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How to think of uniform continuity intuitively?

  1. Oct 22, 2013 #1
    I'm struggling with the concept of uniform continuity. I understand the definition of uniform continuity and the difference between uniform and ordinary continuity, but sometimes I confuse the use of quantifiers for the two.

    The other problem that I have is that intuitively I don't understand why uniform continuity is needed to be defined as a new concept. Now I know that pure mathematicians might be interested in a lot of abstract stuff just because they preserve some sort of structure or they allow them to prove some desirable theorems, but I think the first one who noticed that this concept is an interesting one must have had something in his mind.

    My main questions are:
    How should I think about uniform continuity for real functions?
    Is it possible to generalize the concept of uniform continuity to more general topological spaces?
    Is it possible to visualize uniform continuity in a similar fashion like how we visualize a continuous function as a function that its graph can be drawn without lifting your pen from the paper?
     
  2. jcsd
  3. Oct 22, 2013 #2

    Office_Shredder

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    The difference between uniform continuity and continuity is that continuity is a local condition - for a fixed a, if x is within [itex] \delta[/itex] of a, then f(x) is within [itex] \epsilon [/itex] of f(a) (for appropriate delta and epsilon) Uniform continuity is a global condition - for any a, if x is within [itex] \delta [/itex] of a, then f(x) is within [itex] \epsilon[/itex] of f(a), where delta and epsilon are independent of the choice of a.

    If you restrict yourself to differentiable functions, then uniform continuity is the same as the derivative being bounded - there is some M such that |f'(x)| < M for all x.
     
  4. Oct 22, 2013 #3
    That is just restating the definition I believe. I'm looking for a more visual and intuitive picture of uniform continuity.
     
  5. Oct 22, 2013 #4
    The function ##\sqrt{x}## on ##(0, +\infty)## is a counterexample to that. One implication (bounded derivative implies uniform continuity) is true though.
     
  6. Oct 22, 2013 #5
    Just to clarify, only one direction is true. Bounded derivative implies uniform continuity but there exists uniformly continuous functions with unbounded derivatives. See sqrt(x) on (0, ∞).
     
  7. Oct 22, 2013 #6
    To visualize continuity: given any point (x,f(x)) and any choice of box height ε, there is a box width δ s.t. the graph of the function "leaves" the box centred at (x,f(x)) through it's left and right sides. Uniform continuity tells you that there is a choice of δ that will let you slide this box along the graph of the function without the top or bottom of the box ever intersecting the graph. Visualizing it this way, we can see that a uniformly continuous function is just a continuous function that changes in a controlled way.

    For functions defined on a closed interval, uniform continuity is equivalent to continuity. Specifically, uniform continuity is used to prove that a continuous function on a closed interval is Riemann integrable.
     
  8. Oct 22, 2013 #7
    You can think of Lipschitz continuity (a stronger requirement than uniform continuity) as asking for a bounded slope. You can think of uniform continuity as requiring that a function can be well-approximated by a function with bounded slope. That is, [itex]f[/itex] is uniformly continuous if for any error tolerance [itex]\epsilon>0[/itex], you can find a function [itex]f_\epsilon[/itex] of bounded slope such that every [itex]x[/itex] has a nearby (i.e. within [itex]\epsilon[/itex]) point [itex]x_\epsilon[/itex] with [itex]f_\epsilon(x_\epsilon)=f(x_\epsilon)[/itex]. (I've never seen uniform continuity expressed like this, but I believe it's correct.)
     
  9. Oct 22, 2013 #8
    This is not correct. Consider [itex]f:(0,∞)\rightarrowℝ[/itex] [itex]x\longmapsto\frac{1}{x}[/itex]
    Given ε>0, define [itex]f_ε:(0,∞)\rightarrowℝ[/itex] by:
    [tex]f_ε(x) = \left\{\begin{matrix}
    -\frac{1}{ε^2}(x-ε)+\frac{1}{ε},& &\: x\in(0,ε) \\
    f(x),& &\: x\in[ε,∞)
    \end{matrix}\right.[/tex]

    [itex]f_ε[/itex] has bounded derivative. [itex]f_ε=f[/itex] on [itex][ε,∞][/itex], and for any [itex]x\in(0,ε)[/itex] we can take [itex]x_ε=ε\in(x-ε,x+ε)[/itex], then [itex]f_ε(x_ε)=f(x_ε)[/itex]. But f is not uniformly continuous.
     
    Last edited: Oct 22, 2013
  10. Oct 22, 2013 #9
    The sequence you defined now does not converge uniformly. Economicsnerd never mentioned it specifically, but it's clear from his post he wants uniform convergence.
     
  11. Oct 22, 2013 #10
    That's charitable of you. :)

    I actually just made a mistake; it's a caveat that never occurred to me...

    With the corrected version you suggested, do you think that's a pedagogically nice way to explain uniform continuity, or do you think it just makes it unnecessarily complicated?

