Let $f:[a,b]\longrightarrow\mathbb{R}^2$ be a continuous function such that $$f(a)=(0,0),\ f(b)=(0,1).$$ Is it true that there must exist $t_1,t_2\in [a,b]$ such that $\displaystyle f(t_1)-f(t_2)=(0,\frac{1}{2})?$

If not, please give a counterexample.

  • 1
    $\begingroup$ The obvious generalization to $n \geq 3$ is false. Consider the case $n = 3$: there are paths from $(0, 0, 0)$ to $(0, 0, 1)$ such that no two points are $(0, 0, 1/2)$ apart. Take for, example, $[a, b] = (0, 1)$; then $f(t) = (\cos (2\pi t) - 1, \sin (2 \pi t), t)$ sketches out one revolution of a helix and has no points with the same $x$ and $y$ coordinates except for the endpoints. $\endgroup$ May 15, 2018 at 16:58

6 Answers 6


More of a comment, the particular case $f(t) = (f_1(t), t)$, the graph of a function, is relevant, and somehow well known. In this case, if $f_1(0)=f_1(1)$, there exists $t$, $t+\frac{1}{2}\in [0,1]$ so that $f_1(t) = f_1(t+\frac{1}{2})$. The proof should be clear. It should work with $\frac{1}{n}$ -but not with other distances! ( there are counter examples).So for those proofs that work for any $\delta >0$, they shouldn't.

${Added:}$ Here is a link to an article with more relevant references on "horizontal chords in graphs".

$\bf{Added:}$ Here is an example of a function $f(t) = (f_1(t), t)$ from $[0,1]$ to $\mathbb{R}^2$, $f(0) = (0,0)$, $f(1)= (0,1)$, so that there does not exist on this curve a vertical chord or length $\frac{\pi}{4}$. Take $f_1(t) = \sin(8 t) - t \sin 8$.

  • $\begingroup$ Can you give an example of a counterexample? $\endgroup$
    – Wen
    Sep 11, 2017 at 8:51
  • $\begingroup$ @Wen: added a link $\endgroup$
    – orangeskid
    Sep 11, 2017 at 12:37
  • $\begingroup$ @Wen : added a counterexample $\endgroup$
    – orangeskid
    Sep 15, 2017 at 2:46
  • $\begingroup$ Your added counterexample is very interesting. It raises the question which values of $\alpha\in[0,1]$ must appear as the length of a vertical chord. I initialy thought all of $[0,1]$, but this example shows otherwise. $\endgroup$
    – M. Winter
    Sep 15, 2017 at 9:01
  • $\begingroup$ An easier counterexample would be $f(t)=(\sin(2\pi t),t)$ which has no vertical chord of any length $\alpha\in(1/2,1)$. Still no idea how to avoid chord lengths in $(0,1/2]$. $\endgroup$
    – M. Winter
    Sep 15, 2017 at 10:14

Yes this is true. I'll give a detailed proof based on homotopy invariance of winding number of closed paths in $\mathbb{R}^2-(0,1/2)$. (There's nothing special about $(0,1/2)$ in the theory of winding numbers, of course, but I'll stick with that point since its the relevant center of winding numbers for this problem).

Winding number theory. I'll use the notation $W(\gamma)$ for the winding number around $(0,1/2)$ of any closed path $$\gamma : [0,1] \to \mathbb{R}^2 - \{(0,1/2)\} $$ Just to review the definition of winding number around $(0,1/2)$, one uses polar coordinates $(r,\theta)$ centered at $(0,1/2)$, given by $$(x,y) = (0,1/2) + (r \cos(\theta),r\sin(\theta)) $$ For any path $W : [0,1] \to \mathbb{R}^2 - \{(0,1/2)\}$ one defines its winding number around $(0,1/2)$ to be
$$W(\gamma) = \frac{1}{2\pi} \int_\gamma d\theta \in \mathbb{Z} $$ One can also use covering space theory, which for computations in this problem is easier to use. The universal covering space of $\mathbb{R}^2 - \{(0,1/2)\}$ is identified (using polar coordinates) with the upper half plane $\mathbb{H} = \{(\theta,r) \mid \theta \in \mathbb{R}, r > 0\}$ using the covering map $g : \mathbb{H} \to \mathbb{R}^2 - \{(0,1/2)\}$ given by the formula $$g(r,\theta) = (0,1/2) + (r \cos(\theta), r \sin(\theta)) $$ Then one chooses any lift $\tilde\gamma : [0,1] \to \mathbb{H}$ with coordinate functions $r \circ \tilde\gamma : [0,1] \to (0,\infty)$ and $\theta \circ \tilde\gamma : [0,1] \to \mathbb{R}$ and then $$W(\gamma) = \frac{1}{2\pi}\bigl(\theta\circ\tilde\gamma(1) - \theta\circ\tilde\gamma(0)\bigr) $$ This number is well-defined independent of the choice of the lift $\tilde\gamma$.

