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What would be a example of a function that is continuous, but not uniformly continous? will f(x)=1/x be a example? Give domain from (0,2), why? Strictly, in definitions

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3 Answers

up vote 7 down vote accepted

Clearly $\,\displaystyle{\frac{1}{x}}\,$ is continuous in $\,(0,2)\,$ as it is the quotient of two polynomials and the denominator doesn't vanish there.

Now, if the function was uniformly continuous there then

$$\forall\,\epsilon>0\,\,\exists\,\delta>0\,\,s.t.\,\,|x-y|<\delta\Longrightarrow \left|\frac{1}{x}-\frac{1}{y}\right|<\epsilon$$

But taking $\,\epsilon=1\,$ , then for any $\,\delta>0\,$ we take

$$x:=\min(\delta,1)\,\,,\,y=\frac{x}{2}\Longrightarrow |x-y|=\frac{x}{2}<\delta, \,\text{but nevertheless}$$

$$\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{1}{x}-\frac{2}{x}\right|=\left|\frac{1}{x}\right|\geq 1=\epsilon$$

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+1 for this crystal way. –  B. S. Jan 30 '13 at 12:17
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$f(x)=\frac{1}{x}$ is an example, since its derivative is unbounded on $(0,2)$. This easily follows from definition.

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So what? $\sqrt{x}$ defined on $(0,2)$ has unbounded derivative, yet is uniformly continuous. –  Chris Eagle Jan 30 '13 at 10:52
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Here is a very simple graphical example, where the domain of the function is the interval of the $x$ axis between zero and the last doted line, minus the white points.

enter image description here

The function fails to be uniformly continuous at the two first white points. It is however continuous.

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Its domain is not $(0,2)$, though. –  Henning Makholm Dec 15 '13 at 3:07
    
Right, this only answers the first part of the question. –  Martin Van der Linden Dec 15 '13 at 9:25
    
I could be wrong, but I'm pretty sure this function isn't continuous at the white dot points because it doesn't satisfy Weierstrass for all $\epsilon$. –  René G Jan 5 at 16:29
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