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Suppose that a non-constant function $\gamma : \mathbb{R} \to \mathbb{R}$ is $T$ -periodic for some $T \neq 0$. This exercise shows that there is a smallest positive number $T_0$ such that $\gamma$ is $T_0$-periodic. The proof uses a little real analysis. Suppose for a contradiction that there is no such $T_0$.

  1. Show that there is a sequence $T_1, T_2, T_3, \dots$ such that $T_1 > T_2 > T_3 > \dots > 0$ and that $\gamma$ is $T_r$-periodic for all $r \geq 1$.
  2. Show that the sequence $\{T_r\}$ of $(1.)$ can be chosen so that $T_r \to 0$ as $r \to \infty$.
  3. Show that the existence of a sequence $\{T_r\}$ as of $(1.)$ such that $T_r \to 0$ as $r \to \infty$ implies that $\gamma$ is constant.

At the First step I am facing difficulties to show that such a sequence exist.

Any hints how to do that?

Willie Wong
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    Let $\gamma(x)=0$ if $x$ is rational, and $\gamma(x)=1$ if $x$ is irrational. Every positive rational is a period. I expect you want to put an additional condition on $\gamma$. – André Nicolas Oct 12 '15 at 21:09

2 Answers2

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If there is no such $T_0$, by definition that means that for every $T_0 > 0$ there exists $0 <T^\prime <T_0$ such that $\gamma$ is $T^\prime$-periodic.

Use this to define the sequence iteratively: to get $T_1$, take $T_0=1$ in the above, and let $T_1$ be the corresponding $T^\prime$. Then, "take $T_0=T_1$" in the above, and let $T_2$ be the new "corresponding $T^\prime$. Etc.: given $T_n$, define $T_{n+1}$ to be the "$T^\prime$" corresponding to "$T_0=T_n$."

For the second question, one "trick" is to define the sequence above by ensuring it decreases at least by a factor $2$ every time. I.e., take $T_{n+1}$ to be the $T^\prime$ corresponding to "$T_0=\frac{T_n}{2}$", not to "$T_0=T_n$".

Edit: additionally, as André Nicolas points out, you will not be able to prove 3. without some additional assumption on $\gamma$. Indeed, as stated it is false -- there are non-constant functions without a smallest positive period.

Clement C.
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    If we assume that $\gamma$ is continuous how can we conclude that it is constant? –  Oct 20 '15 at 19:40
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    Basically, it will boil down to the fact that the periods are an additive subgroup of $\mathbb{R}$; and "thus" that they either form a dense subset of $\mathbb{R}$, or admit a least positive element. If the latter holds, you are done: this is the period. If not, then the set is dense; but then continuity of $f$ will come to play, as $f$ will take the same value on each element of this set (and by continuity, on $\mathbb{R}$). – Clement C. Oct 20 '15 at 19:46
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    See also this answer. – Clement C. Oct 20 '15 at 19:47
  • For the first question do we have to pick $T_0=1$ or isn't it necessary? @ClementC. –  Oct 24 '15 at 23:01
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    You can take any $T_0>0$ you want. – Clement C. Oct 24 '15 at 23:02
  • Why do we have to pick a specific $T_0$ ?@ClementC. –  Oct 24 '15 at 23:06
  • Just to start the construction of the sequence somewhere, basically. If you don't have a value for $T_0$, how do you define $T_1$? – Clement C. Oct 24 '15 at 23:07
  • I understand. So we have shown that there is such a sequence. We have also shown in this way that $\gamma$ is $T_r$-periodic for all $r \geq 1$, haven't we? @ClementC. –  Oct 24 '15 at 23:14
  • Or do we have to show it in an other way? –  Oct 24 '15 at 23:17
  • No, this is not immediately clear for all $r\geq 1$. But this will follow from the fact that the set of periods $P_f$ of $f$ is an additive group on $\mathbb{R}$, and does not have a least positive element (since it contains elements arbitrarily close to $0$). "So" $P_f$ must be a dense subset of $\mathbb{R}$. – Clement C. Oct 24 '15 at 23:20
  • I haven't understood this. Isn't there an other way to show it? –  Oct 24 '15 at 23:24
  • You may want to look at André Nicolas's answer, where he added an argument that does not rely explicitly on this result about density of additive subgroups. But basically, I don't understand why you want that statement about all $r\geq 1$ being periods. – Clement C. Oct 24 '15 at 23:27
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We concentrate on 2). Suppose that $\gamma$ is periodic but has no smallest positive period. We show that there is a sequence of periods of $\gamma$ with limit $0$.

