Why growth function little-omega $\omega(m)$ equals $\Theta(\log{m})$ I was trying to prove which is asymptotically larger $\log(\log^*{n})$ or $\log^*(\log{n})$, where $\log^*$ is the iterative logarithm that calculates the number of times before we reach 0.
Definition: We define $\omega(g(n))$ (“little-omega of $g$ of $n$”) as the set
$\omega(g(n))$ = {$f(n)$ : for any positive constant $c>0$, there exists a constant
$n \gt n_0$ such that $0≤cg(n)<f(n)$ for all $n ≥ n_0$}.
Definition: We define $\Theta(g(n))$ = { $f (n) $: there exist positive constants $c_1, c_2$, and $n_0$ such that
$0 ≤c_1g(n) ≤ f(n) ≤ c_2g(n)$ for all $n ≥ n_0$}.
Problem: Why growth function little-omega $\omega(m) = \Theta(\log{m})$ please, which is not clear based on respective definitions above? I am not even sure how $\omega(m) = \Theta(\log{m})$ is connected to prove which is asymptotically larger $\log(\log^*{n})$ or $\log^*(\log{n})$, $m$ is not defined as well, but I guess it's an integer. So, presumably, $m$ is an integer.
 A: This is an old question but I have recently been messing around with the $\log^*$ function so I thought I'd give an answer. I am using $\log^*$ as the number of iterations of $\log$ until we reach $1$, your definition will work with trivial adjustments. I am also using $\log$ as the base $2$ logarithm.
Let $T(x)$ be the tower function $T(0) = 1$ and $T(i+1)=2^{T(i)}$, then $\log^* (T(x))=x$ for all $x$. We have that
$$  T(2^n) > 2^{{{T(n^2 - 1)}}} \qquad (*) $$
for large enough $n$ (Proof given below). Fix $x$, and let $n = n(x)$ be such that
$$T(2^n)< x \leq T(2^{n+1}).$$
Assume that $x$ is so large that $(*)$ holds. Then $\log (\log^* x) \leq  n + 1$. However, $x > T(2^n) > 2^{{T(n^2-1)}}$ so that $\log^* (\log  x) \geq n^2 -1$. As $n(x) \to \infty$ as $x\to \infty$, we conclude that
$$  \frac{ \log^* (\log x) }{ \log(\log^* x)  } \geq \frac{n^2-1}{n+1} \to \infty  $$
as $x\to \infty$.
That is, $\log^* (\log x)$ is asymptotically much greater than $\log (\log^* x)$.
Proof of claim: We have by $2^{n}-1 \gg n^2 -1$ that
$$ T(2^n) = 2^{T(2^n-1)} \gg 2^{T(n^2-1)} .$$
