Conway Functional Analysis text example?

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
1 reply · 3K views
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: