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

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.
