# Self studying little Spivak's, stuck on problem 1-6

1. Jun 22, 2011

### SrEstroncio

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 $$|\int_a^b fg | \leq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2}$$.
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.
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 $$f=0$$.

Last edited: Jun 22, 2011
2. Jun 22, 2011

### micromass

Hi SrEstroncio!

Let's first consider an arbitrary quadratic function. Take

$$ax^2+bx+c$$

Now, if this equation has two distinct roots, then the equation must become negative somewhere (either in between the roots or outside the roots). So we know that

$$ax^2+bx+c<0$$

for some x. So the only way that this function is always greater than zero is that the root is unique.

Now, you have a quadratic function in $\lambda$:

$$\int_a^b(f-\lambda g)^2$$

and you know that this function is always greater than zero. The only way to accomplish this is if the function has only 1 or no distinct real roots. So this is why the lambda is unique.

3. Jun 22, 2011

### SrEstroncio

ok ok ok I get it now, thank you very much ;D