Second adjoint operators Let $X$ be a non-reflexive Banach space. Is it possible to decompose the identity operator $I_X=T+S$ where $T,S$ are bounded operators on $X$ with the property that $T^{**}x^{**}, S^{**}y^{**}\in X$ for some $x^{**}, y^{**}\in X^{**}\setminus X$?
 A: In absolute generality, no.  Recall that Gantmacher's Theorem says that $T:X\rightarrow Y$ is weakly-compact if and only if $T^{**}(X^{**}) \subseteq Y$.  So suppose $X$ has the property that every bounded linear map on $X$ is of the form $\lambda I_X+R$ with $R$ weakly compact.  Then if $I_X=T+S$ then $T=\lambda I_X+R$ say, so $S=I_X-T=(1-\lambda)I_X-R$.  If $x^{**}\in X^{**}\setminus X$, then $T^{**}(x^{**}) = \lambda x^{**} + R^{**}(x^{**})$ will be in $X$ if and only if $\lambda=0$.  But if $\lambda=0$ then $S=I_X-R$ and so as $R(y^{**})\in X$ for any $y^{**}$, it's impossible for $S^{**}$ to map something in $X^{**}\setminus X$ to $X$.
Now apply a big hammer-- the Haydon Argyros space solves the scalar-compact problem, and is non-reflexive.  So it provides an example of such an $X$.
Edit: A similarish question on MO made me revisit this.  For $c_0$ you can find such an operator.  Let $(e_n)$ be the basis of $c_0$, and $(e_n^*)$ the basis of $\ell^1=c_0^*$.  Define $T(e_n) = e_{n/2}$ if $n$ even, or $0$ if $n$ odd.  Then $\langle T'(e_n^*), e_m \rangle = 1$ when $m=2n$, or $0$ otherwise; so $T'(e_n^*) = e_{2n}^*$.  Then $\langle T''(1,0,1,0,1,0,\cdots), e_n^* \rangle=0$ for all $n$, so $T''(1,0,1,0,\cdots)=0$.  Also $\langle (I-T)''(1), e_n^* \rangle = 1 - \langle 1,e_{2n}^*\rangle=0$ for all $n$, so $(I-T)''(1)=0$.  So for $X=c_0$ the answer is "yes".
