• Support PF! Buy your school textbooks, materials and every day products Here!

Inner (dot) products

  • Thread starter babyrudin
  • Start date
8
0
1. Homework Statement

For [tex]x,y \in R^n[/tex], their inner ("dot") product is given by

[tex]<x,y>=\sum_{i=1}^n x_i y_i.[/tex]

Also, we write

[tex]<x,x>=\|x\|^2.[/tex]

2. Homework Equations

Fix p>1. Show that for all [tex]x,y \in R^n[/tex] we have

[tex]< \|x\|^{p-2}x -\|y\|^{p-2}y, x-y> \geq 0[/tex]


3. The Attempt at a Solution

Expanding the left-hand side, we can write

[tex]<\|x\|^{p-2}x,x> -<\|y\|^{p-2}y,x>-<\|x\|^{p-2}x,y>+<\|y\|^{p-2}y,y>[/tex]

which further simplifies to

[tex]\|x\|^p +\|y\|^p -(\|x\|^{p-2} +\|y\|^{p-2})<x,y>.[/tex]

Then I'm stuck. How do I show that the above is nonnegative?
 
Last edited:

Answers and Replies

MathematicalPhysicist
Gold Member
4,138
149
how about using cauchy-schwartz inequality?
something like this
<x,y><=sqrt(<x,x><y,y>)=||x||*||y||
then you get that what you wrote is greater than:
||x||^p+||y||^p-(||x||^p-2+||y||^p-2)(||x||*||y||)=
x^(p-1)(x-y)+y^(p-1)(y-x)=(x^p-1-y^p-1)(x-y)
now if ||x||>=||y|| then ||x||^p-1>=||y||^p-1
I leave you to check this proposition.
 
8
0
I've got it now, many many thanks!
 

Related Threads for: Inner (dot) products

  • Last Post
Replies
15
Views
18K
Replies
3
Views
3K
Replies
1
Views
816
  • Last Post
Replies
1
Views
811
  • Last Post
Replies
6
Views
851
  • Last Post
Replies
3
Views
932
  • Last Post
Replies
2
Views
840
  • Last Post
Replies
10
Views
1K
Top