Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality (or ``Schwarz Inequality'') states that for all $\underline{x}\in{\bf C}^N$ and $\underline{y}\in{\bf C}^N$, we have

$$\zbox {\left\vert\left<\underline{x},\underline{y}\right>\right\vert \leq \Vert\underline{x}\Vert\cdot\Vert\underline{y}\Vert}$$

with equality if and only if $\underline{x}=c\underline{y}$ for some scalar $c$.

We can quickly show this for real vectors $\underline{x}\in{\bf R}^N$, $\underline{y}\in{\bf R}^N$, as follows: If either $\underline{x}$ or $\underline{y}$ is zero, the inequality holds (as equality). Assuming both are nonzero, let's scale them to unit-length by defining the normalized vectors $\underline{{\tilde x}}\isdeftext \underline{x}/\Vert\underline{x}\Vert$, $\underline{{\tilde y}}\isdeftext \underline{y}/\Vert\underline{y}\Vert$, which are unit-length vectors lying on the ``unit ball'' in ${\bf R}^N$ (a hypersphere of radius $1$). We have

$$\begin{eqnarray*} 0 \leq \Vert\underline{{\tilde x}}-\underline{{\tilde y}}\Vert... ... 2 - 2\left<\underline{{\tilde x}},\underline{{\tilde y}}\right> \end{eqnarray*}$$

which implies

$$\left<\underline{{\tilde x}},\underline{{\tilde y}}\right> \leq 1$$

or, removing the normalization,

$$\left<\underline{x},\underline{y}\right> \leq \Vert\underline{x}\Vert\cdot\Vert\underline{y}\Vert.$$

The same derivation holds if $\underline{x}$ is replaced by $-\underline{x}$ yielding

$$-\left<\underline{x},\underline{y}\right> \leq \Vert\underline{x}\Vert\cdot\Vert\underline{y}\Vert.$$

The last two equations imply

$$\left\vert\left<\underline{x},\underline{y}\right>\right\vert \leq \Vert\underline{x}\Vert\cdot\Vert\underline{y}\Vert.$$

In the complex case, let $\left<\underline{x},\underline{y}\right>=R e^{j\theta}$, and define $\underline{{\tilde y}}=\underline{y}e^{j\theta}$. Then $\left<\underline{x},\underline{{\tilde y}}\right>$ is real and equal to $\vert\left<\underline{x},\underline{{\tilde y}}\right>\vert=R>0$. By the same derivation as above,

$$\left<\underline{x},\underline{{\tilde y}}\right>\leq\Vert\underl... ...erline{{\tilde y}}\Vert = \Vert\underline{x}\Vert\cdot\Vert\underline{y}\Vert.$$

Since $\left<\underline{x},\underline{{\tilde y}}\right>=R=\left\vert\left<\underline... ...right>\right\vert=\left\vert\left<\underline{x},\underline{y}\right>\right\vert$, the result is established also in the complex case.