Which one of the following statements is TRUE for all positive functions $f(n)$?

Let us analyze each option for positive functions $f(n)$:

• Option A: Let $f(n)$ be a polynomial of degree $k > 0$, i.e., $f(n) = a_k n^k + a_{k-1} n^{k-1} + \dots + a_0$ with $a_k > 0$. Then, $f(n^2) = a_k (n^2)^k + \dots = a_k n^{2k} + \dots = \Theta(n^{2k})$. Also, $(f(n))^2 = (a_k n^k + \dots)^2 = a_k^2 n^{2k} + \dots = \Theta(n^{2k})$. Hence, $f(n^2) = \Theta(f(n)^2)$ is TRUE.

• Option B: For a constant function $f(n) = c > 0$, $f(n^2) = c$ and $(f(n))^2 = c^2$, so $\lim_{n \to \infty} \frac{f(n^2)}{(f(n))^2} = \frac{1}{c} \neq 0$. Thus, $f(n^2) = o(f(n)^2)$ is false.

• Option C: For an exponential function such as $f(n) = 2^n$, $f(n^2) = 2^{n^2}$ and $(f(n))^2 = (2^n)^2 = 2^{2n}$. Since $2^{n^2}$ grows strictly faster than $2^{2n}$, $2^{n^2} \neq O(2^{2n})$, so this statement is false.

• Option D: For $f(n) = 2^{2^n}$, $f(n^2) = 2^{2^{n^2}}$ and $(f(n))^2 = 2^{2^{n+1}}$. Here $(f(n))^2$ grows faster than $f(n^2)$ for certain slowly growing functions (or consider $f(n) = 2^{-n}$, or $f(n) = 1/n$ if restricted to strictly decaying functions). For $f(n) = 2^{-n}$, $f(n^2) = 2^{-n^2}$ while $(f(n))^2 = 2^{-2n}$, and $\frac{2^{-n^2}}{2^{-2n}} \to 0$, so $f(n^2) \neq \Omega((f(n))^2)$.

Therefore, option A is the only correct statement.