# Complemented subspaces in the dual of James' space $J$

James' space $J$ is subprojective; i.e., every infinite dimensional (closed) subspace of $J$ contains an infinite dimensional subspace which is complemented in $J$. This fact can be found in Corollary 11 of [P.G. Casazza et al. Proc. Amer. Math. Soc. 26 (1977), 294-305].

I would like to know if the dual space $J^*$ is subprojective. It is known that a non-reflexive subspace of $J^*$ contains an infinite dimensional subspace complemented in $J^*$ (Theorem 2 in [A. Andrew. Israel J. Math. 38 (1981), 276-282]), but I do not know what happens for infinite dimensional reflexive subspaces.

Yes, $J^*$ is subprojective.
Suppose that $M$ is a closed subspace of $J^*$. By Mazur's theorem, without loss of generality we additionally suppose that $M$ has a normalised basis $(f_k)_{k=1}^\infty$. Let $\iota\colon M\to J^*$ be the inclusion map and let $Q\colon J^{**}\to M^*$ be its (surjective) adjoint. Denote by $(f^*_k)_{k=1}^\infty$ the coordinate functionals of $(f_k)_{k=1}^\infty$.
Choose a bounded sequence $(z_n)_{n=1}^\infty$ in $J^{**}$ such that $Qz_n = f_n^*$ ($n\in \mathbb{N}$). As $J$ is quasi-reflexive, by Rosenthal's $\ell_1$-theorem, $(z_n)_{n=1}^\infty$ contains a weakly Cauchy subsequence, call it $(z^\prime_n)_{n=1}^\infty$. As $(z^\prime_n)_{n=1}^\infty$ fails to have a convergent subsequence, we may suppose that the sequence $(z^\prime_{2n+1}-z^\prime_{2n})_{n=1}^\infty$ is semi-normalised. However, by a result of Andrew [1, Theorem 2.1], every semi-normalised weakly null sequence in $J$ (so also in $J^{**}$ as $J^{**}\cong J$) contains a sequence spanning a complemented copy of $\ell_2$, so $(z^\prime_{2n+1}-z^\prime_{2n})_{n=1}^\infty$ contains such subsequence, say $(w_n)_{n=1}^\infty$. By reflexivity of $\ell_2$, $(w_n)_{n=1}^\infty$ is shrinking so the adjoint of a projection from $J^{**}$ onto $[w_n]_{n=1}^\infty$ is a projection onto a subspace of $M$. As $M\subset J^* \subset J^{***}$, the conclusion follows.