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Math Help - Uniform Continuous?

  1. #1
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    Uniform Continuous?

    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.
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  2. #2
    MHF Contributor

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    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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