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I came across this question in Stephen Abbott's "Understanding Analysis" 2nd edition.

Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous with $f(0) = f(1)$.

a) Show that there must exist $x,y \in [0,1]$ satisfying $|x-y| = 1/2$ and $f(x)=f(y)$.

b) Show that for each $n \in \mathbb{N}$ there exist $x_n , y_n \in [0,1]$ with $|x_n-y_n|=1/n$ and $f(x_n)=f(y_n)$.

For part a), I thought that I would start by splitting the interval $[0,1]$ into two halves and asserting that if $f(1/2)=c\neq 0$ then there must be some $x \in [0, 1/2]$ and $y \in [1/2,1]$ such that $f(x)=f(y)$ (by the Intermediate Value Theorem) but that did not lead me anywhere.

Michael Wilham
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    Consider $g(x) = f(x)-f(x+1/2 \mod 1)$ – Kenny Lau Oct 24 '17 at 02:28
  • Use the fact that $[0,1]$ is compact, so $f$ attains a maximum value on $[0,1].$ – Chickenmancer Oct 24 '17 at 02:46
  • You could also use a purely topological argument: $g : [0, 1/2]\to \mathbb{R}^2$, $x\mapsto (f(x), f(x+1/2))$ is continuous, and therefore the image of $[0, 1/2]$ under $g$ will be path connected. As $g(0) = (f(0), f(1/2))$ and $g(1/2) = (f(1/2), f(1)) = (f(1/2), f(0))$ are on opposite sides of the diagonal $D = {(x, x) : x\in \mathbb{R}}$ (which separates $\mathbb{R}^2$ into disconnected halves), you'll have that the path between them crosses $D$ at at least one point. – Michael L. Oct 24 '17 at 02:48
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    Consider $\phi(t) = f(t)-f(t+{1\over 2})$ and note that $\phi(0) = -\phi({1 \over 2})$ (or $\phi(0) + \phi({1 \over 2}) = 0$). A similar analysis will work for Part b). – copper.hat Oct 24 '17 at 02:51

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Since $f(0)=f(1)$, we may extend $f$ continuously to all of $\mathbb{R}$ by periodicity, say to $F:\mathbb{R}\longrightarrow\mathbb{R}$.

Now, for any $n\in\mathbb{N}$, consider $F_n:\mathbb{R}\longrightarrow\mathbb{R}$ given by

$$F_n(x)=F(x)-F\left(x+\frac1n\right)$$

If $F_n(0)=0$, we can take $x_n=0$ and $y_n=\frac1n$.
Otherwise, suppose without loss of generality that $F_n(0)>0$.

Claim: $F_n$ may not be positive throughout $\left[0,1-\frac{1}n\right]$.

Proof: Indeed, if that were the case we'd have

$$0< \sum_{i=0}^{n-1}F_n\left(\frac{i}n\right) =\sum_{i=0}^{n-1}F\left(\frac{i}n\right)-F\left(\frac{i+1}n\right) =F(0)-F(1)=0,$$

which is a contradiction. $\square$

Since $F_n$ is continuous, the claim (together with the Intermediate Value Theorem) implies $F_n$ must attain the value $0$ at some point in $\left[0,1-\frac1n\right]$, say $x$. It suffices to check that setting $x_n=x$ and $y_n=x_n+\frac1n$ does the trick.

Fimpellizzeri
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