# Thread: Cauchy Schwaz inequality

1. ## Cauchy Schwaz inequality

$\displaystyle 0 \leq \left\| u-\delta v \right\|^2 = \langle u-\delta v,u-\delta v \rangle = \langle u,u \rangle - \bar{\delta} \langle u,v \rangle - \delta \langle v,u \rangle + |\delta|^2 \langle v,v\rangle. \,$

if $\displaystyle \delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}. \,$ what is $\displaystyle \bar{\delta}$ equal to? is $\displaystyle \langle u,u \rangle=u.u$ (dot product)

thanks for your help.

2. Originally Posted by charikaar
$\displaystyle 0 \leq \left\| u-\delta v \right\|^2 = \langle u-\delta v,u-\delta v \rangle = \langle u,u \rangle - \bar{\delta} \langle u,v \rangle - \delta \langle v,u \rangle + |\delta|^2 \langle v,v\rangle. \,$

if $\displaystyle \delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}. \,$ what is $\displaystyle \bar{\delta}$ equal to? is $\displaystyle \langle u,u \rangle=u.u$ (dot product)

thanks for your help.

$\displaystyle \delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}=\langle u,v\rangle \|v\|^{-1\slash 2} \,\Longrightarrow \bar {\delta}=\overline{\langle u,v \rangle }\|v\|^{-1\slash 2}$ (why?)

Tonio

3. Originally Posted by tonio
$\displaystyle \delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}=\langle u,v\rangle \|v\|^{-1\slash 2} \,\Longrightarrow \bar {\delta}=\overline{\langle u,v \rangle }\|v\|^{-1\slash 2}$ (why?)

Tonio
Does the bar mean complex conjugate?

4. Originally Posted by charikaar
Does the bar mean complex conjugate?

I suppose it does....you wrote it, what did you mean by that?!

Tonio