We consider a Banach space $X$ and its dual $X^\*$. 


Let $Q\colon X^\ast \to X^\ast$ be an idempotent operator. 
Question: 
Can we find an idempotent operator $P\colon X^\ast \to X^\ast$ which is weak${}^\ast$-to-weak${}^\ast$ continuous and with range isomorphic to range of $Q$ and $\mbox{im}P\subseteq \mbox{im}Q$? In fact, I am mostly interested in the case $\mbox{im }Q\cong \ell\_p$ for $p\in [1,\infty)$.


Certainly, $P$ would have to be an adjoint to some idempotent on $X$. My feeling is that in general this is not the case but perhaps it might be true for some well-behaved class of Banach spaces $X$ like Banach lattices? Or Banach lattices without a complemented copy of $\ell\_1$?



Weak*-closed and complemented subspaces of dual Banach spaces
In 


Stegall, C. Banach spaces whose duals contain $\ell\_1(\Gamma)$ with applications to the study of dual $L\_1(\mu)$ spaces. Trans. Amer. Math. Soc. 176 (1973), 463–477 


Stegall proved that $\ell\_2$ is isometrically isomorphic to a norm one complemented subspace of $X^\*$ when $X= (\sum\_{n=1}^\infty \ell\_2^n)\_1$, yet $\ell\_2$ does not embed into $X$.


It would be interesting to have an example of this phenomenon with $X^\*$ separable. 


ADDED 6/25/12: 


Here is a more interesting example because the dual is separable, but the range of the projection is not $\ell\_p$. Take any separable reflexive space $X$ that fails the approximation property and let $(E\_n)$ be an increasing sequence of finite dimensional subspaces of $X$ whose union is dense in $X$. Let $c(E\_n)$ be the space of sequences $(x\_n)$ with $x\_n$ in $E\_n$ and $\lim\_n x\_n$ existing in $X$, normed by $\|(x\_n)\|= \sup \|x\_n\|$.  

Define $c\_0(E\_n)$ to be $(\sum\_{n=1}^\infty E\_n)\_0$. 


Consider the short exact sequence (ses)


$0\to c\_0(E\_n) \to c(E\_n) \to X\to 0$


where the second arrow is the inclusion mapping and the third is the quotient mapping $Q$ defined by 
$(x\_n)\mapsto \lim x\_n$. This sequence locally splits (with constant one), hence $Q^\*$ maps $X^\*$ onto a norm one complemented subspace of $c(E\_n)^\*$.
The ses itself does not split because $ c(E\_n)$ has the approximation property (even a finite dimensional decomposition). 


This example is Proposition 2.4 in


Johnson, William B.; Oikhberg, Timur:
Separable lifting property and extensions of local reflexivity, 
Illinois J. Math. 45 (2001), no. 1, 123–137.


The construction itself is due to W. Lusky


A note on Banach spaces containing $c\_0$ or $C\_\infty$, 
J. Funct. Anal. 62 (1985), no. 1, 1–7.


Lusky was interested in the case that $X$ has the bounded approximation property. Then the resulting ses splits. 


