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Assume $a$ and $b$ are real numbers such that $0<a<b$, and let $\textsf{K}_{a,b}$ the set of all non-negative, monotonically decreasing functions $f$ satisfying $$\int_a^b f(t) dt=1$$ and that $af(a)=bf(b)$ on the interval $[a,b]$.

Find the value of $$\sup \left\{ \int_a^b\max\{f(t),g(t)\}dt \, : \, f,g\in \textsf{K}_{a,b} \right\}$$

Em, I basically have no idea. How to deal with the $\max$ function? Using inequality? I don't know...

Somehow it looks like a variational problem? Everything is thankful.

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    $\begingroup$ Where is this exercise from? $\endgroup$ – Jack Jul 28 at 16:15
  • $\begingroup$ @Jack from a 2019's summer camp test of some university of China $\endgroup$ – Kyle Tao Jul 29 at 12:38
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From $f(b) \leq f(t) \leq f(a)$ and $f(b) = af(a)/b$, we get the inequality: $$f(a)\frac{a(b-a)}{b} =\int_a^bf(b)dt \leq \int_a^b f(t)dt = 1 \leq \int_a^bf(a)dt = f(a)(b-a.)$$

Therefore: $$ m = \frac{1}{b-a} \leq f(a) \leq \frac{b}{a(b-a)} = M.$$

Here's a rough argument given this:

To maximize the integral of the max function, we want two functions with masses distributed with minimal overlap. So let's take $f(x) = M$ for $ b \leq x \leq x_0$ where $x_0 $ is such that $(x_0-a)M = 1$ and $0 $ after that point.

On the other hand, let us take $g(x) = m$. Then: $$\int_b^a \max\{f,g\}dt = \int_b^{x_0} fdt + \int_{x_0}^a gdt = 1+ (b-x_0)(b-a)$$

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