Prove that $h(x)=\|x\|\sin\left(\frac{1}{\|x\|}\right)$ is uniformly continuous on $\mathbb{R}$

Question

Let $h: \mathbb{R}^d \to \mathbb{R}$ be given by $h(x)=\|x\|\sin\left(\frac{1}{\|x\|}\right)$ for $x \in \mathbb{R}^d/\{0\} $, and $h(0)=0$. Prove that $h$ is uniformly continuous on $\mathbb{R}$.

Any idea?

Answer 1

A bit of detail.

Let $d=1$, $x \ge 0$.

$a) h(x)$ is continuos on $[0,\infty)$.

$b) \lim_{x \rightarrow \infty} h(x)=L(=1).$

$\epsilon >0$ be given.

There is a $M(\epsilon) >0$, real, s.t.

$x \ge M$ implies $|h(x)-L|<\epsilon$.

1) Let $x,y \ge M$ :

$|h(x)-h(y)| =$

$|(h(x)-L)+(L-h(y))| \le$

$|h(x)-L|+|h(y)-L| < 2\epsilon$.

2) $x,y \in [0,M]:$ $h$ is uniformly continuos on $[0,M]$.

There is a $\delta(M) >0$ s.t.

$|x-y|<\delta$ implies $|h(x)-h(y)|<\epsilon$.

3) Let $x\le M$, and $y \ge M$.

For $|y-x| < \delta:$

$|(y-M)+(M-x)|=$

$(y-M)+(M-x)<\delta$,

Since $M-x< \delta$, and $M \le y:$

$|h(x)-h(y)|=$

$ |h(x)-h(M)| +|h(M)-h(y)| <$

$ \epsilon +2\epsilon=3\epsilon$.

3) Sum it up:

Given $\epsilon$. Determine $M(\epsilon)$ for the limit.

$h(x)$ is uniformly continuos on $[0,M]$. This determines $\delta (M)$.

$ \epsilon \rightarrow M(\epsilon) \rightarrow \delta (M(\epsilon))$

Finally :

For $x,y \in [0,\infty)$ there is a $\delta(\epsilon)$ s.t.

$|x-y| <\delta$ implies $|h(x)-h(y)|<3\epsilon$.

Answer 2

Hint: The function is continuous and it has finite limit at $0$. It also has a finite limit as $\|x\| \to \infty$. This is enough for uniform continuity.