Equality in the Cauchy-Schwarz inequality for integrals

[Axiomer]

Homework Statement

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

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 0=\int_a^b (f-λg)^2 for some λ\inℝ 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: 0<\int_a^b (f-λg)^2 for all λ\inℝ
\Rightarrow 0<\int_a^b f^2 - 2λ\int_a^b fg + λ^2\int_a^b g^2 for all λ\inℝ
this is a quadratic equation in λ with no real roots, hence the discriminant is < 0:
(2\int_a^b fg)^2 - 4\int_a^b g^2\int_a^b f^2&lt;0 \Rightarrow |\int_a^b fg|&lt;(\int_a^b f^2)^\frac{1}{2}(\int_a^b g^2)^\frac{1}{2}
\squareCase 2: 0=\int_a^b (f-λg)^2 for some λ\inℝ
\Rightarrow 0=\int_a^b f^2 - 2λ\int_a^b fg + λ^2\int_a^b g^2
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:
(\int_a^b fg)^2≥(\int_a^b f^2)(\int_a^b g^2)

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

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

We prove by contradiction that 0 &lt; \int_a^b (f-λk_ε)^2 for all λ\inℝ:
Suppose 0 = \int_a^b (f-λk_ε)^2 for some λ\inℝ. This has at most 2 roots. We have:
\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.
Such that any roots must be λ₁ or λ₂. Without loss of generality, suppose λ₁ is a root. Then:
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 (the first term is 0) = \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 (the first two terms are 0) =\int_\frac{a+b}{2}^b (λ_1ε)^2
\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) a contradiction!

So 0 &lt; \int_a^b (f-λk_ε)^2 for all λ\inℝ \Rightarrow (\int_a^b fk_ε)^2&lt;(\int_a^b f^2)(\int_a^b k_ε^2) by case 1. We rewrite this to get:
(\int_a^b fg)^2 &lt; (\int_a^b f^2)(\int_a^b g^2) + ε*N + ε^2*M for some N,M\inℝ
But we can take ε = min(1,\frac{δ}{2|N|},\frac{δ}{2|M|}) such that εN≤ε|N|≤\frac{1}{2}δ and ε^2M≤ε^2|M|≤ε|M|≤\frac{1}{2}δ
But then (\int_a^b fg)^2&lt;(\int_a^b f^2)(\int_a^b g^2) + δ
a contradiction! \blacksquare
Is there a simpler way to do this?

 

[Axiomer]

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!

 

[Karamata]

Is there a simpler way to do this?

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

0 \le &lt;x+\lambda y, x+\lambda y&gt; = &lt;x,x&gt; + 2\lambda &lt;x,y&gt; + \lambda^2 &lt;y,y&gt;, and then choosing that \lambda = -\dfrac{&lt;x,y&gt;}{&lt;y,y&gt;} you will get |&lt;x,y&gt;|^2 \le &lt;x,x&gt; &lt;y,y&gt;

So, maybe, but maybe, you can use \lambda = -\dfrac{&lt;f,g&gt;}{&lt;g,g&gt;} = -\dfrac{\int_a^b fg}{\int_a^b g^2}

 

[Axiomer]

Yes, that definitely works too. Thanks!