This is part of a proof for the third case in the Master Theorem in [CLRS], 3rd edition. $a\geq 1$, $b>1$ and $c<1$. Also, $f$ is a nonnegative function.
It makes sense for polynomial functions but I don't see how they generalized it. I tried mathematical induction but no luck. Any help would be appreciated!
This is the outline -- to be perfectly formal, there would be annoying details or discussion about floors and ceilings to take care of, for instance, since $\frac{n}{b^i}$ is not an integer in general.
Assumption: for all $n$ greater than some constant $n_0$, we have $$ af\left(\frac{n}{b}\right) \leq c f(n) \tag{$\dagger$} $$
Then, iterate (i.e., perform an induction in disguise): $$\begin{align} a^i f\left(\frac{n}{b^i}\right) &= a^{i-1}\cdot f\left(\frac{(n/b^{i-1})}{b}\right) \operatorname*{\leq}_{(\dagger)} a^{i-1}\cdot c f\left(\frac{n}{b^{i-1}}\right) = ca^{i-2}\cdot af\left(\frac{(n/b^{i-2})}{b}\right) \\ &\operatorname*{\leq}_{(\dagger)} ca^{i-2}\cdot c f\left(\frac{n}{b^{i-2}}\right) = c^2a^{i-3}\cdot a f\left(\frac{(n/b^{i-3})}{b}\right) \\ &\vdots\\ &\operatorname*{\leq}_{(\dagger)} c^{i-1} a^0\cdot cf\left(\frac{n}{b^{0}}\right) = c^i f(n). \end{align}$$