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A Real Analysis question on anti-derivatives

  1. May 20, 2012 #1
    Let f : R to R be a continuous function, and assume anti-derivative of f(x)dx from m to n≤ (n-m)^2 for every closed bounded interval [m,n] in R. Prove that f(x) = 0 for all x in R.

    I tried using fundamental theorem of calculus but got stuck.

    Any help/suggestion would be appreciated.
     
  2. jcsd
  3. May 20, 2012 #2
    Edit:

    I think you mean ##\int_m^nf(x)dx\leq(n-m)^2##. Again, definite integrals are not the same as antiderivatives.
     
    Last edited: May 20, 2012
  4. May 20, 2012 #3

    Fredrik

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    I don't understand what you mean by "assume anti-derivative" or "f(x)dx from m to n≤ (n-m)^2".
     
  5. May 20, 2012 #4
    I am sorry. Here is the exact question:

    Let f : R to R be a continuous function, and suppose that definite integral from m to n |∫(m to n)f(x)dx|≤(n-m)^2 for every closed bounded interval [m, n] in R. Then is it the case that f(x) = 0 for all x in R?
     
  6. May 20, 2012 #5

    Ray Vickson

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    Hint: mean-value theorem.

    RGV
     
  7. May 20, 2012 #6
    So after using Mean Value theorem and FUndamental theorem of Calc, I got to a point where F'(c)=f(c) ≤ m - n. BUt this still doesn't seem to work enough.
    Is there anything crucial that I am missing?

    Thanks!
     
  8. May 20, 2012 #7
    I think he meant Mean Value Theorem for integrals:

    If ##f## is continuous on ##[a,b]##, then there is ##c\in(a,b)## with ##\int_a^bf(x)dx=f(c)(b-a)##.

    Given the FTC, it's just a rewrite of the "regular" mean value theorem.


    The key here is to apply this to your problem, with the assumptions given, and notice what the implications are when ##n-m## is very small.
     
  9. May 20, 2012 #8
    Do you think I should go by contradiction instead? coz I am not really figuring out how to use this assumption here.
    Some elaboration would really help!
     
  10. May 20, 2012 #9
    Yes, I think it would be easiest to assume that ##f\neq0## and show that this gives a contradiction.

    Supposing that ##f\neq0## gives that there exists ##x_0## such that ...

    Since ##f## is continuous, there is ##\delta>0## such that ... on ##(x_0-\delta,x_0+\delta)##.

    Then there is ##c\in(x_0-\delta,x_0+\delta)## such that ##\left|\int_{x_0-\delta}^{x_0+\delta}f(x)dx\right|=\left|f(c)\right|\cdot 2\delta>## ..., contradicting the assumption that ##\left|\int_{x_0-\delta}^{x_0+\delta}f(x)dx\right|\leq##... .

    Now you may need to fidget a bit with the ##\delta## to get the inequalities that you need, but the direction that you need to fidget is a "good" one.

    There is a fairly straightforward way to get around using the Mean Value Theorem. However it requires you to identify that you are in a situation where ##\left|\int_{a}^{b}f(x)dx\right|=\int_{a}^{b} \left| f(x) \right|dx## (i.e. the triangle inequality is actually an equality).
     
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