MathLabs

Problem 2

Let R+\mathbb{R}^+ denote the set of positive real numbers. Find all functions f:R+→R+f:\mathbb{R}^+\to\mathbb{R}^+ such that for each x∈R+x\in\mathbb{R}^+, there is exactly one y∈R+y\in\mathbb{R}^+ satisfying xf(y)+yf(x)≤2xf(y)+yf(x)\le2
Step 5 of 6: Let ε→0\varepsilon\to0 to force f(x)≥1/xf(x)\ge1/x
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f(x)>x+2ε(x+ε)2=1x−ε2x(x+ε)2f(x)>\frac{x+2\varepsilon}{(x+\varepsilon)^2}=\frac1x-\frac{\varepsilon^2}{x(x+\varepsilon)^2}
Detailed analysis

Apply the strict inequality with y=x+εy=x+\varepsilon for small ε>0\varepsilon>0: x/(x+ε)<xf(x+ε)+(x+ε)f(x)x/(x+\varepsilon)<xf(x+\varepsilon)+(x+\varepsilon)f(x), and combining with (x+ε)f(x+ε)≤1(x+\varepsilon)f(x+\varepsilon)\le1 gives f(x)>x+2ε(x+ε)2=1x−ε2x(x+ε)2f(x)>\frac{x+2\varepsilon}{(x+\varepsilon)^2}=\frac1x-\frac{\varepsilon^2}{x(x+\varepsilon)^2}; letting ε→0\varepsilon\to0 yields f(x)≥1/xf(x)\ge1/x.