Determining asymptotics of a function given a series of difference-like inequalities

I have a function $f: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ and I know it satisfies the following properties.

$f(x) \leq \frac{\log{\sqrt{2}}}{2x}$

and for all $A \geq 1$ and $B \geq 0$ there is a $K(A,B) \geq 1$ such that

$f(x) \leq K(A,B) f(Ax + B)$.

I know that there are functions which satisfy this property, for instance: $f(x) = \begin{cases} \frac{\log{\sqrt{2}}}{2x} & \mbox{for } 1 \leq x\\ \frac{\log{\sqrt{2}}}{2} & \mbox{for } 0 \leq x < 1\end{cases}$ ought to satisfy all the relevant properties.

I want to know a bit more about the family of functions that satisfy this property though.

I know some things about my functions. Playing around with various values of $A$ and $B$ and recursion I think you can get that for all $a > 1$ there is some $C$ such that the function $f(x)$ is strictly bounded below by $Ca^{-x}$.

I'm most interested in the asymptotics of my function. In particular, can I get a better lower bound. Is it possible to bound it below by a rational function? Or even better yet by $\frac{C}{x}$?

PS: I have no idea how to tag this question. Please feel free to tag it more appropriately.

Edit: With only this information I don't think a rational lower bound is possible. In particular for all $n$ we have the function $f_n(x) = \begin{cases} \frac{\log{\sqrt{2}}}{2x^n} & \mbox{for } 1\leq x \\ \frac{\log{\sqrt{2}}}{2} & \mbox{for } 0 \leq x < 1 \end{cases}$. These should satisfy all of my properties, and any rational function is eventually beaten out by $f_n(x)$ for large enough $n$.