Product of two uniformly continuous functions is uniformly continuous if either of them is bounded? We know that product of bounded uniformly continuous functions defined on R is uniformly continuous. But is the product of two uniformly continuous functions is uniformly continuous if either of them is bounded?
 A: No. Consider $f(x) = x \sin x$.
Since $f^\prime(x) = \sin x + x \cos x$ , wich can be of order $x$, this is not uniformly continuos.
Edit: added a detailed proof for the last claim.
To see that the last statement is true consider $\varepsilon > 0$. Assume $f$ is uniformly continuous. Then there is $\delta > 0 $ sucht that $|f(x+h)-f(x)|< \varepsilon$ for each $h$ such that $|h| < \delta$. Now, for integer $k$,  look at $x_k=2k\pi$, so $f(x_k) =0$, while, according to Taylor's theorem,
$$f(x_k+h)-f(x_k) =f^\prime(x_k)h +\frac{1}{2} f^{\prime\prime}(c_k)h^2$$
for some $c_k \in (x_k, x_k+h)$ 
Now use the formula for $f^\prime$ to see that the first term on the right hand side is equal to $\pm x_k h=\pm 2\pi k h$. without loss of generality it is positive (consider only even $k$)
Calculate $f^{\prime\prime} (x) = \cos (x) + \cos(x) - x \sin(x) =2\cos(x) -x\sin(x)$
Since $\sin$ is uniformly continuous and $\sin(x_k) =0$ we may assume that $|\sin(c_k)|\le 1/2$, say. Then the second term on the rhs is estimated by (note $\cos $ is bounded by $1$)
$$|\frac{1}{2}c_k \sin(c_k)h^2 -\cos(c_k) h^2|\le \frac{1}{4}(2\pi k +2+2h) h^2$$ So
$$|f(x_k +h) -f(x_k)|>2\pi k h - \frac{\pi k}{2}h^2 -\frac{h^2+h^3}{4}=\pi kh(2-\frac{h}{2})  -\frac{h^2 + h^3}{2}$$
Clearly the last term is smaller than the first one  on the rhs and is constant if $h$ is small and fixed. If $|h|<4$ the factor of $\pi k h$ is positive. Now fix $0 < h< \delta$ and let $k\rightarrow \infty$ to see that the lhs cannot be bounded by $\varepsilon$. 
A: Take $f(x)=\sin(\sqrt{|x|})$. It is a uniformly continuous and bounded function. Then take $g(x)=x$. It is a uniformly continuous function. Over $\mathbb{R}^+$, the derivative of:
$$ f(x)g(x) = x \sin\sqrt{x} $$
is unbounded, and it is easy to check that $f\cdot g$ is not uniformly continuous.
