TITLE: A question about uniform continuity
QUESTION [5 upvotes]: Let $F$ be a continuous function on the real set $\mathbb R$ such that the function $x \mapsto xF(x)$ is uniformly continuous on  $\mathbb R$ . Prove that $F$ is also uniformly continuous on  $\mathbb R$ .

REPLY [3 votes]: Some hints:

Prove that there exists $A,B>0$ such that for all real number $x$, $|xF(x)|\leq A|x|+B$. In particular, $F$ is bounded, say by $M$.  
We write for $x\geq 0$,
$$|F(x)-F(y)|\leq \frac 1x|xF(x)-yF(y)|+\frac 1x|F(y)|\cdot |x-y|.$$
So if $|x|\geq 1$, we have 
$$|F(x)-F(y)|\leq |xF(x)-yF(y)|+M\cdot |x-y|.$$
Conclude, using uniform continuity of $F$ on $[-2,2]$.