Uniform continuity of $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0.$ For the $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0,$ my text book asks the following questions.
(b) Why is $f$ uniformly continuous on any bounded subset of $\mathbb{R}$? 
(c) Is $f$ uniformly continuous on $\mathbb{R}$??
The graph for the function is this.

For the question (b), if I take subset between $[0.2,0.6]$ or the subset where the slope is steep, I don't think the function is uniformly continuous because I think for a given $\epsilon>0$, there is no unique $\delta >0$ for the bounded subset. Therefore, it also cannot be uniformly continuous on $\mathbb{R}.$ However, the questions sounds like the function is uniformly continuous and the book says that it is uniformly continuous. The answer on the book says something but I need more explanation. Thanks. 
 A: This function $f(x)=x\sin (1/x)$ is even so I show uniformness on $[0,∞)$ spilit this as $[0,1]$ and $[1,∞]$ now note that this is a continuous on $[0,1]$ so uniform there. And $f'(x)=\sin(1/x)-\cos (1/x)/x$ and limit $x$ tends to $∞$, $f'(x)$ becomes $0$ so $f'(x)$ is bounded for $x\geq 1$ hence $f$ is uniform on $[1,∞)$ & hence the result.
A: Would anyone be so kind as to critique the following direct demonstration that $\displaystyle x \sin \frac{1}{x}$ is uniformly continuous on $(0,1)$ ?
Let $\epsilon > 0$ and let $x, y \in (0,1)$.
Then
\begin{align*}
x\sin\frac{1}{x} - y \sin\frac{1}{y}
&= x\sin\frac{1}{x} - y\sin\frac{1}{x} + y\sin\frac{1}{x} - y \sin\frac{1}{y} \\
&= (x-y)\sin\frac{1}{x} + y \left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right ),
\end{align*}
so
\begin{equation}
\label{eq: star}
\left | x\sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x - y| + y \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |.
\end{equation}
But
\begin{align}
\label{eq: starstar}
\left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |
&= \left | 2 \cos \left ( \frac{1}{2} \left ( \frac{1}{x} + \frac{1}{y} \right ) \right ) \sin \left ( \frac{1}{2} \left ( \frac{1}{x} - \frac{1}{y} \right ) \right ) \right | \notag \\
&\leq 2 \left | \sin \frac{y-x}{2xy} \right | \notag \\
&\leq \frac{|y - x|}{|xy|} \notag \\
&= \frac{|x - y|}{xy}.
\end{align}
By \eqref{eq: star} and \eqref{eq: starstar} we have
\begin{equation}
\label{eq: potato}
\left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x-y| + \frac{|x-y|}{x}
=|x-y| \left ( 1 + \frac{1}{x} \right ).
\end{equation}
By symmetry, we also have
\begin{equation}
\label{eq: banana}
\left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x-y| \left ( 1 + \frac{1}{y} \right ).
\end{equation}
Now let $\gamma > 0$. \
\noindent
\textbf{Case I:}
Suppose $x \geq \gamma$ or $y \geq \gamma$.
Then \eqref{eq: potato} and \eqref{eq: banana} imply that
\begin{equation}
\left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x-y| \left (1 + \frac{1}{\gamma} \right ),
\end{equation}
so we should let $\displaystyle \delta = \frac{\epsilon}{1 + \frac{1}{\gamma}}$. \
\noindent
\textbf{Case II:}
Suppose instead that $x < \gamma$ and $y < \gamma$.
Then $\displaystyle \left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |$ equals either $\displaystyle x \sin\frac{1}{x} - y \sin\frac{1}{y}$ or $\displaystyle y \sin\frac{1}{y} - x \sin\frac{1}{x}$.
But
\begin{equation}
x \sin\frac{1}{x} - y \sin\frac{1}{y}
\leq x - (-y)
= x + y
< 2 \gamma,
\end{equation}
and similarly,
\begin{equation}
y \sin\frac{1}{y} - x \sin\frac{1}{x} < 2 \gamma,
\end{equation}
hence
\begin{equation}
 \left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right | < 2 \gamma,
\end{equation}
so we should set $\displaystyle \gamma = \frac{\epsilon}{2}$.
