$g ( n ) ∈ ω ( 1 )$ and $f ( n ) ∈ o ( g ( n ) )$ imply $2 f ( n ) ∈ o ( 2 g ( n ) )$

Question

Prove that if $g ( n ) ∈ ω ( 1 )$ and $f ( n ) ∈ o ( g ( n ) )$, then $2 f ( n ) ∈ o ( 2 g ( n ) )$.

I was going over this question in my Algorithms class and could'nt understand why first condition has to be met. How would $g ( n ) ∈ ω ( 1 )$ affect our reasoning? Also, what would happen if instead of $g ( n ) ∈ ω ( 1 )$, it was $g ( n ) ∈ o ( 1 )$?

Answer 1

Considering non negative case we have $$O(f)=\{g: \exists C>0, \exists N \in \mathbb{N}, \forall n>N, g(n) \leqslant Cf(n)\}$$ From this definition for $\forall C>0,$ we have: $$ C \cdot O(f)= O(C \cdot f) = O(f)$$ It can be written also as: $$C \cdot g \in O(f) \Leftrightarrow g \in O(C \cdot f) \Leftrightarrow g \in O(f)$$ without any additional condition.