To me it looks like there can always be values of constants c1, c2, n0 such that $$ c_1 \cdot g(n) \leq f\left(\frac{n}{2}\right) \leq c_2 \cdot g(n) $$
Is there a way to disprove it with a counterexample?
I thought of $$ g(n) = f(n) = 2^n \\ c_1 \cdot 2^n \leq 2^{\frac{n}{2}} \leq c_2 \cdot 2^n $$
But can't we choose any values of c1 and c2 to always satisfy the condition?
Yes, this is a good counterexample. If $c_1 \cdot 2^n \le 2^{n/2}$ for all large enough $n$, then $$ c_1 \le \lim_{n\to\infty}\frac{2^{n/2}}{2^n} =\lim_{n\to\infty} 2^{-n/2} = 0 $$ so $c_1 = 0$, which is not allowed.