In the special case that $\gamma$ is a closed path in $\mathbb{R}^2 - \{(0,1/2)\}$, the number $W(\gamma)$ is always an integer, and invariant under homotopy of closed paths in $\mathbb{R}^2 - \{(0,1/2)\}$.

The formula for $W(\gamma)$ is defined for any path in $\mathbb{R}^2-\{(0,1/2)\}$, not just for closed paths, and there's a few useful little facts which can be easily proved using the covering space definition:

Fact 1: If $(0,1/2)$ is the midpoint of the segment $\overline{\gamma(0),\gamma(1)}$ then $W(\gamma)$ is a half integer, $$W(\gamma) = n + \frac{1}{2} \quad\text{for some $n \in \mathbb{Z}$}$$


Fact 2: Letting $R : \mathbb{R}^2 \to \mathbb{R}^2$ denote any rigid rotation of the plane around $(0,1/2)$, and letting $\gamma : [0,1] \to \mathbb{R}^2 - \{(0,1/2)\}$ be any path, we have $$W(R \circ \gamma) = W(\gamma)$$


Fact 3: $W$ is additive under concatenation. That is, for any concatenation of paths $\gamma = \gamma_1 * \gamma_2$ in $\mathbb{R}^2 - (0,1/2)$ we have $$W(\gamma)=W(\gamma_1)+W(\gamma_2)$$

Working with the path $f$. Allow me to reparameterize the path $f$ to have domain $[0,1]$, so $$f : [0,1] \to \mathbb{R}^2 $$ such that $f(0)=(0,0)$ and $f(1)=(0,1)$.

Denote $$T = \{(x,y) \in [0,1] \times [0,1] \mid x \le y\} $$ which is just the triangle with vertices $(0,0)$, $(0,1)$, $(1,1)$ together with the inside of that triangle. Thus $T$ is homeomorphic to the 2-disc $D^2$, by a homeomorphism that takes the boundary triangle $\partial T$ is homeomorphic to the circle $S^1$.

Define $F : T \to \mathbb{R}^2$ by the formula $$F(x,y) = f(y)-f(x) $$ The goal is to prove that the point $(0,1/2)$ is in the image of $F$.

Case 1: If there exists $(x,y) \in \partial T$ such that $F(x,y)=(0,1/2)$ then we're done.

Case 2: Suppose that $F(x,y) \ne (0,1/2)$ for each $(x,y) \in \partial T$. It follows that $F \mid \partial T$ is a closed curve in $\mathbb{R}^2 - \{(0,1/2)\}$.

Claim: The winding number of $F \mid \partial T$ around $\{(0,1/2)\}$ is nonzero: $$W(F \mid \partial T) \ne 0$$

Once this claim is proved, then by applying homotopy invariance of winding number it follows that $F \mid \partial T$ is not homotopic in $\mathbb{R}^2-\{(0,1/2)\}$ to a constant curve. But $F \mid \partial T$ is homotopic to a constant curve in $\mathbb{R}^2$: the function $F : T \to \mathbb{R}^2$ may be regarded as such a homotopy, considering that the topological pairs $(T,\partial T)$ and $(D^2,S^1)$ are homeomorphic. Therefore, the image of $F$ must contain $(0,1/2)$.

Proving the claim. Let me write $\partial T$ as a closed path $\tau$ which is a concatenation of its three sides $\tau = \tau_v * \tau_h * \tau_d$ where:

  • $\tau_v$ is the vertical side going upwards, $\tau_v(t)=(0,t)$ for $t \in [0,1]$.
  • $\tau_h$ is the horizontal side going rightwards, $\tau_h(t)=(t,1)$ for $t \in [0,1]$.
  • $\tau_d$ is the diagonal side going down and to the left $\tau_d(t)=(1-t,1-t)$ for $t \in [0,1]$.

The path $F | \partial T$ can be written as a concatenation of three paths: $$F | \partial T = (F \circ \tau_v) * (F \circ \tau_h) * (F \circ \tau_d) $$ and so, by Fact 3, we have $$W(F | \partial T) = W(F \circ \tau_v) + W(F \circ \tau_h) + W(F \circ \tau_d) $$ and we have to show this sum is nonzero.

