- #1
maverick280857
- 1,789
- 5
Hi...I've been scratching my head for this one:
If f(x) and g(x) are realvalued functions integrable over an interval (a,b) then prove that
[tex]|\int_{a}^{b} f(x)g(x)dx| \leq \sqrt{\int_{a}^{b}(f(x))^2dx \int_{a}^{b}(g(x))^2dx}[/tex]
I actually don't want the proof of this inequality...I already have it (take F(x) as (f(x)- lambda g(x))^2=>integral of F(x) over (a,b) is positive as is the integrand. Then, we use properties of the quadratic function and get the desired inequality using the fact that the discriminant of the resulting quadratic must not exceed 0). However, an alternative proof would be appreciated.
But what I really want is the physical interpretation of this inequality because that is hard to come up with. The inequality reminds me of the Cauchy Schwarz Inequality and incidentally, its proof is quite similar to Cauchy Schwarz.
I would be very grateful if you could offer an explanation.
Thanks and cheers
Vivek
If f(x) and g(x) are realvalued functions integrable over an interval (a,b) then prove that
[tex]|\int_{a}^{b} f(x)g(x)dx| \leq \sqrt{\int_{a}^{b}(f(x))^2dx \int_{a}^{b}(g(x))^2dx}[/tex]
I actually don't want the proof of this inequality...I already have it (take F(x) as (f(x)- lambda g(x))^2=>integral of F(x) over (a,b) is positive as is the integrand. Then, we use properties of the quadratic function and get the desired inequality using the fact that the discriminant of the resulting quadratic must not exceed 0). However, an alternative proof would be appreciated.
But what I really want is the physical interpretation of this inequality because that is hard to come up with. The inequality reminds me of the Cauchy Schwarz Inequality and incidentally, its proof is quite similar to Cauchy Schwarz.
I would be very grateful if you could offer an explanation.
Thanks and cheers
Vivek