Cauchy-Bunyakovsky-Schwarz Inequality

Semi-Inner Product Spaces

Let $\mathbb K$ be a subfield of $\C$.

Let $V$ be a semi-inner product space over $\mathbb K$.

Let $x, y$ be vectors in $V$.

Then:

$\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 \le \left \langle {x, x} \right \rangle \left \langle {y, y} \right \rangle$

Lebesgue $2$-Space

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $f, g: X \to \R$ be $\mu$-square integrable functions, i.e. $f, g \in \mathcal{L}^2 \left({\mu}\right)$, Lebesgue $2$-space.

Then:

$\displaystyle \int \left\vert{f g}\right\vert \, \mathrm d \mu \le \left\Vert{f}\right\Vert_2^2 \cdot \left\Vert{g}\right\Vert_2^2$

where $\left\Vert{\cdot}\right\Vert_2$ is the $2$-norm.

Cauchy's Inequality

The special case of the Cauchy-Bunyakovsky-Schwarz Inequality in a Euclidean space is called Cauchy's Inequality. It was Cauchy who first published this result in 1821.

It is usually stated as:

$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$

Complex Numbers

$\displaystyle \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$

where all of $w_i, z_i \in \C$.

Definite Integrals

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

Then:

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

Also known as

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

Source of Name

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