Fix a value $x\in(0,1)$ and recursively define the sequences
$$a_0=0,~b_0=1$$
$$c_{n+1}=\frac{\frac{a_n+b_n}2+x}2=\frac{a_n+2x+b_n}4$$
$$a_{n+1}=\begin{cases}a_n,&c_{n+1}>x\\c_{n+1},&c_{n+1}<x\end{cases}$$
$$b_{n+1}=\begin{cases}c_{n+1},&c_{n+1}>x\\b_n,&c_{n+1}<x\end{cases}$$
Supposing that we never have $c_n=x$, what is the rate of convergence of $a_n$ and $b_n$ as they approach $x$, defined by
$$\mu(x)=\lim_{n\to\infty}(b_n-a_n)^{1/n}$$
as a function of $x$? Specifically I am interested in the infinum, supremum, and geometric mean (see integrals below) of $\mu$ over $(0,1)$.
Attempting to plot this in MATLAB for $n=30$ and $1000$ equally spaced points, I get
with a geometric mean of roughly $0.3613$, which matches up reasonably for $n\in[25,30]$ as well as my integral results (see below). The supremum and infinum for these were found to be approximately
$$0.25\le\mu(x)\le0.3862$$
which suggests the upper bound may be given by $\frac{-1+\sqrt{17}}8$ from the fixed-point below.
Trivially we have the easily proven bounds of $1/4\le\mu\le1/2$, and I can compute this at some points by studying fixed-points. For example, we have
$$\mu\left(\frac{-3+\sqrt{17}}4\right)=\frac{-1+\sqrt{17}}8\simeq0.39039$$
$$\mu\left(\frac{-7+\sqrt{89}}{20}\right)=\sqrt{\frac{-5+\sqrt{89}}{32}}\simeq0.37224$$
which were found by solving $a_n=0$ and $1-x=x/b_n$ for $n=1$ and $n=2$ respectively. By taking the limit of these fixed-points, the general term of which can be found explicitly, we get the infinum by
$$\inf_{x\in(0,1)}\mu(x)=\frac14$$
However, I'm not really sure if this $\mu$ is continuous (as it likely exhibits a fractal-like nature), in which case computing $\mu(x)$ is not very useful to me (see context at the end). Hence, I'm also interested in the "average" rate by instead considering $f_n(x)=b_n-a_n$ as a function of $x$ and taking the geometric average as
$$\lim_{n\to\infty}\exp\left(\frac1n\int_0^1\ln(f_n(x))~\mathrm dx\right)$$
which would be more meaningful to me as a sort of "expected rate of convergence".
For $n\in[1,4]$, we have
\begin{align} \exp\left(\int_0^1\ln(f_1(x))~\mathrm dx\right)&=e^{-1}\simeq0.36788\\ \exp\left(\frac12\int_0^1\ln(f_2(x))~\mathrm dx\right)&=e^{-1/2}2^{9/5}3^{-8/5}\simeq0.36418\\ \exp\left(\frac13\int_0^1\ln(f_3(x))~\mathrm dx\right)&\simeq0.36353\\ \exp\left(\frac14\int_0^1\ln(f_4(x))~\mathrm dx\right)&\simeq0.36306\end{align}
which seems to converge somewhat reasonably to $0.36179$ (by Aitken delta squared acceleration), but this is a fairly tedious computation.
For context, this problem arose while studying a bisection-like algorithm I made by averaging the arithmetic mean with a faster converging approximation of the root, such as the secant's root. That is to say, the $x$ used in $c_{n+1}$ is only an approximation of $x$ that becomes more precise on every iteration.