Notice that $F$ is constant on $\tau_d$, namely $F(\tau_d(t))=(0,0)$ for all $t \in [0,1]$. It follows that $W(F \circ \tau_d)=0$, so we just have to show that the sum $$W(F \circ \tau_v) + W(F \circ \tau_h) $$ is not zero.

Now let's examine those two paths separately: \begin{align*} F(\tau_v(t)) &= F(0,t) = f(t)-f(0) = f(t) - (0,0) \\ F(\tau_h(t)) &= F(t,1) = f(1)-f(t) = (0,1)- f(t) \end{align*} A little vector arithmetic gives us $$\frac{1}{2} \bigl(F(\tau_v(t)) + F(\tau_h(t)\bigr) = (0,1/2) $$ To say this in geometric language, for each $t \in [0,1]$, the point $(0,1/2)$ is the midpoint of the segment with endpoints $F(\tau_v(t))$ and $F(\tau_h(t))$. It follows that the path $F \circ \tau_v$ and path $F \circ \tau_h$ differ from each other by a $180^\circ$ rotation of the plane centered on the point $(0,1/2)$. In symbols, letting $\rho : \mathbb{R}^2 \to \mathbb{R}^2$ denote that rotation, we have $$\rho \circ (F \circ \tau_v) = F \circ \tau_h $$ Applying Fact 2 it follows that $$W(F \circ \tau_h) = W(F \circ \tau_v) $$

The path $F \circ \tau_v$ has initial and terminal endpoints $$F \circ \tau_v(0) = (0,0) $$ $$F \circ \tau_v(1) = (0,1) $$ and the point $(0,1/2)$ is the midpoint of the segment between those two endpoints. By applying Fact 1, the winding number of the path $F \circ \tau_v$ is a half-integer: $$W(F \circ \tau_v) = n + \frac{1}{2} \quad\text{for some $n \in \mathbb{Z}$} $$ We therefore have \begin{align*} W(F \circ \tau_v) + W(F \circ \tau_h) &= W(F \circ \tau_v) + W(F \circ \tau_v) \\ &= 2 \cdot W(F \circ \tau_v) \\ &= 2n + 1 \end{align*} which is an odd integer, and therefore nonzero.

  • $\begingroup$ Nice. I did not read the proof carefully yet, although a general proof for $n$i instead of $2$ would imply that winding numbers of simple loops are $1$. So, perhaps the proof should also involve winding numbers somehow. $\endgroup$
    – orangeskid
    May 11, 2018 at 16:49
  • $\begingroup$ Looks very nice. There is another proof using a loop iof index $2$ around $0$ ( $\exp( 4 \pi f(t)) $ that cannot be injective and producing $t_1$,, $t_2$, although it doesn't get $t_1< t_2$. The same argument for any $n$ would produce inductively first some difference $\frac{k_1}{n}$, then $\frac{k_2}{n}$, till we reach $\frac{1}{n}$. But it uses the subtle thing that an injective loop has index $\pm 1$ (wr to $0$)... So maybe there is a direct proof along your approach. $\endgroup$
    – orangeskid
    May 11, 2018 at 17:26
  • $\begingroup$ When you say "$n$ instead of $2$", what exactly do you mean? I'm guessing this refers to something in another answer, but I'm not sure. $\endgroup$
    – Lee Mosher
    May 11, 2018 at 17:30
  • $\begingroup$ Oh, if's $f(t_2)-f(t_1) = (0, 1/n)$. $\endgroup$
    – orangeskid
    May 11, 2018 at 17:35

This answer is wrong, as pointed out by Wen. I won't delete it: maybe someone will find it useful to create a working counterexample

