Self studying little Spivak's, stuck on Schwartz ineq. for integrals

[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:
<br /> \int (f-\lambda g)^{2}&gt;0<br />
Then:
<br /> \lambda^{2}\int g^{2}-2\lambda\int fg +\int f^{2}&gt;0<br />
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:
<br /> \int (f-\lambda g)^{2}=0<br />
The statement means that (f-\lambda g)^{2}=0, so what does that say about f and g?

 

[SrEstroncio]

For the case:
<br /> \int (f-\lambda g)^{2}=0<br />
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
<br /> \int (f-\lambda g)^{2}=0<br />
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
<br /> \int (f-\lambda g)^{2}=0,<br />
then
<br /> {\lambda}^2 \int g^2 -2\lambda \int fg + \int f^2 =0.<br />
Solving for \lambda I get
<br /> \lambda = \frac{2\int fg \pm \sqrt{4{\int fg}^2 -4(\int f^2)(\int g^2)}{2\int g^2}.<br />
Now if the discriminant of the above equation is equal to zero we obtain
<br /> {\int fg}^2 -(\int f^2)(\int g^2) =0<br />
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
<br /> {\int fg}^2 -(\int f^2)(\int g^2) \geq 0<br />
and we would conclude that
<br /> {\int fg}^2 \geq (\int f^2)(\int g^2),<br />
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.