Does g(n) ∈ O(f(n)) imply f(n) ∈ Ω(g(n))?

Question

I am just trying to understand how Big O and Big Omega work. I know that Big O means no better than, and Big Omega means no worse than running times. So if I have a function g(n) such that g(n) = O(f(n)) then can I say that f(n) = Ω(g(n))?

Answer 1

Notation-wise, it is better to write g(n) ∈ O(f(n)), because "O(f(n))" can be seen as the set of all functions that grow no faster than a multiple of f(n).

Let us restate the two relevant formal definitions used in complexity theory:

• g(n) ∈ O(f(n)) ⇔ ∃k>0 ∃N≥0 ∀n≥N [|g(n)| ≤ k·|f(n)|]
• f(n) ∈ Ω(g(n)) ⇔ ∃k>0 ∃N≥0 ∀n≥N [f(n) ≥ k·g(n)]

If we can assume that f and g are non-negative functions (which is almost always the case for functions used in computer science), then we can drop the absolute value signs. Thus:

• g(n) ∈ O(f(n)) ⇔ ∃k>0 ∃N≥0 ∀n≥N [g(n) ≤ k·f(n)]
• f(n) ∈ Ω(g(n)) ⇔ ∃k>0 ∃N≥0 ∀n≥N [f(n) ≥ k·g(n)]

Next, flip the inequality on the second logical statement:

• g(n) ∈ O(f(n)) ⇔ ∃k>0 ∃N≥0 ∀n≥N [g(n) ≤ k·f(n)]
• f(n) ∈ Ω(g(n)) ⇔ ∃k>0 ∃N≥0 ∀n≥N [k·g(n) ≤ f(n)]

Now let's prove that the right-hand side of the first statement implies the right-hand side of the second statement:

1. Assume that ∃k>0 ∃N≥0 ∀n≥N [g(n) ≤ k·f(n)] is true.
2. Instantiate the k>0 that satisfies ∃N≥0 ∀n≥N [g(n) ≤ k·f(n)].
3. Let kʹ = 1/k, which is legal because k ≠ 0.
4. Instantiate the N≥0 that satisfies ∀n≥N [g(n) ≤ k·f(n)].
5. Let n be an arbitrary number such that n≥N.
6. Then we have g(n) ≤ k·f(n).
7. Next we have g(n)/k ≤ f(n).
8. By substitution, we have kʹ·g(n) ≤ f(n).
9. Because n is arbitrary, we derive that ∀n≥N [kʹ·g(n) ≤ f(n)].
10. We derive that ∃N≥0 that satisfies ∀n≥N [kʹ·g(n) ≤ f(n)].
11. We derive that ∃kʹ>0 ∃N≥0 ∀n≥N [kʹ·g(n) ≤ f(n)].
12. We rename kʹ to k, so that ∃k>0 ∃N≥0 ∀n≥N [k·g(n) ≤ f(n)].
13. Thus [∃k>0 ∃N≥0 ∀n≥N [g(n) ≤ k·f(n)]] implies [∃k>0 ∃N≥0 ∀n≥N [k·g(n) ≤ f(n)]].
14. Therefore g(n) ∈ O(f(n)) implies f(n) ∈ Ω(g(n)), as wanted.