Find all $(a,b)$ such that $x^a\sin (x^b)$ uniformly continuous on $(0,1]$.

Question

Let $a,b\in\Bbb R$. There's a question that ask the reader to find all $(a,b)$ such that $x\mapsto x^a\sin (x^b)$ is uniformly continuous on $(0,1]$. I can occasionally find some possible pairs of $a,b$, but is there a systematic way or theorem to discuss such problem?

PS: The sequel problem of that is to find $(a,b)$ such that $x\mapsto e^{ax}\sin(e^{bx})$ is uniformly continuous on $[0,\infty).$

Answer 1

Consider the function $\begin{cases} 0, x=0 \\ x^asin(x^b), x\in (0,1] \end{cases}$

Note $\forall x\in (0,1]$, we have a product and composition of continuous functions so the function is continuous. We now just need to show we can extend continuity to $x=0$. This shows uniform continuity because a continuous function on a compact subset of $\mathbb{R}^n$ is uniformly continuous. I will provide some, but not all details.

for $a\geq 1, b\geq 1$, this is clear. We have several interesting cases. For $a=0$, $b\leq-1$ we cannot extend this continuously to $0$. Note $sin(x^-1)=sin(1/x)$ is not continuous at $x=0$, but $xsin(1/x)$ is continuous (use squeeze theorem here). Use this is a guide.

The remaining cases of interest are $0<a<1$, $-1<b<1$, and $a<0$. Try these out, as the strategy is the same. The goal is to show that the values of $a,b$ determine whether or not the function can be continuously extended to $0$. If it can, then we have uniform continuity, otherwise we don't.

The sequel problem $f(x)=e^{ax}sin(e^{bx})$ can be handled similarly, but it's easier than the above problem. Consider $a,b>0$ and then $a,b<0$. I think this will be sufficient, if not, consider $0<a<1$ and $0<b<1$.

Answer 2

As indicated in the comment by Rigel, we need that the limit exists when $x\to0$ (and the function could be extended to a continuous and hence uniformly continuous function on $[0,1]$, also pointed out in the answer by Kernel_Dirichlet).

I will consider different cases, than in the answer already posted.

Case when $b=0$. Then our function is $x^a\sin (1)$, the limit exists if and only if $a\ge0$.

Case when $b>0$. Represent our function as $x^a\sin (x^b)=\frac {x^{a+b}\sin (x^b)}{x^b}$ and use that $\frac {\sin (x^b)}{x^b}\to 1$, so in this case the limit of $x^a\sin (x^b)$ exists if and only if the limit of $x^{a+b}$ does, which happens exactly when $a+b\ge0$, i.e. $a\ge-b$.

The remaining case is $b<0$, then $\sin (x^b)=\sin (\frac1{x^{-b}})$ oscillates between $-1$ and $1$ as $x\to0$ (and remains bounded). The limit of $x^a\sin (x^b)$ would exist only when $x^a\to0$, i.e. when $a>0$.

A graph would be helpful to visualize. Here is a graph (using Wolfram Alpha) of $g(a,b)= 0.001^a\sin(0.001^b)$, that is just taking a small $x=0.001$ in $x^a\sin (x^b)$, and considering $a$ and $b$ to be independent variables, parameters, in the plane.

Here is also the contour plot, more or less clearly showing the region, which is the right open half-plane $a>0$ union with the closed half-plane above the anti-diagonal $a\ge-b$. The first contour plot goes with $x=0.001$ and the second with $x=0.0000001$.

Answer 3

First notice that $\sin x^b$ is bounded, therefore, when $a>0$ the function $x^a \sin x^b$ is uniformly continuous.

For $a=0$, $\sin x^b$ cannot have a singularity at the origin, so $b\geq0$.

When $a<0$ we need $b>0$. In the regime $b>0$ by the Taylor expansion we have $\sin x^b\sim x^b$ near the origin, therefore $x^a \sin x^b\sim x^{a+b}$ near $0$. Consequently, we need $a+b\geq 0~ \&~ a<0~$

To sum up the admissible set is: $$ \{(a,b)\in \mathbb{R}^2|a> 0\} \cup \{(a,b)\in \mathbb{R}^2|a=0,b\geq 0\}\cup \{(a,b)\in \mathbb{R}^2|a+b\geq 0, a<0\} $$ Or more succintly, $$ \{(a,b)\in \mathbb{R}^2|a> 0\} \cup \{(a,b)\in \mathbb{R}^2|a+b\geq 0, a\leq 0\} $$