What makes positive linear functionals always finite?

  • Context: Graduate 
  • Thread starter Thread starter redrzewski
  • Start date Start date
  • Tags Tags
    Representation Theorem
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
4 replies · 5K views
redrzewski
Messages
115
Reaction score
0
I'm strugging with a portion of Rudin's proof.

Quick statement of the bulk of the theorem:

Let X be a locally compact Hausdorff space. Let A be a positive linear functional on Cc(X) (continous functions with compact support). Then (among other things), there exists a measure u() that represents A:

A(f) = Integral(fdu) for every f in Cc(X).

Now assuming that, he shows that u(K) for any compact set K is finite.

He basically shows that u(K) <= A(f) for some f in Cc(X) with 0 <= f <= 1. This far I follow. But then he immediately concludes that therefore u(K) is finite for any compact K.

Is there some basic property of positive linear functionals that makes them always finite? There is a somewhat similar proof in Rudin's Principles of Mathematical Analysis where he shows that the norm of linear functionals on finite dimensional vector spaces is finite. But that proof assumes a finite dimensional space. So I can't see how to apply that proof here.

Any help is appreciated.
thanks
 
Physics news on Phys.org
I can use the compactness of K to choose an finite open cover {Vi} of K. Then we have:

A(V1) + A(V2) = A(V1+V2), etc by linearity. Showing that this is >= u(K) is straightforward with the countable additivity of the measure. Since there are finite Vi, I get a finite sum very similar to Rudin's proof on the finite vector space in PMA.

However, I'm still assuming that each A(Vi) is finite. But it seems like I could be on the right track. What's the justification that each A(Vi) is finite?

thanks
 
redrzewski said:
He basically shows that u(K) <= A(f) for some f in Cc(X) with 0 <= f <= 1. This far I follow. But then he immediately concludes that therefore u(K) is finite for any compact K.

isn't this obvious? A(f) must be finite by the definition of a linear function, so u(K) <= A(f)< infinity.
 
Let me make sure I understand.

Since A: X -> R is a function, then for any f in X, A(f) must be finite by the definition of a function since A is well defined on its domain.

Is that right?

Also, Royden has a section about extended real-valued functions where apparently f(x)=infinity for some x is a valid definition. So I just need to assume that we aren't dealing with these extended functions here.

thanks
 
redrzewski said:
Since A: X -> R is a function, then for any f in X, A(f) must be finite by the definition of a function since A is well defined on its domain.

Yes, except that it is A:Cc(X)->R and f in Cc(X). The Riesz representation theorem is for real valued linear maps.
Although, u(f) could be infinite if f>=0 doesn't have compact support but then it isn't in the domain of A anyway.