Uniform Continuous?

**tttcomrader** · October 11th 2008, 08:14 PM

Prove that $f(x) = \frac {1}{x^2+1}$ is either uniform continuous or not.

Proof:

I claim that this function is uniform continuous, since its derivative is bounded, here is my proof.

Let $\epsilon > 0$ be given, and pick $0 < \delta < \epsilon$ , then for $x,y \in \mathbb {R} , |x-y|< \delta$ , we have $| f(x) - f(y) | = | \frac {1}{x^2+1} - \frac {1}{y^2+1} | = | \frac {y^2+1-x^2-1}{(x^2+1)(y^2+1)} |$ $\frac { | x^2 - y^2 | }{|(x^2+1)(y^2+1)|} = \frac {|x-y||x+y|}{|(x^2+1)(y^2+1)|} < \frac { \delta |x+y|}{|(x^2+1)(y^2+1)|} < \delta < \epsilon$

Is this right? Thanks.

**Laurent** · October 12th 2008, 02:57 AM

Hi,
this is right, but I think you should explain the inequality $\frac { \delta |x+y|}{|(x^2+1)(y^2+1)|} < \delta$ , using like $\frac { |x+y|}{|(x^2+1)(y^2+1)|} \leq \frac{ |x|+|y|}{|x^2+1||y^2+1|}\leq \frac{ \frac{1}{2}(x^2+1)+\frac{1}{2}(y^2+1)}{|x^2+1||y^2 +1|}$ $=\frac{1}{2}\frac{1}{y^2+1}+\frac{1}{2}\frac{1}{x^ 2+1}\leq \frac{1}{2}+\frac{1}{2}=1$ (where I used $|a|\leq \frac{a^2+1}{2}$ , which results from $a^2+1-2a=(a-1)^2\geq 0$ ). There may be a quicker proof though, that I didn't notice.

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