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\(\displaystyle x = \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y + ( x - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y ) \)is a decomposition into two mutually orthogonal vectors ... ... "I have tried to demonstrate that the two vectors \(\displaystyle \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y\) and \(\displaystyle ( x - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y )\) 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.

 

[math771]

Hi Peter,

The proof of the Cauchy-Schwarz Inequality can be a bit tricky to follow, so don't worry if you're having trouble with it. Let me try to explain why the two vectors in question are orthogonal.

First, let's recall the definition of orthogonality. Two vectors x and y are orthogonal if their inner product, denoted by <x, y>, is equal to 0. In other words, if x and y are orthogonal, then <x, y> = 0.

Now, let's look at the two vectors in question. The first vector, \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y, is a multiple of y. This means that it is in the same direction as y, but with a different magnitude. The second vector, (x - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y), is simply the difference between x and the first vector.

If we take the inner product of these two vectors, we get:

<x - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y, \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} y> = \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} <x, y> - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} <y, y>

Since <x, y> is the inner product of two vectors, it is a scalar. Similarly, <y, y> is also a scalar. This means that we can pull these scalars out of the inner product, giving us:

\frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} <x, y> - \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} <y, y> = \frac{ \langle x, y \rangle }{ \mid \mid y \mid \mid^2} \cdot \langle x, y> - \frac{ \langle x, y \rangle