Binary representation

It's easy to prove that for $x=a_12^{-1}+a_22^{-2}+a_32^{-3}+\dotsc$ we have \[f(x)=a_1b+a_2b^{2-a_1}(1-b)^{a_1}+a_3b^{3-a_1-a_2}(1-b)^{a_1+a_2}+a_4b^{4-a_1-a_2-a_3}(1-b)^{a_1+a_2+a_3}+\dotsc,\]first by induction for all $x$ with finite binary representation and then by continuity for all $x \in [0,1]$. It is then already clear that $f(x)-x<c$ since $0<1-b<b<1$ and hence \[f(x)-x=a_1\left(b-\frac{1}{2}\right)+a_2\left(b^{2-a_1}(1-b)^{a_1}-\frac{1}{4}\right)+\dotsc \le b-\frac{1}{2}+b^2-\frac{1}{4}+b^3-\frac{1}{8}+\dotsc=\frac{b}{1-b}-1=c\]and equality can't hold since in the first summand it would imply $a_1=1$ but in the second one $a_1=0$. However the inequality $f(x)>x$ is not so easy to prove this way since the individual summands $b^{k-a_1-a_2-\dotsc-a_{k-1}}(1-b)^{a_1+a_2+\dotsc+a_{k-1}}-2^{-k}$ might very well be negative. But fortunately, we can proceed more directly. Let us first prove that $f(x) \ge x$ for all $x$. Indeed, by continuity it suffices to prove this for all numbers with finite binary representation. And clearly $f(x)\ge x$ implies $f(x/2)=bf(x)\ge\frac{x}{2}$ as well as $f((x+1)/2)=b+(1-b)f(x)>b+(1-b)x=(x+1)/2+(1-x)(b-1/2) \ge (x+1)/2$. So the claim for $x$ implies the one for $x/2$ and $(x+1)/2$ and hence by induction for all $x$ with finite binary representation and thus $f(x) \ge x$ for all $x$. However, we cannot immediately run the same argument with the strict inequality since we might lose this one in the continuity argument. But there is an easy way out: Suppose that $f(x)=x$ for some $x \in (0,1)$. First suppose that $x \le \frac{1}{2}$. Then $f(2x)=\frac{f(x)}{b}=\frac{x}{b}<2x$ which is a contradiction to what we proved above. Similarly, if $x>\frac{1}{2}$, we have $f(2x-1)=\frac{f(x)-b}{1-b}=\frac{x-b}{1-b}<2x-1$ since $(2x-1)(1-b)-(x-b)=(1-x)(2b-1)>0$ and hence again find a contradiction. Done.