I have two real-valued functions $f,g$ defined over the $N$-dimension Real Euclidean space: $$ f,g: \mathbb{R}^N\to\mathbb{R}. $$ They satisfy this property: $$ \forall x_n \in \mathbb{R}^N: f(x_n)\to 0 \text{ if and only if } g(x_n)\to 0 $$ I am wondering whether there is a mathematical notation/ term for naming such a property? I thought about using $f=O(g)$, but it turns out to mean something different from the above property. I also thought about using terms like continuity, etc.

Any idea? Thank you.

  • $\begingroup$ Maybe $(f,g)(x_n)\to 0$ $\endgroup$
    – Masacroso
    May 28, 2016 at 14:51
  • $\begingroup$ You wrote $\forall x_n \in \mathbb{R}^N: f(x_n)\to 0 \text{ if and only if }g(x_n)\to 0$. This doesn't make sense. Did you mean $\forall (x_n)_{n\in\mathbb{N}} \in \left(\mathbb{R}^N\right)^\mathbb{N}: f(x_n)\to 0 \text{ if and only if }g(x_n)\to 0$ $\endgroup$
    – Xaver
    May 29, 2016 at 16:14
  • $\begingroup$ @Xaver. Thanks. Mathematics is a language for communication, so, I do not see there is such a need to further obscure the notation (as what you proposed) when it can already be well understood. $\endgroup$
    – zell
    May 29, 2016 at 16:57
  • $\begingroup$ Are the functions continuous? $\endgroup$
    – Itai
    May 29, 2018 at 13:14
  • $\begingroup$ Possibly over-complicating, you can say something like "$f^{-1}((-1/n,1/n))$ and $g^{-1}((-1/n,1/n))$ are equivalent as filter bases" (which also implies either both are or both aren't filter bases). If both $f$ and $g$ are continuous the condition is equivalent to $f^{-1}(0) = g^{-1}(0)$. $\endgroup$
    – Itai
    May 29, 2018 at 14:08


You must log in to answer this question.

Browse other questions tagged .