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A question on sequence of functions.

  1. Mar 30, 2007 #1
    i have the follow question:
    Does there exist a continuous function g:R->R that is not periodic, and for every sequence of real numbers {a_k} the sequence of functions g_k(x)=g(x+a_k) has a subsequence converging uniformly on R.

    i think i found a suitable function, i just need to be sure it's fine:
    i defined g_k(x)=g(x)=sin(e^x)
    this function isn't periodic, but if i use Arzela-Ascoli lemma in here, then first of all, this function is uniformly bounded by 1, and i can show that it's equicontinuous, i mean all i need to show is that:
    for every e'>0, there exist d>0, such that for every x,y in R whihc satisfy
    |x-y|<d and for every n then |f_k(x)-f_k(y)|<e'.
    here i can use the fact that |sin(e^(a_k+x))-sin(e^(a_k+y))|<=|e^a_k||e^x-e^y|
    e^a_k converges, and e^x is continuous, so we can prove the equality from this, am i correct?

    thanks in advance.
     
  2. jcsd
  3. Mar 30, 2007 #2
    now when i look at this again, i dont think my example is correct, cause im not guaranteed that e^(a_k) converges at all.
    so can you find an example or the answer that there isnt one function such as this.
     
  4. Mar 30, 2007 #3

    matt grime

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    Let's suppose any continuous function with extra properties will do.

    If the a_k are bounded, then they have a convergent subsequence, so use that. If not, then either a subsequence tends to +infty or -infty, so use that.

    They're the only obviously 'nice' looking subsequences. Now, can we use a nice function like g(x)=exp(-x^2) to do anything? I chose that because it is bounded, which seems like the least you need for g, and tends to zero as |x| tends to infinty.
     
  5. Mar 30, 2007 #4
    well, i think i need to expand the function youve given to include also the point x=0, which is no problem cause we can define it as f(0)=0 and otherwise f(x)=exp(-x^2).
    obviously this function is continuous, the only question is, is it equicontinuous, well let's look at the difference: |e^-(x+a_k)^2-e^-(y+a_k)^2|=|e^(-(a_k)^2)|*|e^-(x^2+2xa_k)-e^-(y^2+2ya_k)|
    basically i dont think i have got matters simpler cause now i have x^2+2xa_k in the exponent in the absolute brackets.
     
  6. Mar 30, 2007 #5

    matt grime

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    Why is exp(-x^2) not defined (and equal to 1) at zero?

    I had to google equicontinuous. It seems you mistyped something above - a sequence of functions is equicontinuous. You made it sound like a function was equicontinuous, which confused me.
     
    Last edited: Mar 30, 2007
  7. Mar 30, 2007 #6
    yes i thought it was e^-1/x^2. my mistake, sorry.

    anyway, back to topic.
    how should i prove that e^-x^2 fits the criteria in my question?
    obvoulsy it's uniformly bounded, but the second is what's bugging me.

    thanks in advance.
     
  8. Mar 30, 2007 #7

    Dick

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    Hey Matt, he's trying to prove that a sequence of FUNCTIONS (arbitrary translates of the original function) converge. Sounds like you are going to have a hard time finding such a function. You can't apply Arzela-Ascoli because R isn't compact. (If g is periodic then you can just work over one period and that's compact).
     
  9. Mar 30, 2007 #8
    the question asks if there exists such a function which isnt periodic which converges uniformly on R, obviously it's hard to find such a function but is it impossible?
     
  10. Mar 30, 2007 #9

    Dick

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    Don't know. Let's think about it...
     
  11. Mar 30, 2007 #10
    btw, i have another question which is obviously an arzela-ascoli question:
    prove that there exists a subsequence {f_k_M} that converges uniformly to on the real line R, where f_k(x)=sin(x+a_k), a_k is a sequence of real numbers.
    i proved that for the interval [0,2pi] there exists such subsequence by using arzela ascoli, now i think i only need to expand it to fit also the real line, so perhaps as we expand the interval but keep it compact, we still have a converging subsequence of f_k, but is it enough in order to prove for the whole real line?
     
  12. Mar 30, 2007 #11

    Dick

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    If the sequence converges uniformly on [0,2pi] then it converges uniformly on the whole line. The values on the rest of the line just repeat the values in the interval [0,2pi].
     
    Last edited: Mar 30, 2007
  13. Apr 2, 2007 #12
    any new insight on the initial question, dick or matt?
     
  14. Apr 2, 2007 #13

    Dick

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    No. And it's still annoying me.
     
  15. Apr 2, 2007 #14

    Dick

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    Ok. The class of 'almost periodic' functions also has this property. I haven't managed to cobble together a proof - maybe you can do that??
     
  16. Apr 3, 2007 #15
    when you say almost periodic, what to do you mean?
    something like my example that sin(e^x), or something like this: e^(ax)*(Asin(wx)+Bcos(wx))?
     
  17. Apr 3, 2007 #16

    Dick

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    Nope, not like those. You could look it up. One definition is that for all e>0 there exists a T such that |f(x+T)-f(x)|<e for all x.
     
  18. May 10, 2007 #17
    Dick, I think that g(x)=c where c is constant will do the job, what do you think?
     
  19. May 10, 2007 #18

    Dick

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    Sure. But constant functions are periodic according to my definition.
     
  20. May 10, 2007 #19
    Im not sure that it's periodic, cause there isn't a small non zero T such that g(x+T)=g(x), cause all of the values of g are the same.
     
  21. May 10, 2007 #20

    Office_Shredder

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    g(x) = c

    g(x+T) = c

    Therefore, g(x+T)=g(x) for all T and x. So given e>0, |g(x+T)-g(x)|=0<e for all T, x. So the constant function not only is periodic, but has infinitely small periods
     
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