Consider the following recurrenc relation:
$f(n) = f(n/2) +nlogn$
Since this does not honor the form of the Master Recurrence, we need to obtain an estimate of the asymptotic order of $f$. According to the book, we can conclude from the recurrence that:
$f(n) = \Omega(nlgn)$
How? If I'm not mistaken, for any $f(n)$ to be $\Omega(g(n))$, $f(n)$ must be greater than or equal to $Cg(n)$ (where $C$ is some constant).
How can $n$ >= $nlgn$? Much less $n$ >= $C*nlgn$? What am I missing?