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show that 0 < f(x) -x < c for every 0 < x < 1

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October 5, 2010
algebra unsolvedalgebra

Problem Statement

Every x,0≤x≤1x, 0 \leq x \leq 1, admits a unique representation x=∑j=0∞aj2−jx = \sum_{j=0}^{\infty} a_j 2^{-j}, where all the aja_j belong to {0,1}\{0, 1\} and infinitely many of them are 00. If b(0)=1+c2+c,b(1)=12+c,c>0b(0) = \frac{1+c}{2+c}, b(1) =\frac{1}{2+c},c > 0, and f(x)=a0+∑j=0∞b(a0)⋯b(aj)aj+1f(x)=a_0 + \sum_{j=0}^{\infty}b(a_0) \cdots b(a_j) a_{j+1} show that 0<f(x)−x<c0 < f(x) -x < c for every x,0<x<1.x, 0 < x < 1.
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