xeno_gear
- 40
- 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?
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?