Show $X$ is reflexive iff $\iota_{X^{**}}=(\iota_X)^{**}$

Question

Exercise 2.96 (The canonical inclusion) Let $X$ be a normed vector space and let $\iota_X\colon X\to X^{**}$ be the canonical inclusion defined by (2.39).

(a) Show that $(\iota_X)^*\iota_{X^*}=\operatorname{id}_{X^*}$ and determine the kernel of the projection $$P:=\iota_{X^*}(\iota_X)^*\colon X^{***}\to X^{***}.$$ (b) Assume $X$ is complete. Show that $X$ is reflexive if and only if $$\iota_{X^*}(\iota_X)^*=\operatorname{id}_{X^{***}}.$$ (c) Lipton's Pullback Let $Y\subset X$ be a closed subspace and let $j\colon Y\to X$ be the obvious inclusion. Then $\iota_X\diamond j=j^{**}\diamond\iota_{Y}\colon Y\to X^{**}$. This map is an isometric embedding of $Y$ into $X^{**}$ whose image is $$\iota_X\diamond j(Y)=j^{**}\diamond\iota_Y(Y)=\iota_X(X)\cap j^{**}(Y^{**})\subset X^{**}.$$ (d) Deduce from Lipton's Pullback that $Y$ is reflexive whenever $X$ is reflexive.

(e) Show that $X$ is reflexive if and only if $\iota_{X^{**}}=(\iota_X)^{**}$.

Note This exercise requires the notion of the dual operator, introduced in Definition 4.1 below.

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I see to show that the identity $\iota_{X^{**}}=(\iota_X)^{**}$ implies that $X$ is reflexive. I have the other direction using only (b). This is part (e) of a five part problems and I have to use parts (c), (d).

It only makes sense that I have to appeal to Linton's pullback using some kind of closed subspace but I cannot possibly think of one. I'm guaranteed the trivial subspace but that's rather unhelpful. I tried assuming that $x\in X^{**}-\iota_X(X)$ and then applying Linton to $\langle x\rangle$ (the span of $x$) to get some kind of contradiction but that makes no sense. I'm very stuck here.

Answer 1

The missing direction can be done elementarily, without the other steps. Show that if $X$ is not reflexive that then $\iota_{X^{**}}$ and $\iota_X^{**}$ must differ, hence if they are the same then $X$ has no choice but to be reflexive.

To that end let $f\in \iota_X(X)^\perp\subset X^{***}$ (by Hahn Banach this space is not zero if $X$ is not reflexive) and $x\in X^{**}$ with $f(x)\neq0$ (as a remark: $x\notin \iota_X(X)$).

Now $$\iota_{X^{**}}(x)\,[f]=f(x)\neq0,\ \ \text{ while }\ \ \ \iota_X^{**}(x)\,[f]= x(\iota_X^*(f)) = x( v\in X\mapsto f(v) ) = x(0)=0.$$

It follows that $\iota_{X^{**}}\neq \iota_X^{**}$.