# Universal Chord Theorem

Let $f \in C[0,1]$ and $f(0)=f(1)$.

How do we prove $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$?

In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$.

For any other non-zero real $r$ (i.e not of the form $\frac{1}{n}$), there is a continuous function $f \in C[0,1]$, such that $f(0) = f(1)$ and $f(a) \neq f(a+r)$ for any $a$.

This is called the Universal Chord Theorem and is due to Paul Levy.

Note: the accepted answer answers only the first question, so please read the other answers too, and also this answer by Arturo to a different question: https://math.stackexchange.com/a/113471/1102

This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.

and here: List of abstract duplicates.

• The statement should seem intuitive to you, and when it does the proof should follow fairly easily. What we have here (can be thought of as) a loop in $\mathbb{R}^2$ parameterized by $t \in [0,1]$. Picture the values of $f(a)$ and $f(a+ 1/2)$ as $a$ varies continuously from 0 to 1/2. Can you visualize why these values must equal each other at some point? Hope this at least makes sense and possibly helps :) – jericson Jan 4 '11 at 23:23
• The function f(x) = x^2 - x satisfies the problem assumptions but isn't periodic. – Qiaochu Yuan Jan 5 '11 at 0:47
• There is a nice exposition of this in an issue of American Mathematical Monthly, circa 1970, where it is called the theorem of the horizontal chord. It gives an example of a piece-wise linear function with no horizontal chord of some value $a.$ ( I don't remember what $a$ was in the example.) – DanielWainfleet Dec 5 '17 at 11:29

## 4 Answers

You want to use the intermediate value theorem, but not applied to $f$ directly. Rather, let $g(x)=f(x)-f(x+1/2)$ for $x\in[0,1/2]$. You want to show that $g(a)=0$ for some $a$. But $g(0)=f(0)-f(1/2)=f(1)-f(1/2)=-(f(1/2)-f(1))=-g(1/2)$. This gives us the result: $g$ is continuous and changes sign, so it must have a zero.

• You define $g(x)=f(x) - f(x+1/2)$ for $x \in [0,1/2]$ and you get that $g(0)=-g(1)$. Why $a \in [0,1/2]$ then? – Ma.H Jan 4 '11 at 23:26
• @Andres: I vaguely recall seeing an extension of this to show what values of a (besides 1/2) you could prove that you could find an x such that f(x)=f(x+a). I think this goes easily to 1/n for any n, but there may be more. – Ross Millikan Jan 4 '11 at 23:28
• @Ma.H: It's a typo, should be $g(0)=-g(1/2)$. – Shai Covo Jan 4 '11 at 23:35
• (Sorry about the typo.) @Ross: I don't know, it doesn't ring a bell but sounds interesting. I'll see if I can think of a non-trivial extension. – Andrés E. Caicedo Jan 5 '11 at 0:39
• @Ross: $1/n$ is right, and those are the only ones. See my answer. – Aryabhata Jan 5 '11 at 6:21

Interestingly,

The numbers of the form $r = \displaystyle \frac{1}{n} \ \ n \ge 1$ are the only positve numbers such that for any continuous function $\displaystyle f:[0,1] \to \mathbb{R}$ such that $\displaystyle f(0) = f(1)$, there is some point $\displaystyle c \in [0,1-r]$ such that $\displaystyle f(c) = f(c+r)$.

For any other positive $r$ we can find such a continuous function for which there is no $c$ such that $f(c) = f(c+r)$.

For a proof that $\displaystyle r = \frac{1}{n}$ satisifies this property, let $\displaystyle g(x) = f(x) - f(x+ \frac{1}{n})$, for $\displaystyle x \in [0, 1-\frac{1}{n}].$

Then we have that $\displaystyle \sum_{k=0}^{n-1} \ g\left(\frac{k}{n}\right) = 0$.

Thus, if none of $\displaystyle g\left(\frac{k}{n}\right)$ are $\displaystyle 0$, then $\displaystyle \exists i,j \in [0, 1, ..., n -1] \ni \displaystyle g\left(\frac{i}{n}\right) \gt 0$ and $\displaystyle g\left(\frac{j}{n}\right) \lt 0$.

