# Cauchy Schwartz Inequality Proof

I found a lot of applications of the Cauchy Schwartz inequality but no proofs, any help will be greatly appreciated!

Prove that

$(\int^{b}_{a} fg)^2 \le \int^{b}_{a} (g)^2.\int^{b}_{a} (f)^2$

We get a quadratic in $\;t\;$ :

$$0\le\int(g(x)t-f(x))^2dx=\left(\int g^2\right)t^2-\left(2\int f g\right)t+\left(\int f^2\right)$$

Since the above is always non-negative its discriminant must be non-positive:

$$\Delta:=4\left(\int fg\right)^2-4\left(\int f^2\right)\left(\int g^2\right)\le0\;\;\ldots$$

HINT:

Let $I(t)$ be given by

$$I(t)=\int_a^b (f(x)-tg(x))^2\,dx$$

Note that the derivative of $I(t)$ is

$$I'(t)=2\int_a^b g(x)(tg(x)-f(x))\,dx$$

and $I'(t)=0$ when $\displaystyle t=\frac{\int_a^b f(x)g(x)\,dx}{\int_a^b g^2(x)\,dx}$.

Finally, note that $I''(t)\ge 0$.

• Are you sure $t$ is the integration variable? – Bernard Nov 4 '16 at 21:03
• @Bungo Thank you for alerting. I've added the definition of $I(t)$ and corrected the erratum. Much appreciative!! -Mark – Mark Viola Nov 4 '16 at 22:19

Hint:

Consider $I(t)$ as a quadratic polynomial in $t$. When does a quadatic polynomial have a constant sign?

By other means: $$0\le\Bigl\|\,\|v\|\,u\pm \|u\|\,v\,\Bigr\|^2=2\|u\|\,\|v\|\,\Bigl(\|u\|\,\|v\|\pm\langle u,v\rangle\Bigr)$$ leads directly to $$|\langle u,v\rangle|\le \|u\|\,\|v\|$$