Uniqueness of λ in Schwarz inequality proof from Spivak

[SrEstroncio]

Homework Statement

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 $f$ and $g$ be integrable functions on $[a,b]$.
Prove that $|\int_a^b fg | \leq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2}$.

Homework Equations

He suggests that you treat the cases $0=\int_a^b (f-\lambda g)^2$ for some $\lambda \in R$ and $0 \lt \int_a^b (f-\lambda g)^2$ for all $\lambda$ separately.

The Attempt at a Solution

My question is: how do I know the $\lambda$ is unique?
Considering the two cases given above I got a cuadratic expression in $\lambda$ whose discriminant gave me the strict inequality when $$0 \lt \int_a^b (f-\lambda g)^2$$ for all $\lambda$ (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 $$|\int_a^b fg | \geq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2}$$, which is obviously wrong), meaning that there is only one root of the equation, or equivalently that the lambda that satisfies $0=\int_a^b (f-\lambda g)^2$ 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 $$0=\int_a^b (f)^2$$ implies $$f=0$$.

 

[hunt_mat]

Suppose that:
$$\int (f-\lambda g)^{2}>0$$
Then:
$$\lambda^{2}\int g^{2}-2\lambda\int fg +\int f^{2}>0$$
Consider this as a quadratic in $\lambda$, what does this statement say? It says that the entire quadratic lies above the x-axis which in turn implies that there are no real roots which is turn puts a condition on the discriminant of this quadratic.

For the case:
$$\int (f-\lambda g)^{2}=0$$
The statement means that $(f-\lambda g)^{2}=0$, so what does that say about f and g?

 

[SrEstroncio]

For the case:
$$\int (f-\lambda g)^{2}=0$$
The statement means that $(f-\lambda g)^{2}=0$, so what does that say about f and g?

We do not know if $f$ and $g$ are continuous, we only assume them to be integrable, so it is not necessarily true that
$$\int (f-\lambda g)^{2}=0$$
implies $(f-\lambda g)^{2}=0$, since $f-\lambda g$ could be zero except at an isolated number of points (it's integral would still be zero but the function won't be zero everywhere).

Maybe I should elaborate on my question. Suppose
$$\int (f-\lambda g)^{2}=0,$$
then
$${\lambda}^2 \int g^2 -2\lambda \int fg + \int f^2 =0.$$
Solving for $\lambda$ I get
$$\lambda = \frac{2\int fg \pm \sqrt{4{\int fg}^2 -4(\int f^2)(\int g^2)}{2\int g^2}.$$
Now if the discriminant of the above equation is equal to zero we obtain
$${\int fg}^2 -(\int f^2)(\int g^2) =0$$
from which we obtain the equality part of the problem. But how do I know there's exactly one lambda that satisfies the equation? What if the equation had two real solutions? so that $\Delta \geq 0$, then we would have
$${\int fg}^2 -(\int f^2)(\int g^2) \geq 0$$
and we would conclude that
$${\int fg}^2 \geq (\int f^2)(\int g^2),$$
which is nonsense. How do I prove the equation has only one solution so that the above explained does not happen?

 

[hunt_mat]

What type of integral are you using? Riemann or Lebesgue? The answer may be different depending on what you choose.