$\sup_t |T(t)|<+\infty$ implies $\sup_t |T(t)^*|<+\infty$? Let $X$ be a Banach space. $T(t)$ a family of bounded operators for $t\in\mathbb{R}$. $T(t)^*$ is the adjoint operator of $T(t)$.
If $\sup_t |T(t)^*|<+\infty$ , then by Hahn-Banach, there's a $x^*\in X ^*$ with $|x^*|=1$ such that $|T(t)x|=\langle x^*,T(t)x \rangle$, we have then
$$|T(t)x|=|\langle x^*,T(t)x \rangle|=|\langle T(t)^* x^*,x \rangle|\leq |T(t)^*||x|$$
and we have $\sup_t |T(t)|<+\infty.$
My question is the converse i.e. if $\sup_t |T(t)|<+\infty$ when do we have $\sup_t |T(t)^*|<+\infty$ ?
Note: As David Mitra mentioned we have even $|T(t)|=|T(t)|^*$.

This was resolved in comments: $\|T^*\|=\|T\|$ for every bounded linear operator between normed spaces. In fact, we only need the easier part here: $\|T^*\|\le \|T\|$, which comes directly from the fact that $T^*$ acts on linear functionals by composing them with $T$.
