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Homework Help: Help with a proof

  1. Mar 7, 2005 #1
    i need some help with this proof. Please do not give me the solution i want to figure this on my own. ALl i need is your hints and steering.
    Also note that i am on a first year level intro to real analysis course. I do however know the upper and lower reimann sums which is probably how this problem is to be solved.

    Suppose f is continuous and non negative on [a,b]. Suppose f(c) > 0 for some c in (a,b). Prove that [tex] \int_{a}^{b} f > 0 . [/tex]

    What worries me here is that f(c) does not include f(a) and f(b) exclusively. however how would i go about proving that? Would i have to use limits to show that for some epsilon [tex] \lim_{\epsilon \rightarrow 0} f(x - \epsilon) > 0 [/tex] and the same would apply for the f(b) paart??
    I know that if i picked a partition P = {a,a+E,b-E,b} i would encounter this problem because of the explicit value (limit rather) of function at a nad b not being greater than zero.

    P.S. i can't find a website that shows how to prove limits formally (although i have learnt it in the past i cannot find it in my notes).
  2. jcsd
  3. Mar 7, 2005 #2


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    Since f is continuous, there's an [itex]\epsilon[/itex] such that [itex]f(x)>0[/itex] for [itex]|x-c|<\epsilon[/itex].
    You only have to deal with this region.
    You need some more results than this, but I don't know what you may or may not assume to know.
  4. Mar 7, 2005 #3
    what do you mena more results... Throw the stuff at me... ill probably know it after all my course went pretty deep into this stuff
  5. Mar 7, 2005 #4


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    1. As Galileo said, you can always choose a small interval around c so that f(x) is positive in that interval- that will give a positive addition to the Riemann sums.

    2. Since you are given that the function is never negative, there will never be any negative terms to cancel that positive number.
  6. Mar 7, 2005 #5
    CLEARLY this is going to be wayyyyyy off but what i understand from what you said is that i pick some delta > 0 in an interval [itex] P:=[ c- \delta, c , c+\delta ] [/itex]. For this interval since F is nonnegative and f(c) > 0 for all c then [itex] f(c - \delta) > 0 \{and} \ f(c+\delta) > 0[/itex]. This is clearly not enough.. some explicit proof would be required... i'm working on that.
    I'm thinking it would go suppose f(c + delta) was not zero then blah blah (right?)
  7. Mar 7, 2005 #6


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    Last edited: Mar 7, 2005
  8. Mar 7, 2005 #7
    damn i'm stupid at this :cry: :confused:

    xanthym how would i go about doing that?? I know that the function is postitive over the interval (a,b) but what does taht say about the endpoints?/
  9. Mar 7, 2005 #8
    i think this problem is in Spivak 3rd edition. Its perhaps knowns as the best and hardest calculus books, specially analysis. I dunno if its helpful but u gota have some resources when doing this course.
    Neways, i remember somestuff when i did this problem.
    Let [c,d] be an interval in [a,b] with f(x)>f(c)/2 for all x in [c,d].
    Now you should be able to get an idea of what the proof should come out too, and look carefully at this partition P={a,c,d,b}.
  10. Mar 8, 2005 #9
    yes i have come across Spivak. It is a very good book. Too bad taht we used Salas Hille Etgen One variable calculus in my course which was totally (TOTALLY) useless because that book has nothing to do with analysis, whatsoever

    but tyhanks for the hint it really helps
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