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I'm a bit confused as to how to interpret $\delta$ and $\varepsilon$ mean in real analysis. My textbook gives an example demonstrating that $\frac{1}{x^2}$ is not uniformly continuous on $(0,1)$.

Definition of not uniformly continuous: $(\exists \varepsilon >0)(\forall \delta > 0)(\exists x)(\exists y)$ such that $|x-y| < \delta \implies |f(x) - f(y)| \geq \varepsilon$.

Let $\varepsilon = 1$ and $y = x + \frac{\delta}{2}$ and show that $\left\vert f(x) - f(x + \frac{\delta}{2}) \right\vert \geq 1.$

With some algebra we see that $\displaystyle 1 \leq \frac{\delta (2x+\frac{\delta}{2})}{2x^2(x+\frac{\delta}{2})^2}$.

Now my book makes the claim that it suffices to prove that the function is not uniformly continuous for $\delta < \frac{1}{2}$. My book picks $x = \delta$ and then $\displaystyle \frac{\delta (2\delta+\frac{\delta}{2})}{2\delta^2(\delta+\frac{\delta}{2})^2} = \frac{5}{9\delta^2} > 1$.

So I see that this shows that if $0 < \delta < \frac{1}{2}$ then $|f(\delta) - f(\delta + \frac{\delta}{2})| > 1$. But the inequality $\displaystyle \frac{5}{9\delta^2} > 1$ is not true if $\delta = 3/4$ and by definition it needs to be true for all $\delta > 0$.

So in general should I think of $\delta$ and $\varepsilon$ as positive numbers very very close to $0$? And why does it suffice to take $\delta < 1/2$?

Student
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1 Answers1

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The variables $x$ and $y$ are bounded by $\exists$ in the written definition. So, if there exist such $x$ and $y$ for some $\delta_0<1/2$, then the same $x$ and $y$ will be good for any $\delta>\delta_0$, as both $|x-y|<\delta$ and $|f(x)-f(y)|\ge\varepsilon$ will hold.

By the way, the implication part ($P\implies Q$) rather goes to 'and' ($P\land\,\lnot Q$) under negation: $$ (\exists \varepsilon >0)(\forall \delta > 0)(\exists x)(\exists y): \ (|x-y| < \delta)\, \land \,(|f(x) - f(y)| \geq \varepsilon)\,. $$

Berci
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  • Thanks for correcting the negation of the implication. But I'm still confused about having $\delta < 1/2$. If $\delta < 1/2$ then he inequality $5/9\delta^2 > 1$ is true. But if $\delta = 3/4$ the inequality is not true. – Student Jun 18 '13 at 23:48
  • Let a $\delta>1/2$ be given. Now find $x$ and $y$ according to the given procedure, but for $\delta_0:=1/4$. I claim that the same $x$ and $y$ will satisfy the required condition: $|x-y|<\delta$ and $|f(x)-f(y)|\ge\varepsilon$. – Berci Jun 18 '13 at 23:49
  • So.. Let $\varepsilon:=1$, and let arbitrary $\delta$ be given. If $\delta<1/2$, do the given procedure to find $x$ and $y$ with the required property. Else, choose $x:=1/4,\ y:=3/8$. – Berci Jun 18 '13 at 23:55
  • Sorry I'm slow. But your comment cleared it up! – Student Jun 18 '13 at 23:56
  • When you see $\delta$ and $\varepsilon$, how do you think of them? – Student Jun 18 '13 at 23:57