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Let $(V, \lVert\,\rVert)$ be a Banach space. I want to produce a non-complete norm $\lVert\,\rVert'$ on it such that $\lVert v\rVert' \leq \lVert v\rVert$ for all $v$ in $V$. Given a continuous injection $\varphi\colon V \to W$ with a non-closed image, taking $\lVert v\rVert' = C \lVert \varphi(v)\rVert$ for some $C$ provided by continuity (preimage of a unit open ball from $W$ is open in $V$, hence contains a smaller ball, so we have an inequality of norms) results in an incomplete norm. However, I wasn't able to find a simple (and non-circular) way to map a space into a non-closed subspace.

The question is inspired by a proof of the fact that $V$ is homeomorphic to $V \backslash K$ for all compact subsets $K$ from a book "Selected Topics in Infinite-Dimensional Topology" (C. Bessaga, A. Pelczynski). The existence of a non-complete norm is an essential part of the proof. In the book it is obtained first for separable spaces (by embedding into $l^2$ as a non-closed subspace) and then in general by combining the new norm on a separable subspace and old norm on the entire space.

That proof is short and quite simple, but not particularly elegant. It would be nice to have a uniform treatment for separable and non-separable spaces. (And it would be really neat if every Banach space could inject as a proper non-closed subspace of itself, but that doesn't sound likely).

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    $\begingroup$ To get such a continuous norm on $V$, you need to produce a balanced, open convex subset $B'$ of $V$ containing the unit ball, but no subspace, while ensuring that the associated gauge of $B'$ is not complete. The Bessaga-Pelczynski approach feels quite natural: do something more-or-less explicit on a separable subspace $W$ and distort the original norm on $V$ with the new norm on the subspace. Looks elegant to me. $\endgroup$ – Martin May 19 '13 at 14:43
  • $\begingroup$ I've read the original proof once again and now agree that it is elegant. $\endgroup$ – Dmitry May 19 '13 at 18:10

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