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## Homework Statement

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]

## Homework Equations

## 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 λ

_{1}and λ

_{2}. 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 λ

_{1}and λ

_{2}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 λ

_{1}or λ

_{2}. Without loss of generality, suppose λ

_{1}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?