$\star$ Show that case 3 of the master method is overstated, in the sense that the regularity condition $af(n/b) \le cf(n)$ for some constant $c < 1$ implies that there exists a constant $\epsilon > 0$ such that $f(n) = \Omega(n^{\log_b{a}+\epsilon})$.

$$ af(n/b) \le cf(n) \\ \alpha f(n/b) \le f(n), \quad \alpha = a/c \\ \alpha f(n) \le f(nb) \\ \alpha^i f(1) \le f(b^i) \\ n = b^i \Rightarrow i = \log_{b}n \Rightarrow f(n) \ge \alpha^{\log_b{n}}f(1) = n^{\log_{b}\alpha} \\ \alpha > a \Rightarrow \alpha = a + d \quad (c < 1, d > 0) \\ \Rightarrow f(n) = n^{\log_b{a} + log_b{d}} = n^{\log_b{a}+\epsilon} \quad (\epsilon = \log_{b}d) $$

I did not think this up. I had some help at StackExchange.