# Continuous version of Cauchy-Schwarz

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?

• Well, aren't integrals defined in terms of sums? – Pedro Tamaroff Aug 18 '12 at 4:01
• Try to use the fact that $\int (a-cb)^2 \geq 0$ for all real numbers $c$ and then choose a clever value for $c$. – Dilip Sarwate Aug 18 '12 at 4:03
• It's Cauchy--Schwarz not Schwartz. I fixed it for you. – Nate Eldredge Aug 18 '12 at 4:07
• @NateEldredge: Good catch, Thanks! – user36739 Aug 18 '12 at 4:10
• @DilipSarwate: I see, clever. – user36739 Aug 18 '12 at 4:11

## 1 Answer

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

• 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) – user36739 Aug 18 '12 at 4:09