Let $X$ be a Banach space , $T$ be a continuous linear operator on $X$ such that $\exists x \in X$ such that $\{S(x): S \in (T)'\}=X $ , where $(T)'$ is the commutant of $T$ , then I can show that either the null space of some nonzero element of $(T)'$ is nonzero or else the range of each noninvertible element of $(T)'$ is not dense ; my question is , does it follow that $T$ has an invariant subspace ?
( Here , $(T)':=\{S \in \mathcal B(X) : S\circ T=T\circ S\}$ i.e. the set of bounded linear maps on $X$ which commutes with $T$ )
Note that if you have an element $S$ of the commutant with non-trivial kernel (meaning kernel not $0$ or the whole space), then the kernel of $S$ is a non-trivial invariant subspace, if $y \in \ker(S)$:
$$(S\circ T)(y)=(T \circ S)(y)=0$$
and $T(y) \in \ker(S)$.
If you have an element $S$ of the commutant where the range is neither the whole space nor $0$, then the range of $S$ is also invariant, let $z=S(y)$:
$$T(z)=(T\circ S)(y)=(S\circ T)(y)$$
and $T(z)$ lies in the range of $S$.
If you want closed invariant subspaces, the kernel of a continuous linear map is always closed (being pre-image of $\{0\}$), and the closure of an invariant sub-space is also invariant. So if you have a non-dense invariant subspace then the closure will not be the entire space and also invariant.