Prove that $ \|x\|_2 = \sup\left \{ x \cdot y \; s.t. \; \|y\|_2 = 1 \right \}$ and that $ x \cdot y = \|x\| $ and $ \|y\|=1$ iff $x = \|x\| y$

Question

In $\mathbb{R}^d$

Question:

a) Prove that $ ||x||_2 = Sup\left \{ x \cdot y \; s.t. \; ||y||_2 = 1 \right \}$
b) Prove that $ x \cdot y = ||x|| $ and $ ||y||=1$ iff $x = ||x|| y$

Answer:

a) First by definition $x \cdot y = ||x|| ||y|| cos( \widehat{ x ; y} )$. Now if $||y||=1 \Rightarrow x \cdot y = ||x|| cos( \widehat{ x ; y} )$. Then $Sup\left \{ x \cdot y \; s.t. \; ||y||_2 = 1 \right \} = Sup\left \{ ||x|| cos( \widehat{ x ; y} ) \; s.t. \; ||y||_2 = 1 \right \}$ and indeed, this is equal to $||x||$ when $cos( \widehat{ x ; y} ) = 1$. In brief we have proved that $||x||_2 = Sup\left \{ x \cdot y \; s.t. \; ||y||_2 = 1 \right \}$

b)
$(\Rightarrow)$
On one hand we have: $x \cdot y = ||x|| ||y|| cos(\widehat{ x ; y})= ||x|| \Rightarrow cos(\widehat{ x ; y}) = 1 \Rightarrow \widehat{ x ; y} = 0$ In other words the vectors $x$ and $y$ arec colinear $\Rightarrow y =\alpha x $.
On the other heand we have that $||y||=1 \Rightarrow || x || | \alpha | =1 \Rightarrow |\alpha| = \frac{1}{||x||}$. Then $x= \frac{y}{ \alpha} = ||x|| y $
$(\Leftarrow )$
First we have that $y = \frac{x}{||x||} \Rightarrow ||y|| = ||\frac{x}{||x||}|| = 1$.
More over $x \cdot y = ||x|| ||y|| cos(\widehat{ x ; y}) = ||x|| \frac{||x||}{||x||} cos(\widehat{ x ; y})=||x||$ as if $y = \frac{x}{||x||}$ $x$ and $y$ are colinear abd thus $cos(\widehat{ x ; y})=1$

Is this prove correct in any dimension $\mathbb{R}^d , d \geq 1$ ?

Thank you.

Answer 1

$$0\leq \|\frac{x}{\|x\|}-\frac{y}{\|y\|}\|^2=2-2\frac{\langle x,y\rangle}{\|x\|\|y\|}$$ implies $$\langle x,y\rangle\leq \|x\|\|y\|.$$ Therefore for any $y$ such that $\|y\|=1$ we have $$\langle x,y\rangle\leq \|x\|.$$ Choosing in particular $y=\frac{x}{\|x\|}$ shows the result $$\sup_{y; \|y\|=1} \langle x,y\rangle\leq \|x\|.$$

Answer 2

Using the https://math.stackexchange.com/users/136544/daw comment.

-Using C.S. inequality $\forall x,y \in \mathbb{R}^d $ we have $x \cdot y = \sum_{i=1} ^d x_i \cdot y_i \leq | \sum_{i=1} ^d x_i \cdot y_i | = |<x | y >| \leq \| x \|_2 \cdot \| y \|_2 $

a)
If $\| y \|_2=1 \Rightarrow x \cdot y \leq \| x \|_2$ and so naturally we have that $Sup \left \{ x \cdot y : \| y \|_2=1 \right \} = \| x \|_2$ and the equality of the supremum can be obtain for $y = \frac{x}{ \| x \|}$

b) $x \cdot y = \| x \|$ and $\| y \| = 1 \Rightarrow x= \| x \| y$
If $x \cdot y = \| x \|$ it means that $y= \lambda x $ in the CS inequality writen above.
Moreover as $|| y || = 1 \Rightarrow || \lambda x || = 1 \Rightarrow |\lambda| = \frac{1}{||x||}$

($\Leftarrow $)
Trivial

Q.E.D.