Show that $ g(x)=(x^{2}-1)f(x)$ is uniformly continuous Let $ f: (-1,1) \rightarrow  \mathbb{R}$ be a bounded and continuous function . Prove that the function $ g(x)=(x^{2}-1)f(x) $ is uniformly continuous on $ (-1,1)$ . $$ $$ My little approach is,  Since $f$ is bounded on $(-1,1)$ , there is positive $M \in \mathbb{R}$ such that 
$$\forall x \in (-1,1)\,|f(x)| \leq M$$
Now, $ |g(x)-g(y)|=|(x^{2}-1)f(x)-(y^{2}-1)f(y)|=|x^{2}f(x)-y^{2}f(y)-(f(x)-f(y))|$
 A: Define a new function on a larger domain $G:[-1,1]\to \Bbb R$ by $$G(x):=\begin{cases}
g(x) &\text{if $x\in(-1,1)$}\\
0 &\text{if $x=1$ or $-1$}
\end{cases}.$$
You can check that $G$ is continuous (specifically, you just need to check this at $-1$ and $1$). This follows from boundedness of $f$ and you apply squeeze theorem.
Now $G$ is defined on a compact set and is continuous, and hence uniformly continuous, while $g$ is the restriction of a uniformly continuous function to a smaller domain. So $g$ is uniformly continuous.
A: Given $\epsilon>0$, there exists $a\in(-1,0)$ and $b\in(0,1)$ such that :
$$\forall x\in(-1,a],\,|g(x)|\le\epsilon$$
and similarly :
$$\forall x\in[b,1),\,|g(x)|\le\epsilon$$
(this is because $\lim_{x\to\pm 1}g(x)=0$)
By Heine theorem, $g$ is uniformly continuous on $[a,b]$ : there exists $\delta>0$ such that for every $(x,y)\in[a,b]$, if $x-y<\delta$ then $|g(x)-g(y)|\le\epsilon$.
It is now easy to show, with the triangle inequality, that $|g(x)-g(y)|\le2\epsilon$, for any $(x,y)\in(-1,1)^2$,
