MHB Cauchy-Schwarz Inequality - Duistermaat and Kolk, CH. 1, page 4 .... ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of the proof of the Cauchy-Schwarz Inequality ...

Duistermaat and Kolk"s proof of the Cauchy-Schwarz Inequality reads as follows:View attachment 7639In the above proof we read the following:

" ... ... Now$$x = \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y + ( x - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y ) $$is a decomposition into two mutually orthogonal vectors ... ... "I have tried to demonstrate that the two vectors $$\frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y$$ and $$( x - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y )$$ are in fact orthogonal by showing that the inner product of these two vectors is zero ... but I failed to make any meaningful progress ...Can someone please demonstrate that these two vectors are in fact orthogonal ...
Help will be much appreciated ...

Peter

 

[GJA]

Hi Peter,

Set $a=\langle x, y\rangle /\|y\|^{2}$. You want to use the bilinearity property of the (real) inner product to get

$\langle ay, x-ay\rangle = a\langle x, y\rangle - a^{2}\langle y, y\rangle = \langle x, y\rangle ^{2}/\|y\|^{2}-\langle x, y\rangle ^{2}/\|y\|^{2}=0.$

Let me know if anything is unclear.