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Self studying little Spivak's, stuck on problem 1-6

  1. Jun 22, 2011 #1
    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 [tex] f [/tex] and [tex] g [/tex] be integrable functions on [tex] [a,b] [/tex].
    Prove that [tex] |\int_a^b fg | \leq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2} [/tex].
    He suggests that you treat the cases [tex] 0=\int_a^b (f-\lambda g)^2 [/tex] for some [tex] \lambda \in R [/tex] and [tex] 0 \lt \int_a^b (f-\lambda g)^2 [/tex] for all [tex] \lambda [/tex] separately.
    My question is: how do I know the [tex] \lambda [/tex] is unique?
    Considering the two cases given above I got a cuadratic expression in [tex] \lambda [/tex] whose discriminant gave me the strict inequality when [tex] 0 \lt \int_a^b (f-\lambda g)^2 [/tex] for all [tex] \lambda [/tex] (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 [tex] 0=\int_a^b (f-\lambda g)^2 [/tex] 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

    micromass

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    Hi SrEstroncio! :smile:

    Let's first consider an arbitrary quadratic function. Take

    [tex]ax^2+bx+c[/tex]

    Now, if this equation has two distinct roots, then the equation must become negative somewhere (either in between the roots or outside the roots). So we know that

    [tex]ax^2+bx+c<0[/tex]

    for some x. So the only way that this function is always greater than zero is that the root is unique.

    Now, you have a quadratic function in [itex]\lambda[/itex]:

    [tex]\int_a^b(f-\lambda g)^2[/tex]

    and you know that this function is always greater than zero. The only way to accomplish this is if the function has only 1 or no distinct real roots. So this is why the lambda is unique.
     
  4. Jun 22, 2011 #3
    ok ok ok I get it now, thank you very much ;D
     
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