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Show that the function $f(x) = \frac{1}{x}$ is not uniformly continuous on the interval $(0,\infty)$ but is uniformly continuous on any interval of the form $(\mu, \infty)$ if $\mu > 0$.

My Work

Referring to the definition of uniform continuity, I have that $f$ is unif. cts. if for each $\epsilon > 0$ there is a $\delta > 0$ so that for all $x, c$ in the domain of $f$ $|x - c| \le \delta \ \Rightarrow \ |f(x) - f(c) | \le \epsilon$.

From this definition, it is clear that if $f$ is uniformly continuous, it will be uniformly continuous on its domain, $(-\infty, \infty) \backslash \{0\}$. So for $\mu > 0$, $(\mu, \infty) \subset \mathrm{Dom}\,(f)$. Additionally, $f$ cannot be unif. cts on $(0, \infty)$ because $0 \notin \mathrm{Dom}\, (f)$. (Sorry about the longwindedness)

Now to find the $\delta$:

\begin{align*} |f(x) - f(c)| = \left|\frac{1}{x} - \frac{1}{c}\right| &= \left|\frac{x - c}{cx}\right| \\ \text{since }x\text{ is within }\delta\text{ of }c \ \Rightarrow \ &\le \frac{\delta}{|cx|}\\ x, c>0 \ \Rightarrow \ &= \frac{\delta}{cx} \end{align*}

This is where I am stuck. Should I use that $x \le c + \delta$, or should I break this up into two cases, one where $cx < 1$ and one where $cx \ge 1$?

Edit (due to Brian M. Scott)

It was pointed out that $0 \notin (0, \infty)$ so my above argument is senseless.

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Additionally, I understand that I cannot simply take $\epsilon = \frac{\delta}{c(c + \delta)}$ because I am trying to prove uniform continuity –  Zvpunry Aug 17 '12 at 17:39
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Use the fact that both $x$ and $c$ are at least $\mu$. So then $\delta/(cx)\le\delta/\mu^2$. –  David Mitra Aug 17 '12 at 17:55
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3 Answers

up vote 2 down vote accepted

Since $0\notin(0,\infty)$, $0$ is completely irrelevant to the question of whether $f$ is uniformly continuous on $(0,\infty)$. To show that $f$ is not uniformly continuous on $(0,\infty)$, you should show that there is some $\epsilon>0$ such that no matter what $\delta>0$ you pick, you can find points $x,y\in(0,\infty)$ such that $|x-y|\le\delta$, but $|f(x)-f(y)|>\epsilon$. HINT: You can take $\epsilon=1/2$. Now consider values of $x$ of the form $\frac1n$ for $n\in\Bbb Z^+$.

To prove that $f$ is uniformly continuous on $(\mu,\infty)$ for $\mu>0$, you need what is really the key insight for both questions: for $x>0$, the graph of $y=\frac1x$ gets steeper and steeper as $x$ gets smaller and smaller. Given $x,y\in(\mu,\infty)$ with $|x-y|\le\delta$, where $\delta$ is some as yet unspecified positive real number, can you find an upper bound on $|f(x)-f(y)|$? How does it compare with $\frac{\delta}{\mu^2}$? (Consider $f'(x)$.)

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I understand your hint w/r/t comparing $|f(x) - f(y)|$ to $\delta/\mu^2$, where by definition $\mu$ is the infimum of the interval. Taking $\epsilon = 1/2$, I have $\delta = xy/2$ so I can always find $x, y \in (0,\infty)$ such that $xy > 1$? –  Zvpunry Aug 17 '12 at 18:00
    
Since $\mathbb R$ is Archimedean I can always choose $\delta < xy\epsilon$? I solved your suggestion using $\epsilon$. Thanks for the great hint! –  Zvpunry Aug 17 '12 at 18:03
    
@jmi4: No matter what $\delta>0$ you pick, you can always find $n\in\Bbb Z^+$ such that $\left|\frac1n-\frac1{n+1}\right|<\delta$; what’s $\left|f\left(\frac1n\right)-f\left(\frac1{n+1}\right)\right|$? –  Brian M. Scott Aug 17 '12 at 18:04
    
Ahhh I see. That is equal to $1$, and it is always possible to pick $\epsilon \in (0,1)$. –  Zvpunry Aug 17 '12 at 18:10
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Notice that:

  1. $f$ is $C^1$.
  2. $f'$ is bounded away from $0$.
  3. If a $C^1$ function (on a connected subset of $\bf R$) has a bounded derivative, it is Lipschitz (this follows from fundamental theorem of calculus).
  4. Any Lipschitz function is uniformly continuous.

Alternatively, you can elementarily show that $f$ is Lipschitz on $(\mu,+\infty)$ for $\mu>0$, using the fact that $\lvert 1/x-1/y\rvert=\lvert x-y\rvert/(xy)^2$ and $xy>\mu^2$.

To show that $f$ is not continuous near zero, best way would be to prove it by contradiction, but the essential thing is that near zero, we get arbitrarily large jumps in arbitrarily small intervals.

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I do not have derivatives or $C^1$ accessible to me as we have not proved $1$-$3$. Lipschitz might be a good approach though –  Zvpunry Aug 17 '12 at 18:04
    
@jmi4: well, it shouldn't be too hard to show that it is Lipschitz using elementary means. Notice that $\lvert 1/x-1/y\rvert=\lvert (x-y)/(xy)\rvert$ and $xy>\mu^2$. –  tomasz Aug 17 '12 at 18:07
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Maybe

you can show that

$\mathrm{abs}\left(\frac{1}{x}-\frac{1}{y}\right)<\mathrm{abs}(x-y)$

cause $0<\frac{1}{x}<1$ and same for $y$.

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Maybe?... –  MattAllegro May 15 at 18:02
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