Limit of a functional equation

Question

Starting from the function $$f(x) = ax+bf(xy)$$ where $x,a>0$, $0<b<1$, and $y>1$, and $y,a,b$ are fixed, I got $$f(x) = ax(1+by+b^2y^2+\ldots+b^{n-1}y^{n-1})+b^nf(xy^n)$$ I want to evaluate the limit of this function as $n \to \infty$. The first term is a geometric series so it's easy. My question is: how do I evaluate the limit of $b^nf(xy^n)$ as $n \to \infty$?

Answer 1

There is no such function, assuming you meant that $y$ can be freely picked.

Let $x=1$. Then $f(1) = a+b f(y)$, so $f(y)$ is constant on $y > 1$, and equal to $\frac{f(1)-a}{b}$.

Note that $f(2) = 2a + b f(2y)$ for all $y > 1$, but $f(2) = \frac{f(1) - a}{b}$ and $f(2y) = \frac{f(1)-a}{b}$, so we have $$\frac{f(1)-a}{b} = 2a+f(1)-a$$ whereupon $$f(1)-a = b f(1)+a b$$ so $$f(1) = \frac{2a}{1-b}$$ That is, $$f(y) = \frac{a(1+b)}{b(1-b)}$$

Then $$f(3) = 3a + b f(3y)$$ so $$\frac{a(1+b)}{b(1-b)} = 3a+\frac{a(1+b)}{1-b}$$ which reduces to $b=\frac{1}{2}$; so $f(y) = 6a$ for $y>1$.

Finally, $$f(4) = 4a+b f(4y)$$ so $6a=4a+3a$, which is a contradiction.