Generalizing the weak derivative

Question

I am wondering about the weak derivative in time. We say f has a weak derivative f' if $$\int_0^T f\phi' = -\int_0^T f'\phi$$ for all $\phi \in C_0^\infty(0,T)$.

This definition uses the $L^2$ inner product. Can I generalise this to some Hilbert space $H$? Is there such a notation of derivative? It would be something like $$(f, \phi')_H = -(f', \phi)_H$$ but how does define $\phi'$? What is the space it lies in? Because it makes no sense to consider a derivative of element of abstract Hilbert space. Any references to this area is appreciated.

Answer 1

[Massive edits to correct the result of an earlier what-on-earth-was-I-thinking moment]

Ilya's comment is correct. In that case, then, there are really two things going on here:

1. The operation $\phi \mapsto \phi'$ defines a densely defined operator $D$ on $H = L^2(0,T)$ with dense domain $\operatorname{Dom}(D) = C_0^\infty(0,T)$. This therefore admits, a priori, an adjoint, that is, an operator $D^\ast$ defined on some subspace of $H$ such that $\left\langle f,Dg \right\rangle = \left\langle D^\ast f,g\right\rangle$ for all $f \in \operatorname{Dom}(D^\ast)$ and $g \in \operatorname{Dom}(D)$.
2. It just happens to be the case, however, that for this specific operator $D$, $\left\langle f,Dg \right\rangle = \left\langle -D f,g\right\rangle$ for $f,g \in \operatorname{Dom}(D)$, so that $\operatorname{Dom}(D) \subseteq \operatorname{Dom}(D^\ast)$ with $D^\ast$ restricting to $-D$ on $\operatorname{Dom}(D)$. Indeed, you have that $-D^\ast$ is skew-adjoint, and in particular is the closure of $D$.

Thus, from an abstract perspective, because of the fact that $D$ is skew-symmetric on its domain, $H^{1}(0,T)$ is just (as a vector space) $\operatorname{Dom}(D^\ast)$, and weak differentiation is the extension of $D$ from $C^\infty_0(0,T)$ to $H^1(0,T)$ given by $-D^\ast$.

I hope I haven't made too much of a mess of things [this time around]!