International Mathematics Competition for University Students

July 31 – August 6 2017, Blagoevgrad, Bulgaria


Ivan is Watching You

Day 1
    Problem 1
    Problem 2
    Problem 3
    Problem 4
    Problem 5

Day 2
    Problem 6
    Problem 7
    Problem 8
    Problem 9
    Problem 10


    Day 1 questions
    Day 1 solutions
    Day 2 questions
    Day 2 solutions
    Closing Ceremony

Official IMC site

Problem 2

2. Let $f\colon\mathbb{R}\to(0,\infty)$ be a differentiable function, and suppose that there exists a constant $L>0$ such that $$ \bigl|f'(x)-f'(y)\bigr| \leq L\bigl|x-y\bigr| $$ for all $x,y$. Prove that $$ \big(f'(x)\big)^2 < 2Lf(x) $$ holds for all $x$.

Proposed by: Jan Ĺ ustek, University of Ostrava

Solution. Notice that $f'$ satisfies the Lipschitz-property, so $f'$ is continuous and therefore locally integrable.

Consider an arbitrary $x\in\mathbb{R}$ and let $d=f'(x)$. We need to prove $f(x)>\frac{d^2}{2L}$.

If $d=0$ then the statement is trivial.

If $d>0$ then the condition provides $f'(x-t)\ge d-Lt$; this estimate is positive for $0\le t<\tfrac{d}{L}$. By integrating over that interval, $$ f(x) > f(x) - f(x-\tfrac{d}{L}) = \int_0^{\tfrac{d}{L}} f'\big(x-t\big) \mathrm{d}t \ge \int_0^{\tfrac{d}{L}} (d-Lt) \mathrm{d}t = \frac{d^2}{2L}. $$

If $d<0$ then apply $f'(x+t)\le d+Lt=-|d|+Lt$ and repeat the same argument as $$ f(x) > f(x) - f(x+\tfrac{|d|}{L}) = \int_0^{\tfrac{|d|}{L}} \big(-f'(x+t)\big) \mathrm{d}t \ge \int_0^{\tfrac{|d|}{L}} (|d|-Lt) \mathrm{d}t = \frac{d^2}{2L}. $$

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