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Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function and $f(0)=f(1)$. Show that for all $n \in \mathbb{N}$ there is $x, 0 \leq x < x+\frac{1}{n} \leq 1,$ such that $f(x)=f(x+\frac{1}{n})$.

I define $g(t)=f(t+\frac{1}{n})-f(t)$ but I couldn't find appropriate $a,b$ such that $g(a)g(b)<0$.

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marked as duplicate by Andres Caicedo, T. Bongers, Lord_Farin, Peter Taylor, tetori Oct 31 '13 at 8:52

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1 Answer 1

Since the function is continuous and f(0)=f(1) there there exist at least one c in [0,1] such that f'(c)=0.

It means the function has to have a critical point in [0,1]. Consider x < c. Then we will have some c+h(h>0) which will be equal to x.enter image description here

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