Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals

Theorem

Let $f$ and $g$ be real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

Then:

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

Proof

 $\displaystyle \forall x \in \R: \ \$ $\displaystyle 0$ $\le$ $\displaystyle \left({x f \left({t}\right) + g \left({t}\right)}\right)^2$ $\quad$ $\quad$ $\displaystyle 0$ $\le$ $\displaystyle \int_a^b \left({x f \left({t}\right) + g \left({t}\right)}\right)^2 \mathrm d t$ $\quad$ Relative Sizes of Definite Integrals $\quad$ $\displaystyle$ $=$ $\displaystyle x^2 \int_a^b \left({f \left({t}\right)}\right)^2 \mathrm d t + 2 x \int_a^b f \left({t}\right) g \left({t}\right) \ \mathrm d t + \int_a^b \left({g \left({t}\right)}\right)^2 \mathrm d t$ $\quad$ Linear Combination of Integrals $\quad$ $\displaystyle$ $=$ $\displaystyle A x^2 + 2 B x + C$ $\quad$ $\quad$

where:

$\displaystyle A = \int_a^b \left({f \left({t}\right)}\right)^2 \mathrm d t$
$\displaystyle B = \int_a^b f \left({t}\right) g \left({t}\right) \ \mathrm d t$
$\displaystyle C = \int_a^b \left({g \left({t}\right)}\right)^2 \mathrm d t$

The quadratic equation $A x^2 + 2 B x + C$ is non-negative for all $x$.

It follows (using the same reasoning as in Cauchy's Inequality) that the discriminant $\left({2 B}\right)^2 - 4 A C$ of this polynomial must be non-positive.

Thus:

$B^2 \le A C$

and hence the result

$\blacksquare$

Also known as

This theorem is also known as the Cauchy-Schwarz inequality.

Some sources give it as the Cauchy-Schwarz-Bunyakovsky inequality.

Source of Name

This entry was named for Augustin Louis Cauchy, Karl Hermann Amandus Schwarz and Viktor Yakovlevich Bunyakovsky.

It was first stated in this form by Bunyakovsky in 1859, and later rediscovered by Schwarz in 1888.