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