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Show that ##\frac{1}{x^2}## is not uniformly continuous on (0,∞).

  • #1
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25

Homework Statement


Show that ##f(x)=\frac{1}{x^2}## is not uniformly continuous at ##(0,\infty)##.

Homework Equations


N/A

The Attempt at a Solution


Given ##\epsilon=1##. We want to show that we can compute for ##x## and ##y## such that ##\vert x-y\vert\lt\delta## and at the same time ##\vert f(x)-f(y)\vert\gt 1##. Note that
\begin{align}
\vert x-y\vert\lt\delta &\Leftrightarrow y-\delta\lt x \lt y+\delta
\end{align}
So let ##x=y+\delta/2##. Now to link the desired conditions on ##x## and ##y##, we compute for ##\vert \frac{1}{x^2}-\frac{1}{y^2}\vert \gt 1 \Leftrightarrow \vert \frac{1}{(y+\frac{\delta}{2})^2}-\frac{1}{y^2}\vert \gt 1 \Leftrightarrow \vert \frac{y\delta+\frac{\delta^2}{4}}{y^4+y^3\delta+(\frac{\delta}{2})^2}\vert \gt 1##.
Since ##\lim_{y\to 0}(\frac{y\delta+\frac{\delta^2}{4}}{y^4+y^3\delta+(\frac{\delta}{2})^2})=\infty##, since ##\delta## is fixed, then we can find ##x## and ##y## such that ##\vert f(x)-f(y)\vert \gt 1##.
 

Answers and Replies

  • #2
Delta2
Homework Helper
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1) The denominator should be ##y^4+y^3\delta+y^2(\frac{\delta}{2})^2## ok this is just a typo I guess...

2) ##\delta## is not exactly fixed, as we want to compute specific ##x(\delta),y(\delta)## for any ##\delta## but yes that limit is ##+\infty## for any ##\delta>0##...
 
  • #3
Math_QED
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Alternatively, take ##a_n:= \frac{1}{n}, b_n := \frac{1}{n+1}##. Then ##a_n - b_n \to 0## but it is not true that ##f_(a_n)-f(b_n) = n^2 - (n+1)^2 \to 0##. Hence, ##f## can't be uniformly continuous.
 
  • #4
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Alternatively, take ##a_n:= \frac{1}{n}, b_n := \frac{1}{n+1}##. Then ##a_n - b_n \to 0## but it is not true that ##f_(a_n)-f(b_n) = n^2 - (n+1)^2 \to 0##. Hence, ##f## can't be uniformly continuous.
slick! Thank you!
 

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