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I have trouble with this question:

What does it mean that a function is uniformly continuous in a domain $D\subset\mathbb{R}^n$ ?

I know the definition for uniform continuity , but im not sure if this is enough. Since our domain can be a subset of $\mathbb{R}^n$ ?

AugSB
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Biggiez
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1 Answers1

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In the ordinary definition of uniform continuity, just use the norm of a vector instead of the absolute value to get the generalized notion of uniform continuity for a function defined on a domain in Euclidean space. The point is that for two points at distance less than $\delta$ the difference in the value of the function should be less than $\epsilon$ where $\delta$ only depends on $\epsilon$ but not on the points (with the usual epsilon-delta preamble, of course).

Mikhail Katz
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  • So you mean that i replace our 2 points x and c with vectors x and c? – Biggiez Jun 01 '16 at 16:13
  • ∀ϵ>0,∃δ>0 s.t ||x||,||c||∈D and ||x−c||<δ⇒||f(x)−f(c)||<ϵ.. Like this? (I should have bars over x and c also, but dont know how to do that) – Biggiez Jun 01 '16 at 16:20
  • Yes, $x,c\in\mathbb{R}^n$ though I am a bit worried about the fact that you imply an asymmetry between the two points by denoting one by $x$ and the other by $c$ as if $c$ is more of a constant than $x$. Remember the whole point about uniform continuity is that neither of the points is preferred (uniform continuity is not a point-wise notion but rather a global notion). – Mikhail Katz Jun 01 '16 at 16:20
  • Yes, your second comment is better. – Mikhail Katz Jun 01 '16 at 16:21
  • Ah okay, so both x and c plays equally important roles in the definition? Is that what your saying? – Biggiez Jun 01 '16 at 16:24
  • Im also not sure of the real reason on why we are replacing x and c as vectors instead of points? – Biggiez Jun 01 '16 at 16:26
  • Think this through in the case of ordinary continuity first. We are not doing anything different here in that respect. To define ordinary continuity at a point in Euclidean space you will also have to work with norm in place of absolute value. – Mikhail Katz Jun 01 '16 at 16:33
  • Okay , it makes more sense now. – Biggiez Jun 01 '16 at 16:34
  • So the final definition should be in this style: ∀ϵ>0,∃δ>0 s.t ||x||,||c||∈D and ||x−c||<δ⇒||f(x)−f(c)||<ϵ Where x and c should have bars above them ( to denote that they are vectors) ? – Biggiez Jun 01 '16 at 16:35
  • You should put parentheses around the two conditions before $\Rightarrow$. Otherwise the logical structure of the definition is not clear. – Mikhail Katz Jun 01 '16 at 16:37
  • ∀ϵ>0, ∃δ>0 s.t ( ||x||,||c||∈D and ||x−c||<δ ) ⇒||f(x)−f(c)||<ϵ ? – Biggiez Jun 01 '16 at 16:39
  • Much better. To write a more detailed version, you could explicitly outfit both $x$ and $c$ by a universal quantifier, and then write the implication. – Mikhail Katz Jun 01 '16 at 16:59