I copied my question from cs.stackexchange because I highly doubt it's going to get an answer there. In Introduction to Algorithms, Lemma 4.4 of the proof of the master theorem goes like this. $a\geq1,b>1,f$ is a nonnegative function defined on exact powers of b. The recurrence relation for $T$ is $T(n) = a T(n/b) + f(n)$ for $n=b^i, \,i>0$
For the third case, we have $f(n) = \Omega(n^{log_ba +\epsilon})$ for some fixed $\epsilon>0$ and that $ af(n/b)\leq cf(n)$ for fixed $c<1$ and for all sufficiently large $n$. In this case, $T(n) =\Theta(f(n))$ since $f(n) = \Omega(n^{log_ba +\epsilon})$.
I was wondering if the condition that $f(n) = \Omega(n^{log_ba +\epsilon})$ is unnecessary since the regularity condition $ af(n/b)\leq cf(n)\,\forall n>n_0$ for fixed $c<1$ and for some $n_0$ implies that $$f(n)\geq m(\frac{a}{c})^{log_b(n/n_0)} \,where \,m=min_{1\leq x\leq n_0}{f(x)}\\=(\frac{n}{n_0})^{log_b(a/c)}=\Theta(n^{log_ba +log_b(c^{-1})})=\Theta(n^{log_ba +\epsilon})$$ This will hold as long as f(n) is non-zero. Hence $f(n)=\Omega(n^{log_ba +\epsilon})$. Therefore we merely need to add the condition that f(n) is positive for all but finitely many values of n for case 3. Am I correct about this?