I'm reading Linear Chaos by Karl-G. Grosse-Erdmann and Alfred Peris Manguillot. I'm and having trouble connecting the dots in the "equivalently" part of proof.
$\textbf{Theorem}$ Let $T$ be a hypercyclic operator. Then its adjoint $T^{*}$ has no eigenvalues.
Equivalently Every operator $T-\lambda I$, $\lambda \in \mathbb{K}$, has dense range implies that $\lambda$ not an eigenvalue of $T^{*}$."
Whats written for the "equivalent" part is:
"Moreover, by the Hahn-Banach theorem (see Appendix A), $T-\lambda I$ has dense range precisely when $$\langle x,T^{*}x^{*}-\lambda x^{*}\rangle=\langle(T-\lambda I)x,x^{*}\rangle=0$$ for all $x\in X$ entails that $x^{*}=0$ which is equivalent to $\lambda$ not being an eigenvalue of $T^{*}$.
With regards to what the appendix has in terms of the Hahn-Banach theorem:
"(Hahn–Banach theorem). Let $X$ be a vector space, $M$ a subspace of $X$, $p$ a seminorm on $X$ and $u : M \to K$ a linear functional such that $|u(x)| ≤ p(x)$ for all $x ∈ M$. Then $u$ has a linear extension $\tilde{u}$ to $X$ such that $|\tilde{u}(x)| ≤ p(x)$ for all $x \in X$.
We will mostly apply the Hahn–Banach theorem through one of the following corollaries:
(i) if $p$ is a seminorm on $X$ and $x_0 ∈ X$ then there exists a linear functional $u$ on $X$ such that $u(x_0) = p(x_0)$ and $|u(x)| ≤ p(x)$ for all $x \in X$;
if $X$ is a Fréchet space then
(ii) every continuous linear functional on a subspace of $X$ extends to a continuous linear functional on $X$ (with preservation of norm, if $X$ is a Banach space);
(iii) if $M$ is a closed subspace of $X$ and $x \notin M$ then there exists a continuous linear functional $x^{*}$ on $X$ that vanishes on $M$ with $\langle x, x^{*}\rangle \neq 0$;
(iv) a subspace $M$ of $X$ is dense in $X$ if and only if every continuous linear functional that vanishes on $M$ also vanishes on $X$;
(v) for any $x\in X$, if $\langle x,x^{*} \rangle = 0$ for all $x^{*} \in X^{*}$ then $x=0$."
The part that I'm not understanding is how Hahn-Banach theorem applies in particular which of the corollaries is being using above and how? I suspected it was part (v) but there its stated that $x=0$ not $x^{*}=0$ so my only other guess is part (iv), but I still don't quite see how this applies.
Thank you.
What is being used here is part (iv). Let us refer to the condition that $$\langle(T-\lambda I)x,x^{*}\rangle=0$$ for all $x\in X$ as $(*)$ and let $M$ denote the range of $T-\lambda I$. Note that $(*)$ says exactly that the functional $x^*$ vanishes on $M$. So, by (iv), $M$ is dense iff $(*)$ implies $x^*$ vanishes on $X$, i.e. $x^*=0$.