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Let $f$ be continuous on $[0,1]$ and $f(0)=f(1)$. Also, $n\in \mathbb{N}$. Prove that there exists some $x$ s.t. $f(x)=f(x+\frac{1}{n})$.

I think I need to assume towards a contradiction that if $g(x)=f(x)-f(x+\frac{1}{n})$, then $g(x)\neq 0$ $\forall x$.

So, $g(0)=f(0)-f(\frac{1}{n})$, $g(1)=f(1)-f(1+\frac{1}{n})$. Since $f(0)=f(1)$,

$g(0)-g(1)=f(1+\frac{1}{n})-f(\frac{1}{n})$. For context, I am only up to Ch. 8 in Spivak, so I cannot use any derivative-based theorems which is why I am stuck here.

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You can't assume anything on $f(1+1/n)$, you must limit your $x$ to $0 \le x \le 1- 1/n$ – vonbrand Feb 24 '14 at 1:11
See also: Universal Chord Theorem. (You can find further links there.) – Martin Sleziak Feb 24 '14 at 6:45
up vote 4 down vote accepted

Assume the contrary that $g(x)$ is not $0$ on $[0,1-\frac{1}{n}]$, which means either $g(x)>0$ or $g(x)<0$ on $[0,1-\frac{1}{n}]$ (since, g is continuous). If, $g(x)>0 \implies f(0)>f(\frac{1}{n})>f(\frac{2}{n})>\cdots>f(1-\frac{1}{n})>f(1)$, contradiction !! Similarly, for $g<0$, we get a contradiction. Therefore, $g(x)$ has a zero in $[0,1-\frac{1}{n}]$.

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Thank you for the answer. – user113525 Feb 25 '14 at 0:21
@user113525 You are welcome :) !! – r9m Feb 25 '14 at 0:43

Hint: What is $$g(0)+g\left(\frac 1 n\right)+...g\left(\frac{n-1}n\right)?$$

How does that help prove that $g(x)=0$ for some $x$?

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