Verify that the Riesz vector is unique

  • Thread starter Thread starter rhobymic
  • Start date Start date
  • Tags Tags
    Vector
rhobymic
Messages
5
Reaction score
0

Homework Statement


The Riesz representation theorem gives us that forall f in V* there exists a unique R_f in V such that f(x) = <x, R_f >. (<,> is my attempt to type inner product angle brackets) Verify that R_f is unique.

Homework Equations



If I knew the relevant equations I think I could get this done on my own

The Attempt at a Solution



No attempts at a solution because i don't even know where to begin.

Usually when i prove uniqueness I start with the assumption that two solutions exist and then prove that they end up being the same solution... but i don't see how to work that angle here.

Any help would be great
 
Physics news on Phys.org
rhobymic said:

Homework Statement


The Riesz representation theorem gives us that forall f in V* there exists a unique R_f in V such that f(x) = <x, R_f >. (<,> is my attempt to type inner product angle brackets) Verify that R_f is unique.

Homework Equations



If I knew the relevant equations I think I could get this done on my own

The Attempt at a Solution



No attempts at a solution because i don't even know where to begin.

Usually when i prove uniqueness I start with the assumption that two solutions exist and then prove that they end up being the same solution... but i don't see how to work that angle here.

Any help would be great

You already have the correct approach. Assume there are two solutions. Write down the equations the two vectors in V would have to satisfy.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
6
Views
2K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
14
Views
2K
Replies
1
Views
2K
Replies
9
Views
4K
Back
Top