Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrals and continuity

  1. May 20, 2012 #1
    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?

    I tried using fundamental theorem of calculus but got stuck, since I only got that F'(x)=f(x)≤ 0.

    Any help/suggestion would be appreciated.
     
    Last edited: May 20, 2012
  2. jcsd
  3. May 20, 2012 #2
    Why do you think f(x) would be 0? Doesn't make much sense.
     
  4. May 20, 2012 #3
    If f is non-zero at a point a then there exists some b, K > 0 s.t. f(x) > K for all x in [a-b,a+b]. Can you finish the proof?
     
  5. May 20, 2012 #4
    At first I thought it was wrong, so was trying to find a counterexample. But I saw this question in a book asks to show that f(x)=0 for all x in R, which means it should be true. I was using Mean Value theorem and fundamental theorem of calc but I am not really heading anywhere with those.

    Thanks and any further help will be appreciated!
     
  6. May 20, 2012 #5
    I guess I don't really get what this means (are you using the sing-change property of continuity?). I am really stuck!

    Thanks
     
  7. May 20, 2012 #6

    Using continuity of f is that the condition is fulfilled, so [tex]\,f(x)>K \,\,\forall\,x\in [a-b,a+b]\Longrightarrow \int_{a-b}^{a+b}f(x)dx>K\int_{a-b}^{a+b}dx = 2Kb[/tex] Can you see now how to finish with a contradiction?

    DonAntonio
     
  8. May 20, 2012 #7
    note also if 0<b<c then f(x) >K on [a-c,a+c].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integrals and continuity
Loading...