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 $X$.