Let $W$ be the infimum of all the periods. Suppose that $W$ is positive, and $W$ is not a period. Then there exists a strictly decreasing sequence of $W_1,W_2,W_3,\dots$ of periods, with limit $W$. In particular, for any integer $m$ there exist periods $W_i\lt W_j$ such that $W_j-W_i\lt \frac{1}{m}$. But $\gamma(x+W_j-W_i)=\gamma(x-W_i)=\gamma(x)$, We conclude that for any positive integer $m$, $\gamma$ has a positive period $\lt \frac{1}{m}$.

So we have shown that if there is no smallest positive period, there is sequence of periods with limit $0$.

For 3), as pointed out in a comment, we will need to make a further assumption on $\gamma$. Continuity will do it.

Added: We show that if there is a point $a$ at which the function $\gamma(x)$ is continuous, then $\gamma$ is a constant function. We include more or less full detail.

Suppose that $\gamma(a)=c$, and that there exists a $b$, and a $d\ne c$ such that $\gamma(b)=d$. We will derive a contradiction. It then will follow that $\gamma(x)=c$ for all $x$.

By the result of 2), there is a strictly decreasing sequence $(W_i)$ of periods, with limit $0$. In particular, we may choose the $W_i$ so that $W_i\lt \frac{1}{i}$ for any positive integer $i$.

Since $\gamma(b)=d$, we have $\gamma(b+kW_i)=\gamma(b)=d$ for every integer $k$, positive, negative, or $0$. Since $W_i\lt \frac{1}{i}$, it follows that for any positive integer $i$ there is a $b_i$ such that $|b_i-a|\lt \frac{1}{i}$ and $\gamma(b_i)=d$.

So we have found a sequence $(b_i)$ such that $\lim_{i\to\infty} b_i=a$ and $\gamma(b_i)=d$ for all $i$. Since $f(a)=c\ne d$, this contradicts the continuity of $\gamma(x)$ at $x=a$.

André Nicolas
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  • Also how do we know that there is a $W_i$ so that $W_i\lt \frac{1}{i}$ for any positive integer $i$?

    $$$$ And could you explain me how we get to the following? $$$$ Since $W_i\lt \frac{1}{i}$, it follows that for any positive integer $i$ there is a $b_i$ such that $|b_i-a|\lt \frac{1}{i}$ and $\gamma(b_i)=d$.

     –  Nov 04 '15 at 23:42
  • Why do we know that there is a period $W_i$ of length $< \frac{1}{i}$ ? Also do we know that for some $m \in \mathbb{Z}$ , $\gamma(b + m W_i)= \gamma(a)$ since $\gamma$ is $W_i$-periodic? –  Nov 05 '15 at 17:54
  • We have gone through both of these. In particular, the first part of the answer, before the Added part, shows that under the assumption there is no smallest period, there are arbitrarily small positive periods. By the way, the sequence of comments has gotten out of control long, and the system keeps objecting. I will delete most of my comments and suggest you do the same. – André Nicolas Nov 05 '15 at 18:05
  • I see. But how do we conclude that the distance from $a$ is $< \frac{1}{i}$ ? –  Nov 05 '15 at 18:10
  • I had explained that already, now deleted. There is a period of length $W_i\lt 1/i$. Now suppose $\gamma(b)=d$. Then $\gamma(b+\pm kW_i)=d$ for any integer $k$. But $W_i$ has length $\lt 1i$. So by taking steps of length $W_i$, we can get from $b$ to within $1/i$ of $a$. – André Nicolas Nov 05 '15 at 18:14
  • I have understood this. I was wondering why there is necessarily a $k \in \mathbb{Z}$ such that $b+kW_i=a$. –  Nov 05 '15 at 18:17
  • Let $a=2$, and let $0\lt W\lt 1/1000$. Let $b=17.131313\dots$. Can we find an integer $k$ (here negative) such that $|a+kW|\lt 1/1000$? Sure, keep stepping backwards from $b$ in steps of length $W$. There is a smallest $m$ such that $b-mW\le a$.. Then $b-(m-1)W\gt a$, and $a-(m-1)W\le W\lt 1/1000$. More informally, a frog starts hopping backwards from $b$ taking hops of tiny length $W\lt 1/1000$. Then after a while she will be at distance $\le W$ from $a$. If she continues hopping, she will then get further from $a$. But the point is that at some time she will be close to $a$. – André Nicolas Nov 05 '15 at 18:37