Weak convergence equivalent to weak convergence at $0$

Question

Let $(X, \lVert \cdot \rVert)$ be a Banach space and $(T(t))_{t\geq0}$ a semigroup of linear operators , i.e for all $t, s \geq 0$ we have $T(t):X \to X$ is a linear operator, $T(t+s)=T(t)T(s)$ and $T(0)=$ Id.
For arbitrary but fixed $x\in X,~l \in X^{'}$ consider the mapping $$ F_{l, x}: \mathbb{R}_+ \to \mathbb{R}, t \mapsto F_{l, x}(t):=l(T(t)x) $$ Is continuity of $F_{l,x}$ at $t=0$, i.e. $$ \underset{t\downarrow 0}{\lim} l(T(t)x) = l(x) $$ enough for continuity everywhere?
[The motivation is 1. that weak continuity implies strong continuity, and 2. that there's the equivalent statement for strong continuity.]

Answer 1

Yes because $$ l\big (T_{t+h}(x)\big ) - l\big (T_t(x)\big ) = l\Big (T_{t}\big (T_h(x) - x\big )\Big ) = l'\big (T_h(x) - x\big ), $$ where $ l'=l\circ T_t. $