# How to prove Cauchy-Schwarz integral inequality?

The Cauchy-Schwarz integral inequality is as follows:

$$\displaystyle \left({\int_a^b f \left({t}\right) g \left({t}\right) \ \mathrm d t}\right)^2 \le \int_a^b \left({f \left({t}\right)}\right)^2 \mathrm d t \int_a^b \left({g \left({t}\right)}\right)^2 \mathrm d t$$

How do I prove this using multivariable calculus methods, preferably with double integrals?

• This isn't what you asked for, but to me it seems most clear to prove Cauchy-Schwarz in an abstract inner product space (guided by our intuition from $\mathbb R^n$), then get this inequality as a special case. – littleO Jan 3 '15 at 3:14

## 3 Answers

If by double-integral you mean the identity: $$\frac{1}{2}\int_a^b\int_a^b (f(x)g(y) - g(x)f(y))^2\,dx\,dy \\= \int_a^b f^2(x)\,dx\int_a^b g^2(x)\,dx - \left(\int_a^b f(x)g(x)\,dx\right)^2$$

Then note that the integrand $\displaystyle (f(x)g(y) - g(x)f(y))^2 \ge 0$, hence the inequality follows.

HINT: Use the polynomial function $$P(x) = \int_a^b \left(f(t) + xg(t)\right) ^2dt$$

And you will have two cases to prove : either $g = \Theta_{[a, b] \to \mathbb{R}}$ or not (the first case is pretty easy).

So we take $P(\lambda)=\int_a^b (|f|+\lambda|g|)^2$

Then $$P(\lambda)=\int_a^b |f|^2 +2\lambda\int_a^b |fg| +\lambda^2 \int_a^b |g|^2$$

P is a 2nd degre polynome and beacause he never cancel his discriminant is negative then we have the inegality ! If I don't make a mistake sure...

Shadock

• It's actually not always negative tho, that's why you need $\leq$ and not $<$ – servabat Jan 3 '15 at 3:06
• @servabat If I write "his discriminant is negative" is it right? – Hexacoordinate-C Jan 3 '15 at 3:08
• Well I guess yes. I usually say negative or zero to be more clear for $\leq 0$, but that's just some habit. – servabat Jan 3 '15 at 3:09
• @servabat I agree but by definition ... :) – Hexacoordinate-C Jan 3 '15 at 3:11
• Yes, you are right. – servabat Jan 3 '15 at 3:11