# 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.

• The $D^{\alpha}$ need to be re-interpreted as weak derivatives for this to make sense. So, strictly speaking, they are not the same norms. They just look the same by abuse of notation. Commented Oct 1, 2022 at 9:20

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 $$X$$.
• Is that unique extension obtained by extending the composition $\lVert\cdot\rVert_X\circ h^{-1}$? $h$ is an isometric isomorphism from $X$ to a dense subspace of the completion $Y$. Thank you.
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.