Time complexity analysis of 2 arbitrary algorithms - prove or disprove
It is possible. Example $g_A(n)=1$, $g_B(n)=2$, and $f(n)=1$.

It is also necessary, since $g_B(n) = 2 g_A(n) \in\Omega(f(n))$.

To see that $ 2 g_A(n) \in\Omega(f(n))$ you can use the definition of $\Omega(\cdot)$.

From $g_A(n) = \Omega(f(n))$ you know that here is some $n_0$ and some $c&gt;0$ such that, $\forall n \ge n_0$, $g_A(n) \ge c f(n)$.
This implies that, for the same value of $n_0$ and $c$, $2 g_A(n) \ge 2 c f(n) \ge c f(n)$, i.e.,  $ 2 g_A(n) \in\Omega(f(n))$.