A monotonically nondecreasing function $ f(n) $ s.t $ f \in O(n^2) $ and $ f \notin o(n^2) $ but also $ f \in \Omega(n) $ and $ f \notin \omega(n) $
I may be wrong, but it seems to me that the function defined by:
$f(n) = \left\{\begin{array}{rl}1 &amp; \text{if }n=1\\2^{2^{\left\lfloor \log_2\log_2 n \right\rfloor+1}}&amp;\text{otherwise}\end{array}\right.$
satisfy the conditions.
To explain why, the values taken by $f(n)$ will be $1, 4, 4, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 256, 256, …$
The idea is that when $n=2^{2^k}$, then $f(n) = 2^{2^{k+1}} = n^2$, and when $n=2^{2^k}-1$, then $f(n) = 2^{2^{k}} = n +1$. That way, you will guarantee the 5 conditions: $(O(n^2) \setminus o(n^2)) \cap (\Omega(n) \setminus \omega(n))$ and increasing.