I think this should work as a counterexample.
Take $a=0$, $b=7$. We will define $f$ piecewisely.
For $t\in[0,1]$, $\ \ f(t)=(\frac{t}{2},0) $;
for $t\in[1,2]$, $\ \ f(t)=(\frac{1}{2},\frac{t-1}{4})$;
for $t\in[2,3]$, $\ \ f(t)=(\frac{1}{2}-(t-2),\frac{1}{4})$;
for $t\in[3,4]$ $\ \ f(t)=(-\frac{1}{2},\frac{1}{4}+\frac{3}{8}(t-3))$;
for $t\in[4,5]$, $\ \ f(t)=(-\frac{1}{2}+\frac{3}{2}(t-4),\frac{5}{8})$;
for $t\in[5,6]$, $\ \ f(t)=(1,\frac{5}{8}+\frac{3}{8}(t-5))$;
for $t\in[6,7]$, $\ \ f(t)=(1-(t-6),1)$.
Since it is possible that I made some mistake in the parametrization, the idea is the following:
Start from $(0,0)$; go along the x-axis to the point $(\frac{1}{2},0)$; go up to the point $(\frac{1}{2},\frac{1}{4})$; now "turn back" to the point $(-\frac{1}{2},\frac{1}{4})$; now go up to the point $(-\frac{1}{2},\frac{5}{8})$. At this point you are above the points with y-coordinate $\frac{1}{2}$. To add $\frac{3}{8}$ vertically, without having two points $t_1,t_2$ s.t. $f(t_1)-f(t_2)=(0,\frac{1}{2})$, you need to go back "beyond" $\frac{1}{2}$ in the x-direction, then you can reach the points with y-coordinate 1 "safely".

enter image description here

  • $\begingroup$ Sorry, but can you provide a picture? I don't quite get the 'go back beyond' 1/2 $\endgroup$
    – Wen
    Sep 11, 2017 at 22:08
  • $\begingroup$ @Wen Is it clear now? $\endgroup$
    – Uskebasi
    Sep 12, 2017 at 7:36
  • 1
    $\begingroup$ I believe that (1/2,1/8) an (1/2,5/8) both lie on your graph, so I don't think this works. $\endgroup$
    – Wen
    Sep 12, 2017 at 9:00
  • $\begingroup$ Well, I think you're right $\endgroup$
    – Uskebasi
    Sep 12, 2017 at 9:13

I'll use the well-known concept of Brouwer topological degree. Here is an brief intuitive idea (roughly speaking) of the construction of such tool..

Given a (at least) $C^2$ function $g\colon\Omega\subseteq\mathbb R^n\to\mathbb R^n$ (or, more generally, between two oriented manifolds of same dimension) and $y\in\mathbb R^n$ a regular value for $g$ such that $g^{-1}(y)$ is compact (therefore finite, by the inverse function theorem), one defines the Brouwer degree of $g$ at $y$ as the integer number $$\deg(g,y) = \sum_{x\in g^{-1}(y)} {\rm sign}\det g'(x),$$ where $g'(x)$ is the jacobian matrix of $g$ at $x$ and ${\rm sign}\ t$ is $1$ for $t>0$ or $-1$ for $t<0$.

One can prove that this number is homotopy invariant, that is, if $f$ and $g$ are homotopic and $y$ is a regular value for both $f$ and $g$, then $$\deg(f,y) = \deg(g,y).$$

Using Sard's theorem to define the degree of maps at critical points, we have that, given a $C^2$ function $h$, $\deg(h,\cdot)\colon\mathbb R^n\to\mathbb Z$ is a continuous locally constant function.

Now, given a continuous function $f\colon\Omega\to\mathbb R^n$, we can choose a smooth function $g$ "close enough from $f$" (using Weierstrass approximation theorem) and define the degree of $f$ at some value $y$ by $$\deg(f,y) = \deg(g,y),$$ where the right side is already defined.

Briefly, the Brouwer degree can be thought as a map taking a continuous function $f\colon\Omega\to\mathbb R^n$ and some value $y\in\mathbb R^n$ and giving a integer number.

In addition, notice that if $\deg(f,y) \neq 0$ then the equation $f(x) = y$ has at least one solution.

Now we're ready to think about the problem.

Let ${\rm I} = [0,1]$ and $q=(0,1/2)$. Consider $\alpha\colon{\rm I}\to\mathbb R^2$ be a continuous function such that $\alpha(0) = (0,0)$ and $\alpha(1) = (0,1)$ and .

Let $\gamma\colon{\rm I}\to\mathbb R^2$ be the smooth function $$\gamma(t) = (\sin(\pi t), t).$$ Here is the plot of $\gamma$: plot of $\gamma$

Define $F, G\colon {\rm I}^2\to\mathbb R^2$ (note the same dimension) by $$F(x,y) = \alpha(x) - \alpha(y)$$ and $$G(x,y) = \gamma(x) - \gamma(y).$$

We have that $F$ and $G$ are homotopic since it assumes values on $\mathbb R^2$ (we can take the linear homotopy). Therefore, $\deg(F,q) = \deg(G,q)$.

