Given $v\in L^p(\Omega)$, $v$'s norm is defined by $$ \lVert v\rVert_{L^p(\Omega)} = \left[\int_\Omega |v|^p dx \right]^{1/p} $$
How is $v$'s norm defined if $v\in L^p(\Omega)^d$?
In other words, $v$ is not a scalar, but a vector field over $\Omega$. I saw this denoted as $L^p(\Omega)^d$ in some publications, where $d$ is the dimension.
EDIT:
I did a bit of thinking, and following @herb steinberg's comment, came up with something. I don't have a mathematician's education, and I could use some constructive criticism.
Following Marsden and Hughes' Mathematical Foundations of Elasticity, Definition 2.2,
Let $\Omega\subset\mathbb{R}^n$ be an open set and let $x\in\Omega$. The tangent space to $\Omega$ at $x$ is simply the vector space $\mathbb{R}^n$ regarded as vectors emanating from $x$; this tangent space is denoted $$T_x\Omega = \{\mathbb{R}^n, x \in \Omega \}$$ The tangent bundle of $\Omega$ is the product $$T\Omega = \Omega\times\mathbb{R}^n \quad(\text{or}\;\; \Omega\times T_x\Omega)$$ consisting of pairs $(x,v)$ of base points $x$ and tangent vectors at $x$.
So when I say a vector field over $\Omega$, I mean $v\in T\Omega$.
Also, let
$$\langle \cdot,\cdot\rangle: T_x\Omega\times T_x\Omega\to \mathbb{R}$$
denote the inner product at the tangent space of $x$. The tangent space is then equipped with the norm $$|v|= v \mapsto \sqrt{\langle v,v\rangle}$$
Thus I can extend the original definition of the Lp spaces to account for vector fields. For vector fields $v \in T\Omega$, the vectors at every point belong to a tangent space such that $v(x)\in T_x\Omega$. Then $L^p(T\Omega)$ is the space equipped with the norm
$$ \lVert v\rVert_{L^p(T\Omega)} = \left[\int_\Omega |v(x)|^p dx \right]^{1/p} = \left[\int_\Omega {\langle v(x),v(x) \rangle}^{p/2} dx \right]^{1/p} $$
Definitions of Sobolev norms can follow analogically. Do you see a problem with this derivation?
Naively, we have
\begin{align} \| v\|_{L^p(\Omega)^d} = \sum^d_{j=1} \|v_j\|_{L^p(\Omega)}. \end{align} Since $d$ is finite, then we could also use \begin{align} \| v\|_{L^p(\Omega)^d} = \left(\sum^d_{j=1} \|v_j\|_{L^p(\Omega)}^p\right)^{1/p}. \end{align} since all finite dimensional norms are equivalent.