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

[Terrell]

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 $\vert f(x)-f(y)\vert \gt 1$.

 

[Delta2]

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 $\delta>0$...

 

[member 587159]

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.

 

[Terrell]

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!