# How to find: $~\min\limits_{f\in E}(\int_0^1f(x) \,dx)$

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

• Please try to limit your edits. While editing in general is appreciated, repeated minor edits can be perceived as noise. – quid Nov 8 '17 at 20:27
• 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. – quid Nov 15 '17 at 21:04
• Out of curiosity where did you find this question? I'm sure I've seen it before. Might have been IMC. – user85798 Jan 15 '18 at 13:00
• @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. – 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 $\int_0^1 f(x)dx= \frac 14$