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Let $\mathcal{T}$ the set of sequences $t\mapsto at^\alpha$, with $t\in\mathbb{N}$, $\alpha>0$ and $a\in\mathbb{R}$. If necessary, it is possible to add boundedness assumptions, such that $\alpha\geq \alpha_{\min}>0$ and $0<a_{\min}\leq |a|\leq a_{\max}$.

I'm looking for a metric $d$ on $\mathcal{T}$ such that $$\forall\varepsilon>0,\, \exists \alpha(\varepsilon)>0,\, M(\varepsilon)>0,\, t(\varepsilon)\in\mathbb{N} : \left(d(u,v)>\varepsilon\right) \implies \left(\forall t\geq t(\varepsilon),\, |u(t)-v(t)|\geq M(\varepsilon)t^{\alpha(\varepsilon)}\right).$$

In other words, as soon as $d(u,v)>\varepsilon$ the rate of divergence of $u$ and $v$ is bounded from below by some function of $\varepsilon$ only.

I tried to use the metric $d(at^\alpha,bt^\beta) = \max(|a-b|,|\alpha- \beta)$, but it fails, as for example $d\left((1+2\varepsilon)t,t^{1+\frac{1}{n}}\right)>\varepsilon$, yet these two functions don't satisfy the desired property, as their rate of divergence can be made as small as possible by letting $n$ go to $\infty$.

So I'd like to either find a metric that works, or at least to prove that it does or does not exist.

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  • $\begingroup$ There is another maybe useful metric on the space of monomials $at^\alpha$, $a>0$, $\alpha\in\mathbb R$. Put $d(at^\alpha,bt^\beta)=\max(|\ln a-\ln b|,|\alpha-\beta|)$. Then $d(at^\alpha,bt^\beta)=\mu$ implies $\frac{at^\alpha}{b t^\beta}\leq e^\mu t^\mu$ for $t>1$. Maybe taking quotients of the monomials makes more sense than taking differences. $\endgroup$ – Helmut Mar 27 '18 at 14:11
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I'll write your $t(\epsilon)$ as $t_0(\epsilon)$ to avoid ambiguity. The condition implies that if $u(t)=v(t)$ for some $u,v,t,$ then $t<t_0(\tfrac12 d(u,v)).$ In particular $d(nt,t^2)\to 0$ as $n\to\infty,$ because the two functions are equal at $t=n.$ (To clarify: I mean $d(u_n,v)\to 0$ where $u_n$ is the monomial $nt^1$ and $v$ is the monomial $t^2.$) This means $nt\to t^2,$ but by the same logic $nt\to t^3,$ which means $d(t^2,t^3)=0,$ so $d$ is not a metric.

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  • $\begingroup$ Could you elaborate on the triangle inequality? Did you mean "is implied by both"? Also, would it be possible to take $M(\varepsilon)=1$? $\endgroup$ – Augustin Mar 26 '18 at 13:48
  • $\begingroup$ @Augustin: that answer was wrong (the triangle inequality didn't work, I'd made a mistake when checking). The true answer seems to be that there is no such metric. $\endgroup$ – Dap Mar 26 '18 at 18:51
  • $\begingroup$ That seems correct to me. It means I spent a significant amount of time looking for something that doesn't exist... Anyway, thanks! $\endgroup$ – Augustin Mar 27 '18 at 9:59

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