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Let $f$ be the real valued function 0n $[0,\infty]$ defined by $f(x)=x^{\frac{2}{3}}\ln x$ for $x>0$ and $0$ for $x=0$. Is it uniform continuous on $[0,\infty)$?

I think that if we break $[0,\infty)$ in $[0,1]\cup(1,\infty)$ then being continuous on $[0,1]$ it will be uniform there and by showing that its derivative is bounded on $[1,\infty)$ we can collectively say that it is uniform on $[0,\infty)$ , Is it correct?

Mhenni Benghorbal
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Mathronaut
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    That's a good point. Show uniform continuity for the intervals $[0,2]$ and $[1,\infty)$. In this case, when $x,y$ are close, they are either in the first interval or in the second. If you splitted into $[0,1]$ and $[1,\infty )$ then you can't decide whether both $x$ and $y$ belong in the same interval. – Dimitris Dec 12 '13 at 03:39
  • Yes, you can bound the derivative. I was considering the wrong function. – Mhenni Benghorbal Dec 12 '13 at 04:02
  • By graphing it I don't think its uniformly continuous on $[0,\infty)$ especially if you see $lim_{x\to 0^+} x^{\frac{2}{3}}ln x$. – user60887 Dec 12 '13 at 06:01
  • that limit would come to be $0$ – Mathronaut Dec 12 '13 at 14:25

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