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2016 Taiwan TST Round 1 Quiz 1 Problem 1: Function

Source: 2016 Taiwan TST Round 1 Quiz 1 Problem 1

March 29, 2016
functionalgebraTaiwanFunctional inequality

Problem Statement

Suppose function f:[0,∞)→[0,∞)f:[0,\infty)\to[0,\infty) satisfies (1)∀x,y≥0,\forall x,y \geq 0, we have f(x)f(y)≤y2f(x2)+x2f(y2)f(x)f(y)\leq y^2f(\frac{x}{2})+x^2f(\frac{y}{2}); (2)∀0≤x≤1,f(x)≤2016\forall 0 \leq x \leq 1, f(x) \leq 2016. Prove that f(x)≤x2f(x)\leq x^2 for all x≥0x\geq 0.
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