Let's calculate $\deg(G,q)$. First, we have to verify that $q$ is in fact a regular value for $G$ to then apply the degree formula. We are going to take a look at the set $G^{-1}(q)$.. with some straight forward calculations, we can see that $$G(x,y) = q \Leftrightarrow \left\{ \begin{matrix} x = k + 3/4 \\ y = k + 1/4 \end{matrix}\right. , k\in\mathbb Z$$ and $k=0$ gives the solution $(x,y)=(3/4,1/4)$, that lies in ${\rm I}^2$. So $G^{-1}(q) = \{(3/4,1/4)\}$.

Note that $$G'(x,y) = \begin{bmatrix} \pi\cos(\pi x) & -\pi\cos(\pi y) \\ 1 & -1 \end{bmatrix},$$ thus $$\det G'(x,y) = -\pi(\cos(\pi x) - \cos(\pi y))$$ and $$\det G'(3/4,1/4) = \pi\sqrt{2} > 0.$$ Therefore, \begin{equation*} \begin{split} \deg(G,q) & = \sum_{(x,y)\in G^{-1}(q)} {\rm sign} \det G'(x,y) \\ & = {\rm sign} \det G'(3/4,1/4) \\ & = {\rm sign} \pi\sqrt{2} \\ & = 1. \end{split} \end{equation*}

Since $\deg(F,q) = \deg(G,q) = 1 \neq 0$, we conclude that there exists some $(x_0,y_0)\in{\rm I}^2$ such that $F(x_0,y_0) = q$, i.e., $$\alpha(x_0) - \alpha(y_0) = (0,1/2).$$

  • 1
    $\begingroup$ I added in my answer an example of a curve without vertical chords of length $\frac{\pi}{4}$. $\endgroup$
    – orangeskid
    Sep 15, 2017 at 2:48
  • 1
    $\begingroup$ @orangeskid I confess I didn't think a lot in the final of my answers (have a lot of calculations to do). So I'll remove it and be fine with the case when $T=1/2$. But it's bothering me that I can't change the $\alpha$ in my answer for the $f$ in your answer.. I'll think about where the argument do not work. $\endgroup$ Sep 15, 2017 at 2:55
  • $\begingroup$ It is indeed a very puzzling problem. I am not even sure what the answer is. The example that I wrote it's an old classical example that I got from the paper provided in the link. Still thinking about it.. $\endgroup$
    – orangeskid
    Sep 15, 2017 at 3:02

This is more like a comment extending on an observation in orangeskid's answer which is also discussed in his linked paper (Proposition 4). It turns out that any (non-zero) vertical chord length can be avoided except $1/n$ with $n\in\Bbb N$. This means that any proof of the initial statement cannot use a general argument that fits any vertical chord length.

Check orangekid's answer to see how the folowing functions are used. Let $$f(x)=\sin(2\pi/\omega\cdot x)-x\sin(2\pi/\omega).$$

Note that $f(0)=f(1)=0$. Consider $g(x):=f(x-\omega)$. We have

\begin{align} g(x)&=\sin(2\pi/\omega\cdot x-2\pi)-(x-\omega)\sin(2\pi/\omega)\\ &=\sin(2\pi/\omega\cdot x)-x\sin(2\pi/\omega)+\omega\sin(2\pi/\omega)\\ &=f(x)+\omega\sin(2\pi/\omega). \end{align}

Clearly $f$ and $g$ never intersect except when $\omega=1/n$ for $n\in\Bbb N$. If they do not intersect then this is equivalent to say that the curve $(f(t),t)$ has no (vertical) chord of length $\omega$.

The paper shows (using the intermediate value theorem on the slope of chords) that OP's statement (generalized to chord lengths $1/n$) is indeed true for curves given via $(f(t),t)$ for some function $f:[0,1]\to\Bbb R$ with $f(0)=f(1)=0$ like the one above.


I just came across a curious theorem in Rolfsen's book "Knots and Links", and was reminded of this question.

Theorem B.15 (p. 16). Suppose $X$ is a path-connected subset of $\Bbb R^2$, and $C$ is a chord (that is, a line-segment) with endpoints in $X$. Suppose $0<\alpha<1$. Then among all chords with endpoints in X and parallel to $C$, there is either one of length $\alpha|C|$ (where $|C|$ is the length of the chord $C$) or one of length $(1-\alpha)|C|$.

Let $X$ be the curve in your question. Let $C$ be the chord from $(0,0)$ to $(0,1)$. Set $\alpha=1/2$. Theorem B.15 yields the existence of a chord with end points $x,y\in X$ and $x-y=(0,1/2)$.


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