I came across to the following problem:

Let $E$ be the set of all continuous function $f:[0,1]\to \mathbb{R}$ such that $$f(x)+f(y)\ge |x-y|\qquad\forall\,x,y\in [0,1]$$

Then find $$\min_{f\in E}\left(\int_0^1f(x) dx\right)$$

My attempt: I took the double integral on both side which yields $$ 2\int_0^1f(x)dx =\int_0^1\int_0^1f(x) +f(y)dydx \ge \int_0^1\int_0^1|x-y|dxdy =\frac{1}{3} $$ Thus, $$~\min\limits_{f\in E}(\int_0^1f(x) \,dx) \ge \frac{1}{6}$$ Unfortunately I don't know How to get the minimizer. Please give help me with a hint or an answer.

Minimize $\min_{f\in E}\left(\int_0^1f(x) dx\right)$

  • 2
    $\begingroup$ Please try to limit your edits. While editing in general is appreciated, repeated minor edits can be perceived as noise. $\endgroup$ – quid Nov 8 '17 at 20:27
  • 2
    $\begingroup$ I do not understand the point of your comment. If you want to imply that I have downvoted this question, please note, first, it is poor style to do this, second it is factually incorrect. $\endgroup$ – quid Nov 15 '17 at 21:04
  • $\begingroup$ Out of curiosity where did you find this question? I'm sure I've seen it before. Might have been IMC. $\endgroup$ – user85798 Jan 15 '18 at 13:00
  • $\begingroup$ @bwv869 you may be right but that was actually a problem I encountered with a friend. I did not ask him where he took it from. he wanted the challenge the groups after some lunch. $\endgroup$ – Guy Fsone Jan 15 '18 at 13:04

Setting $y=1-x$, $f(x)+f(1-x)\geq |2x-1|$, and since $\int_0^1 f(1-x) dx =\int_0^1 f(x) dx$, integrate to get $$2\int_0^1 f(x)dx\geq \int_0^1 |2x-1| dx = \frac 12$$ hence $\int_0^1 f(x) dx \geq \frac 14$.

This bound is attained for $f:x\mapsto |x-\frac 12|$. The triangle inequality trivially yields $f(x)+f(y)\ge |x-y|$ and $\int_0^1 f(x)dx= \frac 14$

  • 1
    $\begingroup$ Ingenious +1 :) $\endgroup$ – Maria Mazur Nov 7 '17 at 19:12
  • $\begingroup$ Ingenius solution! +1 $\endgroup$ – Hans Nov 11 '17 at 0:40
  • $\begingroup$ Would you care to check out a variation on this theme math.stackexchange.com/q/2514441/64809? $\endgroup$ – Hans Nov 13 '17 at 16:57

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.