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

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


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?

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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
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.

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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


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 $[a,\infty)$

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