Prove that if $f(n)=\omega(g(n))$ then $f(n)-g(n)$ = $\Theta f(n).$

Question

I'm a beginner in the field of asymptotic analysis and I'm having trouble proving that if $f(n)=\omega(g(n))$ then $f(n)-g(n)$ = $\Theta f(n).$ A hint how to solve this problem will much be appreciated!

Answer 1

You have to prove that there exists $a,b>0$ such that $$ af(n)\le f(n)-g(n) \le bf(n) $$ for all $n \in \mathbf{N}$, where $f,g: \mathbf{N} \to (0,\infty)$ are functions such that $f(n)=\omega(g(n))$. So, it is enough to set $b=1$ and any $a \in (0,1)$. Indeed, $af(n) \le f(n)-g(n)$ if and only if $$f(n) \ge \frac{1}{1-a}g(n),$$ which is eventually true.

Answer 2

Note that $f(n)-g(n) < f(n)$ therefore $f(n)-g(n) = O(f(n))$. Now you need to prove $f(n) - g(n) = \Omega(f(n))$.

Since $f(n) = \omega(g(n))$, how do you compare $f(n)-g(n)$ to $0.9f(n)$?