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Let the metric space $X=(0,\infty)$ and determine whether the following are uniformly continuous on $X$:
(1) $f(x)=\sqrt{x}$
(2) $f(x)=1/x$
(3) $f(x)=\ln(x)$
(4) $f(x)=x\ln(x)$

Since this isn't $\mathbb{R}$ I don't think I can use the usual method: showing that $|f'|$ is bounded.

Any tips on how to solve these problems?

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Are these homework problems? –  fpqc Feb 8 '12 at 9:24
    
Wikipedia says that "Every Lipschitz continuous map between two metric spaces is uniformly continuous. In particular, every function which is differentiable and has bounded derivative is uniformly continuous." –  Matt N. Feb 8 '12 at 9:29
    
@BenjaminLim Yes. Matt N. Unfortunately the lecturer specifically pointed out that we cannot use that property since we have not proven it. –  Emir Feb 8 '12 at 9:33
    
In the first three cases the functions are monotone. Therefore (if you are trying to show uniform continuity) it suffices to explicitly compute $f(x+\delta) - f(x)$ and show that it is bounded by a function $g(\delta)$ independent of $x$ with $g(\delta)\to 0$ as $\delta \to 0$. –  Willie Wong Feb 8 '12 at 9:47
    
OTOH, if you don't have bounded derivatives, do you know how to demonstrate a contradiction to uniform continuity? In particular, do you know how to explicitly construct a sequence $(x_n)$ such that the $\delta_n$ neighborhoods whose image are in $\epsilon$ neighborhoods necessarily have $\delta_n \to 0$ as $n\to\infty$? –  Willie Wong Feb 8 '12 at 9:50
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1 Answer

up vote 3 down vote accepted

You can prove it directly. For example, for (1) use that $|\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} + \sqrt{y}|$:

Then you have $|\sqrt{x} - \sqrt{y}|^2 \leq |\sqrt{x} + \sqrt{y}||\sqrt{x} - \sqrt{y}| = |x - y|$

So for $\delta := \varepsilon^2$ you get $|\sqrt{x} - \sqrt{y}| < \varepsilon$.

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