# almost invariant half space for a dual of a restricted operator

Let $X$ be an infinite-dimensional Banach space and $T\in\mathcal{L}(X)$ a continuous linear operator acting on $X$. A closed subspace $Y$ of $X$ is said to be an almost-invariant halfspace (hereafter, AIHS) under $T$ just in case the following two conditions hold:

(i) $TY\subseteq Y+E$ for some error subspace $E$ of $X$ with $\text{dim}(E)<\infty$ (i.e., $Y$ is almost-invariant under $T$); and

(ii) $Y$ has both infinite dimension and infinite codimension (i.e., $Y$ is a halfspace).

Equivalently, a halfspace $Y$ is almost-invariant under $T$ whenever it is invariant under $T+F$ for some finite-rank operator $F\in\mathcal{L}(X)$.

Suppose $W$ is a finite-codimensional $T$-invariant subspace of $X$ and write $S:=T|_W\in\mathcal{L}(W)$ for the restriction of $T$ to $W$. It is clear that every AIHS for $S$ is an AIHS for $T$, with the same error.

Question. Suppose $S^*\in\mathcal{L}(W^*)$ admits an AIHS. Does this imply that $T^*\in\mathcal{L}(X^*)$ also admits an AIHS?

This is probably a simple exercise, but I can't see how to prove it just yet.

Note that every dual operator acting on an infinite-dimensional complex Banach space admits an AIHS by a result of Popov and Tcaciuc, so when trying to answer the above question we can restrict our attention to real Banach spaces $X$.

• Sure. Since $W$ is finite codimensional, WLOG $X=W \oplus_\infty F$ for some finite dimensional space $F$ and $X^*= W^*\oplus_1 F^*$. – Bill Johnson Jun 9 '16 at 1:59
• Thanks Bill. Adi used the same approach in an email he sent me. I posted his answer below. – Ben W Jun 9 '16 at 3:11

Let us decompose $X=W\oplus E$ for some finite-dimensional subspace $E$ of $X$. Denote by $P_W\in\mathcal{B}(X)$ the bounded linear projection onto $W$ along $E$, and by $E^\perp\subseteq X^*$ the annihilator of $E$. Define the bounded linear map $j:E^\perp\to W^*$ by the rule $$jx^*=x^*|_W\;\;\;\text{ for all }x^*\in E^\perp,$$ which admits a bounded linear inverse $j^{-1}:W^*\to E^\perp$ given by the rule $$j^{-1}w^*=w^*\circ P_W\;\;\;\text{ for all }w^*\in W^*.$$
Set $R:=j^{-1}\circ S^{*}\circ j:E^{\perp}\to E^{\perp}$, and let $H$ be an AIHS for $S^{*}$ with error $M$. Put $G:=j^{-1}H$ and $N:=j^{-1}M$, which are closed subspaces of $E^{\perp}$ and hence also of $X^{*}$. Then for any $x^{*}\in E^{\perp}$ and any $x\in X$ we now have \begin{eqnarray*} ((P_W^{*}T^{*})x^{*})(x) & = & x^{*}(TP_Wx) \hskip 4cm (\mbox{note } P_Wx\in W)\\ &=& x^{*}(SP_Wx)=(jx^{*})(SP_Wx) \hskip 1cm (\mbox {note } (jx^{*})\in W^{*} )\\ &=& (S^{*}(jx^{*}))(P_Wx) = ((j^{-1}\circ S^{*}\circ j)(x^{*}))(x) \\ &=& (Rx^{*})(x) \end{eqnarray*} so that $P_W^{*}T^{*}\equiv R$ on $E^\perp$. Since also $G\subseteq E^{\perp}$, this gives us \begin{multline*}(P_W^{*}T^{*})G=RG=(j^{-1}\circ S^{*}\circ j)(j^{-1}H)\\=j^{-1}(S^{*}H)\subseteq j^{-1}(H+M)=G+N,\end{multline*} and hence $$\tag{1}T^{*}G=(P_W^{*}T^{*})G+((1-P_W^{*})T^{*})G\subseteq G+N+((1-P_W^{*})T^{*})G.$$ Now, $I-P^*$ is a projection onto $W^{\perp}$, hence $(I-P^*)T^*(G)\subseteq W^{\perp}$. As well, $T^*(W^{\perp})\subseteq W^{\perp}$, since $W^{\perp}$ is a (finite dimensional) invariant subspace for $T^*$ . Put now $G_0:=G+W^{\perp}$ Then: \begin{eqnarray*} % \nonumber % Remove numbering (before each equation) T^{*}(G+W^{\perp}) &\subseteq & T^*(G)+T^*(W^{\perp})\subseteq G+N+((I-P^{*})T^{*})(G)+W^{\perp} \\ & \subseteq & G+W^{\perp}+N=G_0+N \end{eqnarray*} As $\text{dim}(N)=\text{dim}(M)$, the space $G_0$ is an AIHS under $T^*$ with defect $\leq d$.