Is $$\log \log n \times \log \log \log n = \Omega(\log n) $$ How can we prove it.
Actually I'm trying to prove that $f(n) = \lceil(\log \log n)\rceil !$ is polynomially bounded. It means
$$c_1 n^{k_1} \leq f(n) \leq c_2 n^{k_2} \quad \forall n > n_0$$ $$m_1 \log n \leq \log [f(n)] \leq m_2 \log n$$ $$\log [f(n)]=\theta(\log n) \text{ i.e. } \log [f(n)]=\Omega(\log n) \text{ and }\log [f(n)]=O(\log n) $$
I've proved that $\log [f(n)] = O(\log n)$, But I'm having trouble proving $\,\log \left[f(n)\right] = \Omega\left(\log n\right)$. Can anybody tell me how can we do it.
You will not be able to prove this. $f(n)$ grows asymptotically slower than any polynomial, i.e. $f(n) = n^{o(1)}$. Indeed, for any $c > 0$ $$ n^c = 2^{c\log n} $$ while $$f(n) = 2^{\Theta(\log\log n\cdot \log\log\log n)}$$ as you showed. But $$ \frac{\log\log n\cdot \log\log\log n}{\log n} < \frac{(\log \log n)^2}{\log n} \xrightarrow[n\to\infty]{} 0 $$
Edit: this does not contradict the fact that $f(n)$ is polynomially bounded. It is: it is asymptotically bounded (above) by any polynomial.