Let
- $\lambda$ denote the Lebesgue measure on $\mathbb R$
- $I\subseteq\mathbb R$ be an interval
- $X,Y$ be a $\mathbb R$-Banach space such that $X$ can be embedded into $Y$
- $1\le p,q\le\infty$
- $u\in L^p(I,X)$
We say that $u$ has a weak derivative in $L^q(I,Y)$ $:\Leftrightarrow$ $\exists v\in L^q(I,Y)$ with $$\int_I\varphi'u\;{\rm d}\lambda=-\int_I\varphi v\:{\rm d}\lambda\;\;\;\text{for all }\varphi\in C_c^\infty(I)\tag 1\;.$$ In that case, we write $$\frac{{\rm d}u}{{\rm d}t}=v\tag 2\;.$$
The question is: Does the notation introduced in $(2)$ makes sense?
To be precise: I've read that we can prove the following: Let $Z$ be another $\mathbb R$-Banach space such that $Y$ is densely embedded into $Z$ and such that $Z'$ is separable (which condition on $Z$ would imply the seperability of $Z'$?). Moreover, let $1\le r\le\infty$ and $w$ be the weak derivative of $u$ in $L^r(I,Z)$. Then $v=w$ $\lambda$-almost surely.
How can we prove this statement?
To prove the statement note that if $$\int_I v\varphi=\int_I w\varphi\tag{A}$$ for all $\varphi\in C_c^\infty$, then $v=w$ a.e. by the Du Bois Reymond Lemma. Note that the density of the embedding $Y\subset Z$ and the separability of $Z'$ were not used.
With respect to the other question: if $Z$ is separable and reflexive, then $Z'$ is separable (and reflexive). If $Z$ is separable but not reflexive, then $Z'$ can be not separable ($Z=\ell^1$ is an example).
Edit.
The above argument works if the embedding $\iota$ from $X$ to $Y$ and the embedding $\kappa$ from $Y$ to $Z$ are linear and continuous. In this case, the precise meaning of $(A)$ is $$\int_I \kappa(v)\varphi=\int_I w\varphi\tag{A}$$ and the precise conclusion is $\kappa(v)=w$ a.e. However this is not the case that the OP have in mind (see the comments below).