- #1
EquationOfMotion
- 22
- 2
I'm trying to prove that the set of all square integrable functions f(x) for which ∫ab |f(x)|^2 dx is finite is a vector space. Everything but the proof of closure is trivial.
To prove closure, obviously we should expand out |f(x)+g(x)|^2, which turns our integral into one of |f(x)|^2 (finite), |g(x)|^2 (finite), and the two cross terms f*(x)g(x) and f(x)g*(x).
This question has been posted before, and it's resolved by using the Cauchy Schwarz inequality. Since <f|g> ≤ √(<f|f><g|g>) the two cross terms will be finite and we are done.
But why are we allowed to use the Cauchy Schwarz inequality in the first place? Isn't the inequality only true given that we are in an inner product (and thus vector) space? So since I'm trying to prove that what I have is a vector space, it seems using the inequality is illogical.
The way I tried to work around this problem is by arguing that f(x) and g(x) belong to the vector space of all functions, and my inner product is defined as ∫ab f*(x)g(x) dx. But then the inner product could be infinite! Which doesn't make much sense.
To prove closure, obviously we should expand out |f(x)+g(x)|^2, which turns our integral into one of |f(x)|^2 (finite), |g(x)|^2 (finite), and the two cross terms f*(x)g(x) and f(x)g*(x).
This question has been posted before, and it's resolved by using the Cauchy Schwarz inequality. Since <f|g> ≤ √(<f|f><g|g>) the two cross terms will be finite and we are done.
But why are we allowed to use the Cauchy Schwarz inequality in the first place? Isn't the inequality only true given that we are in an inner product (and thus vector) space? So since I'm trying to prove that what I have is a vector space, it seems using the inequality is illogical.
The way I tried to work around this problem is by arguing that f(x) and g(x) belong to the vector space of all functions, and my inner product is defined as ∫ab f*(x)g(x) dx. But then the inner product could be infinite! Which doesn't make much sense.