Equip the space $C([0,1])$ with the usual supremum norm. Show that this space is not reflexive.
I already saw a lot of solutions to this kind of question, but I just don't know how to start this.
Many people are starting with the fact "if $E$ is a reflexive Banach space then each linear continuous functional attains its norm." What does this mean? Or why is that true?
I just know that $X$ is reflexive if $J(X) = X''$ and $J$ is defined as $J\colon X \to X''$ by $J(x)=F_x$ and $F_x$ is the functional from $X \to F$.
Best Answer
If you like cannon shooting note that $C[0,1]$ is separable by the (Stone-)Weierstaß theorem. On the other hand, $C[0,1]^*$ is not reflexive as the total variation of $\delta_x-\delta_y$ (which is the norm in $C[0,1]^*$), $\|\delta_x - \delta_y\|=2$ for distinct $x,y\in [0,1]$. This means that $C[0,1]^*$ is not separable as $\{\delta_x\colon x\in [0,1]\}$ is an uncountable discrete subset of $C[0,1]^*$.
By the Hahn-Banach theorem, dual of a non-separable space is non-separable, hence $C[0,1]^{**}$ cannot be isomorphic to $C[0,1]$ (note that this is stronger that non-reflexivity).
Moral: a separable space with non-separable dual cannot be reflexive.