Cauchy-Schwarz inequality

In mathematics, the Cauchy-Schwarz inequality is a fundamental and ubiquitously used inequality that relates the absolute value of the inner product of two elements of an inner product space with the magnitude of the two said vectors. It is named in the honor of the French mathematician Augustin-Louis Cauchy and German mathematician Hermann Amandus Schwarz.

The inequality for real numbers

The simplest form of the inequality, and the first one considered historically, states that

$\left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}\leq \left(\sum _{i=1}^{n}x_{i}\right)^{2}\left(\sum _{i=1}^{n}y_{i}\right)^{2}$

for all real numbers x₁, …, x_n, y₁, …, y_n (where n is a arbitrary positive integer). Furthermore, the inequality is in fact an equality

$\left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}=\left(\sum _{i=1}^{n}x_{i}\right)^{2}\left(\sum _{i=1}^{n}y_{i}\right)^{2}$

if and only if there is a number C such that $x_{i}=Cy_{i}$ for all i.

The inequality for inner product spaces

Let V be a complex inner product space with inner product $\langle \cdot ,\cdot \rangle$. Then for any two elements $x,y\in V$ it holds that

$|\langle x,y\rangle |\leq \|x\|\|y\|,\quad (1)$

where $\|a\|=\langle a,a\rangle ^{1/2}$ for all $a\in V$. Furthermore, the equality in (1) holds if and only if the vectors $x$ and $y$ are linearly dependent (in this case proportional one to the other).

If V is the Euclidean space Rⁿ, whose inner product is defined by

$\langle x,y\rangle =\sum _{i=1}^{n}x_{i}y_{i},$

then (1) yields the inequality for real numbers mentioned in the previous section.

Another important example is where V is the space L²([a, b]). In this case, the Cauchy-Schwarz inequality states that

$\left(\int _{a}^{b}f(x)g(x)\,\mathrm {d} x\right)^{2}\leq \int _{a}^{b}{\big (}f(x){\big )}^{2}\,\mathrm {d} x\cdot \int _{a}^{b}{\big (}g(x){\big )}^{2}\,\mathrm {d} x$

for all real functions f and g for which the integrals on the right-hand side exist.

Proof of the inequality

A standard yet clever idea for a proof of the Cauchy-Schwarz inequality for inner product spaces is to exploit the fact that the inner product induces a quadratic form on V. Let $x,y$ be some fixed pair of vectors in V and let $\phi (x,y)$ be the argument of the complex number $\langle x,y\rangle$. Now, consider the expression $f(t)=\langle x+te^{i\phi (x,y)}y,x+te^{i\phi (x,y)}y\rangle$ for any real number t and notice that, by the properties of a complex inner product, f is a quadratic function of t. Moreover, f is non-negative definite: $f(t)\geq 0$ for all t. Expanding the expression for f gives the following:

${\begin{aligned}f(t)&=\langle x+te^{i\phi (x,y)}y,x+te^{i\phi (x,y)}y\rangle \\&=\|x\|^{2}+te^{i\phi (x,y)}\langle y,x\rangle +te^{-i\phi (x,y)}\langle x,y\rangle +t^{2}\|y\|^{2}\\&=\|x\|^{2}+2t|\langle x,y\rangle |+t^{2}\|y\|^{2}.\end{aligned}}$

Since f is a non-negative definite quadratic function of t, it follows that the discriminant of f is non-positive definite. That is,

$4|\langle x,y\rangle |^{2}-4\|x\|^{2}\|y\|^{2}=4(|\langle x,y\rangle |^{2}-\|x\|^{2}\|y\|^{2})\leq 0,$

from which (1) follows immediately.