Sobolev Norm: Making sense of integral over $\mathbb{R}^{n^k}$ for $k \ge 1$ The Sobolev space $W^{k,p} (\Omega)$ is defined as
\begin{equation}
W^{k,p}(\Omega)=\{u \in L^{p}(\Omega) | D^{\alpha}u \in L^{p}(\Omega), \forall \alpha \in \mathbb{N}^{n} \mid |\alpha| \leq k\}
\end{equation} and norm is defined by
$$ ||u||_{W^{k,p}(\Omega)} = \Big({\sum_{|\alpha| \leq  k} 
\int_{\Omega} |D^{\alpha} u|^p}\Big)^\frac{1}{p} $$ for $1\le p< \infty $
How does one make sense of integral when $|\alpha| \ge 1$?
For $|\alpha|= 1$, the integrand is $ |D^{1} u|^p= |(D_1u, D_2u,...,D_nu)|^p$, i.e the gradient, a vector in $\mathbb{R}^n$.
For $|\alpha|= 2$, the integrand is $ |D^{2} u|^p= |(D_iD_ju)_{1\le i,j \le n}|^p$, i.e the Hessian, a vector in $\mathbb{R}^{n^2}$.
How are they supposed to be integrated considering the fact that we are integrating over a subset of $\mathbb{R^n}$?
Any hints will be appreciated. Thanks!
 A: First, to expand on the comment that Hans Engler left, the $\alpha$ are multiindices. So $D^1 u, D^2 u$ don't strictly make sense; $\alpha$ should be an $n$-tuple with entries in the nonnegative integers. For example, if we let $\alpha_i = (0, \ldots, 1, \ldots, 0)$ be the multiindex with a 1 in the $i$th entry and zeros elsewhere, then
$$
D^{\alpha_i}u = \frac{\partial}{\partial x_i} u
$$
is the usual $i$th partial of $u$. So the $D^\alpha u$ are not vectors, but scalars (as long as $u : \mathbb{R}^n \to \mathbb{R}$ is a scalar function). Thus, in the expression
$$
\|u\|_{W^{k,p}} = \left(\sum_{|\alpha| \leq k} \int_\Omega |D^\alpha u(x)|^p\, dx\right)^{1/p}
$$
is just a sum of integrals of scalar functions. This can be confusing with the notation (found in Evans, and I'd imagine in many other places) of $Du = \nabla u$. In this case Evans uses $Du$ to denote the vector $(\frac{\partial}{\partial x_1} u, \ldots, \frac{\partial}{\partial x_n} u)$ i.e. the gradient. But $D^\alpha u$ is just one mixed partial derivative, not a vector consisting of all of them.
