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An example of a continuous function in L1 space with no limit at infinity

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

    I am trying to come up with a continuous function in L1[0,infinity) that doesn't converge to 0 as the function goes out to infinity.

    2. Relevant equations

    I am trying to show an example of an f in L1[0,infinity) (i.e. ∫abs(f) < infinity) where the limit as the function goes to infinity does not exist.

    3. The attempt at a solution

    My professor said it could be done in the lecture, but I have had no success... I have been experimenting with decreasing oscillating functions (e.g sinx/x - where I can't stop it converging to 0) and rapid oscillating functions (e.g. sin(e^x) - can't seem to get an integral on the absolute value) and haven't figured it out yet. Anyone have any ideas about a function that would work?
     
  2. jcsd
  3. May 24, 2012 #2
    I don't think you can do it with smooth functions. Perhaps something discontinuous might do the trick!
     
  4. May 24, 2012 #3
    My professor said clearly that you can do it with a continuous function. With a discontinuous function, you could put in pieces with measure zero that would mess up the limit, while still being in L1...

    Any more ideas?
     
  5. May 24, 2012 #4
    Yeah sorry; I typed my answer horribly wrong :D What I meant is that you might want to try to find a function which is continuous, but behaves more and more like a discontinuous function when you go to high x
     
  6. May 24, 2012 #5
    Thanks, that sounds more like it.
    I have been trying some functions like sin(e^x) which is continuous and has no limit, but I can't prove it is in L1. Is it in L1? Do I need to change the function in some way to get it into L1?
     
  7. May 24, 2012 #6

    Dick

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    Sum a series of bumps of height 1 but smaller and smaller area centered on points running off to infinity.
     
  8. May 24, 2012 #7
    Thanks for the reply...
    With abs(sin(e^x)) I have a function that does that, but I am getting a limit of 0 according to wolfram alpha... am I doing something wrong?
     
  9. May 24, 2012 #8

    Dick

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    It doesn't have limit 0, but it's also not in L1. Define the function piecewise. Can you define a function f(x) such that f(0)=1 and f(x)=0 for x outside of the interval [-1,1]? That's what I mean by a 'bump'. What's the area under your f(x)? What would be the area under f(cx) for some constant c? Now think about making an infinite sum of them.
     
  10. May 24, 2012 #9
    The function f=abs(cos x) satisfies f(0)=1 and f(pi/2)=0. The area under the curve is 2 for every bump (-pi/2 to pi/2). If you take f(2x), the area under the curve is 1 for the bump (-pi/4 to pi/4). I see how the area and interval are cut by the value of c, so if the c increases each bump has progressively less area under the curve over a progressively smaller interval.
    But, what is the proper notation for setting up an infinite sum of these bumps and how can I prove this doesn't converge to 0?
     
  11. May 24, 2012 #10

    Dick

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    f(k(x-c)) is going to be a bump centered at c. Pick a sequence of points c going to infinity. Arrange the values of the k so the infinite series that sums the areas converges. You could express the final result using sigma notation, but describing it in words should be fine too. All you have to do to show the sum doesn't converge to 0 as x->infinity is notice that your sum assumes the value 1 for arbitrarily large values of x.
     
  12. May 24, 2012 #11
    As I said earlier, you can't really use a smooth function like that (or you have to choose a very weird smooth function..) Easiest thing to do would be just to have a saw-tooth function which is nonzero on a sequence of decreasing intervals between each natural number. So for example it's nonzero in [itex] \lbrace [0,1] \cup [1,1+1/2] \cup [2,2+1/4] \cup [3, 3+1/8] \cup ... \rbrace [/itex]
     
  13. May 24, 2012 #12
    Thanks, this should help, but I meant how do I show that that the limit as x-> doesn't exist?
     
  14. May 24, 2012 #13

    Dick

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    Probably should have been more careful to say explicitly, "define the function f(x)=cos(x) for x in [-pi/2,pi/2] and f(x)=0 otherwise". That's what I was thinking was intended, but perhaps not. Does that make it clearer why a function defined as an infinite sum of bumps in the right way doesn't approach 0 as x->infinity?
     
  15. May 24, 2012 #14
    Totally; I think I was momentarily confusing the value of the function (will always fluctuate between 1 and 0) and the area under the curve. So I guess the final thing I don't know how to prove is that the integral for the entire absolute value of the function is less than infinity so that the function remains in L1[0,infinity). Is it enough to say that the bumps get infinitely thin (area under each bump approaches 0)?
     
  16. May 24, 2012 #15

    Dick

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    Yes. You have to specify how the bumps get thin and where they are centered. Then make sure the sum of the area under all of the bumps is a convergent series.
     
  17. May 24, 2012 #16
    So I figured out a function that starts with an area of 2 under the first bump and each successive area is cut in half...

    f(t) = [itex]\sum^{\infty}_{k=0} [/itex] abs{cos([itex]2^{k}[/itex](t - [itex]\frac{3(2^{k-1})\pi }{2^{k+1}}[/itex]} where [itex]\frac{3(2^k)\pi - 4 \pi}{2^{k+1}}[/itex] [itex]\leq[/itex] t [itex]\leq[/itex] [itex]\frac{3(2^k)\pi - 2 \pi}{2^{k+1}}[/itex]

    The widths diminish: pi,pi/2,pi/4,pi/8.... and integrals diminish: 2,1,1/2,1/4....
    How does it look to you?
    Can I just use the standard proof that Ʃ 1/(2^n) converges to 1? It would give a total area under the curve as 3 (from 0 to infinity).
     
    Last edited: May 24, 2012
  18. May 24, 2012 #17

    Dick

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    It's a good try. But the center point of your bumps isn't going to infinity. They are piling up at a single point, the limit of 3*2^(k-1)pi/2^(k+1) as k->infinity. I think you should just define it in terms of of the function I defined in post 13. You could write it as the sum of f(2^k*(x-ck)). Pick a sequence for ck that goes to infinity.
     
  19. May 24, 2012 #18
    But if I pick some sequence for ck that goes to infinity, how will the bumps line up to create a continuous function? I may be totally missing something here - could you please give me an example of a function for ck that goes to infinity and how the whole function would look? At this point I am lost...
     
    Last edited: May 24, 2012
  20. May 24, 2012 #19

    Dick

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    I'm not sure why you think the bumps should 'line up'. Each function in your sum is continuous all by itself. How about just ck=k? Then you have an infinite sequence of bumps of height 1 getting narrower and narrower as the centers go to infinity (well except maybe at the beginning, the first few might overlap).
     
  21. May 25, 2012 #20
    I see... I was confusing continuity as lining up everything perfectly. Even if the terms overlap, the bumps will just look a little wobbly over the overlapping interval. One last thing, for L1[0,infinity) do the centers always need same distance from each other, or can the distance get larger? (From the earlier example, the sum converges to quickly if the distance gets smaller).
     
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