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Let $f:[0,1]\to\mathbb R^k$ be a continuous function with $f(1) = \overrightarrow 0$. Is it true that there always exist $k$ points $0 \le a_1 \le a_2 \le \ldots \le a_k \le 1$ such that $\sum_{i=1}^k (-1)^{k+1} f(a_k) = f(a_1) - f(a_2) + f(a_3) - \ldots = \frac{f(0)}{2}$?

When $k=1$ we have to find one point $a_1 \in [0,1]$ with $f(a_1) = \frac{f(0)+f(1)}{2}$, which is the intermediate value theorem.

When $k=2$ the situation is more complicated. We believe the statement is still true. However, we noted that if we are not finding $\frac{f(0)}{2}$ but, say, $.49 f(0)$ instead, there exist counter-examples. A counter-example looks like a sine function $f(t) = (t, \sin 100t)$, rotated in the 2-d plane a little bit. (This is not the example, but you can imagine that changing the constants $.49$ and $100$ a bit makes it work.)

For bigger $k$ we have no idea.

Are there any similar results known before?

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up vote 15 down vote accepted

The statement is true. It is almost precisely Lemma 2 in the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210. The proof is short but not easy to invent. The paper can be read on Google Books here.

Notes on the text: the intervals in the formulation of Lemma 2 are in fact disjoint, the term "antipodal map" means "odd map" (i.e. $\varphi(-x)=-\varphi(x)$).

For $k=2$, this fact is classic and one can replace 0.5 by any inverse integer ($1/3, 1/4,\dots$) but for every other value there is a counter-example.

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I knew the k=2 result for functions (ie. f(x)=(x,g(x)), but not this one. More sadly, I cannot even understand it. Why can it not happen that the sum of the segments is 1/2(r(b)+r(a))? If x_1 and x_{n+1} have opposite signs, this seems to be the case. – domotorp Jan 16 '13 at 18:51
Everything in the proof is translation-invariant, there is no way terms like $r(b)+r(a)$ can appear. What he proves is that you can subdivide $[a,b]$ into $n$ intervals $[y_i,y_{i+1}]$ and equip them with signs $\varepsilon_i=\pm 1$ so that $\sum \varepsilon_i (r(y_{i+1})-r(y_i))=0$. This means that the sum of the terms equipped with pluses equals the sum of the terms equipped with minuses. But the total sum of all differences equals $f(b)-f(a)$. So the sum of "positive" terms equals $(f(b)-f(a))/2$. And by changing sign you may assume that there are at most $n/2$ of "positive" terms. – Sergei Ivanov Jan 16 '13 at 19:37
Yes, clear now, thx. Is anything known about how many dividing points are needed for the inverse integer generalization in higher dims? – domotorp Jan 17 '13 at 4:36

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