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Given the Gelfand triple $V \subset H \subset V^*$, we for for $y \in L^2(0,T;V)$ also $y \in L^2(0,T;V^*)$ and thus $y \in L^1(0,T;V^*)$.

I understand this because if $y(t) \in V$ then $y(t) \in V^*$ since $V$ is a subset. The compactness of $[0,T]$ gives us the $L^1$ result.

Therefore, it makes sense to require that $y$ has weak derivative $y_t \in L^1(0,T;V^*)$, and to impose the additional condition $y_t \in L^2(0,T;V^*)$.

Why does it therefore make sense? Please can someone explain. Thanks.

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