Cauchy-Schwarz inequality for Integrals

• January 16th 2009, 12:24 PM
Cauchy-Schwarz inequality for Integrals
Suppose that the functions $f,g,f^2,g^2, fg$ are integrable on the closed, bounded interval [a,b]. Prove that:

$\int ^b _a fg \leq \sqrt { \int ^b_a f^2 } \sqrt { \int ^b_a g^2 }$.

Proof.

Now, $0 \leq \sqrt { \int _a ^b (f- \lambda g)(f- \lambda g) } ^2$ $= \int ^b_a (f^2- \lambda fg - \lambda fg + \lambda ^2 g^2 )$
$= \int ^b_a f^2 - \lambda \int ^b _a fg - \lambda \int ^b _a fg + \lambda ^2 \int ^b _a g^2$

Choose $\lambda = \int _a ^b fg [ \int ^b _a g^ {2}]^{-1}$

then the previous equation becomes $\int ^b _a f^2 - \int ^b _a (fg)^2 [ \int ^b _a g^2 ]^{-1} - \int ^b _a (fg)^2 [ \int ^b _a g^2 ] ^{-1} + \int ^b _a (fg)^2$
which implies $0 \leq \int ^b _a f^2 - \int ^b _a (fg)^2 (2 [ \int ^b _a g^2 ]^{-1})$
$\int ^b _a (fg)^2 (2 [ \int ^b _a g^2 ] ^{-1} \leq \int ^b _a f^2$
$\int ^b _a (fg)^2 \leq \frac { \int ^b _a f^2 \int ^b _a g^2 } {2}$

How do I get rid of the 2 at the bottom?

Thanks.
• January 16th 2009, 12:50 PM
ThePerfectHacker
The Cauchy-Schwarz inequality is a general inequality for inner product spaces.
Thus, what you wrote is a consequence of this general inequality found here*.

*)I am just a little worried because the space of all integral functions is not an inner product space with norm $\left< f,g\right> = \smallint_a^b fg$.
However, the proof on that page seems to work.
• January 16th 2009, 11:34 PM
Moo
Hello,

A more general view on this inequality is Hölder's inequality.
Cauchy-Schwarz is then a specific situation, with p=q=2.

Hölder's inequality - Wikipedia, the free encyclopedia