Is the Integral of (f(x)-q(g(x)))^2 Greater Than Zero?

[mruncleramos]

Lets say we have two functions f and g that are riemann integrableon the interval [a,b] If (f(x)-q(g(x)))^2 is greater than zero for all real q, is the integral from a to b of (f(x)-q(g(x)))^2 greater than zero? Also, let's say we have a function h(x) that is defined from [0,1] as follows: f(x) is zero if x is not 0.5 and if x is 0.5, then f(x) is a number, let's say 15. What is the integral from a to b of this function? If it is zero, we can continue to add points. When does this process yield a nonzero integral?

 

[philosophking]

For the first question, just think of what such a function would look like. If a function is greater than zero for all real q, then what does this say about the area.

For the second function, do you know Lebesgue's condition for integrability? This sounds like an analysis-type class, so you might have covered it. Basically, Lebesgue's condition would tell you that such an integral exists and is equal to the lower/upper sum (here, the lower sum is easier, since every neighborhood of a point on [a,b] contains an x such that f(x)=0).

I think the answer to your final question would follow from just a little more thought on what I said above.

I'm sorry if you don't know Lebesgue's condition for integrability. I'm not trying to pull some higher math on you-- I did learn that last semester in Real Analysis I.

 

[mruncleramos]

Just for contextualization. I just finished Calculus on Manifolds, and I'm coming back to some of the problems that i thought could have had better solutions to. Problem 6 of chapter one asks you to prove the schwarz ineqaulity for integrals. Spivak inserts a cryptic (in my opinion) hint: Consider seperately the cases $\int_{a}^{b} f(x) - \lambda g(x) dx$ = 0 and $\int_{a}^{b} f(x) - \lambda g(x) dx$ > 0


I suppose this is supposed to include all possibilites of f and g. So far, I've used the latter inequality to derive part of the schwarz inequality, but i cannot find a use for the former. Oh yeah. Excuse me for being picky, but i have to figure out what spivak meant by this hint. I don't want to use riemann sums.

 

[Galileo]

Just for contextualization. I just finished Calculus on Manifolds, and I'm coming back to some of the problems that i thought could have had better solutions to. Problem 6 of chapter one asks you to prove the schwarz ineqaulity for integrals. Spivak inserts a cryptic (in my opinion) hint: Consider seperately the cases $\int_{a}^{b} f(x) - \lambda g(x) dx$ = 0 and $\int_{a}^{b} f(x) - \lambda g(x) dx$ > 0


I suppose this is supposed to include all possibilites of f and g. So far, I've used the latter inequality to derive part of the schwarz inequality, but i cannot find a use for the former. Oh yeah. Excuse me for being picky, but i have to figure out what spivak meant by this hint. I don't want to use riemann sums.

I have the book as well. The LateX seems somewhat mixed up, so I`ll just copy the problem:
Spivak asks to prove that:
\left| \int_a^b f \cdot g \right| \leq (\int_a^b f^2)^{\frac{1}{2}}\cdot (\int_a^b g^2)^{\frac{1}{2}}
for integrable functions f,g on [a,b],
and hints to consider seperately the cases:

0 = \int_a^b (f-\lambda g)^2
for some \lambda \in \mathbb{R} and
0 < \int_a^b (f-\lambda g)^2
for all \lambda \in \mathbb{R}.

This covers all cases, since for any two integrable functions f,g on [a,b] there either exists a \lambda such that the integral is zero, or for all \lambda the integral is not equal to zero. (The proof that is must be greater than zero is actually the point of this thread).

You cannot mimic Theorem 1-1 (2)exactly, since he uses the argument that x and y are either linearly dependent or not and uses the fact that if they are linearly independent, then \lambda y-x \neq 0 for all \lambda and so 0<|\lambda y-x|^2 for all \lambda (by theorem 1-1 (1)).
You cannot use this argument for the integral, because even if f and g are linearly independent, it is not necessarily the case that:

\int_a^b (f-\lambda g)^2 > 0 for all \lambda. (||f||=(\int_a^b f^2)^{\frac{1}{2}} does not constitute a norm on the given space).