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Let $ \Psi$ be a set of functions $\chi :\mathbb{R}^+\rightarrow [0,1)$ satisfying $$\chi(t_n)\rightarrow 1\Rightarrow t_n\rightarrow 0$$

I want to find some functions that belongs to $ \Psi$, other than

$\chi_1 (t)=\left\{\begin{matrix} e^{-2t} & \text{if } t>0\\ 1 & \text{if } t=0 \end{matrix}\right.$

$\chi_2 (t)=\left\{\begin{matrix} \frac{1}{1+t} & \text{if } t>0\\ 1 & \text{if } t=0 \end{matrix}\right.$


I'm working on this kind of functions, and i want to know more about it.

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    $\begingroup$ Well you can take a function of this form $f(t) = 0$ if $t=1 $ and $f(t)=a$ if $t\not =1$ for any $a\in(0,1)$. In this case $f(t_n)\rightarrow 0$ if and only if $t_n=0$ for all $n>N$. More generally pick any function $g(t)$ which is bounded below by some $c>0$ and change it's value in $t=1$ to be zero. $\endgroup$ – Yanko Dec 7 '18 at 14:45
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    $\begingroup$ Neither of your sample functions has the property you describe. Both satisfy $\chi(t_n) \to 0 \implies t_n \to \infty$ instead of $1$. And neither of them require a piece-wise definition, since $$e^{-2\cdot 0} = \frac 1{1 + 0} = 1$$Did you mean something else? $\endgroup$ – Paul Sinclair Dec 8 '18 at 0:47
  • $\begingroup$ @ Paul Sinclair To be honest I find it like that in a paper, and I didn't understand why the authors use piece-wise definition $\endgroup$ – Motaka Dec 8 '18 at 10:16
  • $\begingroup$ @Paul Sinclair: I'm so sorry I made an error in the definition of $\chi$.. It's correct now $\endgroup$ – Motaka Dec 10 '18 at 9:13
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    $\begingroup$ It's alright. I've modified my answer to address the corrected condition. Now I understand the piecewise definition (though it still seems a poor way to state it): $0$ is not actually in the domain of $\chi$, instead they are using this to indicate that the limit of the function at $0$ is in fact $1$. $\endgroup$ – Paul Sinclair Dec 11 '18 at 1:20
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Edit: I've updated this post to match the corrected condition. As I had predicted, the updated version is very similiar to the original.

The condition $\chi(t_n) \to 1 \implies t_n \to 0$ is equivalent to (for functions $[0,\infty) \to [0,1)$):

For every $\epsilon > 0$ there is some $m_\epsilon > 0$ such that $\chi(t) \le 1- m_\epsilon$ for all $t \in [\epsilon, \infty)$.

I.e., the function is bounded away from $1$ everywhere except possibly at $t = 0$.

Your condition does not require any behavior for $\chi$ near $0$. It could be that $\lim_{t\to 0} \chi(t) = 1$, but it could also be true that $\chi$ is bounded away from $1$ everywhere. In that case, there are no sequences $\{t_n\}$ such that $\chi(t_n) \to 1$, so your condition is vacuously true.

Any function bounded away from $1$ is an example:

  • $\chi(t) = c$ for any constant $c < 1$.
  • $\chi(t) = a + b\sin t$, with $1 > a + b$ and $ a \ge b$.
  • $\chi(t) = \begin{cases} \frac{\arctan t}\pi&t \text{ rational}\\ 0.5& t\text{ irrational}\end{cases}$

Functions that satisfy the condition, but are not bounded away from $1$ near $0$ include the two you've already given, and

  • $\chi(t) = \begin{cases} 1-t & t < 1\\0& t > 1\end{cases}$
  • $\chi(t) = e^{-(t-1)^2}$
  • $\chi(t) = \frac{2\arctan (1/t)}\pi$

The boundedness condition is clearly sufficient to imply yours. To show that it is also necessary, first note that if $\chi$ is not bounded away from $1$ at some $p > 0$, then one can always find a $t_n \in (p - 1/n, p + 1/n)$ with $\chi(t_n) > 1 - 1/n$, and therefore $\chi(t_n) \to 1$ while $t_n \to p$. A similar argument holds for $p = \infty$. Thus every point in $[\epsilon, \infty]$ has a neighborhood bounded away from $1$. By compactness, a finite number of those neighborhoods covers it. The highest bound among the neighborhoods in the finite cover serves as $1-m_\epsilon$.

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