For any positive $\displaystyle r$, consider the following example, due to Paul Levy.

$\displaystyle f(x) = \sin^2\left(\frac{\pi x}{r}\right) - x \ \sin^2\left(\frac{\pi}{r}\right)$. Clearly, $f$ is continuous and $f(0)=0=f(1).$

If $\displaystyle f(x) = f(x+r)$, then, $\displaystyle r\ \sin^2\left(\frac{\pi}{r}\right) = 0$ and hence, $\displaystyle r = \frac{1}{m}$ for some integer $\displaystyle m$.

Apparently this is called the Universal Chord Theorem (due to Paul Levy!).

• I like this example! – Andrés E. Caicedo Jan 5 '11 at 6:21
• @Chandru: Thanks, but credit goes to Paul Levy :-) – Aryabhata May 25 '11 at 4:29
• Credit goes to you for knowing his theorem as well :) – user9413 May 25 '11 at 4:36

Hint: consider $g(x)$, defined on $[0,1/2]$ by $g(x)=f(x+1/2)-f(x)$

I have first encountered this result in the book Van Rooij, Schikhof: A Second Course on Real Functions. I will copy here the text of Exercise 9.P

Let $0<\alpha<1$. Let $f\colon[0,1]\to\mathbb R$, $f(0)=f(1)$.
(i) Show that if $\alpha$ is one of the numbers $\frac12,\frac13,\frac14,\dots$ and if $f$ is continuous, then the graph of $f$ has a horizontal chord $\alpha$, i.e., there exists $s,t\in[0,1]$ with $f(s)=f(t)$ and $|s-t|=0$.
(ii) The proof you gave probably relies on Darboux continuity. Prove, however, that the given continuity condition on $f$ may not be weakened to Darboux continuity. (Take $\alpha:=\frac12$ and start with a function on $(0,\frac12]$ that maps every subinterval of $(0,\frac12]$ onto $\mathbb R$.)
(iii) Now let $\alpha\notin\{\frac12,\frac13,\dots\}$. Define a continuous function on $[0,1]$ with $f(0)=f(1)$ whose graph has no horizontal chord of length $\alpha$. (Choose $f$ such that $f(x+\alpha)=f(x)+1$ for $x\in[0,1-\alpha]$.)

For this questions only the first and the third part are relevant. And the first part has been already solved in other answers.

Let us spell out in details construction following the hint from the third part. (Although the hint given there already gives quite a good idea how to proceed.) This is slightly different from the examples given in other answers.

We want to have $f(x+\alpha)=f(x)+1$. Notice that this also implies $f(x+k\alpha)=f(x)+k$.

If we define the function on the interval $[0,\alpha]$, then the above condition determines the function $f$ uniquely on the rest of the interval $[0,1]$. We want to have $f(0)=0$ and $f(\alpha)=f(1)$.

Let us denote $n=\left\lfloor\frac1\alpha\right\rfloor$, i.e., $n$ is the largest integer such that $n\alpha<1$. Then we have $1-n\alpha\in(0,\alpha)$. We need to choose $f(1-n\alpha)=-n$ in order to get $f(1)=0$. (Notice that this cannot be done if $n\alpha=1$. It is possible only if $0<1-n\alpha<\alpha$, since the values $f(0)$ and $f(1)$ are already prescribed.)

Then arbitrary function defined as above (i.e., with the prescribed values in the points $0$, $1-n\alpha$, $\alpha$ and extended from $[0,\alpha]$ to the whole interval using $f(x+\alpha)=f(x)+1$) satisfies the required conditions.

Such function for a specific choice of $\alpha$ is illustrated in this picture: For comparison, here is plot of Lévy's function mentioned in Aryabhata's answer for the same value of $\alpha$ can be checked on WolframAlpha.

## protected by AryabhataFeb 26 '12 at 10:41

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