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I am referring to the introduction of this paper: G. Flores - On an identification of the Lipschitz-free spaces of $\mathbb{R}^n$

Here, the authors describe the predual space of the space

$$ Lip_0(M) := {\left\{ f: M \to \mathbb{R} ~ | ~ f(0_M)=0, ||f||_{Lip} < + \infty \right\} }$$

where $(M,d)$ is a "pointed" metric space (this just means that we fix a point $0_M$), and $||f||_{Lip} = { \sup_{x \neq y} \frac{|f(x)-f(y)|}{||x-y||}}$ is the Lipschitz norm.

Then, they define the space

$$ \mathcal{F}(M) = \overline{ \text{span}\left\{ ev_x | x\in M \right\} } $$

where $ev_x \in Lip_0(M)^*$ is the evaluation function at a point $x$, i.e. $ev_x(f) = f(x)$, and they state that

$$ \mathcal{F}(M)^* = Lip_0(M) $$

My question is: it is always true that we can embed a space in its double dual, therefore

  • $ \mathcal{F}(M)^* = Lip_0(M) \subset (Lip_0(M)^*)^* $

Moreover, by definition, $\mathcal{F}(M) \subset Lip_0(M)^*$, from which follows immediately that

  • $ \mathcal{F}(M)^* \supset (Lip_0(M)^*)^* $

But then I would conclude that $Lip_0(M) = Lip_0(M)^{**}$, i.e. $Lip_0(M)$ would be reflexive, which is not stated anywhere and seems too good to be true.

Can anyone spot the mistake in this reasoning, if there is any?

rod
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    You prove that the bidual is isomorphic to itself. However, this is not the definition of a reflexive Banach space. For this this you would need that the isomorphism is given by $$ X \rightarrow X^{**}, x\mapsto (\varphi \mapsto \varphi(x)). $$ The James' space is an example of a space which is isomorphic to its bidual, but not reflexive. – Severin Schraven Nov 13 '23 at 23:29
  • I found an answer: the reason lies exactly in this question: https://math.stackexchange.com/questions/655718/why-is-this-inclusion-of-dual-of-banach-spaces-wrong – rod Nov 13 '23 at 23:31

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