I wonder if the following is true
$$ L_{\mathrm{loc}}^{\infty} (\mathbb{R}_+ ; H^1 (\mathbb{R}^3)) \cap C(\mathbb{R}_+ ; H^{-1} (\mathbb{R}^3)) \subset C_{\mathrm{w}} (\mathbb{R}_+ ; H^1 (\mathbb{R}^3)). \hspace{2cm} (1) $$
For me $\psi \in C_{\mathrm{w}} (\mathbb{R}_+ ; H^1 (\mathbb{R}^3))$ means $\forall L \in H^{-1} (\mathbb{R}^3)$,
$$ \left| \langle \psi (t_n) - \psi (t) , L \rangle_{H^1 , H^{-1}} \right| \xrightarrow{n \rightarrow \infty} 0 ~~~~ \mathrm{when} ~~~~ t_n \xrightarrow{n \rightarrow \infty} t. $$
I can think I can conclude (1) via the following little argument.
For $L \in H^{-1}$, let $v_L$ denote the $H^1$-function given by the Riesz Representation theorem for $H^1$, i.e.,
$$ L (u) = \langle u , L \rangle_{H^1 , H^{-1}} = \langle u , v_L \rangle_{H^1}. $$
Similarly, for $v \in H^1$ let $L_v$ denote the functional determined by $v$ via the Riesz Representation theorem. Then, since $\psi \in C (\mathbb{R}_+ ; H^{-1} )$, we have
$$ \left| \langle \psi (t_n) - \psi (t) , L \rangle_{H^1 , H^{-1}} \right| = \left| \langle v_L , L_{\psi (t_n)} - L_{\psi (t)} \rangle_{H^1 , H^{-1}} \right| \leq \| L_{\psi (t_n)} - L_{\psi (t)} \|_{H^{-1}} \| v_L \|_{H^{1}} \rightarrow 0. $$
Is my reasoning correct, or am I misunderstanding the spaces and convergences involved?
Let $u$ be given as $u\in L^\infty(I, X)$ and $u\in C(\bar I,Y)$ for an intervall $I$, reflexive Banach space $X$ and normed space $Y$ with continuous embedding $X\hookrightarrow Y$. Then $u\in C_w(\bar I,X)$.
Take $t_n\to t$. Then $(u(t_n))$ is bounded in $X$, hence contains a weakly converging subsequence $u(t_{n_k})$ converging weakly to some $\tilde u$ in $X$. Due to the continuity in $Y$, we have $\tilde u= u(t)$. This means the weak subsequential limit does not depend on the subsequence. Moreover, each subsequence of $(u(t_n))$ contains another subsequence converging to $u(t)$ weakly in $X$. This implies $u(t_n)\rightharpoonup u(t)$ in $X$, which is the weak continuity.