Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I seem to have hit a dead-end in the following proof.

Define $f:\mathbb{R}\to\mathbb{R}$ by:

$f(x)=\frac{1}{1+x^2}$

Show that $f$ is uniformly continuous.

My proof:

Let $x_{0}\in \mathbb{R}$.

Also let $\epsilon >0$

Choose $\delta = ?$

Then, for $x\in \mathbb{R}$, such that $|x-x_{0}|<\delta$, we have:

|$f(x)-f(x_{0})|=|\frac{1}{1+x^2}-\frac{1}{1+x_{0}^2}|=|\frac{x_{0}^2-x^2}{(1+x^2)(1+x_{0}^2)}|=\frac{|x_{0}-x||x_{0}+x|}{(1+x^2)(1+x_{0}^2)}\le \delta|x+x_{0}|$

The last line uses the fact that $x^2, x_{0}^2\ge 0$. How can I finish the proof?

share|improve this question
2  
Start by showing that $f'(x)$ is bounded. Then show in general that functions with bounded derivative are Lipschitz and thus uniformly continuous. –  Henning Makholm Mar 13 '12 at 18:13
add comment

2 Answers

up vote 6 down vote accepted

Write $$|f(x)-f(y)|=\left|\frac 1{1+x^2}-\frac 1{1+y^2}\right|=\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|\leq |x-y|\frac{|x|+|y|}{(1+x^2)(1+y^2)}\\\ \leq \frac{|x-y|}2\frac{x^2+1+y^2+1}{(1+x^2)(1+y^2)}\leq \frac{|x-y|}2\left(\frac 1{1+y^2}+\frac 1{1+x^2}\right)\leq |x-y|$$ and now the $\delta$ you have to choose is clear. In fact, $f$ is lipschitz continuous.

share|improve this answer
    
I used $x^2-y^2=(x-y)(x+y)$ and $|x+y|\leq |x|+|y|$ and $2ab\leq a^2+b^2$. –  Davide Giraudo Mar 13 '12 at 19:48
add comment

Hint:

Prove this theorem and use it,

THEOREM If $f$ continuous on $[a,\infty)$ and $\lim_{x\to\infty}f(x)$ exist, then $f$ uniformly continuous on $[a,\infty)$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.