Question about projections acting on dual space Let $X$ be a complex Banach space, and let $P$ be a bounded linear operator acting on the dual $X^{*}$ such that that $P^2=P$.
I research for a bounded linear operator $Q$  acting on $X$ such that its adjoint $Q^{*}=P?$
 A: Let $\varphi $ denote the invariant mean on $\ell^\infty(\mathbb{N})$ (or any continuous extension of the limit on the space of convergent sequences). Define $$Px^*=\varphi(x^*)\mathbf{1},\quad x^*\in \ell^\infty$$ where $\mathbf{1}$ denotes the  sequence constantly equal $1.$ Then $P$ is a projection operator on $\ell^\infty(\mathbb{N}).$ However the operator $P$ is not adjoint to any projection on $\ell^1(\mathbb{N}).$ Indeed any such projection $Q$ should be one-dimensional. Hence $Qx=\psi(x)y,$ where $\psi$ is a bounded linear functional on $\ell^1(\mathbb{N})$ and $y\in \ell^1(\mathbb{N}).$ Thus $\psi(y)=1.$ We have $$(Q^*x^*)(x)=x^*(Qx)=\psi(x)x^*(y)$$ and $$(Px^*)(x)=\varphi(x^*)\sum_{n=1}^\infty x_n$$ Substituting $x^*=\delta_n$ gives $P\delta_n=0$ for every $n.$ On the other hand $\delta_n(y)=y_n,$ therefore $\delta_n(y)\neq 0$ for some $n.$ Then $$(Q^*\delta_n)(y)=y_n\neq 0,\quad (P\delta_n)(y)=0$$
A: Note that $Q$ must be a projector: ${Q^2}^*={Q^*}^2 = P^2 = P = Q^*$, hence $Q^2=Q$. The space $X$ must decompose into the image of $Q$ which is the space on which $Q$ projects, and the kernel of $Q$, which is the space parallel to which $Q$ projects.
$$
X = \ker(Q) \oplus \ker(id -Q), \quad \ker(id-Q)= im(Q).
$$
For a projector $Q$, there is a relation between its invariant subspaces and the invariant subspaces of its dual operator: $\ker(Q^*) =(im(Q))^\perp$ etc that should allow you to find the desired subspaces.
