Does there exist a function $f(n)$ that $f(n^2) = O(f(n))$
In other words I need to show that there exists $,\exists n_0 ,c \gt 0 , \forall n \gt n_0$ that will makes this true:
$$f(n^2) \le c \cdot f(n)$$
I thought about taking $\frac{1}{n}$, Because then:
$\cfrac{1}{n^2} \le c \cfrac{1}{n} \rightarrow \cfrac{1}{n} \le c \rightarrow 1 \le c\cdot n$ Which is true, for example $c = 1, n = 2$