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To Prove

  1. Jun 26, 2005 #1
    hello all

    well iv been workin on this problem but no matter how many ideas cross my mind i cant prove it, iv tried playing around with the properties of continous functions on closed intervals but didnt get me anywhere

    let f be a contiuous function on [a,b] which is non-negative, suppose that

    show that
    [tex]f(x)=0 \ \forall x\epsilon [a,b][/tex]

    this is probably the best of my proofs but i dont think its correct because i think i have to prove contradiction for all functions that are more than zero, please help any advice would be helpful

    thank you

    lets assume that f(x)>0 let f(x)=c>0 by FTC
    since f(x)>0 then F(x)=cx where c>=0
    0=c(b-a) since x is an element of [a,b] for all a,b an element of R then c=0
    therefore a contradiction and f(x)=0
  2. jcsd
  3. Jun 26, 2005 #2

    matt grime

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    if f is non-zero at some point in [a,b] it is non-zero in some interval around that point. and that's sufficient to show the answer.

    if f(x)=c then it is a constant function. but you aren't told f(x) is constant, are you?
    Last edited: Jun 26, 2005
  4. Jun 26, 2005 #3


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    That certainly doesn't follow!

    If, for any x, f(x)> 0 then there exist some small closed interval around that x on which f(x) is positive (since f is continuous). In fact, f must have a minimum on that interval which is positive. The integral of f on that small interval must be a positive number and, since f is non-negative the integral on the entire interval cannot be less than that positive number.
  5. Jun 26, 2005 #4


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    Why do I find this stuff so interesting?

    I hear whispers of Cauchy in the hallways of Analysis. :smile:
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