Question

Given f(n) = o(g(n)), show that it is not necessary that  log (f(n)) = o(log (g(n)))

[Hint, use a counterexample]

Answer 1

A simple contradiction would be n² and n³.

While n² = o (n³),

Log(n²) = 2 Log n,  Log(n³) = 3 Log n , Hence Log(n²) = θ ( Log(n³))

Answer 2

lim(n->infinity)(log(f(n))/log(g(n))) = lim n->infinity (1/(f(n)*ln2) * f'(n)) / (1/(g(n) * ln2)) * g'(n))

                                                     = lim n->infinity (g(n)/f(n) * f'(n)/g'(n))

lim n->infinity (g(n)/f(n)) = infinity

lim n->infinity (f'(n)/g'(n)) = 0

let f(n) = n, g(n) = n^2

lim n-> infinity (f(n)/g(n)) =  lim n-> infinity (1/n)

                                       = 0

f(n) = o(g(n))​

In this condition

lim(n->infinity)(log(f(n))/log(g(n)))  = lim n->infinity (n * 1/2n)

                                                      = 1

so it is not necessary that  log (f(n)) = o(log (g(n))) when f(n) = o(g(n))

Comments

I don't know what the first 3 lines add to this proof by counter example.  Wouldn't you just want to start with details of f(n) and g(n)?