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Let $f:[0,1]\to \mathbb{R}$ be continuous, and suppose that $f(0)=f(1).$ Show that there is a value $x\in [0, 1998/1999]$ satisfying $f(x)=f(x+1/1999)$.

Raheem Najib
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1 Answers1

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Hint: examine the sign of the function$$g(x)=f(x+1/1999)-f(x)$$ on that domain and use the intermediate value theorem.

Arthur
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  • Nice hint! +1 :) – RFZ Nov 15 '15 at 16:08
  • I considered the same function. But how to check that this function changes sign? – Raheem Najib Nov 15 '15 at 16:09
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    I consider the sum $\sum \limits_{k=0}^{1998}g\left(\dfrac{k}{1999}\right)=f(1)-f(0)=0$. Hence we have the following cases: 1) $x_j=0$ or 2) $x_i>0$ and $x_j<0$ and we use Intermediate value theorem. Right? – Raheem Najib Nov 15 '15 at 16:45
  • @RaheemNajib Exactly. If any of the $g(k/1999)$ are zero, we are done. If none of them are zero, then your sum says that there must be both positive and negative values for $g$. In that case, the intermediate value theorem says that there is a $y$ somewhere (between a positive and a negative point) with $g(y)=0$. – Arthur Nov 15 '15 at 17:27
  • @Arthur, Excellent hint! Thank for help! +1 – Raheem Najib Nov 15 '15 at 17:39