# Does the product of two functions equal the product of their Big-O's?

let's say $$f(n) = O(g(n))$$ and $$l(n) = O(m(n))$$

is it always true that $$f(n) \cdot l(n) = O(g(n)) \cdot O(m(n))$$ ?

By definition, $$f(n) \leq c_1 \cdot g(n)$$ and $$l(n) \leq c_2 \cdot m(n)$$, for some $$c_1$$ and $$n \geq n_0$$, and for some $$c_2$$ and $$n \geq n_0'$$ respectively. Suppose we set $$n_0^* = \max (n_0, n_0')$$, then both inequalities are satisfied for $$n \geq n_0^*$$. Then obviously $$f(n) \cdot l(n) \leq c_1 \cdot g(n) \cdot c_2 \cdot m(n)$$ for $$n \geq n_0^*$$. So indeed the claim holds. We can go take one additional step, however. Note that we can replace $$c_1 \cdot c_2$$ by a new constant, and thus we conclude that $$f(n) \cdot l(n) = O(g(n) \cdot m(n))$$.