Seems nice

Killer · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 200

Let $f:R->R$ a continuous function.Show that if $f$ admits a local maximum and local minimum then there exist $(a,b) \in R^2$ ,distinct numbers,such that $f( \frac {a+b}{2})=\frac{f(a)+f(b)}{2}$ .

**Posted:** Wed Nov 17, 2004 1:46 pm

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Kill'em all!!!

grobber

It follows from a few basic facts:

Since $f$ is continuous, if the conclusion doesn't hold, then $\frac{f(x)+f(y)}2-f(\frac{x+y}2)$ must have a constant sign. Let's assume it's always $>0$ . In this case, using, again, the continuity of $f$ , we can show that $f$ must be convex, but not constant on any interval, and it's easy to show now that such a function cannot have a local maximum (we do pretty much the same thing if $f$ is concave).

**Posted:** Wed Nov 17, 2004 2:10 pm

LevonNurbekian

The problem is already solved by grobber,so I suggest a little bit other solution,but the basic ideas are same.Let $x_{1},x_{2}$ are the points where function f approaches it�s local minimum and local maximum,respectively.Then there exist ${\delta}_{1}>0$ and ${\delta}_{2}>0$ ,such that $f(x_{1})\leq{f(x)},\forall x\in{[x_{1}-{\delta}_{1},x_{1}+{\delta}_{1}]}$ ,and $f(x_{2})\geq{f(x)},\forall x\in{[x_{2}-{\delta}_{2},x_{2}+{\delta}_{2}]}$ .So $f(x_{1})\leq\frac{f(x_{1}-{\delta}_{1})+f(x_{1}+{\delta}_{1})}{2}$ and $f(x_{2})\geq\frac{f(x_{2}-{\delta}_{2})+f(x_{2}+{\delta}_{2})}{2}$ .So we found four points $a,b,c,d$ such that $f(\frac{a+b}{2})\leq\frac{f(a)+f(b)}{2}$ ,and $f(\frac{c+d}{2})\geq\frac{f(c)+f(d)}{2}$ .Now it is easy to prove that as $f$ is continious there exist(as grobber noticed) $u,v\in{R}$ such that $f(\frac{u+v}{2})=\frac{f(u)+f(v)}{2}.$

**Posted:** Thu Nov 18, 2004 4:02 am