Cauchy-Schwarz Inequality/Inner Product Spaces

Theorem

Let $V$ be a semi-inner product space over $\mathbb K$ where $\mathbb K$ is a subfield of $\C$.

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

Then:

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

Proof

Let $\lambda \in \mathbb K$. Since an inner product is generated by a norm on the underlying normed linear space we may expand as follows:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 0\)	\(\le\)	\(\displaystyle \left\Vert {x - \lambda y}\right\Vert^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left \langle {x - \lambda y, x - \lambda y} \right \rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left \langle {x, x} \right \rangle - \left \langle {x, \lambda y} \right \rangle - \left \langle {\lambda y, x} \right \rangle + \left \langle {\lambda y, \lambda y} \right \rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left \langle {x, x} \right \rangle - \lambda^* \left \langle {x, y} \right \rangle - \lambda \left \langle {y, x} \right \rangle + \left \langle {\lambda y, \lambda y} \right \rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

where $\lambda^*$ is the complex conjugate of $\lambda$.

(If $\mathbb K = \R$, then $\lambda^* = \lambda$.)

When $y = 0$ the equality is the trivial statement that $0 \leq 0$, so assume that $y \neq 0$.

If we let $\lambda = \left \langle {x, y} \right \rangle \times \left \langle {y, y} \right \rangle^{-1}$ then we obtain:

$0 \le \left \langle {x, x} \right \rangle - \left|{\left \langle {x, y} \right \rangle}\right|^2 \times \left \langle {y, y} \right \rangle^{-1}$

Solving this for $\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 $, we see that:

$\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 \le \left \langle {x, x} \right \rangle * \left \langle {y, y} \right \rangle = \left\Vert{x}\right\Vert^2 \times \left\Vert{y}\right\Vert^2$

as desired.

$\blacksquare$

Alternative names

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

Source of Name

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

Sources

• John B. Conway: A Course in Functional Analysis (1990)... (previous)... (next) $I.1.4$