Cauchy-Schwarz Inequality

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

$$\zbox {\left\vert\left<\underline{u},\underline{v}\right>\right\vert \leq \Vert\underline{u}\Vert\cdot\Vert\underline{v}\Vert}$$

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

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

$$\begin{eqnarray*} 0 \leq \Vert\underline{\tilde{u}}-\underline{\tilde{v}}\Vert^2... ...=& 2 - 2\left<\underline{\tilde{u}},\underline{\tilde{v}}\right> \end{eqnarray*}$$

which implies

$$\left<\underline{\tilde{u}},\underline{\tilde{v}}\right> \leq 1$$

or, removing the normalization,

$$\left<\underline{u},\underline{v}\right> \leq \Vert\underline{u}\Vert\cdot\Vert\underline{v}\Vert.$$

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

$$-\left<\underline{u},\underline{v}\right> \leq \Vert\underline{u}\Vert\cdot\Vert\underline{v}\Vert.$$

The last two equations imply

$$\left\vert\left<\underline{u},\underline{v}\right>\right\vert \leq \Vert\underline{u}\Vert\cdot\Vert\underline{v}\Vert.$$

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

$$\left<\underline{u},\underline{\tilde{v}}\right>\leq\Vert\underli... ...derline{\tilde{v}}\Vert = \Vert\underline{u}\Vert\cdot\Vert\underline{v}\Vert.$$

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