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Prove that f = 0 integral

  1. Mar 24, 2006 #1
    "Suppose f is continuous on [a,b] and for all k = 0,1,2,3,4,5,... [tex]\int_a^b x^kf(x) = 0[/tex]. Prove that f = 0."

    What I know so far:

    f(c)=0 for some c in [a,b]

    [tex]\int_a^b f=0[/tex]

    [tex]\int_a^b x^k \int_a^b f=0[/tex]

    Any hint on how to proceed would be appreciated.
     
  2. jcsd
  3. Mar 24, 2006 #2

    AKG

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    Use Weierstrass approximation, and consider [itex]\int f^2[/itex].
     
  4. Mar 25, 2006 #3
    I can't see the solution. Can you elaborate a little more?
     
    Last edited: Mar 25, 2006
  5. Mar 25, 2006 #4

    AKG

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    First, do you know the Weierstrass approximation theorem? Next, if p is any polynomial, what is [itex]\int pf[/itex]? Putting those two facts together, what can you conclude about [itex]\int f^2[/itex]?
     
  6. Mar 25, 2006 #5
    Weierstrass approximation theorem tells me that f(x) can be uniformly approximated by polynomials. So give e>0, for some p(x), |p(x)-f(x)|<e for all x. x^k * f(x) is apprximately x^k * p(x). But I don't understand how this has to do with f^2.
     
  7. Mar 25, 2006 #6

    AKG

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    f is bounded, so there exists M such that M > |f(x)| for all x in [a,b]. For all e > 0, there exists a polynomial p(x) such that |p(x) - f(x)| < e/M for all x. Thus, for such a polynomial, |f(x)p(x) - f²(x)| < e

    [tex]|\int f^2| = |\int f^2 + 0| = |\int f^2 - \int fp| = |\int f^2 - fp| \leq \int |fp - f^2| \leq (b-a)e[/tex]

    Since we can make e arbitarily small, [itex]|\int f^2| = 0[/itex], but f² is non-negative continuous, so f² is zero, hence so is f.
     
  8. Mar 26, 2006 #7
    At no point did you use the hypothesis of 'x^k for all k=0,1,2,...' . This would imply that if f is continuous on [a,b] and [tex]\int_a^b f=0[/tex], then f=0.
     
  9. Mar 26, 2006 #8

    shmoe

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    Try to justify each of AKG's steps carefully.
     
  10. Mar 26, 2006 #9
    I understand now. Thanks for the help.
     
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