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Self studying little Spivak's, stuck on Schwartz ineq. for integrals

  1. Jun 22, 2011 #1
    1. The problem statement, all variables and given/known data

    In an effort to keep me from spending all summer lying on the couch, I recently started reading Michael Spivak's Calculus on Manifolds; while working on problem 1-6 I got stuck on a technical detail and I was wondering if anyone could provide a little insight.

    Problem 1-6 says:
    Let [itex] f [/itex] and [itex] g [/itex] be integrable functions on [itex] [a,b] [/itex].
    Prove that [itex] |\int_a^b fg | \leq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2} [/itex].

    2. Relevant equations

    He suggests that you treat the cases [itex] 0=\int_a^b (f-\lambda g)^2 [/itex] for some [itex] \lambda \in R [/itex] and [itex] 0 \lt \int_a^b (f-\lambda g)^2 [/itex] for all [itex] \lambda [/itex] separately.


    3. The attempt at a solution

    My question is: how do I know the [itex] \lambda [/itex] is unique?
    Considering the two cases given above I got a cuadratic expression in [itex] \lambda [/itex] whose discriminant gave me the strict inequality when [tex] 0 \lt \int_a^b (f-\lambda g)^2 [/tex] for all [itex] \lambda [/itex] (since there are no real roots of the equation), but in order to conclude that [\tex] |\int_a^b fg | \leq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2} [/tex] I am forced to assume that the discriminant of the equation is equal to zero (otherwise I get [tex] |\int_a^b fg | \geq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2} [/tex], which is obviously wrong), meaning that there is only one root of the equation, or equivalently that the lambda that satisfies [itex] 0=\int_a^b (f-\lambda g)^2 [/itex] is unique, fact that I feel must be proven, not assumed).

    How do I know said lambda is unique? Keep in mind that since f and g are integrable (but may not be continuous) one cannot assume that [tex] 0=\int_a^b (f)^2 [/tex] implies [tex] f=0 [/tex].
     
    Last edited: Jun 22, 2011
  2. jcsd
  3. Jun 22, 2011 #2

    hunt_mat

    User Avatar
    Homework Helper

    Suppose that:
    [tex]
    \int (f-\lambda g)^{2}>0
    [/tex]
    Then:
    [tex]
    \lambda^{2}\int g^{2}-2\lambda\int fg +\int f^{2}>0
    [/tex]
    Consider this as a quadratic in [itex]\lambda[/itex], what does this statement say? It says that the entire quadratic lies above the x-axis which in turn implies that there are no real roots which is turn puts a condition on the discriminant of this quadratic.

    For the case:
    [tex]
    \int (f-\lambda g)^{2}=0
    [/tex]
    The statement means that [itex](f-\lambda g)^{2}=0[/itex], so what does that say about f and g?
     
  4. Jun 22, 2011 #3
    We do not know if [itex] f [/itex] and [itex] g [/itex] are continuous, we only assume them to be integrable, so it is not necessarily true that
    [tex]
    \int (f-\lambda g)^{2}=0
    [/tex]
    implies [itex](f-\lambda g)^{2}=0[/itex], since [itex] f-\lambda g [/itex] could be zero except at an isolated number of points (it's integral would still be zero but the function wont be zero everywhere).

    Maybe I should elaborate on my question. Suppose
    [tex]
    \int (f-\lambda g)^{2}=0,
    [/tex]
    then
    [tex]
    {\lambda}^2 \int g^2 -2\lambda \int fg + \int f^2 =0.
    [/tex]
    Solving for [itex] \lambda [/itex] I get
    [tex]
    \lambda = \frac{2\int fg \pm \sqrt{4{\int fg}^2 -4(\int f^2)(\int g^2)}{2\int g^2}.
    [/tex]
    Now if the discriminant of the above equation is equal to zero we obtain
    [tex]
    {\int fg}^2 -(\int f^2)(\int g^2) =0
    [/tex]
    from which we obtain the equality part of the problem. But how do I know there's exactly one lambda that satisfies the equation? What if the equation had two real solutions? so that [itex] \Delta \geq 0 [/itex], then we would have
    [tex]
    {\int fg}^2 -(\int f^2)(\int g^2) \geq 0
    [/tex]
    and we would conclude that
    [tex]
    {\int fg}^2 \geq (\int f^2)(\int g^2),
    [/tex]
    which is nonsense. How do I prove the equation has only one solution so that the above explained does not happen?
     
    Last edited: Jun 22, 2011
  5. Jun 23, 2011 #4

    hunt_mat

    User Avatar
    Homework Helper

    What type of integral are you using? Riemann or Lebesgue? The answer may be different depending on what you choose.
     
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