How to interpret $|\nabla u|$ when using distributions? In Sobolev spaces, to define the respective norm, we need to be able to interpret $||\nabla u||_{L^2(\Omega)}=\left( \int_{\Omega} |\nabla u|^2 \ dt\right)^{1/2}$
However, how do we interpret $|\nabla u|$ or even $|\dot u|$ when we are derivatives in the sense of distributions/generalised functions?
Supposedly, a distribution is a functional, and their norms can be calculated like any usual operator.
$$||F||_{B(H_1,H_2)}=\sup_{||h_1||_{H_1}=1} ||Fh_1||_{H_2}$$
which in our case would be $$||\dot u||=\sup_{||h_1||_{\mathcal{D}(\Omega)}=1} |Fh_1|_{\mathbb{R}}$$,
where $\mathcal{D}(\Omega)$ is the space of test functions. What would be the norm in $\mathcal{D}(\Omega)$?
 A: May I suppose you are working with $\Omega \subset \mathbb{R}^{n}$? I don't know if this is the most correct interpretation, but it is very useful in a lot of applications. You know how to define, given $u \in L^{2}(\Omega)$ the distribution 
$T_{u} \in \mathcal{D}'(\Omega)$ by $$T_{u}(\varphi) = \int{u(x)\varphi(x)}dx$$ and then you define $\frac{\partial T_{u}}{\partial x_{i}}$ in the distributional sense.  It may happen to exist $g \in L^{2}(\Omega)$ such that $\frac{\partial T_{u}}{\partial x_{i}} = T_g$, in this case, we say $g = \frac{\partial u}{\partial x_{i}}$ in the distributional sense and $$\left\|\frac{\partial u}{\partial x_{i}}\right\|_{L^{2}} = \|g\|_{L^{2}}$$ with this norm being the usual in $L^{2}$.
In this way, if $\nabla u = \left(\frac{\partial u}{\partial x_{1}}, ..., \frac{\partial u}{\partial x_{n}} \right) \in [L^{2}(\Omega)]^{n}$ we define
$$ \|\nabla u\|_{L^{2}}^{2} = \sum_{i=1}^{n}{\left\|\frac{\partial u}{\partial x_{i}}\right\|_{L^{2}}^{2}}$$
You find more about this on books dealing with Sobolev spaces.
(Edit: I hope this answer your question, I am not sure if I got what is your doubt)
