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Let $f$ and $g$ be two periodic functions over $\Bbb{R}$ with the following property: If $T$ is a period of $f$, and $S$ is a period of $g$, then $T/S$ is irrational.

Conjecture: $f+g$ is not periodic.

Could you give a proof or a counter example? It is easier if we assume continuity. But is it true for arbitrary real valued functions?

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i should not be periodic, but at the moment I cannot find a proof :-( –  mau Jul 29 '10 at 15:12
    
I think if f and g are continuous then their sum is never periodic. Are you more interested in the general case? –  Qiaochu Yuan Jul 29 '10 at 17:42
    
Qiaochu: That's exactly the case. Sorry about the confusion, it was my fault. I have updated the question. –  AgCl Jul 29 '10 at 17:55
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4 Answers

up vote 14 down vote accepted

Here is a counterexample. Let $a, b, c \in \mathbb{R}$ be linearly independent over $\mathbb{Q}$. Let $\text{span}(x, y, z, ...)$ be the $\mathbb{Q}$-vector space in $\mathbb{R}$ spanned by $x, y, z, ...$. Let $AB = \text{span}(a, b), BC = \text{span}(b, c), AC = \text{span}(a, c)$. And for a subset $S$ of $\mathbb{R}$, let $\chi_S$ denote the characteristic function of $S$. Now define

$\displaystyle f(x) = \chi_{AB} - 2 \chi_{BC}$

and

$\displaystyle g(x) = 3 \chi_{AC} + 2 \chi_{BC}.$

Then $f$ has period set $\text{span}(b)$, $g$ has period set $\text{span}(c)$, and $f + g$ has period set $\text{span}(a)$. (I am not sure if the coefficients are necessary; they're just precautions.)

Are you still interested in the continuous case?


(Old answer below. I slightly misunderstood the question when I wrote this.)

Here is a simpler example. I claim that the function $h(x) = \sin x + \sin \pi x$ cannot possibly be periodic. Why? Suppose an equation of the form

$\sin x + \sin \pi x = \sin (x+T) + \sin \pi (x+T)$

held for all $x$ and some $T > 0$. Take the second derivative of both sides with respect to $x$ to get

$\sin x + \pi^2 \sin \pi x = \sin (x+T) + \pi^2 \sin \pi(x+T).$

This implies that $\sin x = \sin (x+T)$ and that $\sin \pi x = \sin \pi(x+T)$, which is impossible.

(Or is the question whether the sum can be periodic?)

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I've removed the comment that was here about your older answer, but I wish I could upvote this post again. :-) –  ShreevatsaR Jul 31 '10 at 5:08
    
Thanks Qiaochu! That's a very nice counterexample. Also, I'd really like to see your proof for the continuous case. –  AgCl Jul 31 '10 at 12:17
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@AgCl: It's sort of annoying to write out in full, so here's a sketch. First show that continuous functions have minimal periods. If f+g were periodic with period T then you could find multiples of T which approximate multiples of the period of f. In particular, you could find multiples of T such that f(nT) approaches its maximum on a period. If you do this, then g(nT) should wander around its period and in particular be taking different values, so f(nT)+g(nT) can't be constant. –  Qiaochu Yuan Jul 31 '10 at 17:24
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Pick a basis $B$ of $\mathbb R$ as a $\mathbb Q$ vector space, and split it into two non-empty disjoint parts $B_1$ and $B_2$. Define $\mathbb Q$-linear maps $f,g:\mathbb R\to\mathbb R$ such that $f(x)=x$ and $g(x)=0$ if $x\in B_1$, $f(x)=0$ and $g(x)=x$ if $x\in B_2$. Then $f(x)+g(x)=x$ for all $x\in B$, so that in fact $f+g=\operatorname{id}_{\mathbb R}$, which is not a periodic function. Morever $f$ and $g$ are periodic, and their sets of periods are precisely $B_1$ and $B_2$. Since $B_1\cup B_2$ is linearly independent over $\mathbb Q$, it is easy to see that $x/y\not\in\mathbb Q$ whenever $x\in B_1$ and $y\in B_2$.

This is then an example where the sum is not periodic.

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If each function has a smallest period, and otherwise fits the conditions, then a proof may be forthcoming by attempting to compute the smallest period of the sum and failing. However, things become unclear if there is no smallest period, as in the case of the characteristic function of the rationals. Progress might be made in this case by decomposing such a function as an infinite sum of periodic functions, or at least give more counterexamples to study. (e.g. Write the characteristic function of the rationals as an infinite sum of functions of smallest period 1. )

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is it possible to build two different function with no smallest period and such that all periods are irrational? (yes, I think it should) –  mau Jul 29 '10 at 20:25
    
Yes. For any irrational number a, the characteristic function of the rational multiples of a works. –  Qiaochu Yuan Jul 29 '10 at 20:31
    
Take c(x), the above characteristic function of the rationals, and form d(x) = c(x * sqrt(2) ). However, you should be able to prove that any function which has uncountably many periods (and may also need some weak condition resembling continuity) must be constant. [signature removed by moderator] –  G. Paseman Jul 29 '10 at 20:36
    
@Gerard: no dice without a continuity assumption. Take, for example, the characteristic function of the Q-vector space spanned by all but one element of a Hamel basis for R over Q. –  Qiaochu Yuan Jul 30 '10 at 21:14
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as a first start, if f+g is periodic the period cannot be rational wrt to the period of f and g.

Let us suppose that T is the smallest period of f(x), that is ∀x, f(x+T) = f(x); similarly S is the smallest period of g(x), that is ∀x, g(x+S) = g(x). If f+g had a period Q, and Q/T = m/n, we have that ∀x, f(x+nQ)+g(x+nQ) = f(x)+g(x). But f(x+nQ)=f(x+mT)=f(x), thus ∀x, g(x+nQ)=g(x) and therefore nQ is a period of g, which is impossible since it would mean that T/S is rational.

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This is correct even if T and S aren't the smallest periods of f and g, BTW. This matters because there may be no smallest period. –  ShreevatsaR Jul 30 '10 at 18:08
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