# Examples of functions characterized by sequence

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.

• 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. – Yanko Dec 7 '18 at 14:45
• 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? – Paul Sinclair Dec 8 '18 at 0:47
• @ 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 – Motaka Dec 8 '18 at 10:16
• @Paul Sinclair: I'm so sorry I made an error in the definition of $\chi$.. It's correct now – Motaka Dec 10 '18 at 9:13
• 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 $1$. – Paul Sinclair Dec 11 '18 at 1:20

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 $$1-m_\epsilon$$.