My Math Forum D[f] is bounded, show that f is uniformly continuous

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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 $f : X \to Y$ is differentiable between metric spaces, and that f'(x) is bounded. Then f is uniformly continuous. Proof: $f'(x) \leq K$ for all x. $f \leq K*x+C$. $d_Y(f(x), f(a)) < \varepsilon$. $d_Y(f(x), f(a))= |f(x)-f(a)| \leq |(K\,x) + (K\,a) + C|$, where in the last step, the signs of K were changed to ensure the operation to be true. $|(K\,x) + (K\,a) + C| \leq |(K\,x) + (K\,a)| + |C|$. $|x-a|=< \delta,\,\,\, x= a + {\delta \over 2}$. $2|K||a+{\delta \over 2} + C| \Longrightarrow \delta\,<\,\frac{2\varepsilon}{|K|} - 4|y|$ 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 $|f(x)-f(y)|\le |f'(t)||x-y|\le M|x-y|$, choose $\delta = \frac{\epsilon}{M}$

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