2
$\begingroup$

Proving the discrete version of Cauchy-Schwarz is easy:

$ \left(\sum_{i} a_i^2\right) \left(\sum_i b_i^2\right) \geq \left(\sum_i a_ib_i\right)^2 $

can be done via the determinant of the quadratic formula.

Now, however, I want to prove the continuous version, which states:

$ \int a^2 \int b^2 \geq \left(\int ab\right)^2$

How do I prove this?

  • 0
    @user1311390 Yes, clever. Try it with $c = \frac{\int ab}{\int b^2}$ (assuming that \int b^2 > 0), and work out a separate proof for the case when $\int b^2 = 0$.2012-08-18

1 Answers 1

3

First note that $ab \leq \frac{a^2}{2} + \frac{b^2}{2}$. Then take $a = \frac{f}{(\int{f^2})^{\frac{1}{2}}}, b = \frac{g}{(\int{g^2})^{\frac{1}{2}}}$.

  • 0
    I like this solution. It's much better than what I ended up with (expressing the integrals as reinmen sums, and arguing that the approximations are within epsilon)2012-08-18