The problem is as follows:

Suppose that $f$ is continuous on $[0,1]$ and $f(0) = f(1)$. Let $n$ be any natural number. Prove that there is some number $x$ such that $f(x)=f(x + \frac{1}{n}).$

I was wondering if my proof was logically sound, especially the last bit. I tried to apply the logic of the Intermediate Value Theorem, but was curious as to whether the contradiction in the end actually gives me the result I want:

Let $g$ be a function such that $g(x) = f(x+\frac{1}{n}) - f(x)$. We want to show that there exists an $x \in [0,1]$ such that $g(x) = 0$. We prove by contradiction.

  1. Assume $\forall x \in [0,1], g(x) < 0.$ Then for $i = 0, 1, 2, ..., n$ we have that $g(\frac{i}{n}) < 0$. Therefore, $f(0) > f(\frac{1}{n}) > f(\frac{2}{n}) > ... > f(1),$ which means that $f(0) \neq f(1)$.
  2. By similar logic, it follows that $g(x) > 0$ cannot hold for all $x \in [0,1]$.

Hence, there must exist $a,b \in [0,1]$ such that $a$ and $b$ have different sign, i.e. $g(a) \leq 0 \leq g(b)$. By the Intermediate Value Theorem, there exists $c$ such that $g(c) = 0$, and we are done.

Thank you for your help!

  • $\begingroup$ The idea is completely sound, yet I cannot grasp the reasoning in (1): why if $\;g(x)<0\;$ for all $\;x\in[0,1]\;$ , we must have thet $\;f(0)>f\left(\frac1n\right)>\ldots etc.$ ? Where does this sequence of inequalities come from? $\endgroup$ – DonAntonio Mar 26 at 1:27
  • $\begingroup$ @DonAntonio For $f(1/n)-f(0)<0$, $f(2/n)-f(1/n)<0$, etc. $\endgroup$ – fantasie Mar 26 at 1:33
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    $\begingroup$ @DonAntonio . If $g<0$ then for $0\le j \le n-1$ we have $0>g(j/n)=f((j+1)/n)-f(j/n)$. $\endgroup$ – DanielWainfleet Mar 26 at 1:37
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    $\begingroup$ Very nice. An alternate is that $\sum_{j=0}^{n-1}g(j/n)=\sum_{j=0}^{n-1}f((j+1)/n)-f(j/n)=f(1)-f(0)=0$, so the values $g(j/n)$ for $0\le j \le n-1$ cannot be all $+$ nor all $-.$ This is called the Horizontal Chord Theorem. $\endgroup$ – DanielWainfleet Mar 26 at 1:52

I think your argument is both neat and sound.

Here is a way to avoid the whole contradiction argument, at the cost of using another well-known theorem about continuous functions. Because $f$ is continuous on the closed interval $[0,1]$, it attains its maximum, say $f(x_0)=M$. Now $$g(x_0-\tfrac1n)=M-f(x_0-\tfrac1n)\geq0.$$ And $$ g(x_0)=f(x_0+\tfrac1n)-M\leq0. $$ As $g$ is continuous, there exists $x$ between $x_0-\tfrac1n$ and $x_0$ such that $g(x)=0$.

The argument above does not work in principle if $x_0-\tfrac1n<0$, but that can be solved by defining $g$ as $f(0)$ outside of $[0,1]$.

  • $\begingroup$ What is the intuition behind this idea? Just curious as to how someone could come up with something like this. $\endgroup$ – alexion Mar 26 at 3:22
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    $\begingroup$ I thought of the curve climbing up from $f(0)$ up to the max (think of this as the first part) and the climbing down from the max to $f(1)$ (think of this as the second part). The horizontal distance between the two curves is $1$ at the bottom and $0$ at the max, so all distances are achieved in between. I struggled for a bit to make this into someting analytic, until I noticed that your $g$ is exactly what one needs. $\endgroup$ – Martin Argerami Mar 26 at 3:32

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