prove that $\log( f(n) )=o(\log(g(n)))$ implies $f(n)=o(g(n))$ (small o)

Question

I think it's not true. I disprove it by assuming $f(n) = 1$, $g(n) = 2$.

so $$\lim \frac{\log\left(f(n)\right)}{\log\Big(g\big(g(n)\big)\Big)} = 0$$

but $$\lim \frac{f(n)}{g(n)} = 0.5$$

which disprove it, nevertheless, I am not sure if $\log(1) =O(0)$ have any meaning.

thanks

Answer 1

For the definition take a look here prove that $\log f(n) = o(\log g(n))$ implies $f(n)=o(g(n))$.

Your choice seems indeed a bit strange since $\log g(n)\not \to 0$ but it is correct by the definition $\log( f(n) )=0=o(\log(g(n)))$.

As an alternative let consider

• $f(n)=e^{\frac 1 {n^2}}$
• $g(n)=e^{\frac 1 {n}}$