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a) Prove that every Lipshitz function is uniformly continuous.

b)Let g : [1,∞) −→ R s.t g(x) = √x. Prove that g is uniformly continuous.
What about if we take g : (0,∞) −→ R?

*I did part a) which is direct using the definition of uniformly continuous function.

*Concerning part b):

From the definition if we choose δ=ϵ^2 |√x−√y|^2 ≤ |√x−√y||√x+√y|=|x−y|<ϵ^2⟹|√x−√y|<ϵ.

And from Lipschitz |f(x)−f(y)|=|x−y|/|√x+√x| if we choose L=1/|√x+√y|

then |f(x)−f(y)|≤L|x−y|.

I found similar question already posted but not on these intervals:[1,∞) &(0,∞).

How can I continue the proof on both intervals? Should I proof that L exists on [1,∞) but doesn't on (0,∞)?

Elio
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  • Yes, on $[1,\infty)$ you have to see if the $L$ that you have shown, can be bounded above (your hint is that $\frac{1}{\sqrt x}$ is smaller than $1$ whenever $x>1$). In the case of $(0,\infty)$, you have to see that $L$ cannot be bounded above. – Sarvesh Ravichandran Iyer Nov 16 '16 at 11:32
  • Essentially duplicate: http://math.stackexchange.com/questions/569928/sqrt-x-is-uniformly-continuous. – Martín-Blas Pérez Pinilla Nov 16 '16 at 11:43

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