# quasi-nilpotent part of a dual operator

Definitions and notation.

Let $X$ be a complex Banach space and $T\in\mathcal{L}(X)$ a continuous linear operator on $X$. We define the quasi-nilpotent part of $T$ as \begin{equation*}H_0(T):=\left\{x\in X:\lim_{n\to\infty}\|T^nx\|^{1/n}=0\right\}.\end{equation*} Note that $H_0(T)$ is a $T$-invariant linear subspace of $X$, although perhaps not a closed subspace. Trivially, we have the inclusion \begin{equation*}\mathcal{N}^\infty(T):=\bigcup_{n=1}^\infty\mathcal{N}(T^n)\subseteq H_0(T),\end{equation*} where $\mathcal{N}(\cdot)$ denotes the null space AKA kernel.

I am given an operator $T\in\mathcal{L}(X)$ satisfying the following (somewhat bizarre) conditions:
(i) $0\in\partial\sigma(T)$,
(ii) $\partial\sigma(T)\subseteq\sigma_p(T)$ (where $\sigma_p(\cdot)$ denotes the point spectrum),
(iii) $\sigma_p(T^*)=\emptyset$,
(iv) $\sigma(T)$ is connected and has infinite cardinality,
(v) the null spaces $\mathcal{N}(T^n)$, $n\in\mathbb{N}$, have finite and strictly increasing dimension,
(vi) $\mathcal{N}^\infty(T)$ is dense in $X$,
(vii) for every $x\in X$ the space $[T^nx]_{n=0}^\infty$ is either finite-dimensional or all of $X$, and
(viii) for every linearly independent set $(u_n)_{n=1}^\infty$ of eigenvectors under $T$ we have $[u_n]_{n=1}^\infty=X$.

Question. Could we use (i)-(viii) to obtain a nonzero $x^*\in H_0(T^*)$?

Condition (iii) means that any such $x^*$ will have a linearly independent orbit. Condition (vi) guarantees that $H_0(T)$ is dense in $X$. It seems plausible that this could generate a nonzero functional $x^*\in H_0(T^*)$. However, $H_0(T)\neq X$ since otherwise $T$ would be quasinilpotent, contradicting (iv), so the construction may not be easy.

Thanks!

EDIT: Regarding $H_0(T)$ being dense in $X$: It is easy to find examples of operators $T$ for which $\overline{\mathcal{N}^\infty(T)}=\overline{H_0(T)}=X$ but $H_0(T^*)=\{0\}$. Just take, for instance, an unweighted unilateral left-shift operator on $c_0$ or $\ell_p$ ($1\leq p<\infty$). So, condition (vi) is not enough by itself to ensure that $H_0(T^*)\neq\{0\}$.