A question about continuous curves in $\mathbb{R}^2$

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.

Solution 1:

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 $f_1(t) = \sin(8 t) - t \sin 8$.

Solution 2:

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.