When is $x^{\alpha}\sin(x^{\beta})$ uniformly continuous for $\alpha, \beta > 0$? Consider a function $f_{\alpha, \beta}\colon (0, \infty) \longrightarrow \mathbb{R}$ defined in the following way:
$$f_{\alpha, \beta} = x^{\alpha}\sin(x^{\beta}) \quad \alpha, \beta > 0$$
Then we can pose the questions: 


*

*For which pairs $\alpha, \beta$ is this function uniformly continuous?

*For which sets $(\alpha, \beta)$ in $(0, \infty)^2$ is the family equicontinuous?


I am baffled as to how to go about answering these questions in a clear and concise way. I think that it is possible to produce an answer by considering many cases and lots of tedious estimates. Is there a better way to approach the problem?
Any help will be appreciated. 
 A: Fact 1. A uniformly continuous function on $(0,\infty)$ is Lipschitz for large distances, which can be expressed concisely by the inequality 
$$|f(x)-f(y)|\le L|x-y|+M,\quad x,y>0 \tag{1}$$
where $L$ and $M$ are constants. This is  a general and useful fact, so it's worth recording regardless of this problem. Proof: there is $\delta>0$ such that $|f(x)-f(y)| \le 1$ whenever $|x-y|\le \delta$. Divide $[x,y]$ into intervals of size at most  $\delta$; we need no more than  $1+|y-x|/\delta$ of these. Then $|f(x)-f(y)|\le 1+|y-x|/\delta$, which gives (1). $\quad \Box$
In particular, (1) implies that $(f(x)-f(1))/(x-1)$ remains bounded as $x\to \infty$. This excludes $\alpha>1$ from consideration. 
Furthermore, consider the sequence $x_n=(\pi/2+\pi n)^{1/\beta}$. Observe that 
$|x_{n+1}-x_n|  \le C n^{\frac{1}{\beta}-1}$ and $|f(x_n)-f(x_{n+1})|\ge c n^{\alpha}$ where $C$ and $c$ are positive constants independent of $n$. By (1), $\alpha\le \frac{1}{\beta}-1$.  
Summarize the necessary conditions obtained so far: 
$$0<\alpha\le 1, \qquad 0<\beta \le \frac{1}{\alpha+1}\tag{2}$$
Here's an interesting fact: under conditions (2), the second derivative of $f$ is bounded at infinity. Indeed, the largest term in $f''$ has $x^{\alpha+2\beta-2}$, and 
$$\alpha+2\beta-2 \le \frac{\alpha^2-\alpha}{\alpha+1}\le 0$$
Fact 2. If $f$ is  uniformly continuous  on $(0,\infty)$ and $f''$ is bounded at infinity, then $f'$ is also bounded at infinity. 
Proof:  there is $\delta>0$ such that $|f(x)-f(y)| \le 1$ whenever $|x-y|\le \delta$. By the mean value theorem, this implies $|f'|\le 1/\delta$ at some point within the interval $[x,x+\delta]$. But then $|f'(x)| \le 1/\delta+ \delta  \sup_{[x,x+\delta]} |f''|$, which is a uniform bound on $f'$. $\quad \Box$  
It is not hard to see that   $f'$ is bounded at infinity if and only if 
$$ \alpha +\beta \le 1  \tag{3}  $$
So, (3) is necessary for uniform continuity. It is also sufficient, since bounded derivative implies Lipschitz implies uniformly continuous on $[1,\infty)$; the interval $[0,1]$ is compact, so it's not a problem. 

Concerning equicontinuity: on $[1,\infty)$, the family described by (3) is equicontinuous because the derivative $f'$ is uniformly bounded. There is an issue at $0$, where $f$ behaves like $x^{\alpha+\beta}$. You don't want to allow this exponent to be arbitrarily small. So, 
$$  \epsilon \le \alpha +\beta \le 1  \tag{4}  $$
gives an equicontinuous family, for every $\epsilon>0$. 
