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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.

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Well, I think that this is an example of the adjoint operator. Think of the derivative as being just a linear operator. – Ilya Jan 16 '13 at 21:08

[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]!

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I'm confused: I suppose $C_{0}^\infty(0,1)$ denotes the functions with compact support in $(0,1)$. Aren't these functions dense in $L^2(0,1)$ with the usual norm? So where does this repeated strict inclusion $H^{0}_0(0,1) \subsetneq L^2(0,1)$ come from? And what do you mean by saying that $D^\ast$ a priori densely defined? It seems you need some sort of symmetry property or you need to know that $D$ is closable in order to say that, no? The domain of the adjoint of an densely defined operator can be zero. – Martin Jan 17 '13 at 6:47
@Martin Gah, of course you're write on all accounts---I trust what I wrote is now correct. I've no idea what I was thinking before... – Branimir Ćaćić Jan 17 '13 at 6:53
Where by "write" I mean "right." [sigh] – Branimir Ćaćić Jan 17 '13 at 7:00
Yes, this looks more like what I expected. Thanks for the edit. – Martin Jan 17 '13 at 7:01
@BranimirĆaćić Thanks for the answer. My hope was to avoid these Lebesgue, Sobolev, and $C_0^\infty$ spaces and have some further generalization to abstract Hilbert space. Apart from writing the differentiation as a linear operator I guess there is not much more I can do? – pde_lover Jan 17 '13 at 11:14

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