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

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 $$g ( n ) ∈ o ( 1 )$$?

• Please use MathJax to type equations. The first condition is completely redundant. For any $f,g$: $f = o(g) \implies 2f = o(2g)$. – Dmitry Oct 13 '20 at 16:14

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.