Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined by $f\left( x\right) =1 / \left( 1+x^{2}\right)$. Prove that $f$ uniformly continous.

My proof. Let $\varepsilon >0$. Pick $\delta=$min{$1,\varepsilon(\dfrac { ( 2+2\left| a\right| +\left| a^2\right|) \left| 1+a^2\right| } {(1+\left| a\right|) })$}

If $\left| x-a\right|<\delta$ and $a,x\in\mathbb{R}$ then $\left| f\left( x\right) -f\left( a\right) \right| =\dfrac {\left| x-a\right| \left| x+a\right| } {\left| 1+x^2\right| \left| 1+a^2\right| }$. We need to show this is smllar than $\varepsilon$.

Note that $\left| x-a\right| < 1 \Rightarrow \left| x\right| -\left| a\right| \leq \left| x-a\right| < 1 \Rightarrow \left| x\right| < 1+\left| a\right| \Rightarrow \left| x^{2}\right|+1\leq \left| x^{2}+1\right| < 2+2\left| a\right| +\left| a^2\right| \Rightarrow$ Also, $\left| x\right| < 1\Rightarrow \left| x\right| < 1\Rightarrow \left| x\right|+a < 1+a \Rightarrow \left| x+a\right| < 1+\left| a\right| $.

Now, we can do this,

$\dfrac {\left| x-a\right| \left| x+a\right| } {\left| 1+x^2\right| \left| 1+a^2\right| } \leq \dfrac {\delta (1+\left| a\right|) } {( 2+2\left| a\right| +\left| a^2\right|) \left| 1+a^2\right| }\leq \varepsilon$

Can you check my proof?

  • 1
    $\begingroup$ you define $\delta$ before defining $a$, that's not good $\endgroup$ – fonfonx Jan 17 '17 at 19:20

It is better to use this: \begin{eqnarray} \left| f\left( x\right) -f\left( a\right) \right|&=&\dfrac {\left| x-a\right| \left| x+a\right| } {(1+x^2)(1+a^2)}\\ &=&\left| x-a\right|\frac{1}{1+a^2}\dfrac { \left| x+a\right| } {1+x^2}\\ &\le&\left| x-a\right|\frac{1}{1+a^2}\dfrac {|x|+|a|} {1+x^2}\\ &\le&\left| x-a\right|\frac{1}{1+a^2}(\dfrac {|x|} {1+x^2}+\dfrac {|a|} {1+x^2})\\ &\le&\left| x-a\right|\frac{1}{1+a^2}(\frac12+|a|)\\ &\le&\left| x-a\right|(\frac12+\frac{|a|}{1+a^2})\\ &\le&\left| x-a\right|(\frac12+\frac12)\\ &=&|x-a|. \end{eqnarray} Here $2x\le 1+x^2$ is used.

  • $\begingroup$ So, does my delta work? $\endgroup$ – James Ensor Jan 17 '17 at 19:30
  • 1
    $\begingroup$ @Kahler, $\delta=\epsilon$. $\endgroup$ – xpaul Jan 17 '17 at 19:37
  • $\begingroup$ So, where did I go wrong? $\endgroup$ – James Ensor Jan 18 '17 at 7:33
  • $\begingroup$ @Kahler, your $\delta$ depends on $a$. $\endgroup$ – xpaul Jan 18 '17 at 15:12
  • $\begingroup$ Why $\delta$ should not depend on $a$? $\endgroup$ – James Ensor Jan 19 '17 at 14:41

Continuous + Infinitesimal as $|x|\to +\infty$ imply Uniformly Continuous.

If we consider some $\varepsilon>0$, there is some interval $I=[-M,M]$ such that $|f(x)|\leq\frac{\varepsilon}{2}$ on $\mathbb{R}\setminus I$.
Then $f(x)$ has bounded variation on $\mathbb{R}\setminus I$, and since $I$ is a compact interval, $f(x)$ is uniformly continuous on $I$. It follows that there is some $\delta$ ensuring $$ |x-y|\leq\delta\quad\Longrightarrow\quad \left|f(x)-f(y)\right|\leq\varepsilon.$$

Now just notice that $\frac{1}{x^2+1}$ is continuous and infinitesimal as $|x|\to +\infty$.


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