I know the answer is f(n) is Big O of g(n).
This is the graph for f(n) and g(n). Here when C is 1, this is the case and f(n) is upper bounded by g(n).
When I'm giving 0.1 as the C value, as shown below, we see that for sufficiently large values of n, f(n) is actually lower bounded by g(n).
Does this mean f(n) is actually big theta(g(n)) or am I missing something?
in the equivalent definition, $f(n)$ is $\theta(g(n))$ iff $lim_{x\to \infty} \frac {f(n)}{g(n)}=c, c\in \Bbb R\setminus \{0\}$. but, according to L'Hôpital's rule, $lim_{x\to \infty} \frac {\log\log(x)}{0.1\log(x)}=lim_{x\to \infty} \frac {\frac {1}{xlogx}}{\frac {1}{10x}} = lim_{x\to \infty} \frac{10x}{x\log(x)}= lim_{x\to \infty} \frac {10}{log(x)}=0$, and therfore, even if its looks like $f(n)$ lower bounded by $g(n)$, at the very very end (much greater values that we checked in those graphs) $f(n)$ is upper bounded by $g(n)$.