Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Hi SrEstroncio! :smile:

    Let's first consider an arbitrary quadratic function. Take


    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook