Let $E$ be a (complete) topological vector space, and $u:E\to E$ be continuous. Is it always true that if ${\rm Im}(u)$ is of finite codimension in $E$, then it is closed in $E$ or do we have to assume something on $E$? (It is OK if $E$ is Frechet by the open mapping theorem applied to ${\rm id}\oplus u:F\oplus E\to E$, where $F$ is a supplementary subspace to $E$.)
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2 Answers
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No. For $E$ take $X\oplus \ell_2$, where $X$ is the direct sum of continuum many copies of the scalar field under the direct sum topology. This is the largest locally convex topology on $X$ and any linear mapping from $X$ into a locally convex space is continous. Write $X=X_1 \oplus X_2$ with each $X_i$ isomorphic to $X$ and $\ell_2 = Y_1 \oplus Y_2$ with $Y_1$ and $Y_2$ closed and infinite dimensional. Let $Y_0$ be a dense codimension one subspace of $Y_1$. Define $T$ by having $T$ map $X_1$ one to one onto $X$, $T$ maps $X_2$ one to one onto $Y_0$, and $T$ maps $\ell_2$ isometrically onto $Y_2$. Then $T$ is one to one, continuous, and maps $E$ onto a dense codimension one subspace of $E$.
answered Aug 31, 2011 at 20:29
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$\begingroup$ Very nice, thanks! One could still wonder whether it is possible to construct such a counterexample without using axiom of choice as in your map from $X_2$ to $Y_0$... $\endgroup$ Commented Sep 1, 2011 at 6:34
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$\begingroup$ Something is needed in order to get a discontinuous linear functional on $\ell_2$ (which is equivalent to having a dense subspace of codimension one). $\endgroup$ Commented Sep 1, 2011 at 14:49
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$\begingroup$ Are you basically saying that one cannot produce non closed finite codimensional subspaces of complete TVS without using axiom of choice? That would be perfect as far as I am concerned (it would, in practice, be stronger than what I was hoping for). $\endgroup$ Commented Sep 1, 2011 at 19:45
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$\begingroup$ No, but IIRC you need something beyond ZF to prove the existence of a discontinuous linear functional on $\ell_2$. To prove the existence of Hamel bases in vector spaces (from which you get a discontinuous linear functional on $\ell_2$) it is enough to have the product of every family of two point sets non empty, and this axiom is strictly weaker than choice. For just $\ell_2$, less is needed. but I think not just ZF. Where are the logicians when we need them? There are least a zillion of them on MO. :) $\endgroup$ Commented Sep 1, 2011 at 22:37
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Bill: The existence of Hamel basis in (all) vector spaces is equivalent to the axiom of choice (see Blass, Andreas "Existence of bases implies the axiom of choice". Contemporary mathematics 31, 1984). For the existence of a Hamel basis in $l_2$ it is enough to have a well-ordering of the reals. Also, it is consistent with ZF that all linear functionals on $l_2$ are continuous (for instance, a model in which every set of reals has the property of Baire).
answered Sep 15, 2011 at 17:35
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$\begingroup$ I intended this to be a comment on Bill´s answer, but I don´t know how to do it. $\endgroup$ Commented Sep 15, 2011 at 17:38
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1$\begingroup$ Ramiro, you need 50 reputation to be able to comment on answers (mathoverflow.net/faq#reputation). $\endgroup$– B RCommented Sep 15, 2011 at 17:46
$L^p$ to itself whose range is a proper, finite codimensional subspace.$\endgroup$