Question: Let $f:[0,1]\to\mathbb{R}$ be a continuous function. The cord joining the points $(\alpha, f(\alpha))$ and $(\beta, f(\beta))$ of the curve $y=f(x)$ is said to be horizontal if $f(\alpha) = f(\beta)$. Suppose that the chord joining the points $(0,f(0))$ and $(1,f(1))$ is horizontal. By considering the function $g$ defined on $[0,\frac{1}{2}]$ by $g(x) = f(x+\frac{1}{2}) - f(x)$ or otherwise, show that the curve $y=f(x)$ has a horizontal chord of length $\frac{1}{2}$ in $[0,1]$. Show, more generally, that it has a horizontal chord of length $\frac{1}{n}$ for each positive integer $n$.
Is my solution valid?
Solution: We know that $f(0)=f(1)$
For the first part I want to show that $g$ has a root in $[0,\frac{1}{2}]$. Note that $g(0) = f(\frac{1}{2}) - f(0)$ and $g(\frac{1}{2}) = f(1) - f(\frac{1}{2}) = f(0) - f(\frac{1}{2})= - g(0)$. First, if $g(0) = 0$ then $f(0) = f(\frac{1}{2})$ and so we have found our horizontal chord. Otherwise, if $g(0) \neq 0$ then $g(0)$ and $g(\frac{1}{2})$ have opposite signs, so $0 \in (g(0), g(\frac{1}{2}))$ (or $0\in (g(\frac{1}{2}), g(0))$, just dependent on whether $g(0) < g(\frac{1}{2})$ or not). So by Intermediate Value Theorem $\exists c \in (0,\frac{1}{2})$ such that $g(c) = 0 \implies f(c+\frac{1}{2}) = f(c)$. So we have found our horizontal chord.
For the generalised $n$ part, consider the function $g: [0,1- \frac{1}{n}]\to\mathbb{R}$ defined by $g(x) = f(x + \frac{1}{n}) - f(x)$. Again we'd like to show $\exists c\in (0, 1-\frac{1}{n})$ such that $g(c) = 0$. If $ g(x) \neq 0 \quad \forall x\in [0, 1- \frac{1}{n}]$ then either $g>0$ or $g<0$ over the whole interval. Let us suppose that $g(x) > 0 \quad \forall x\in [0, 1- \frac{1}{n}]$. This means we must have $f(x+\frac{1}{n}) > f(x)$, and noting that $\frac{1}{n}, \frac{2}{n}, \ldots, \frac{n-1}{n}\in [0,1-\frac{1}{n}]$ we have that $f(0) < f(\frac{1}{n}) < f(\frac{2}{n}) < \ldots < f(\frac{n-1}{n}) < f(1)$, but this contradicts the fact that $f(0) = f(1)$. A similar argument shows that we cannot have $g(x) < 0 \quad \forall x\in [0,1-\frac{1}{n}]$. Applying Intermediate Value Theorem tells us that $\exists c\in (0,1-\frac{1}{n})$ such that $g(c) = 0 \implies f(c+ \frac{1}{n}) = f(c)$. So we have found our horizontal chord.
