Recently, I have seen the following argument:
$$f(x) < Dx + f\left(\frac{x}{2} \right)$$
$$\Rightarrow f(x) < Dx + f \left( \frac{x}{2} \right ) < Dx + \frac{Dx}{2} + f \left(\frac{x}{4} \right)$$
$$\Rightarrow f(x) < Dx (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^n}) + f\left ( \frac{x}{2^n} \right)$$ Since $f(0) = 0$, then taking the limit as $n \rightarrow \infty$ gives:
$$f(x) < 2 Dx$$
I was wondering if this could be generalized. My goal is to perform a similar trick on the system of inequalities:
$$f(x) > D x - f\left(\frac{x}{a} \right) $$
$$f(x) < Dx - f \left( \frac{x}{a} \right ) + f \left( \frac{x}{b} \right ) + f \left( \frac{x}{c} \right)$$
The first iteration isn't too bad. Replacing the negative terms above by the lower bound and the positive terms above by the upper bound, we get:
$$f(x) < Dx - \frac{Dx}{a} + f \left( \frac{x}{a^2} \right ) + \frac{Dx}{b} - f \left ( \frac{x}{ab} \right ) + f \left ( \frac{x}{b^2}\right) + f \left( \frac{x}{cb} \right ) + \frac{Dx}{c} - f \left ( \frac{x}{ac} \right ) + f \left ( \frac{x}{bc} \right ) + f \left ( \frac{x}{c^2} \right) $$
As you can see, it gets complicated quickly. Again, the goal here is to end up with a statement like $ f(x) < ACx$ for some constant A, presumably in terms of a, b and c $ \in \mathbb{Z}$. Has anyone seen anything like this before / tried to make a similar argument? Any advice, info, suggestions would be very much appreciated. thanks