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As part of a question of several parts I have proved the following:

1) that every Lipshitz function is uniformly continuous.

2) that every function that is differentiable and whose derivative is bounded, is Lipshitz.

3)Considering the function $g: [1, \infty) ~~; ~~g(x) = \sqrt{x} $, I have proved that it is uniformly continuous by first proving that it's differentiable and that $g'(x)$ is bounded, therefore it is Lipshitz, and therefore, it is uniformly continuous.

The last part is where i'm stuck on. if we changed g so that $g: (0, \infty)$ then is it uniformly continuous? I have showed that it is unbounded, but I don't know if that means anything.

i'd appreciate any help.

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    Are you familiar with Heine-Cantor theorem? –  Nov 17 '16 at 13:18
  • not really, sorry –  Nov 17 '16 at 13:18
  • I've looked it up, yes I know this! but (0, inf) is not compact, so it doesn't hold. am I wrong? –  Nov 17 '16 at 13:22
  • If a function is uniformly continuous on $A$ and on $B$ then it is uniformly continuous on $A \cup B$. (Just take your $\delta$ that works on $A \cup B$ to be $\min { \delta_A,\delta_B }$.) Accordingly, you can prove that $\sqrt{x}$ is uniformly continuous on $[0,1]$ using any method (Heine-Cantor would be the easiest), and then you can show it is uniformly continuous on $(1,\infty)$ by showing it is Lipschitz there. – Ian Nov 17 '16 at 13:22
  • But $[0,2]$ is compact. – Daniel Fischer Nov 17 '16 at 13:22
  • but the set is open at 0, doesn't that make a difference? –  Nov 17 '16 at 13:25
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    Nope; if a function is uniformly continuous on $A$ and $B \subseteq A$ then it is uniformly continuous on $B$. (Just reuse the same $\delta$.) – Ian Nov 17 '16 at 13:28

3 Answers3

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Any continuous function over a compact (ie. closed, bounded) interval is uniformly continuous (Heine Cantor theorem). Therefore the function is uniformly continuous over $[0,1]$. Deriving the result for $[0, +\infty[$ is then easy given what you've already proved.

Ian
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Ulysse Mizrahi
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You can also use that the function is Hölder continuous, for if $0\le x\le y$ one has that \begin{align} \sqrt{y}-\sqrt{x}&=\sqrt{x+(y-x)}-\sqrt{x} \\ &\le\sqrt{x+2\sqrt{x}\sqrt{y-x}+(y-x)} \\ &=(\sqrt{x}+\sqrt{y-x})-\sqrt{x}=\sqrt{y-x} \end{align} so that in general $$ |\sqrt{y}-\sqrt{x}|\le\sqrt{|y-x|} $$ which is Hölder with index $1/2$ and thus uniformly continuous on $[0\infty)$.

Lutz Lehmann
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On $[1,+\infty [$, using mean value theorem, $|\sqrt{x}-\sqrt y|=|x-y|$ and thus is uniform continuous. On $[0,1]$, the uniform continuity is obvious (continuous function on a compact). Now, if $x\leq 1\leq y$, you have that $$|\sqrt x-\sqrt y|\leq |\sqrt x-\sqrt 1|+|\sqrt y-\sqrt 1|,$$ and thus, the claim almost follow using previous results.

Surb
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