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Does there exists a function $f(x)$, which is differentiable in $\mathbb{R}$, and the middle value point is unique, that is, the $\xi$ in formula


is unique, having following property:

Connect pairs of arbitrary point $x_0$ and $x_1$ (here $x_0 < x_1$), get $\xi_1$, with connecting $x_0$ and $\xi_1$, get $\xi_2$, with connecting $x_0$ and $\xi_2$, get $\xi_3$, $\dots\,$, satisfy: $\inf\{\xi_n\}>x_0$.

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I edited the text a bit, but I still do not quite understand what is being asked. – Raskolnikov Apr 28 '11 at 10:04
You've edited the question to redefine what "bounded from below" means. That's not a good idea, since that's a well-established term. My suggestion would be to remove the "bounded from below" part and just write $\inf\{\xi_n\}>x_0$, which speaks for itself. ($\xi_n\ge \inf\{\xi_n\}$ is redundant). – joriki Apr 28 '11 at 11:27
@joriki Thanks for point out. – abc Apr 28 '11 at 11:32
@Gerry: Yes, that comment was about an earlier version of the question that just said "is bounded from below" -- I've deleted it now to avoid further confusion. – joriki Apr 28 '11 at 13:17
up vote 0 down vote accepted

Call $d(a,b) = \frac{f(a)-f(b)}{a-b}$ and $\delta(a,b) =$ the unique $\xi \in ]a;b[$ such that $f'(\xi) = d(a,b)$.

For any $a \in \mathbb{R}$, the function $d_a : b \mapsto d(a,b)$, extended on $b=a$ with $d_a(a) = f'(a)$ is continuous and injective. To prove the injectiveness, if you have $d(a,b) = d(a,c)$, then $d(a,b) = d(a,c) = d(b,c)$, and if you look at the bigger of the three intervals, you will find one value for $\xi$ in each of the smaller intervalls, which contradicts your hypothesis.

Therefore, for every $a$, $d_a$ is either strictly increasing or strictly decreasing. From this and from the continuity of $d$, you get that either they are all strictly increasing, either they are all strictly decreasing.

Without loss of generality, suppose they are increasing. Then you get that $f'$ is also strictly increasing : if $a \lt b$, then $f'(a) = d_a(a) \lt d_a(b) = d_b(a) \lt d_b(b) = f'(b)$.

From here, choose $x_0, x_1$ and build the sequence $\xi_n$ as you described. $\xi_n$ is a strictly decreasing sequence, bounded below by $x_0$ so it converges to some limit $\xi$. If $\xi \gt x_0$ then $f'(\xi) = f'(\lim \xi_n) \le \lim f'(\xi_n) = \lim d(x_0, \xi_{n-1}) = d(x_0, \xi) = f'(\delta(x_0, \xi)) \lt f'(\xi)$, which is impossible. Thus $\lim \xi_n = x_0$.

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Do you mean, if the middle value point is unique in whole field, the function is convex up or down? – abc Apr 29 '11 at 1:28
@abc, yes under this hypothesis, the function is convex or concave, and you can also prove that f' is continuous (which I didn't need) – mercio Apr 29 '11 at 9:04

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