MathDB
There exists a function u

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September 5, 2010
functionalgebra proposedalgebra

Problem Statement

Let f:(0,+)Rf : (0,+\infty) \to \mathbb R be a function having the property that f(x)=f(1x)f(x) = f\left(\frac{1}{x}\right) for all x>0.x > 0. Prove that there exists a function u:[1,+)Ru : [1,+\infty) \to \mathbb R satisfying u(x+1x2)=f(x)u\left(\frac{x+\frac 1x }{2} \right) = f(x) for all x>0.x > 0.