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Continuous functions

  1. Aug 12, 2006 #1
    This is a simple question: given a continuous function, f(x), in a closed interval, how do we show that there is value "a" small enough such as for arbitrary x:

    f(x+a) - f(x) < e

    Where e lower bound is 0.

    ???

    Thxs in advance.
     
  2. jcsd
  3. Aug 12, 2006 #2

    HallsofIvy

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    If [itex]x_0[/itex] is in the closed interval then the definition of "continuous" is that [itex]\lim_{x\rightarrow x_0} f(x)= f(x_0)[/itex]. That in turn means "given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x-x_0|< \delta[/itex] then [itex]|f(x)-f(x_0)|< \epsilon[/itex].
    To get your result, let [itex]x_0= x[/itex] and [itex]x= x_0+ \delta[/itex]. Then your "a" is "[itex]\delta[/itex]", "e" is "[itex]\epsilon[/itex] and [itex]|f(x)- f(x_0)|<\epsilon[/itex] becomes |f(x+a)-f(x)|< e. Of course, it's still true without the absolute value since if f(x+a)- f(x) is negative it's still less than the positive e.
     
  4. Aug 12, 2006 #3

    StatusX

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    If the function is uniformly continuous, the proof is easy. This would mean that for any e>0, there is a d>0 such that |x-y|<d implies |f(x)-f(y)|<e for all x,y in the domain. So just pick your a as d/2.

    It's a little more complicated to show this from just the assumption of continuity (which, remember, is a local property, where as you're looking for an a with a ceratin global property), but the Heine-Cantor Theorem states that these definitions are equivalent if the domain is compact, which a closed interval is. If you can't just cite this theorem, there is a link to a proof on the wikipedia page.
     
  5. Aug 12, 2006 #4

    quasar987

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    Mmh, yes, the existence of the 'a' for all x is assured by definition of uniform continuity + Heine-Cantor in the case [itex](X,d(x,y))=(\mathbb{R},|x-y |)[/itex].


    P.S. Halls, do you know how to write in italic w/o using Latex?
     
    Last edited: Aug 12, 2006
  6. Aug 12, 2006 #5

    mathwonk

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    just subdivide your interval into unit length subintervals, then subintervals of length 1/10, then 1/100, and so on....


    assuming no subdivision ever has intervals on all of which the variation is less than epsilon, resulots in the existence of an infinite decimal, i.e. a real number, in our interval where the function is not continuous.
     
  7. Aug 16, 2006 #6
    My question has been misunderstood... sorry the lack of clarity. What I meant is:

    Let e and c be constants, for all x. Is there such a constant for which, regardless of x;

    If within the interval [a,b]

    (</= stands for inferior or equal)

    f(x+e) </= a

    then

    f(x+e) - f(x) </= c


    Additionally,

    f(x+l) - f(x) ; if l < k < e

    then

    f(x+l) - f(x) < f(x+k) - f(k) < c


    ???

    If there is, what is the proof?


    This all has to do with a proof that all Riemann sums converge towards the same value when the span of its subdivision tends to 0. Thanks in advance.
     
    Last edited: Aug 16, 2006
  8. Aug 17, 2006 #7

    StatusX

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    I can't make any sense out of your last post, werg. If what you're trying to do is prove all Riemann sums converge to the same limit for continuous functions on a closed interval, you would want to start with the question we thought you asked and then answered.
     
    Last edited: Aug 17, 2006
  9. Aug 17, 2006 #8
    Okay, so after a little thought I solved it. Sorry for the lack of clarity again, I tried my best to explain. N/P now :smile:
     
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