if $\log(f(n))=\omega(\log(g(n))$ so then $(f(n))=\omega(g(n)$?

Question

I want to show that $$f(n)=\omega(g(n)$$ . for $$f(n)=(\log(n))^n , g(n)=n^{(1/2)\log n}$$ and it is easier for me to show that** $$\log(f(n))=\omega(\log(g(n))$$. is it correct to do that?is it right to claim that if $\log(f(n))=\omega(\log(g(n))$ so then $(f(n))=\omega(g(n)$?

**

with $$\log(f(n))=n\log(\log n),\log(g(n))=1/2*\log^2n$$ I show that $$\lim_{n\to\infty}(\log f(n)/\log(g(n))=\infty $$ and conclude that $(f(n))=\omega(g(n)$

Answer 1

I have corrected your codes here.

I want to show that $f(n)=\omega(g(n))$ for $$f(n)=\log^n n,g(n)=n^{\frac{1}{2}\log n},$$ and it is easier for me to show that $$\log f(n)=\omega(\log g(n)).$$ Is it correct to do that ? Is it right to claim that if $$\ \log f(n)=\omega(\log g(n)),$$ then $$f(n)=\omega(g(n))\ ?$$ With $$\log f(n)=n\log\log n,\log g(n)=\frac{1}{2}\log^2 n,$$ I show that $$\lim_{n\to \infty}\frac{\log f(n)}{\log g(n)}=\infty,$$ and conclude that $f(n)=\omega(g(n))$.

The answer is presented below.

Solution. For $x\geqslant e$, we have $$e^x\log x\geqslant e^x>\frac{1}{2}x^2,$$ Now letting $x=\log n$ can get $$n\log\log n>\frac{1}{2}\log^2 n.$$ It shows that for $n\geqslant 16$, $$\log^n n>n^{\frac{1}{2}\log n}.\tag{*}$$ We can verify that (*) holds for $n\geqslant 4$. It shows that $f(n)=\omega(g(n))$.$\quad\square$