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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 $\mathbb R$ .

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What did you try? Where is this failing? –  Did Oct 27 '12 at 12:44

1 Answer 1

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 $[-2,2]$.

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could you please tell me how to prove the first statement? –  Une Femme Douce 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

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