Let $\{f_{a}(x)=x^{-a},\hspace{1mm} a\in\mathbb{R}^{+}\}$ for $x\in I=(1,+\infty)$. Then we have $f_{a}(x)\in (0,1)\subset[0,1]$ thus for any sequence $(x_{n})$ there exists $(x_{n_{k}})$ and $y\in[0,1]$ such that:
$$f_{a}(x_{n_{k}})\longrightarrow y$$
So that $f_{a}(x_{n})$ admits converging subsequences. We can not directly apply Ascoli Arzelà as $I$ is not compact. But on any compact $K\subset I$ this result holds and $f_{a}$ is equicontinous on $K$. From such information can we derivate the applicability of Ascoli-Arzelà on $I$ as numerable union of compact subsets?
Can we show this family be equicontinous via Ascoli Arzelà
arzela-ascolicompactnessequicontinuityreal-analysis
Best Answer
The family is not uniformly equicontinuous on $(1, +\infty)$. If it were, since each $f_a$ has a continuous extension to $[1, +\infty)$, the family of extensions would be uniformly equicontinuous on $[1, +\infty)$. (For every $\varepsilon > 0$ the same $\delta$ would work on $[1, +\infty)$ as on $(1, +\infty)$.) A fortiori we would obtain a uniformly equicontinuous family on $[1,2]$ by restriction, and the Ascoli-Arzelà theorem says every sequence in that family would have a uniformly convergent subsequence. But we have $$\lim_{n \to \infty} f_n(x) = \begin{cases} 1 &\text{if } x = 1,\\ 0 &\text{if } x > 1, \end{cases}$$ thus a sequence whose pointwise limit is discontinuous. It follows that the sequence has no uniformly convergent subsequence, hence the family is not equicontinuous on $[1,2]$. (Note: on a compact domain the concepts of uniform equicontinuity and pointwise equicontinuity coincide.)
As for pointwise equicontinuity, we can show that the family is pointwise equicontinuous on $(1, +\infty)$ using the Ascoli-Arzelà theorem.
Consider a sequence $(f_{a_n})_{n \in \mathbb{N}}$ and the sequence $(a_n)$ of indices. Either $(a_n)$ has a convergent subsequence - in which case we may assume the whole sequence converges, say to $\alpha \in [0, +\infty)$ - or we have $a_n \to +\infty$.
Consider any compact interval $[u,v] \subset (1, +\infty)$. In the former case, $(f_{a_n})$ converges to $f_{\alpha}$ uniformly on $[u,v]$, in the latter case it converges uniformly to $0$ on $[u,v]$.
Thus by the Ascoli-Arzelà theorem the family of the restrictions $\{ f_a\rvert_{[u,v]} : a \in \mathbb{R}^+\}$ is equicontinuous, and that implies the original family is equicontinuous at all $x \in (u,v)$. Since every $x \in (1, +\infty)$ lies in some $(u,v)$ with $1 < u < v < +\infty$, the family is pointwise equicontinuous on $(1, +\infty)$.