Uniform continuity proof problem Let $f$ be continuous on $\mathbb R$ and assume that there is a constant $C$ such that
$$|f(x)| \leq C$$
for all $x \in \mathbb R$.
Let $g$ be uniformly continuous on $\mathbb R$.  Show that $fg$ is uniformly continuous on $\mathbb R$.
I just need a hint on how to start this! I'm drawing a blank. 
 A: It is not true: you need more. In particular, let us understand what is missing. 
$fg$ is uniformly continuous on whole $\mathbb R$ if
$$\forall \epsilon>0 ~\text{there exists} ~\delta=\delta(\epsilon)~ \text{s.t.}~
\forall x_1,x_2\in\mathbb R : |x_1-x_2|<\delta\Rightarrow |f(x_1)g(x_1)-f(x_2)g(x_2)|<\epsilon~~(*)$$
Let us study conditions on $f$ and $g$ to ensure $(*)$. We can write
$$|f(x_1)g(x_1)-f(x_2)g(x_2)|=|f(x_1)g(x_1)+f(x_1)g(x_2) - f(x_1)g(x_2) -f(x_2)g(x_2)|\leq |f(x_1)g(x_1)-f(x_1)g(x_2)  |+
|f(x_1)g(x_2)-f(x_2)g(x_2)|=\\|f(x_1)||g(x_1)-g(x_2)|+ 
|g(x_2)||f(x_1)-f(x_2)|; $$
Now, in the first term in the r.h.s. of the last equality, i.e. $|f(x_1)||g(x_1)-g(x_2)|$ we can use that $f$ is bounded and $g$ is uniformly continuous. What about the second term, i.e.
$$|g(x_2)||f(x_1)-f(x_2)| ~~(**)$$
? Can you find additional hypothesis on $f$ and $g$ to estimate $(**)$ and arrive at the thesis, i.e. $(*)$?
A: Here is a start,
$$ |g(x)f(x)-g(y)f(y)| = |g(x)f(x)-g(y)f(x)+g(y)f(x)-g(y)f(y)|  $$
$$  \leq |g(x)f(x)-g(y)f(x)| + |g(y)f(x)-g(y)f(y)| . $$
