Okay I need to show this using calculus and mean value theorem.

My try :

Let $D$ be a convex and compact set in $R^2$ Now let $R$ be a compact closed rectangle such that $D \subset R$. Draw two lines parallel to the axis such that $D$ is now composed of four subsets name them $R_1 \cup R_2 \cup R_3 \cup R_4 = D $

Okay now create these functions :

$$F(p)= 1 \ if\ p\epsilon R_1 \\ =0 \ else\\\\G(p)= 1 \ if\ p\epsilon R_2 \\ =0 \ else \\\\ H(p)= 1 \ if\ p\epsilon R_3 \\ =0 \ else \\\\ T(p)= 1 \ if\ p\epsilon R_4 \\ =0 \ else$$

Then we have $$\int\int_{R}F(p)+\int\int_{R}G(p)+\int\int_{R}H(p)+\int\int_{R}T(p)=A(D)$$

$A(D)$ denotes the area of this compact convex set $D$

Now since $$\int\int_{R}F(p)= A(R_1)=\sum_{R_{ij}: \bigcup R_{ij}=R_1} A(R_{ij})$$

We can say that $\exists R_{ij}: \bigcup R_{ij}=R_1$ such that $$\sum_{R_{ij}: \bigcup R_{ij}=R_1} A(R_{ij})> \frac{A(D)}{4}$$$

Also $\exists R_{ij}: \bigcup R_{ij}=R_1$ such that $$\sum_{R_{ij}: \bigcup R_{ij}=R_1} A(R_{ij}) < \frac{A(D)}{4}$$$

Then by the intermediate value theorem $\exists R_{ij}: \bigcup R_{ij}=R_1$ such that $$\sum_{R_{ij}: \bigcup R_{ij}=R_1} A(R_{ij}) = \frac{A(D)}{4}$$$

Then we can find $R_{ij}$ which form the $R_1$ such so that above is the case.

I can apply the same procedure to other functions. What do you think about this approach? Is it correct/incorrect what kind of correction does it need?

  • $\begingroup$ Where have you used convexity? $\endgroup$ – copper.hat Dec 5 '16 at 18:17
  • $\begingroup$ Used convexity in assuming that those areas of each subsets are the sums of the areas of rectangles $\endgroup$ – Xenidia Dec 5 '16 at 18:18
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    $\begingroup$ @Xenidia: But in general you can't choose the lines parallel to the axis, as copper.hat's example shows. $\endgroup$ – Dominik Dec 5 '16 at 18:25
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    $\begingroup$ @Xenidia: You are missing my point. Since your approach has fixed axes, it cannot work. Also, I don't understand what you are doing above. What continuous function are you applying the mean value theorem to? $\endgroup$ – copper.hat Dec 5 '16 at 18:30
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    $\begingroup$ Sorry, your approach & responses are too vague for me to understand. $\endgroup$ – copper.hat Dec 5 '16 at 18:35

Let me sketch the proof (it is going to be relatively brief, so a lot of details need to be checked...).

Say $S$ is your compact convex set in the plane, take a huge circle that contains $S$. Fix a point on the circle, and let $D_p$ be the corresponding diameter of $C$. A simple exercise shows that we can find a line $L_p$ parallel to $D_p$ and cutting $S$ in two pieces of the same area, and similarly we can find a line $L_p'$ orthogonal to $D_p$ and cutting $S$ in two pieces of the same area. These two lines $L_p$ and $L_p'$ determine a division of $S$ into $4$ pieces (number them say counterclockwise) $S_1(p)$, ... , $S_4(p)$ and it is clear that $S_1(p)$ and $S_3(p)$ have the same area, and similarly for $S_2(p)$ and $S_4(p)$. So we just need to ensure that there is a point $P$ on the circle $C$ for which $S_1(p)$ and $S_2(p)$ have the same area. Now study the continuous function given by$$A(p) := \text{area}(S_1(p)) - \text{area}(S_2(p))$$and apply the intermediate value theorem (it is easy to find two points $p$, $p'$ such that $A(p) = -A(p')$).


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