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By definition, $X$ is reflexive if canonical injection $J:E \to E^{**}$ is surjective, where $\langle Jx,f\rangle_{E^{**},E^*}=\langle f,x\rangle_{E^*,E},~\forall x \in E,~\forall f \in E^*.$ In order to show $E$ is reflexive, it is not enough to show the existence of linear surjective isometry from $E$ to $E^{**}$.

I'd like to know if it is possible to show $L^p$ is reflexive for $1<p<\infty$ by using Riesz representation theorem. By Riesz representation theorem, $(L^p)^* =L^{p'}$, where $1/p+1/p'=1.$

Usually people say "since $(L^p)^{**} =(L^{p'})^*=L^p$, $L^p$ is reflexive"

It seems right, but I'd like to prove it in detail. Would you give me any comment for this question? Thanks in advance!

04170706
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  • What kind of detail would you like to see? The one line proof is detailed enough, given Riesz representation theorem. – Cave Johnson Jan 11 '18 at 05:29
  • @Cave Johnson I'd like to show $L^p$ is reflexive by showing the canonical injection is surjective. Is it enough by the one line proof? – 04170706 Jan 11 '18 at 05:34
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    If you write explicitly the isomorphism from $(L^{p})^{}$ to $L^{p'}$ and the isomorphism from $(L^{p'})^{}$ to $L^{p}$ , the inverse of their composition becomes exactly the canonical embedding of $L^{p}$ onto $(L^{p})^{**}$. No tricks needed. – Kavi Rama Murthy Jan 11 '18 at 05:48
  • @ Kavi Rama Murthy I don't get it. Sorry. Do we need adjoint operator?Could you write it in detail by answer? – 04170706 Jan 11 '18 at 06:03

1 Answers1

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Note that the isomorphism $(L^p)^* = L^{p'}$ is given by

$$\Phi_p : L^{p'}\to (L^p)^*,\ \ \Phi_p (f) (g) = \int_X fg d\mu.$$

Now we check that the canonical homomorphism is isomorphism: indeed, for all $f\in L^p$ and $\ell \in (L^p)^*$,

$$\begin{split} J (f) (\ell) &= \ell (f) \\ &= (\Phi_p\Phi_p^{-1})(\ell) (f) \\ &= \int_X \Phi_p^{-1}(\ell) f d\mu \\ &=\int_X f \Phi_p^{-1}(\ell) d\mu \\ &= \Phi_{p'} (f) (\Phi^{-1}_p \ell) \\ &= ((\Phi_p^{-1})^*\circ \Phi_{p'} (f))(\ell). \end{split}$$

So $J(f) = (\Phi_p^{-1})^*\circ \Phi_{p'}(f)$ and thus $J$ is an isomorphism.