Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb R$ .

share|cite|improve this question
What did you try? Where is this failing? – Did Oct 27 '12 at 12:44
up vote 3 down vote accepted

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 $[-2,2]$.

share|cite|improve this answer
could you please tell me how to prove the first statement? – Un Chien Andalou Feb 17 '13 at 14:30
@CityOfGod I'm pretty sure it has been asked recently in the site. But I don't find the link. – Davide Giraudo Feb 17 '13 at 16:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.