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Nov 2, 2018 at 2:13 vote accept Jeremy
Nov 2, 2018 at 2:13 answer added Jeremy timeline score: 0
Oct 30, 2018 at 20:05 comment added Mateusz Kwaśnicki In general, the set $A$ that minimizes $\int_A f$ given $\int_A g = p$ is the set $A_p = \{f / g < c(p)\}$ for an appropriate $c(p)$. To see this, simply write $$\begin{aligned}\int_A f - \int_{A_p} f &= \int_{A\setminus A_p} f - \int_{A_p\setminus A}f \\&\ge c(p)\int_{A\setminus A_p} g - c(p) \int_{A_p\setminus A}g \\&= c(p)\int_A g - c(p)\int_{A_p} g \\&= c(p) \times p - c(p) \times p = 0.\end{aligned}$$
Oct 30, 2018 at 19:27 history edited Jeremy CC BY-SA 4.0
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Oct 30, 2018 at 18:18 history asked Jeremy CC BY-SA 4.0