# Thread: A problem involving integrable and continuous functions

1. ## A problem involving integrable and continuous functions

Suppose that the functions $h,g:[a,b]\mapto\mathbb{R}$ be continuous. Prove that $\int^b_{a}\leq\sqrt{\int^b_{a}h^2}\sqrt{\int^b_{a} g^2}$

Suppose that the functions $h,g:[a,b]\mapto\mathbb{R}$ be continuous. Prove that $\int^b_{a}\leq\sqrt{\int^b_{a}h^2}\sqrt{\int^b_{a} g^2}$
There's a problem here. What's that first integral?

-Dan

Suppose that the functions $h,g:[a,b]\mapto\mathbb{R}$ be continuous. Prove that $\int^b_{a}\leq\sqrt{\int^b_{a}h^2}\sqrt{\int^b_{a} g^2}$
Are you asking for a proof of the Cauchy-Schwartz inequality for continuous real functions on [a,b]?

[int(h(x) g(x) dx, a,b)]^2<=int([h(x)]^2 dx, a,b) int([g(x)]^2 dx, a,b)

RonL

4. I believe the way your prove it is by taking the limit of the Riemann sum. And since for all finite sums this inequality is true then it is true also for an infinite sum.*

*)If a_n<= s then lim a_n <= s if it exists.

5. Originally Posted by ThePerfectHacker
I believe the way your prove it is by taking the limit of the Riemann sum. And since for all finite sums this inequality is true then it is true also for an infinite sum.*

*)If a_n<= s then lim a_n <= s if it exists.
You don't have to, the proof works out of the box.

RonL

6. Originally Posted by CaptainBlack
You don't have to, the proof works out of the box.
But! If you accept the standard Cauhy-Swartzh inequality for inner product spaces in general then there is nothing to it. However, he might want to use the algebraic Cauchy-Swartzh inequality and apply it ot integration.

7. I don't know what is going on with the coding, did I do something wrong? Why isn't my inequality showing up as it is?