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Difference between continuity and uniform continuity

  1. May 7, 2015 #1
    I noticed that uniform continuity is defined regardless of the choice of the value of independent variable, reflecting a function's property on an interval. However, if on a continuous interval, the function is continuous on every point. It seems that the function on that interval must be uniformly continuous. Is this correct? Is there a counterexample for the statement?
     
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  3. May 7, 2015 #2

    jbunniii

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    What you said is almost true. If, on a closed, bounded interval, a function is continuous at every point, then the function is uniformly continuous on that interval.

    Counterexample on a non-closed interval: ##f(x) = 1/x## on the interval ##(0,1)##.

    Counterexample on a closed but unbounded interval: ##f(x) = x^2## on the interval ##[0,\infty)##.
     
  4. May 8, 2015 #3

    Svein

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    Even more general: If a function is continuous at every point in a compact set, it is uniformly continuous on that set.
    The proof is simple: Let the compact set be K and take an ε>0. Since the function is continuous at every point x in the set, there is a δx for every x∈K such that |f(w)-f(x)|<ε for every w in <x-δx, x+δx>. Let Ox=<x-δx, x+δx>, then K is contained in [itex]\bigcup_{x\in K}O_{x} [/itex]. Since K is compact, it is contained in the union of a finite number of the Ox, say [itex] \bigcup_{n=1}^{N}O_{n}[/itex]. Take δ to be the minimum of the δn and |f(w)-f(x)|<ε whenever |w-x|<δ.
     
  5. May 8, 2015 #4

    jbunniii

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    And here's the proof that a closed, bounded interval is compact, so you can apply Svein's proof.

    Let ##[a,b]## be a closed, bounded interval, and let ##\mathcal{U}## be any collection of open sets which covers ##[a,b]##. Let ##S## be the set of all ##x \in [a,b]## such that ##[a,x]## can be covered by finitely many of the sets in ##\mathcal{U}##. Clearly ##a \in S##. This means that ##S## is nonempty and is bounded above (by ##b##), so it has a supremum, call it ##c##. Since ##c \in [a,b]##, it is contained in some ##U_c \in \mathcal{U}##, hence there is some interval ##(c - \epsilon, c + \epsilon)## contained in ##U_c##. Since only finitely many sets from ##\mathcal{U}## are needed to cover ##[a,c - \epsilon/2]##, those sets along with ##U_c## form a finite cover of ##[a,c+\epsilon/2]##. This shows that ##c \in S## and moreover, that ##c## cannot be less than ##b##. Therefore ##c=b##, so all of ##[a,b]## can be covered by finitely many sets in ##\mathcal{U}##.
     
    Last edited: May 8, 2015
  6. May 8, 2015 #5

    Svein

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    Excellent! And, of course, since one closed and bounded interval is compact, the union of a finite number of closed and bounded intervals is again compact.
     
  7. May 8, 2015 #6

    jbunniii

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    I don't think this will work if ##w## and ##x## are not contained in the same ##O_n##. I think you need to take ##O_x = (x - \delta_x/2, x + \delta_x/2)## and ##\delta## to be ##\min\{\delta_n/2\}## in order to ensure that ##|w-x| < \delta## implies ##|f(w) - f(x)| < \epsilon##.
     
  8. May 8, 2015 #7

    Svein

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    Possibly. I need to think about that. The crux of the matter is that there is a finite number of intervals that cover K, which means that the minimum of the (finite number of) δ's exist and is greater than 0.
     
  9. May 8, 2015 #8

    HallsofIvy

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    Here is the fundamental difference between "continuous" and "uniformly continuous":

    A function is said to be continuous at a point, x= a, if and only if, given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|f(x)- f(a)|<\epsilon[/itex].

    A function is said to be continuous on a set, A, if and only if, given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|f(x)- f(a)|<\epsilon[/itex] for all a in set A.

    That is, a function is uniformly continuous on a set A if and only if it is continuous at every point in A and, given [itex]\epsilon> 0[/itex] the same [itex]\delta[/itex] can be used for ever point in A.
     
  10. May 8, 2015 #9

    Svein

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    Agree. I was sloppy and overlooked the simple fact that that the length of the interval Ox is 2⋅δx.
     
  11. May 13, 2015 #10
    The defition of uniformly continuous is perplexing for me. Is it possible for you to illustrate it with a proof about a function which is not uniformly continuous but continuous on a set so that I can compare the two? Thank you.
     
  12. May 14, 2015 #11

    Svein

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    OK. Take the function [itex]f(x)=\frac{1}{x} [/itex] on the open interval <0, 1>. It is continuous (and even differentiable) on that interval, but not uniformly continuous.
     
  13. May 17, 2015 #12

    Svein

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    After using pencil an paper for a bit, I came to the conclusion that I should have used ε/2 and δ/2. But - I recall a proof in "Complex analysis in several variables" that ended up in "... less than 10000ε, which is small when ε is small".
     
  14. May 17, 2015 #13

    HallsofIvy

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    First note an important point about "uniform continuity" and "continuity" you may have overlooked: we define "continuous" at a single point and then say that a function is "continuous on set A" if and only if it is continuous at every point on A. We define "uniformly continuous" only on a set, not at a single points of a set.

    A function is uniformly continuous on any closed set on which it is continuous and so on any set contained in a closed set on which it is continuous.
    To give an example of a function that is continuous but not uniformly continuous, we need to look at f(x)= 1/x on the set (0, 1).

    To show that it is continuous on (0, 1), let a be a point in (0, 1) and look at [itex]|f(x)- f(a)|= |1/x- 1/a|= |a/ax- x/ax|= |(a- x)/ax|= |x- a|/ax< \epsilon[/itex].
    We need to find a number, [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex], then [itex]|f(x)- f(a)|> \epsilon[/itex]. We already have [itex]|x- a| ax\epsilon[/itex] so we need an upper bound on ax. If we start by requiring that [itex]\delta< a/2[/itex] then [itex]|x- a|< \delta< a/2[/itex] so that [itex]-a/2< x- a< a/2[itex] or [itex]a/2< x< 3a/2[/itex] so an upper bound on ax is [itex]3a^2/2[/itex]. If [itex]|x- a|< a/2[/itex] and [itex]|x- a|< 3a/2[/itex] then [itex]|f(x)- f(a)|< |x- a|/ax< |x- a|/(3a^2/2)= 2|x- a|/3a^2[/itex] which will be less than [itex]\epsilon[/itex] as long as [itex]|x- a|< 3a^2\epsilon/2[/itex]

    So we can take [itex]\delta[/itex] to be the smaller of [itex]a/2[/itex] and [itex]3a^2/\epsilon[/itex]. Therefore, 1/x is continuous at any point a in (0, 1) so continuous on (0, 1).

    Now the point is that this [itex]\delta[/itex] depends on a. It is a decreasing function of a and will go to 0 as a goes to 0. That's fine for "continuity" but for uniform continuity we must be able to use the same [itex]\delta> 0[/itex] for a given [itex]\epsilon[/itex] no matter what the "a" is and we cannot do that. If the problem were to prove uniform continuity on the set [p, 1), which is contained in the closed set [p, 1], we could use, for any a in that set, the [itex]\delta[/itex] that we get for a= p, the smallest x value in the set. But with (0, 1), the smallest [itex]\delta[/itex] is 0 which we cannot use since we must have [itex]\delta> 0[/itex].
     
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