# Studying uniform continuity.

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!

• Yeah why not? If you find one I can take the half of it and have another one. – Shashi Nov 14 '17 at 16:56
• The $0 <$ in $0 < |x^\prime - x^{\prime\prime}|< \delta$ is not needed. – Gribouillis Nov 14 '17 at 17:06
• 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. – David C. Ullrich Nov 14 '17 at 17:29
• I meant definition , my bad. – Raku Nov 14 '17 at 17:35
• You may be interested in Fig in the part Visualization. – user2820579 Jul 31 '19 at 18:32

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.
• @Raku I understood your question. Yes, two distinct appoaches may well lead to two distinct but equally valid $\delta$'s. – José Carlos Santos Nov 14 '17 at 18:06