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.

Statement of the Cauchy-Schwarz inequality

Let V be a complex inner product space with inner product $\langle \cdot ,\cdot \rangle$. Then for any two elements $x_{1},x_{2}\in V$ it holds that

$|\langle x_{1},x_{2}\rangle |\leq \|x_{1}\|\|x_{2}\|,\quad (1)$

where $\|y\|=\langle y,y\rangle ^{1/2}$ for all $y\in V$. Furthermore, the equality holds if and only if the vectors $x_{1}$ and $x_{2}$ are linearly dependent.

Proof of the inequality

A standard yet clever idea for a proof of the Cauchy-Schwarz inequality 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, if 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 by the substitution $x_{1}\rightarrow x$ and $x_{2}\rightarrow y$.