Apparently weaker form of big O notation

Question

If instead of saying there exists some $C \in \mathbb{R}^{+}$ and $N \in \mathbb{N}$ such that for any $n \geq N$ we have $f(n) \leq Cg(n)$ we say that there is some $C \in \mathbb{R}^{+}$ and some subsequence $(f(n_{k}))_{k \in \mathbb{N}}$ with $(n_{k})_{k \in \mathbb{N}}$ strictly increasing such that for any $k \in \mathbb{N}$ $f(n_{k}) \leq Cg(n_{k})$, Is it a weaker condition? Or is it equivalent to $f(n) = O(g(n))$? Let it be or not, it reminds me of some theorems about sequences in mathematical analysis. Technique of change of variables made me ask this question.

Answer 1

Let's get the terminology straight: when we have $A \Leftrightarrow B$, then we call $A$ and $B$ equivalent and its natural to say, that they are equally stronger. When we have $A \Rightarrow B$, but not $B \Rightarrow A$ ($B\nRightarrow A$), then we call $A$ more strong (look here or here).

Outgoing from stated, your suggestion gives more weaker condition, because, $f\in O(g)$ in usual sense implies $f\in O(g)$ in new sense, but not reverse, as shows example: $$f(n)=\begin{cases}1, n \text{ is odd}\\ n, n \text{ is even} \end{cases}$$ this function will be in "new" $O(1)$, but not in usual $O(1)$.