# Numerically Evaluate the limit of the solution of a functional equation

I need to evaluate the limit of $$f(x)$$ as $$x\to0$$, where the function $$f$$ solves the following equation: $$f(x)=\left\{ \begin{array}{ll} g(x) & \text{if } x\geq \frac{1}{2};\\ \frac{1}{2} f(\alpha x) + \frac{1}{4} (1-x) f\left(\frac{\alpha x}{1-x}\right) + \frac{1}{4} (1+x) f\left(\frac{\alpha x}{1+x}\right) & \text{otherwise } (0

where $$\alpha\in (1,\frac{3}{2})$$ is a given constant, and the function $$g$$ is known over the interval $$[\frac{1}{2},\alpha]$$.

I wonder if there is a nice numerical trick to compute $$\lim_{x\to 0} f(x)$$ ? Or maybe even a formula ?

I tried to solve the recurrence equation by discretizing $$f$$ over $$[0,\frac{1}{2}]$$ (which reduces to solving a linear system of size $$1/2\epsilon$$ for a discretization size of $$\epsilon$$ ), but the solution looks "fractal-like", so I don't know if my computation can be trusted, and if there are not numerical issues around $$x=0$$...

• just an observation: taking derivatives and then sending $x\rightarrow 0$ I find $f^{(n)}(0)=\alpha^n f^{(n)}(0)$, hence $f^{(n)}(0)=0$ for any $n\geq 1$. – Carlo Beenakker Aug 14 '19 at 19:55
• Since g is known on [1/2,α], we can determine f(α/3) putting x=1/2 in the functional equation. But then, which other values can be found? Putting x=α/3 again we may find f(α/(3+α)), unless α<√(3/2), which is not excluded. In any case, I don't see how f(x) can possibly be determined for x< α-1... – Pietro Majer Aug 14 '19 at 20:46