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Suppose $(B,\|\cdot\|)$ is a Banach space, $V\subset B$ a dense subspace, and $V$ is equipped with a norm $\|\cdot\|_V$ such that $\|x\|_V = \|x\|$ for all $x\in V$.

Is $(B,\|\cdot\|)$ a completion of $(V,\|\cdot\|_V)$ with respect to the $\|\cdot\|_V$ topology? I.e., can the spaces be considered the same/identical up to some isometry?

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closed as too localized by Yemon Choi, Alain Valette, Bill Johnson, Andreas Blass, quid Jan 15 '13 at 20:15

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Yes, $B$ is a completion of $V$. Note that the completion does not make sense. – Jochen Wengenroth Jan 11 '13 at 9:44
By definition, $(Id,(B,\lVert\cdot\rVert))$ is a completion. All the other completions can be obtained by an isometry. What do you mean by "the norms are identical"? – Davide Giraudo Jan 11 '13 at 9:44
I agree wirh Jochen. Perhaps the OP had in mind a particular construction of a completion, e.g. via equivalence classes of Cauchy seuences? – Yemon Choi Jan 11 '13 at 10:07
Thanks, yes, I meant a completion by Cauchy sequences, or at least that the completion and $B$ can be considered the same space via isometry. – Nemis L. Jan 11 '13 at 10:32
At first, I thought I must be misunderstanding the question, as the hypothesis says $\Vert\cdot\Vert_V$ is the restriction to $V$ of the norm $\Vert\cdot\Vert$ of $B$. So the question asks whether a Banach space is the completion of each dense subspace (with the induced norm). Since that's practically the definition of completion, I didn't see a real question. According to the comments, it seems the question is whether the definition I had in mind agrees with the construction by Cauchy sequences. But that's surely covered in standard textbooks. So I vote to close. – Andreas Blass Jan 12 '13 at 16:09
up vote 2 down vote accepted

This is standard, and the answer has been been indicated in the comments. Recall that in a normed space $X,$ the triangle inequality easily yields that $| \|x\| - \| y \| |\leq \|x-y\|$ for all $x,y \in X.$

Now let's turn to your dense subspace $V$ of the Banach space $B.$ Take an element $b \in B.$ There is a sequence $(v_{n})$ of elements of $V$ such that $v_{n} \to b$ (with respect to the norm on $B).$ Then by the above remark, the sequence $( \|v_{n} \|)$ is a Cauchy sequence of real numbers whose limit is the real number $\| b \|$. Hence $\| b \|$ is uniquely specified in terms of $ \| \|_{V} $ since you assume that $\| \|$ and $\| \|_{V}$ agree on $V.$ Also, $\|b\|$ is the same as would be assigned if considering $b$ as an element of the completion of $V.$ Hence $B$ embeds isometrically as a susbpace of the completion of $V.$ However, it is clearly dense in that completion, as $V$ already is, and it is a closed subspace of that completion, since any Cauchy sequence of elements of $V$ has a limit which lies in $B.$ Hence $B$ is indeed isometrically isomorphic to the completion of $V.$

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Great answer, thanks! – Nemis L. Jan 11 '13 at 10:40

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