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.

  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.