If $f,g$ are uniform continuous with $f$ bounded, then $fg$ is uniform continuous. I heard that with two uniform continuous functions $f,g$ where one of them is bounded, then their product $fg$ is uniform continuous. But I can't prove it why. Can you give me some advice or hints to solve this?
 A: It is not true. You can fairly easily prove that the product of two bounded uniformly continuous functions is uniformly continuous, but only one being bounded doesn't suffice.
Consider for a counterexample the bounded uniformly continuous function $\sin \colon \mathbb{R}\to \mathbb{R}$. See whether you can find a uniformly continuous $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)\sin x$ is not uniformly continuous.
A: Let $f$ and $g$ uniform continuous. That is, any fixed $\epsilon>0$ then any $a$ varying common domain of $f$ and $g$ we have that there is a $\delta>0$ (that depends only on $\epsilon>0 $ ) such that $0 <| x - a | <\delta$ implies 
$
|f(x)-f(a)|<\epsilon \quad\mbox{ and } \quad |g(x)-g(a)|<\epsilon.
$
To prove that $fg$ is uniformly continuous we must show that any fixed $\epsilon>0$ then any $a$ varying we have that there is a $\delta>0$ (that depends only on $\epsilon>0 $ ) such that $0 <| x - a | <\delta$ implies $ | f(x)g(x)-f(a)g(a)|<\epsilon$.
Use the equality
\begin{align}
f(x)g(x)-f(a)g(a) = &\Big([ f(x)-f(a)]+f(a)  \Big)\cdot \Big([g(x)-g(a)]+g(a)\Big)-f(a)g(a)
\\
=
&
\Big(f(x)-f(a) \Big)\Big( g(x)-g(a)\Big)
\\
+
&f(a)\Big( g(x)-g(a)\Big)
\\
+
&g(a)\Big( f(x)-f(a)\Big)
\end{align}
to conclude
\begin{align}
|f(x)g(x) - f(a)g(a)| = & |f(a)-f(x)|\cdot |g(a)-g(x)|\\ +&|g(a)|\cdot|f(a)-f(x)|\\+&|f(a)|\cdot |g(a)-g(x)|
\\ <&\epsilon^2 +|f(a)|\cdot \epsilon + |g(a)|\cdot\epsilon
\end{align}
and to prove that $f\cdot g$ is uniformly continuous ( if $f$ and $g$ are limited).
Suppose say that $g$ is not limited. Now, to see this situation in general without necessarily using a counter example we can use the reverse triangular inequality 
$
|a+b+c| \geq  |a|-|b|-|c|
$
to obtain the inequality
\begin{align}
|f(x)g(x)-f(a)g(a)| \geq &
\Big|g(a)\Big|\Big| f(x)-f(a)\Big|
\\
-
&\Big|f(a)\Big|\Big| g(x)-g(a)\Big|
\\
-
& \Big|f(x)-f(a) \Big|\Big| g(x)-g(a)\Big|.
\end{align}
Now what can you say about $g(x)f(x)$?
