Let $V$ be a separable Banach space and $H$ be a separable Hilbert space such that \begin{equation} V \subset H \subset V' \end{equation} and the inclusion maps are continuous with dense images. Here $V'$ is the dual space of $V$.
Then, the notion of Gelfand triple as in Section 7.2 of "Nonlinear Partial Differential Equations" (2000) by Tomas Roubicek implies that if $u(t) : [0,1] \to V$ satisfies \begin{equation} u \in L^p(0,1 ; V) \text{ while } \partial_t u \in L^{q}(0,1; V') \text{ where } p,q \in (1,\infty) \text{ with } \frac{1}{p}+\frac{1}{q}=1 \end{equation} then, $t \to \lVert u(t) \rVert_H^2$ is absolutely continuous with \begin{equation} \lVert u(t_2) \rVert_H^2 - \lVert u(t_1) \rVert^2_H = 2\int_{t_1}^{t_2} \langle [\partial_tu](t), u(t) \rangle_{V' \times V}dt \text{ for all } t_1, t_2 \in [0,1] \end{equation}
Now, I wonder if a "converse" of this Gelfand triple also holds true.
That is. for a given mapping $u(t) : [0,1] \to V$, assume that $t \to \lVert u(t) \rVert_H^2$ is absolutely continuous and $\partial_t u$ exists as an element of $L^{q}(0,1; V')$ for some $q \in (1,\infty)$. Then, \begin{equation} \text{ Is it necessarily true that } u(t) \in L^{p}(0,1 ; V) \text{ for } p \text{ satisfying } \frac{1}{p}+\frac{1}{q}=1? \end{equation}
I think this is quite plausible but cannot see if it is indeed true.
Could anyone please help me?