    [I'm not just seeking validation. I'd love to figure out a good way to make uniform continuity intuitive to neophytes.]
     
    Last edited: Oct 22, 2013
  12. Oct 22, 2013 #11
    Uniform convergence of what? There aren't any sequences of functions defined in mine or economicsnerd's posts.
     
  13. Oct 22, 2013 #12
    I think it's pretty nice.

    To be honest, I always see uniform continuity as a technical condition which is useful at times. Given a function ##\mathbb{R}\rightarrow \mathbb{R}##, I usually have an intuition for when it's uniform continuous or not. But I can't seem to give a formal condition that explains the intuition. Your characterization of Lipschitz functions is very close to my intuition though.

    One sad point is that apparently it doesn't generalize to arbitrary metric spaces. See http://en.wikipedia.org/wiki/Modulus_of_continuity But for stuff like ##\mathbb{R}##, I think it's very nice.
     
  14. Oct 22, 2013 #13
    You have a function ##\varepsilon\rightarrow f_\varepsilon##. The idea is that if ##\varepsilon\rightarrow 0##, then ##f_\varepsilon\rightarrow f## uniformly. So this can be seen as some kind of net.
     
  15. Oct 26, 2013 #14
    I really don't know if there is a good intuitive approach to uniform continuity. I'll try...

    The difference between the two is that with regular continuity, the ball of radius delta about some point c depends on how far you are away from c in the domain (delta depends on x and and epsilon). In other words, after fixing an epsilon precision in the codomain, you fix an x some distance away from some center c, and then from this epsilon and this x you determine a delta precision in the domain so that you end up with a ball: |x-c|<delta and x being inside this ball will guarantee that all points f(x) in the codomain lie in the ball (or coball?) |f(x)-f(c)|<epsilon. Since this delta is determined from x, after fixing x, you're still stuck with that delta. It can then be seen that once epsilon and c are fixed, the radius delta may become infinite if x is far enough from c and if f is unbounded that far from c.

    On the other hand, for uniform continuity, is a tool used to separate the average continuous function from a better behaving function. It is a tool that refers to pairs of points rather than just individual points. In the definition of uniform continuity, the radius of the delta ball is not dependent on the distance from the center (delta is not dependent on x) but it obviously must still be dependent on your fixed precision epsilon. In words, after fixing an epsilon precision, you choose some FINITE delta radius and construct a delta ball. The function is well enough behaved so that, if the interval [x,c] is in this delta ball, then the precision in the codomain is guaranteed.

    In this sense, uniform continuity is a tool used to determined how well behaved a continuous function is. Most functions that we deal with from day to day are uniformly continuous on some closed set. If you fix any epsilon precision, and then you can choose some delta precision so that when you drag this interval through some portion of the domain (in R^2 we are looking at a rectangle of height epsilon and width delta) the graph of f enters the rectangle on the left side and exits on the right side (if it enters through the bottom or exits through the top it violates our precision epsilon). The function is then uniformly continuous on the interval through which you drug it. In other words, there is some minimal and finite delta such that you can do this over an interval for any function. If you can do this for all of R then the function is uniformly continuous on R.

    Note that through the context here, we have subtly constructed at a geometric proof that every function defined only on a closed bounded interval is uniformly continuous.
     
  16. Oct 31, 2013 #15
    The really neat thing about uniform continuity is that it allows the exchange of limits. If a function is uniformly continuous, you can move limits in and out of integral signs.
     
  17. Oct 31, 2013 #16

    WannabeNewton

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    That's uniform convergence.
     
  18. Nov 1, 2013 #17
    Silly me, you're right. But still. A cool thing about uniform convergence, then.
     
  19. Nov 2, 2013 #18

    HallsofIvy

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    One important difference between "continuous" and "uniformly continuous" is that we define "continuous" at individual points: "f(x) is continuous at x= a if and only if ...". We can talk about a function being "continuous on a set" but what we mean is "continous at every point of that set". In order to talk about a function being uniformly continuous we must say it is uniformly continuous on a set. Being "uniformly continuous" at a single point is meaningless.

    A function is continous at a point, x= a, if and only if, as x approaches a, f(x) approaches f(a). A function is uniformly continuous on a set if it is continous at every point of the set and the rate at which f(x) approach f(a) is uniform over the set.
     
  20. Nov 8, 2013 #19

    Erland

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    One situation where uniform continuity is needed, and ordinary continuity does not suffice, is in the following theorem:

    Let X and Y be metric spaces, Y complete, let S be a dense subset of X, and let f: S → Y be a uniformly continuous function.
    Then there exists a contionous function g : X → V which extends f (i.e. f i s the restriction of g to S). (Actually, this g is unique and uniformly continuous too.)


    The conclusion of this theoren need not hold if f is only assumed to be continuous. As an example, consider the function f(x)=sin(1/x) defined on R - {0}. This function is bounded and continuous, but not uniformly continuous on R - {0}. This function cannot be extended to a continuous function on R, since it has non-removable discontinuity at 0.
     
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