# 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 am trying to look for an example of a monotonically non-decreasing function $$f(n)$$ such that:

$$f(n) \in O(n^2)$$ and $$f(n) \notin o(n^2)$$ but also $$f(n) \in \Omega(n)$$ and $$f(n) \notin \omega(n)$$. The domain could also be all real numbers.

I considered different variations of piecewise functions, for example I considered the function $$f(n)= \begin{cases} n^{2-\frac{1}{n}} &\text{if}\, n =2k+1\\ n^{1+\frac{1}{n}}&\text{if}\, n=2k\\ \end{cases}$$

which satisfies the growth rate conditions, but is clearly not monotonically non-decreasing.

I would be happy to hear your ideas, thank you.

$$f(n) = \left\{\begin{array}{rl}1 & \text{if }n=1\\2^{2^{\left\lfloor \log_2\log_2 n \right\rfloor+1}}&\text{otherwise}\end{array}\right.$$
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.