Does Big Omega imply little-oh

Question

"If f= BigOmega(g) then g=o(f)"

Is this true? My understanding is that f is Big Omega bounded by g. So it's at least g(n) on a graph or more. So then examining g, if it is little-oh of f - then it should be at most but not inclusive bounded by f. Seems true to me?

Answer 1

Not necessarily. Let f(x) = g(x) = x.

Then f = BigOmega(g). Proof: let k = 1/2, n_0=1, then for all n > n_0, f(n) >= k * g(n) (since x >= x/2 when x > 1).

However, g != o(f). If you let k=1/2, then |g(n)| <= k * |f(n)| simply isn't true for all n.