Why is the completion $Y$ of a normed space $X$ equipped with the norm $\lVert\cdot\rVert_X$? This is a question arising from my study on the Sobolev space, but it may be applicable to other spaces that are also defined by completing a normed space. Let $\Omega$ be a nonempty open subset of $\mathbb{R}^n$. Then the Sobolev space $H^{k,p}(\Omega)$ is defined as the completion of
$$S:=\{u\in C^\infty(\Omega):\lVert u\rVert_{k,p}<\infty\}$$
in the norm
$$\lVert u\rVert_{k,p}=\left(\sum_{|\alpha|\leq k}\lVert D^\alpha u\rVert_{L^p(\Omega)}^p\right)^\frac{1}{p}.$$
My question is, why does $H^{k,p}$ share the same norm $\lVert\cdot\rVert_{k,p}$ that is initially equipped on the vector space $S$ in some literature?
Surely, I know exactly how to completing a normed space using Cauchy sequences, and I'm familiar with the statement that every normed space is isometrically isomorphic to a dense subspace of a Banach space. But somehow I don't know why we can evaluate the $H^{k,p}$ norm for the functions in the Sobolev space. Isn't there a different norm for the Sobolev space that results from the completing process? Why is that? Thank you.
 A: This is by definition of completion.  $(Y,\|\cdot\|_Y)$ is a completion of $(X,\|\cdot\|_X)$ iff $(Y,\|\cdot\|_Y)$ is complete (duh!) and $(X,\|\cdot\|_X)$  is (or can naturally be viewed as) a dense subspace of it. More specifically subspace means not just subset but rather subobject in the category of normed spaces, that is, $\|\cdot\|_Y$ restricted to $X$ coincides with $\|\cdot \|_X$.
Therefore, if the given norm on $\|\cdot\|_X$ has an established name and/or symbol, it is common to keep the name/symbol for the norm on the completed space, even though perhaps some of the originally definiing properties do not immediately make sense in the completed space.
A: On a normed space $X$, the norm function $x \mapsto \|x\|_X$ is uniformly continuous on $X$.  This is a consequence of the triangle inequality.  But any (real) uniformly continuous function extends uniquely to a continuous function on the completion $Y$.  In this case, the norm on $Y$ given by the same formula is of course that (unique continuous) extension of the original norm on $X$.
