Second adjoint of the canonical embedding

Question

Suppose that $X$ is a Banach space. Denote by $\kappa_X$ the canonical embedding of $X$ into $X^{**}$. Do we always have $$(\kappa_X)^{**} = \kappa_{X^{**}}? $$

Answer 1

No. Not that that makes any sense to me, but no. In fact it turns out that $\kappa_{X^{**}}=\kappa_X^{**}$ if and only if $X$ is reflexive.

In fact the truth behind my erroneous feeling that the two must be equal in general is this: $$\kappa_{X^{**}}\kappa_X=\kappa_X^{**}\kappa_X.$$

Proofs: Writing $x\in X$, $x^*\in X^*$, etc.:

If you unpack two definitions you see the question is whether $$x^{***}(x^{**})=x^{**}(\kappa_X^*x^{***}).$$

If $X$ is not reflexive, choose $x^{***}\ne0$ so $x^{***}(\kappa_X x)=0$ for all $x$. That says precisely that $\kappa_X^*x^{***}=0$. But there exists $x^{**}$ with $x^{***}(x^{**})\ne0$.

Of course it must be true if $X$ is reflexive. But I thought it must be true in general, and actually writing down a proof took me a few minutes, so here it is:

The proof that $\kappa_{X^{**}}\kappa_X=\kappa_X^{**}\kappa_X$ in general is just unpacking definitions:

$$\kappa_{X^{**}}\kappa_X(x)(x^{***})=x^{***}(\kappa_X(x)),$$ while

$$\kappa_X^{**}\kappa_X(x)(x^{***})=\kappa_X(x)(\kappa_X^*(x^{***})) =\kappa_X^*(x^{***})(x)=x^{***}(\kappa_X(x)).$$

And now if $X$ is reflexive then $\kappa_X$ is an isomorphism, so it follows that $\kappa_{X^{**}}=\kappa_X^{**}$.

If there are no typos here it's a friggin miracle.

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Or do this:

Exercise Whether $X$ is reflexive or not, $\kappa_X^*\kappa_{X^*}x^*=x^*$. Show that this implies $\kappa_{X^{**}}=\kappa_X^{**}$ if $X$ is reflexive.