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

  • #1

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

Homework Equations



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


The Attempt at a Solution



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

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,745
26
Suppose that:
[tex]
\int (f-\lambda g)^{2}>0
[/tex]
Then:
[tex]
\lambda^{2}\int g^{2}-2\lambda\int fg +\int f^{2}>0
[/tex]
Consider this as a quadratic in [itex]\lambda[/itex], 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:
[tex]
\int (f-\lambda g)^{2}=0
[/tex]
The statement means that [itex](f-\lambda g)^{2}=0[/itex], so what does that say about f and g?
 
  • #3
For the case:
[tex]
\int (f-\lambda g)^{2}=0
[/tex]
The statement means that [itex](f-\lambda g)^{2}=0[/itex], so what does that say about f and g?

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

Maybe I should elaborate on my question. Suppose
[tex]
\int (f-\lambda g)^{2}=0,
[/tex]
then
[tex]
{\lambda}^2 \int g^2 -2\lambda \int fg + \int f^2 =0.
[/tex]
Solving for [itex] \lambda [/itex] I get
[tex]
\lambda = \frac{2\int fg \pm \sqrt{4{\int fg}^2 -4(\int f^2)(\int g^2)}{2\int g^2}.
[/tex]
Now if the discriminant of the above equation is equal to zero we obtain
[tex]
{\int fg}^2 -(\int f^2)(\int g^2) =0
[/tex]
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 [itex] \Delta \geq 0 [/itex], then we would have
[tex]
{\int fg}^2 -(\int f^2)(\int g^2) \geq 0
[/tex]
and we would conclude that
[tex]
{\int fg}^2 \geq (\int f^2)(\int g^2),
[/tex]
which is nonsense. How do I prove the equation has only one solution so that the above explained does not happen?
 
Last edited:
  • #4
hunt_mat
Homework Helper
1,745
26
What type of integral are you using? Riemann or Lebesgue? The answer may be different depending on what you choose.
 

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

Replies
1
Views
2K
  • Last Post
Replies
16
Views
7K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
10
Views
4K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
11
Views
1K
Top