Conway Functional Analysis text example?

Click For Summary
SUMMARY

The discussion focuses on John Conway's example 1.5 from "A Course in Functional Analysis," specifically regarding the multiplication operator \( M_\phi : L^p(\mu) \to L^p(\mu) \) and its adjoint \( M_\phi^* : L^q(\mu) \to L^q(\mu) \). It establishes that \( M_\phi^* = M_\phi \) under the condition \( 1 \le p < \infty \) and \( 1/p + 1/q = 1 \). The key to understanding this example lies in recognizing the dual relationship between \( L^p \) and \( L^q \) spaces, which is facilitated by Hölder's inequality. The discussion emphasizes the importance of precise definitions and the potential confusion arising from Conway's notation.

PREREQUISITES
  • Understanding of \( L^p \) spaces and their properties
  • Familiarity with multiplication operators in functional analysis
  • Knowledge of dual spaces and the relationship between \( L^p \) and \( L^q \)
  • Proficiency in applying Hölder's inequality
NEXT STEPS
  • Study the properties of \( L^p \) spaces, focusing on duality and isomorphisms
  • Explore the concept of multiplication operators in functional analysis
  • Review Hölder's inequality and its applications in functional analysis
  • Read about the implications of adjoint operators in the context of \( L^p \) spaces
USEFUL FOR

Mathematicians, graduate students in functional analysis, and anyone seeking to deepen their understanding of operator theory and dual spaces in the context of \( L^p \) spaces.

xeno_gear
Messages
40
Reaction score
0
Hello, I'm reading through John Conway's A Course in Functional Analysis and I'm having trouble understanding example 1.5 on page 168 (2nd edition):

Let [tex](X, \Omega, \mu)[/tex] and [tex]M_\phi : L^p(\mu) \to L^p(\mu)[/tex] be as in Example III.2.2 (i.e., sigma-finite measure space and [tex]M_\phi f = \phi f[/tex] is a multiplication operator, where [tex]\phi \in L^\infty(X, \Omega, \mu)[/tex]). If [tex]1 \le p < \infty[/tex] and [tex]1/p + 1/q = 1[/tex], then [tex]M_\phi^* : L^q(\mu) \to L^q(\mu)[/tex] is given by [tex]M_\phi^*f = M_\phi f[/tex]. That is, [tex]M_\phi^* = M_\phi[/tex].

I think I see how to do it if it's L^2 since there we have an inner product to work with. But otherwise I'm just not sure and don't really get the example. Any ideas?
 
Physics news on Phys.org
The essense is that the dual of L^p is L^q, via

[tex]L^q\to (L^p)^*[/tex]
[tex]g\mapsto (f\mapsto \int fg)[/tex]

Use good old Hölder's inequality to show that this mapping makes sense. That this is an isomorphism is not trivial, but it is probably explained in your measure theory book.

Once you have proven (or accepted) this result, Conway's claim is not hard anymore; just write out the definitions. Of course, he is being a bit sloppy, piling identifications on top of each other (e.g. it doesn't make sense to say M*=M, as they have totally different domain and codomain).
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K