose some integer d <=2 and let
f(ia + jb + kc) = j sign(b) min(|b|,d) + k sign(c) min(|c|,d)

>
>
> David C. Ullrich

.

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