Let $(X,d)$ is a metric space, $T:X\rightarrow X$ is a continuous self-map.
Def1. $T$ is strongly sensitive if there exist $\epsilon>0$, for any $x\in X$ and any positive number $\delta>0$, we can find a point $y\in B(x,\delta)$ and $n\in\mathbb{N}$, such that $d(T^nx,T^ny)>\epsilon$.
Def2. $T$ is (weakly) sensitive if there exist $\epsilon>0$, for any $x\in X$ and any positive number $\delta>0$, we can find a point $y\in B(x,\delta)$ and $n\in\mathbb{N}\cup\{0\}$, such that $d(T^nx,T^ny)>\epsilon$.
Of course, Def1 implies Def2, my quesiton is Def2 implies Def 1, too? If not, does there exist a counterexample?
I believe Def 2 does imply Def 1.
Suppose $T$ is (weakly) sensitive.
Then, there exists an $\epsilon > 0$ such that [rest of Def 2].
Let $x \in X$ and $\delta > 0$ be chosen arbitrarily.
Pick a number $0 < \delta' < \min\{\epsilon,\delta\}$.
Then by our choice of $\epsilon$, there exists a $y \in B(x,\delta')$ and $n \in \mathbb{N}\cup\{0\}$ such that $d(T^nx,T^ny) > \epsilon$.
If $n = 0$, then $d(x,y) = d(T^0x,T^0y) > \epsilon \ge \min\{\epsilon,\delta\} > \delta'$, a contradiction since $y \in B(x,\delta')$.
Thus, $n \in \mathbb{N}$. Also, since $\delta' < \min\{\epsilon,\delta\} \le \delta$, we have $y \in B(x,\delta') \subseteq B(x,\delta)$.
Hence, $\epsilon > 0$ is such that for any $x \in X$ and any $\delta > 0$, there exists a $y \in B(x,\delta)$ and $n \in \mathbb{N}$ such that $d(T^nx,T^ny) > \epsilon$.
Therefore, $T$ is strongly sensitive.