# Find the value of $\sup \int_a^b \max\{f,g\}$

Assume $$a$$ and $$b$$ are real numbers such that $$0, and let $$\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.

• Where is this exercise from? – Jack Jul 28 at 16:15
• @Jack from a 2019's summer camp test of some university of China – Kyle Tao Jul 29 at 12:38

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: $$m = \frac{1}{b-a} \leq f(a) \leq \frac{b}{a(b-a)} = M.$$
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: $$\int_b^a \max\{f,g\}dt = \int_b^{x_0} fdt + \int_{x_0}^a gdt = 1+ (b-x_0)(b-a)$$