I'm trying to figure out a new operator compared to the Big O.
Suppose we have two positive functions, $f(n)$ and $g(n)$ then we say that $f(n) = O^*(g(n))$ if there exists a constant $ c > 0 $ such that $f(n) \le c(g(n)) $ for every integer $ n \ge 1 $
It is very similar to the BigO definition but you are a little bit more restricted here because you can't choose $n_0$.
I'm trying to prove that if $f(n) = O(g(n))$ then $f(n) = O^*(g(n))$
Here is what I got so far:
By the definition of BigO we know that if $f(n) = O(g(n))$ then there exist $n_0, c > 0$ such that $f(n) \le c(g(n))$ for every integer $n \ge n_0$
Now I set this $n_0$ to be 1 to apt the definition of $O^*$, but I can't figure out how to determine my $c$.
Can please someone give me an hint? Thanks!