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Prove that f is a constant function

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose that f:R->R satisfies the inequality ##|\sum\limits_{k=1}^n3^k[f(x+ky)-f(x-ky)]|<=1## for every positive integer k, for all real x, y. Prove that f is a constant function.

    2. Relevant equations

    None
    3. The attempt at a solution
    I tried taking f(x)=sinx and then using sinC-sinD=2cos(C/2+D/2)sin(C/2-D/2). Then you will get 2cosx.sinky. |cosx| is always less than or equal to one, but greater than zero. So I ignored cosx from the sum since it is just like a positive constant. and you will be left with
    $$2\sum3^ksinky=2.Im\sum3^ke^{iky}$$
    I am not sure about this procedure. Is there a general method?
     
  2. jcsd
  3. Apr 28, 2015 #2

    Dick

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    To prove a function is constant you often prove that the derivative is zero.
     
  4. Apr 29, 2015 #3
    There are 2 variables x and y. Should I differentiate the expression with respect to x or with respect to y?
     
  5. Apr 29, 2015 #4

    haruspex

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    As far as f is concerned, it is a function of a single variable.
     
  6. Apr 29, 2015 #5
    If I differentiate w.r.t t, then you will have $$\sum\limits_{k=1}^n3^k[f'(x+ky). (\frac{dx}{dt}+k\frac{dy}{dt})-f'(x-ky). (\frac{dx}{dt}-k\frac{dy}{dt})]$$
     
  7. Apr 29, 2015 #6

    haruspex

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    I didn't mean to imply endorsement of differentiating. Since Dick suggested it, I'm prepared to believe that helps, but I don't see it yet.
    I think there's something not quite right in the OP. Should it say "for every positive integer n, for all real x, y"?
     
  8. Apr 29, 2015 #7
    It does.
     
  9. Apr 29, 2015 #8
    I thinks its just <sum expression> for all real x, y.
     
  10. Apr 29, 2015 #9

    haruspex

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    But n is also a variable for that sum, and the terms inside the sum can change sign as k changes, so I think it's important to include "for all n".
     
  11. Apr 29, 2015 #10

    Dick

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    Don't differentiate with respect to either one. Remember that the derivative is the limit of a difference quotient. Let y approach zero and take the limit.
     
  12. Apr 29, 2015 #11

    Dick

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    Actually letting y actually approach zero doesn't cut it. I think you need some sort of mean value theorem type of thing to get the difference to actually equal ##f'(c) 2ky## for some value of ##c## and nonzero ##y##. Or you need to make some sort of approximation argument about the difference in terms a derivative. THEN you can let ##k## approach infinity to conclude ##f'(c)=0##. Maybe I haven't thought this through enough, so you try and fill in the details for me.
     
    Last edited: Apr 29, 2015
  13. Apr 29, 2015 #12

    haruspex

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    Fix on some x to begin with. For simplicity I'll take x=0.
    Consider f(a)-f(-a). By suitable choices of n and y, what bounds can you put on it?
    Next, consider n=2, but changing y such that the extreme ends of the sum are still at -a and +a, you should now be able to put tighter bounds on that difference. And so on.
     
  14. Apr 29, 2015 #13
    For x=0, then you get f(a) for n=1, y=a.
    For n=2, the expression has two pairs: f(y)-f(-y) + f(2y)-f(-2y)
    then there is no value of y for which y=a and -2y=-a.
     
  15. Apr 29, 2015 #14
    According to MVT, $$f'(c)=\frac{f(x+ky)-f(x-ky)}{2ky}$$
    Hence ##f'(c)2ky=g(k)## where g(k)=f(x+ky)-f(x-ky)
     
  16. Apr 29, 2015 #15
    If I take a point x, f(x) on the curve Y=f(X), then the two points x-ky and x+ky on the X-axis are aymmetrically placed about X=x. So can I say that the arbitary point c will remain same for any value of k? That is, can I say that the lines formed by connecting (x-2y,f(x-2y)) and (x+2y, f(x+2y)) will be parallel to that formed by (x-y, f(x-y)) and (x+y,f(x+y))?
     
  17. Apr 29, 2015 #16

    Dick

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    Good point, and no you can't say that. So this won't work in the way I was thinking.
     
  18. Apr 29, 2015 #17

    haruspex

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    Yes, but what inequality do you get for |f(a)-f(-a)|?
    Sorry, a small error in what I posted may have misled you:
    I meant the extreme top end of the sum, singular. I.e. a = ny. What inequality do you get now? How can you combine this with the n=1 inequality to get even tighter bounds on f(a)-f(-a)?
     
  19. Apr 30, 2015 #18
    For n=1, y=a, the inequality will be |f(a)-f(-a)|<=1. So f(a)-f(-a) is greater than -1 and less than 1.
    for n=2,y=a, you get f(a)-f(-a)+f(2a)-f(-2a) and this should again lie between -1 and 1. Let f(a)-f(-a)=A. So---> -1-A<=f(2a)-f(-2a)<=1-A.
     
  20. Apr 30, 2015 #19

    haruspex

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    No, the given inequality has 3k, not 3k-1, and this is the k=n term.
    I meant you to pick y such that the k=n (=2) term corresponds to f(a), but I don't think it matters.
    What does matter is that you've left out the power-of-three factors again.
     
  21. Apr 30, 2015 #20
    Sorry.
    For n=2, y can be taken as a/2
    So 3[f(a/2)-f(-a/2)] + 9[f(a)-f(-a)]
    And ##-1/3\le f(a)-f(-a) \le 1/3##
     
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