if $\displaystyle f$ is unif. cont. on $\displaystyle (0,\infty)$ and $\displaystyle g$ is positive and bounded on $\displaystyle (0,\infty)$, then $\displaystyle f \cdot g$ is unif. cont. on $\displaystyle (0,\infty)$
$\displaystyle \left( {\exists B > 0} \right)\left( {\forall x} \right)\left[ {\left| {g(x)} \right| \leqslant B} \right]$. $\displaystyle g$ is bounded.
If $\displaystyle \varepsilon > 0$ then use $\displaystyle \frac{\varepsilon}{B+1}$ in the uniformity of $\displaystyle f$.
I understand that M>0. $\displaystyle |g(x)| \le M, \forall x$ since g is bounded. We know f is unif. cont. so for any $\displaystyle \epsilon > 0$ there exists delta > 0 such that $\displaystyle |f(x)-f(a)| < \frac{\epsilon}{M+1}, \forall x,a \in A$ with $\displaystyle |x-a| < \delta$. Here is where I think I go wrong and do not understand. Do I simplify $\displaystyle |fg(x) - fg(a)|$ to be $\displaystyle \le |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)|$?
g not being uniformly continuous is really throwing me off. I believe I need to use something from g being continuous?