# Is best case complexity big Omega of worst case complexity?

I need to prove or disprove the following claim:

Given that the best case complexity of the algorithm A is $$O(f(n))$$ and the worst case complexity of A is $$Ω(g(n))$$, it follows that $$f(n) ∈ Ω(g(n))$$.

I know that obviously $$A_b \leq A_w$$. I can also conclude obviously that for constants $$c_1 , c_2$$ and for all $$n_0 \leq n$$ for some $$n_0$$ the following happens: $$A_b \leq c_1f(n)$$ and $$A_w \leq c_2g(n)$$. I can see that it doesn't mean that $$f(n) \geq cg(n)$$ for some constant $$c$$ but I am having trouble disproving it. Any help will be welcomed.

Consider the following algorithm $$A$$:
• Otherwise, count from $$1$$ to $$n^2$$, where $$n$$ is the length of the input.
The best case complexity of this algorithm is $$O(n)$$, and the worst-case complexity of this algorithm is $$\Omega(n^2)$$. Does $$n = \Omega(n^2)$$ hold?