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## Linear maps on incomplete inner-product spaces

Let $(M,g)$ be a compact Riemannian manifold without boundary. Let $V$ be the vector space $C^4(M) \times C^4(M)$, where addition and scalar multiplication are defined in the obvious way. Suppose we turn $V$ into an incomplete inner-product space with a inner product given by the second variation of a $C^2$ functional $E$ on $V$, $E=E(w,v)$, with respect to the second argument $v$. (It is symmetric and positive definite in this case.) Now let $X$ be a nowhere dense subset of $V$. Given this, can there exist a linear map $B$ from $X$ to $V$ that is one-to-one and onto?

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Are you sure you don't want more conditions on $B$, e.g. continuity? As it stands the answer is pretty trivially yes. You are just asking whether $V,X$ are isomorphic as vector spaces, i.e. have the same Hamel dimension. Every infinite-dimensional normed space $V$ has a nowhere dense subspace $X$ with the same Hamel dimension. Just take $X$ to be the kernel of any nonzero continuous linear functional. Since $X$ has codimension 1, it has the same dimension as $V$. – Nate Eldredge Nov 13 2011 at 13:42
@Nate Eldredge: Thanks. When I wrote this I didn't believe that $B$ was continuous, but I have been able to establish this now. This, of course, completely alters the situation. – Viktor Bundle Nov 13 2011 at 22:51
Vote to close as no longer relevant, in view of the comments. Or, if it is still relevant I suggest to edit/correct the question. – quid Dec 26 2011 at 1:08