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.
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 $X$.