[THIS IS HOMEWORK so please, don't just provide solutions, help me reasoning]
I'm dealing with the following problem:
Let $f(n)$ be the worst case time complexity of an algorithm A. Assume that there
exist $c > 0$ and $n_{0}$, such that for each $n > n_{0}$ there exists an input of size $n$ for which A runs in exactly $c\cdot{g(n)}$ steps.
a. Is it necessarily true that $f(n) = \Omega(g(n))$?
b. Assume $f(n) = O(g(n))$. Is it necessarily true that $f(n) = \Theta(g(n))$?
For a. I think it must be true since the $\Omega$ notation has on it all the elements that are either $f(n) = c\cdot g(n)$ or $f(n) > c\cdot g(n)$ so following the assumption leads to $f(n) = c\cdot g(n)$ being always true, therefore making the whole sentence true. (hope I'm right).
For b. I assume $f(n) = O(g(n))$ and $f(n)=c\cdot{g(n)}$ ($\forall{n>n_{0}}$... of course) then I concluded that since (a) is true then $f(n) = \Theta(g(n))$.
any help will be appreciated, thanks.
