Let $S:X\to X$ be a bounded operator with $\sigma(S)\subseteq \{z\in\mathbb{C}: Re(z)>0\}$. Define $T(t)= e^{tS}$ and $\mathcal{T}=\{T(t)\}_{t\geq 0}$.
Define $$D_\mathcal{T}(x)= \{y: \text{ there exist } \{x_n\}\subseteq X \text{ with } x_n\to x \text{ and a strictly increasing sequence} \{t_n\}_{n\geq 0}\text{with } T(t_n)x_n\to y\}.$$
I claim that if $y\in D_\mathcal{T}(0)$, then $y=0$.
Proof of claim. By $y\in D_\mathcal{T}(0)$, there exist $\{ x_n\}$ and $\{t_n\}$ with $x_n\to 0$ and $T(t_n)(x_n)\to y$. Take $t_n= k_n-s_n$ where $k_n\in\mathbb{N}$ and $s_n\in [0, 1)$.
Since $e^{t_nS}x_n\to y$, we have $e^{k_nS}x_n\to e^{\delta S}y$, where $s_n\to\delta$
By $\sigma(S)\subseteq \{z\in\mathbb{C}: Re(z)>0\}$, $\sigma{(e^S)}\subseteq \mathbb{C}\setminus \overline{D}$. This implies that there is $A>1$ with $A^{k_n}||x_n||\leq ||e^{k_nS}x_n||$. By $||e^{k_nS}x_n||\to ||e^{\delta S}y||$, we have $y=0$.
I dont sure that this proof is true. Please help me to know it.