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Equality in the Cauchy-Schwarz inequality for integrals

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Regarding problem 1-6 in Spivak's Calculus on Manifolds: Let [itex]f[/itex] and [itex]g[/itex] be integrable on [itex][a,b][/itex]. Prove that [itex]|\int_a^b fg| ≤ (\int_a^b f^2)^\frac{1}{2}(\int_a^b g^2)^\frac{1}{2}[/itex]. Hint: Consider seperately the cases [itex]0=\int_a^b (f-λg)^2[/itex] for some [itex]λ\inℝ[/itex] and [itex]0 < \int_a^b (f-λg)^2[/itex] for all [itex]λ\inℝ[/itex]

    2. Relevant equations

    3. The attempt at a solution
    I can prove the inequality using Riemann sums and the regular Cauchy-Schwarz inequality, however I didn't see a way to prove that equality holds iff [itex]0=\int_a^b (f-λg)^2[/itex] for some [itex]λ\inℝ[/itex] using this method. Using the hint gave me a bit of trouble, I think i'm doing something wrong/there's an easier way to do it:

    Case 1: [itex]0<\int_a^b (f-λg)^2[/itex] for all [itex]λ\inℝ[/itex]
    [itex]\Rightarrow 0<\int_a^b f^2 - 2λ\int_a^b fg + λ^2\int_a^b g^2 [/itex] for all [itex]λ\inℝ[/itex]
    this is a quadratic equation in λ with no real roots, hence the discriminant is < 0:
    [itex](2\int_a^b fg)^2 - 4\int_a^b g^2\int_a^b f^2<0 \Rightarrow |\int_a^b fg|<(\int_a^b f^2)^\frac{1}{2}(\int_a^b g^2)^\frac{1}{2}[/itex]
    [itex]\square[/itex]


    Case 2: [itex]0=\int_a^b (f-λg)^2[/itex] for some [itex]λ\inℝ[/itex]
    [itex]\Rightarrow 0=\int_a^b f^2 - 2λ\int_a^b fg + λ^2\int_a^b g^2[/itex]
    This is a quadratic equation in λ (otherwise we can show easily that the result holds) with a real root, hence the discriminant is ≥ 0 and we proceed as before to get:
    [itex](\int_a^b fg)^2≥(\int_a^b f^2)(\int_a^b g^2)[/itex]

    We prove this case by contradiction. Suppose that [itex](\int_a^b fg)^2>(\int_a^b f^2)(\int_a^b g^2)[/itex] such that [itex](\int_a^b fg)^2=(\int_a^b f^2)(\int_a^b g^2) + δ[/itex] for some δ>0. Then there are exactly two roots λ1 and λ2. It follows that at least one of [itex]\int_a^\frac{a+b}{2} (f-λg)^2[/itex] or [itex]\int_\frac{a+b}{2}^b (f-λg)^2[/itex] has only λ1 and λ2 as roots. Suppose that it is [itex]\int_\frac{a+b}{2}^b (f-λg)^2[/itex], with the argument being similar otherwise.

    Consider the function [tex]k_ε = \left\{\begin{matrix}
    g& &on\: [a,\frac{a+b}{2}) \\
    g+ε& & on\: [\frac{a+b}{2},b]
    \end{matrix}\right.[/tex]

    We prove by contradiction that [itex]0 < \int_a^b (f-λk_ε)^2[/itex] for all [itex]λ\inℝ[/itex]:
    Suppose [itex]0 = \int_a^b (f-λk_ε)^2[/itex] for some [itex]λ\inℝ[/itex]. This has at most 2 roots. We have:
    [itex]\int_a^b (f-λk_ε)^2 = \int_a^\frac{a+b}{2} (f-λg)^2 + \int_\frac{a+b}{2}^b (f-λ(g+ε))^2[/itex].
    Such that any roots must be λ1 or λ2. Without loss of generality, suppose λ1 is a root. Then:
    [itex]0 = \int_a^b (f-λ_1k_ε)^2 = \int_a^\frac{a+b}{2} (f-λ_1g)^2 + \int_\frac{a+b}{2}^b (f-λ_1(g+ε))^2[/itex] (the first term is 0) [itex]= \int_\frac{a+b}{2}^b (f-λ_1g)^2 - 2λ_1ε\int_\frac{a+b}{2}^b (f-λ_1g) + \int_\frac{a+b}{2}^b (λ_1ε)^2[/itex] (the first two terms are 0) [itex]=\int_\frac{a+b}{2}^b (λ_1ε)^2[/itex]
    [itex]\Rightarrow λ_1 = 0 \Rightarrow \int_a^b f^2 = 0 \Rightarrow \int_a^b fg = 0 \Rightarrow (\int_a^b fg)^2 = (\int_a^b f^2)(\int_a^b g^2)[/itex] a contradiction!

    So [itex]0 < \int_a^b (f-λk_ε)^2[/itex] for all [itex]λ\inℝ \Rightarrow (\int_a^b fk_ε)^2<(\int_a^b f^2)(\int_a^b k_ε^2) [/itex] by case 1. We rewrite this to get:
    [itex](\int_a^b fg)^2 < (\int_a^b f^2)(\int_a^b g^2) + ε*N + ε^2*M[/itex] for some [itex]N,M\inℝ[/itex]
    But we can take [itex] ε = min(1,\frac{δ}{2|N|},\frac{δ}{2|M|}) [/itex] such that [itex]εN≤ε|N|≤\frac{1}{2}δ[/itex] and [itex]ε^2M≤ε^2|M|≤ε|M|≤\frac{1}{2}δ[/itex]
    But then [itex](\int_a^b fg)^2<(\int_a^b f^2)(\int_a^b g^2) + δ[/itex]
    a contradiction! [itex]\blacksquare[/itex]
    Is there a simpler way to do this?
     
  2. jcsd
  3. Jun 26, 2012 #2
    Oh wait.. once I prove the inequality using Riemann sums i'd just have to use the discriminant argument to show that equality holds in the second case!
     
  4. Jun 26, 2012 #3


    Maybe you can prove it like standard proof of Cauchy–Schwarz inequality:

    [itex]0 \le <x+\lambda y, x+\lambda y> = <x,x> + 2\lambda <x,y> + \lambda^2 <y,y>[/itex], and then choosing that [itex]\lambda = -\dfrac{<x,y>}{<y,y>}[/itex] you will get [itex]|<x,y>|^2 \le <x,x> <y,y>[/itex]

    So, maybe, but maybe, you can use [itex]\lambda = -\dfrac{<f,g>}{<g,g>} = -\dfrac{\int_a^b fg}{\int_a^b g^2}[/itex]
     
  5. Jun 26, 2012 #4
    Yes, that definitely works too. Thanks!
     
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