International Mathematics Competition
for University Students
2016

IMC2016: Day 2, Problem 7

7. Today, Ivan the Confessor prefers continuous functions $f:[0,1]\to\mathbb{R}$ satisfying $f(x)+f(y)\geq |x-y|$ for all pairs $x,y\in [0,1]$. Find the minimum of $\int_0^1 f$ over all preferred functions.

Proposed by Fedor Petrov, St. Petersburg State University

Solution. The minimum of $\int_0^1 f$ is $\tfrac14$.

Applying the condition with $0\le x\le \tfrac12$, $y=x+\tfrac12$ we get $$ f(x) + f(x+\tfrac12) \ge \tfrac12. $$ By integrating, $$ \int_0^1 f(x) \,\mathrm{d}x = \int_0^{1/2} \big(f(x)+f(x+\tfrac12)\big) \,\mathrm{d}x \ge \int_0^{1/2} \tfrac12 \,\mathrm{d}x = \tfrac14. $$

On the other hand, the function $f(x)=\big|x-\tfrac12\big|$ satisfies the conditions because $$ |x-y| = \Big|\big(x-\tfrac12\big)+\big(\tfrac12-y\big)\Big| \le \big|x-\tfrac12\big|+\big|\tfrac12-y\big| = f(x)+f(y), $$ and establishes $$ \int_0^1 f(x) \,\mathrm{d}x = \int_0^{1/2} \big(\tfrac12-x\big) \,\mathrm{d}x + \int_{1/2}^1 \big(x-\tfrac12\big) \,\mathrm{d}x = \frac18+\frac18 = \frac14. $$