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The definition of uniform continuity states that a function $f$ defined over an interval $I$ is uniform continuous if $\forall \epsilon > 0, \quad\exists \delta> 0$ such that $ \forall x' , x'' \in I : 0 < |x'-x''| < \delta \implies |f(x')-f(x'')|<\epsilon$ Is it possible to find multiple solutions for $\delta$ while using different approaches?

Thanks for the help!

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    $\begingroup$ Yeah why not? If you find one I can take the half of it and have another one. $\endgroup$ – Shashi Nov 14 '17 at 16:56
  • $\begingroup$ The $0 <$ in $0 < |x^\prime - x^{\prime\prime}|< \delta$ is not needed. $\endgroup$ – Gribouillis Nov 14 '17 at 17:06
  • $\begingroup$ What you state is not a theorem, it's the definition of uniform continuity. A theorem that might be called "the uniform continuity" theorem states that if $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous. $\endgroup$ – David C. Ullrich Nov 14 '17 at 17:29
  • $\begingroup$ I meant definition , my bad. $\endgroup$ – Raku Nov 14 '17 at 17:35
  • $\begingroup$ You may be interested in Fig in the part Visualization. $\endgroup$ – user2820579 Jul 31 '19 at 18:32
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Since we're after a $\delta$ such that$$|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<\varepsilon,$$if you take $\delta'$ such that $0<\delta'\leqslant\delta$, then$$|x-y|<\delta'\implies|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<\varepsilon.$$So, yes, of course that it is possible to have more than one $\delta$. In fact, there are always infinitely many $\delta$'s that will do.

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  • $\begingroup$ What i meant by the question is : imagine we had an exercice where we have to find delta and we found : delta < epsilon/10 using the 1st method. Using a 2nd method we found delta < epsilon/5 Are both values acceptable ? $\endgroup$ – Raku Nov 14 '17 at 18:02
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    $\begingroup$ @Raku I understood your question. Yes, two distinct appoaches may well lead to two distinct but equally valid $\delta$'s. $\endgroup$ – José Carlos Santos Nov 14 '17 at 18:06
  • $\begingroup$ Alright , glad to have been able to get this question out of my mind, Thanks for the help! $\endgroup$ – Raku Nov 14 '17 at 18:08

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