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Prove that $f(x) = \frac {1}{2x-1}$ is not uniformly continuous on $[0,\frac {1}{2})$

So I understand the process I am just having sue trouble picking two values that are valid that make the proof work.

So I am working with the definition for not uniformly continuous as:

$$|x_1 - x_2|\lt \delta \space and \space |f(x_1) - f(x_2)| \ge \epsilon^*$$

Given $\epsilon^* \gt 0$ show that for any positive number $\delta$ there are numbers $x_1 and \space x_2$ in $[0,\frac {1}{2})$ the above property is satisfied.

So I have to pick my $\epsilon^* , x_1, and \space x_2$

So considering my function I chose $x_1=\frac {1}{2} - \delta$ and then I chose $x_2=\frac{1}{2} - \frac{\delta}{2}$ but because of my restrictions on $\delta$ it is possible that $x_1$and $x_2$ could both be $\frac{1}{2}$ which is not in the domain.

So I need help picking two numbers that will satisfy all the different properties and then I should not have a problem picking my epsilon from there.

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    $\begingroup$ The definition of uniform continuity of a function on domain $D$ is that $\forall\epsilon\exists\delta$ such that for all $x_1,x_2$ in $D$, $|x_1-x_2|<\delta\to|f(x_1)-f(x_2)|<\epsilon$. So there is no need to worry about values going out of range. $\endgroup$ Oct 21 '15 at 19:41
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    $\begingroup$ This is good to know: math.stackexchange.com/questions/1441637/… $\endgroup$ Oct 21 '15 at 19:46
  • $\begingroup$ but the $x_1$ and $x_2$ i chose are not $in D$ and I cannot think of any other values that make the function work that satisfy the conditions $\endgroup$
    – B ry
    Oct 21 '15 at 19:47
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Suppose the function is uniformly continuous. So for $\epsilon=1$, there exists some $\delta$ such that if $|x_2-x_1|<\delta$, then $$ \left|\frac1{2x_1-1}-\frac1{2x_2-1}\right|<1,\qquad(\star) $$

Suppose now that $\delta\ge1/2$. But letting $x_1 = 0$ and $x_2=1/2-1/100$ contradicts $(\star)$.

On the other hand if $\delta<1/2$, letting $x_1=1/2-\delta$ and $x_2=1/2-\delta/2$ contradicts $(\star)$. So the function is not uniformly continuous.

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For $n \ge 1$ integer: $$\left\vert f(\frac{1}{2}-\frac{1}{4n})-f(\frac{1}{2}-\frac{1}{2n})\right\vert=n$$ Hence if you pickup $\epsilon = 1$, you won't be able to find $\delta > 0$ such that for all $\vert x_1-x_2 \vert < \delta$ you have $\vert f(x_1) - f(x_2) \vert < 1$.

Why? Because for $n > \frac{1}{2\delta}$, you have $\left\vert (\frac{1}{2}-\frac{1}{4n})-(\frac{1}{2}-\frac{1}{2n})\right\vert=\frac{1}{2n} < \delta$ but $\left\vert f(\frac{1}{2}-\frac{1}{4n})-f(\frac{1}{2}-\frac{1}{2n})\right\vert=n \ge 1$

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Sometimes when working on a problem it might not be best practise to just work with the definition itself to prove that some property does not hold. A more general approach can be easier than the gritty details of a particular example (not that this one was troublesome).

Prove that a uniformly continuous function on a bounded interval is necessarily bounded. (Not difficult). Then simply announce that this example and any similar example cannot be uniformly continuous because it is transparently unbounded. [It is no harder to prove this general statement than to handle this particular problem since, as the hints show, you are merely exploiting what happens near the right-hand endpoint where the given function is unbounded.]

If the next problem is to show that the function $f(x)=\sin x^{-1}$ is not uniformly continuous on $(0,1)$ then you will need a different approach.

I could always spot the talented students from the drudges because they seldom went directly at a problem from a naive approach---they could spin any problem in a different direction. Don't suppress your inner creative nature.

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  • $\begingroup$ Unfortunately, our task is to use this definition approach to answer this question. $\endgroup$
    – B ry
    Oct 25 '15 at 17:55

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