(This is my attempt to simplify or generalize my previous unanswered question Proving technique for exchanging lim and (infinite) sum where the sum converges very slowly.)
Consider a function defined on positive real numbers $f: \mathbb{R_+}\to\mathbb{R}$ with the following properties
- for all $x>a$, $f(x) = kf(x-a) + (1-k) f(x-b)$, for some constants $k$, $a$, and $b$ where $0<k<1$ and $0<b<a$
- $a / b \notin \mathbb{Q}$ (otherwise a counter-example would be $f(x) = \sin(2\pi x/\gcd(a,b))$))
- $f(x)$ is bounded
- $f(x)$ is continuous (otherwise a counter-example would be the indicator function $f(x) = \mathbb{1}_{a\mathbb{Z}+b\mathbb{Z}}$)(can we weaken this to "discontinuity is nowhere dense"?)
Can we conclude that the limit $$\lim_{x\to\infty} f(x)$$ exists? If not, what other condition we need to add to ensure the limit exists?
Intuitively, this is true to me. The first functional equation means that each function value is a weighted average of some previous values, so overall it should converge to some sort of average of the "seeding region" $x\in(0,a]$. But I can't really convert this intuition to a concrete proof.
Another attempt I had was to try solving the functional equation. I can find the following solution basis $$f(x) = e^{tx},$$ where $t$ is a (potentially complex) solution to the equation $$ke^{-at} + (1-k) e^{-bt} = 1.$$ I don't know if this is the entire solution space, nor how to solve the coefficient for each term. Nevertheless, this equation always has a solution $t = 0$, whose corresponding term in $f(x)$ would be a constant, which is the limit if it exists. All other solutions $t$ have negative real components (but can be arbitrarily close to 0), meaning their corresponding term in $f(x)$ shrinks and has individual limit of 0. However, I don't think I can directly say the limit of the function is the sum of the limit of each term in this case.
Another path I tried to explore is to consider the function on a grid $F: \mathbb{Z}^2\to\mathbb{R}$, $F(x, y) = f(ax+by)$. It would be a great first step to prove that $\lim_{\mathbf{v}\to(+\infty,+\infty)}F(\mathbf{v})$ exists, and then we can "fill in" other values for $f(x)$ using continuity.
On the grid, it can be seen that what contributes to $F(x, y)$ is a structure similar to binomial expansion: \begin{aligned} F(x, y) &= kF(x-1, y) + (1-k) F(x,y-1)\\ &= k^2 F(x-2, y) + 2k(1-k) F(x-1,y-1) + (1-k)^2 F(x,y-2)\\ &= k^3 F(x-3, y) + 3k^2(1-k) F(x-2,y-1) + 3k(1-k)^2 F(x-1, y-2) + (1-k)^3 F(x,y-3)\\ ... \end{aligned} As good as it looks, it doesn't help me with the question.