Re: Sum of periodic functions with incommensurate periods

On 14 Oct 2005 09:14:00 -0700, "David M Einstein" <Deinst@xxxxxxxxx>
wrote:

>David C. Ullrich 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.
>>
>> First, you should note that a periodic function need not have
>> a smallest positive period! For example the characteristic function
>> of the rationals is periodic... it's not clear to me whether
>> "with least positive periods T1 and T2 respectively" was
>> supposed to be an additional hypothesis or just a declaration
>> of a name for these things that need not exist.
>>
>> It's easy to see that there exist periodic f and g with no
>> common period, such that f + g is periodic (which may or
>> may not answer your question). Note that R is a vector
>> space over Q. Although people don't usually think of it
>> that way, a "period" of a linear operator is precisely
>> a non-zero element of the kernel. If you look at bases
>> it's easy to "construct" examples of Q-linear f and g
>> with non-trivial kernels, with kernels disjoint except
>> for 0, such that f + g also has a non-trivial kernel,
>> and that gives a counterexample, _except_ that these
>> f and g do not _have_ smallest positive periods. You
>> can get an example where f and g have smallest positive
>> periods by a similar construction, with functions
>> that are not quite Q-linear:
>>
>> Let B be a basis for R as a Q-vector space,
>> and say b1, b2, b3 are distinct elements of B.
>> "b" will always denote an element of B. Say
>> c_b(x) is the coefficient of b in when x is
>> written as a Q-linear combination of elements of B.
>>
>> Let
>>
>> f(sum_b c_b(x) b) = b1 + sum_{b <> b1} c_b(x) b,
>>
>> unless c_b1(x) is an integer, in which case
>>
>> f(sum_b c_b(x) b) = sum_{b <> b1} c_b(x) b.
>>
>> Then f has period b1, and the zero set of f
>> is precisely the integer multiples of b1, so
>> f has no smaller positive period.
>>
>> Let b' = b unless b = b3; let (b3)' = -b3.
>> Let
>>
>> g(sum_b c_b(x) b) = b2 + sum_{b <> b2} c_b(x) b',
>>
>> unless c_b2(x) is an integer, in which case
>>
>> g(sum_b c_b(x) b) = sum_{b <> b2} c_b(x) b'.
>>
>> Then g has smallest positive period b2, and
>> f+g has period b3.
>>
>> I believe that's right - I can fill in details
>> for any of the assertions that are not clear,
>> I think.
>>
>> In that last f+g has arbitrarily small positive
>> periods. Probably that's more interesting than
>> getting f+g to have a smallest positive period;
>> if you'd prefer that f+g also have a smallest
>> positive period then you could probably modify
>> the above, jiggling the way f and g depend on
>> c_b3(x).
>
>I do not see how arbitrarily small periods occur.
>If we perform a parallel construction in
>Q(sqrt(2),sqrt(3)), with b1=1, b2=sqr(2),
>b3=sqrt(3), f+g has minimal period sqrt(3).
>Am I missing something?

I don't know exactly what you mean by "parallel
construction", unless it's the same construction...

Anyway, in the example above any rational multiple
of b3 is a period of f+g. This is because f+g
is depends on all the c_b _except c_b3: if
y = x + r b3 where r is rational then c_b(y)
= c_b(x) for b <> b3, and hence (f+g)(y) = (f+g)(x).

You should pay no attention to this construction
anyway - it's been vastly simplified and improved
(see parallel post).

>> I don't know whether f and g can be measurable
>> (They can't be tempered distributions...)
>>
>> ************************
>>
>> David C. Ullrich
>


David C. Ullrich
.

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