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

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

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 $\underline{{\tilde x}}\isdef \underline{x}/\Vert\underline{x}\Vert$, $\underline{{\tilde y}} \isdef \underline{y}/\Vert\underline{y}\Vert$, 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,

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

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

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

The complex case can be shown by rotating the components of $\underline{x}$ and $\underline{y}$ such that re$\left\{\left<\underline{{\tilde x}},\underline{{\tilde y}}\right>\right\}$ becomes equal to $\left\vert\left<\underline{{\tilde x}},\underline{{\tilde y}}\right>\right\vert$.