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

• SrEstroncio
In summary, the conversation discusses problem 1-6 from Michael Spivak's Calculus on Manifolds, which asks to prove the inequality |\int_a^b fg | \leq (\int_a^b f^2)^{1/2}(\int_a^b g^2)^{1/2}. The speaker was stuck on how to prove the uniqueness of lambda in the equation 0=\int_a^b (f-\lambda g)^2 and sought clarification. The responder explains that the uniqueness can be proven by considering the quadratic function \int_a^b(f-\lambda g)^2 and the fact that it is always greater than zero. This shows that the lambda is unique.
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:
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.

micromass said:
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.

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

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The best way to approach self studying "little Spivak's" is to start with the basics. Make sure you have a strong understanding of the concepts and principles covered in the earlier chapters before moving on to more challenging problems. It can also be helpful to read through the solutions for the previous problems to gain a better understanding of the thought process behind them.

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