1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    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]

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