\noindent
To summarise, let
\begin{equation}
\delta = \frac{\epsilon}{1 + \frac{1}{\epsilon/2}}
= \frac{\epsilon}{1 + \frac{2}{\epsilon}}
= \frac{\epsilon^2}{\epsilon + 2}.
\end{equation}
Then
\begin{equation}
|x-y| < \delta \Rightarrow \left | x \sin \frac{1}{x} - y \sin \frac{1}{y} \right | < \epsilon.
\end{equation}
\noindent
Therefore $\displaystyle h(x) = x \sin \frac{1}{x}$ is uniformly continuous on $(0,1)$.
A: I think it is also uniformly continuous on $(0,1)$. If $(x_{n})$ and $(y_{n})$ are sequences that converge to 0, then since $\sin$ is bounded, $x_{n}\sin(1/x_{n})$ 
and $y_{n}\sin(1/y_{n})$ also converge to 0. Then the absolute value of their 
difference converge to 0 and this is sufficient for uniform continuity. 
Now, if $x_{n}\longrightarrow x$ and $y_{n}\longrightarrow y\Rightarrow |x_{n}-y_{n}|\longrightarrow |x-y|\underset{\text{if}}{=}0\Rightarrow x=y$. If those sequences do not converge to 0, then $x>0\Rightarrow [x-1/m,1-1/m]$ for an appropriate $m$. Since $x\sin(1/x)$ is continuous on that interval, then it's uniformly continuous there.  
A: For b:
1) Show that $f$ is continuous at $0$. To do so, notice that $|f(x) - f(0)| = |x \sin(1/x)| \leq |x|$, so $\lim_{x \to 0} |f(x)-f(0)| \leq \lim_{x \to 0}|x| = 0$. Therefore $\lim_{x \to 0} f(x) = f(0)$ and this tells us that $f$ is continuous at $0$. 
2) Now argue that $f$ is continuous on all $\mathbb{R}$ since it is continuous at $0$ (from 1) and on $\mathbb{R} \backslash\{0\}$ (as the product and composition of continuous functions there). 
3) Since $f : \mathbb{R} \to \mathbb{R}$ is continuous, then on any bounded subset it is uniformly continuous. Why? Let $U \subset \mathbb{R}$ be continuous. Then for some $R > 0$, $U \subset [-R, R]$. Since $f$ restricted to $[-R,R]$ is uniformly continuous (a continuous function restricted to a compact set is uniformly continuous), then $f$ is uniformly continuous for any subset of $[-R,R]$ (in particular, $U$). 
For c: 
$f$ is uniformly continuous on $\mathbb{R}$. Why? You know that $f$ is uniformly continuous on $[-1, 1]$, say. Outside of $[-1,1]$, notice that the derivative of $f$ is $f'(x) = \sin(1/x) - \cos(1/x)/x$ and (since we're restricted away from the orgin), this means that $f'(x)$ is bounded. In particular $|f'(x)| \leq 2$ for every $|x| \geq 1$. This means that $f$ is Lipschitz continuous with Lipschitz constant at most $2$ on the compliment of $[-1,1]$. That means that if $x, y\in [-1,1]^c$ then $|f(x)-f(y)|\leq 2 |x-y|$. To put these pieces together, you can say the following:
Let $\epsilon > 0$. Find $\delta_1 > 0$ such that if  $x,y \in [-2, 2]$ then $|f(x)-f(y)|<\epsilon$ (which you can do by part b). Let $\delta = \min(\delta_1, \epsilon/2, 1)$. Now, if $x, y \in \mathbb{R}$ such that $|x-y| < \delta$, then either $x,y \in [-2,2]$ or $x,y \in [-1,1]^c$ (since we chose $\delta \leq 1$). If $x,y \in [-2,2]$ then $|x-y|<\delta \leq \delta_1$, so $|f(x)-f(y)|<\epsilon$. Otherwise, if $x,y \in [-1,1]^c$ then $$|f(x) - f(y)|\leq 2 |x-y| < 2 \delta \leq 2 (\epsilon/2) = \epsilon$$
In either case, $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$ showing that $f$ is uniformly continuous.
