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February 26th, 2014, 12:29 PM   #1
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Joined: Sep 2010
From: Oslo, Norway

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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.

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,
king.oslo is offline  
July 7th, 2014, 05:30 AM   #2
Joined: Jul 2014
From: jordan

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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
$|f(x)-f(y)|\le |f'(t)||x-y|\le M|x-y|$, choose $\delta = \frac{\epsilon}{M} $
Orlicz is offline  

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bounded, continuous, show, uniformly

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