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Let $f(x)$ continuous function on $R$ wich can be in different signs. Prove, that there is exists an arithmetic progression $a, b, c (a<b<c)$, such that $f(a)+f(b)+f(c)=0$.

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What you done so far? –  Aryabhata Nov 12 '10 at 23:24
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This is not really a functional equation. –  Qiaochu Yuan Nov 12 '10 at 23:55
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I think that your teacher means that $f$ has different signs and not that it can have - else $f$ might be positive. –  AD. Nov 13 '10 at 5:22

3 Answers 3

Try defining the following function on $R \times R^+$

$ g(x,h) = f(x-h) + f(x) + f(x+h) $

Argue that if $f$ can take positive and negative values, then so can $g$. Then, use the intermediate value theorem.

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up vote 0 down vote accepted

Let's ponder like this:

At some point $x$ $f(x)>0$, therefore, in the vicinity of this point there is an increasing arithmetic progression $a_{0}, \ b_{0}, \ c_{0}$ that, $f(a_{0})+f(b_{0})+f(c_{0})>0$.

Like this one will be found increasing arithmetic progression of $a_{1}, \ b_{1}, \ c_{1}$ that, $f(a_{1})+f(b_{1})+f(c_{1})<0$.

For all values of parametr $t[0,1]$ сonsider the arithmetic progression $a(t), \ b(t), \ c(t)$, where

$a(t)=a_{0}(1-t)+a_{1}t$,
$b(t)=b_{0}(1-t)+b_{1}t$,
$c(t)=c_{0}(1-t)+c_{1}t$.

Function $F(t)=f(a(t))+f(b(t))+f(c(t))$ continuously depends on $t$, at $t=0 \ F(t)>0$, and at $t=1 \ F(t)<0$. It means that in some $t \ F(t)=0$ and the corresponding progress $a(t), \ b(t), \ c(t)$ is required.

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HINT:

1) Think about the intermediate value theorem.

2) Think about some $x$ and some $y$ with $f(x)\gt0$ and $f(y)\lt0$.

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You mean, think about the intermediate value theorem. –  user1119 Nov 13 '10 at 12:14
    
@George S.: Yes thanks, fixed it. –  AD. Nov 13 '10 at 14:30

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