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Show $f(x):=\sqrt{x}$ is uniformly continuous on $[0,1]$.

What I did: Need to show that $\forall \varepsilon>0: \exists\delta>0$ such that
$\forall x,y\in(0,1): |x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon$

Choose $\delta=\varepsilon^2$.

then for $x,y\in(0,1)$ with $|x - y| < \delta$,

$|f(x)-f(y)|=|\sqrt{x}-\sqrt{y}|$

$|\sqrt{x}-\sqrt{y}|\le|\sqrt{x}+\sqrt{y}|$

$|\sqrt{x}-\sqrt{y}|^2\le|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}| =|x-y|< \varepsilon^2$

$|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}<\sqrt{\varepsilon^2}=\varepsilon$

Martin Sleziak
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Lost
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  • What is your question? – Vladimir Vargas Nov 11 '14 at 01:36
  • Surely Lost just wants to check the proof is correct. – Suzu Hirose Nov 11 '14 at 01:40
  • I thought it was correct, but received no credit for it. And I'm having trouble typing it in correctly in here. – Lost Nov 11 '14 at 01:44
  • I'll fix the typesetting. Hang on a sec – Ben Grossmann Nov 11 '14 at 01:45
  • Interestingly, thanks to this I found an error in my own solution to the problem. – Suzu Hirose Nov 11 '14 at 01:49
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    $\sqrt{x}-\sqrt{y} = \frac{x-y}{\sqrt{x}+\sqrt{y}}$. Fix $\varepsilon > 0$. Assume without loss of generality that $1 \geq x \geq y \geq 0$. If $y>0$ then you are done, because you have the Lipschitz estimate $|\sqrt{x}-\sqrt{y}| \leq \frac{|x-y|}{\sqrt{\delta}}$ whenever $y \in [\delta,1]$ and $\delta > 0$. If $y=0$ then you can simply check that you need $\delta < \varepsilon^2$. Blindly plugging that into the Lipschitz estimate, you get $|\sqrt{x}-\sqrt{y}| \leq \frac{\varepsilon^2}{\varepsilon} = \varepsilon$ whenever $|x-y|<\varepsilon^2$. – Ian Nov 11 '14 at 01:49
  • For future reference, try to keep the entire mathematical expression between the $'s. – Ben Grossmann Nov 11 '14 at 01:50
  • I think your solution is fine. In general, you can also invoke the Heine-Cantor Theorem. – Daniele A Nov 11 '14 at 01:52
  • (Cont.) The derivation in the OP is fine. What I wrote is essentially a way of seeing where one might come up with $\delta = \varepsilon^2$. – Ian Nov 11 '14 at 01:52

1 Answers1

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Your solution is correct. I would probably not give you full marks for it for not explaining why the inequalities you use are true.

Also, you were required to prove the uniform continuity on $[0,1]$, but only used $(0,1)$ in your argument (it still works for $x=0$ or $y=0$, but you didn't say so).

Martin Argerami
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