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February 26th, 2014, 12:29 PM  #1 
Senior Member Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2  D[f] is bounded, show that f is uniformly continuous
I recently found a more elegant method to prove the following proposition using the mean value theorem, however, I would really like it if someone could check my proof, because I want to learn. This is the proposition: Proposition: Assume is differentiable between metric spaces, and that f'(x) is bounded. Then f is uniformly continuous. Proof: for all x. . . , where in the last step, the signs of K were changed to ensure the operation to be true. . . Does it look correct? Thank you for your time. Kind regards, Marius 
July 7th, 2014, 05:30 AM  #2 
Newbie Joined: Jul 2014 From: jordan Posts: 7 Thanks: 0 
Simply, since $f$ is differentiable and $f'$ is bounded then there exits $M>0$ s.t. $f'(t)\le M$ so that by MVT we have $f(x)f(y)\le f'(t)xy\le Mxy$, choose $\delta = \frac{\epsilon}{M} $ 

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bounded, continuous, show, uniformly 
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