Timeline for Surjectively isometric normed spaces: Hamel vs (extended) Schauder dimension
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| Oct 28, 2011 at 11:33 | comment | added | Salvo Tringali | As a consequence, you're decidedly right with your last remark too, and the original question is no longer that interesting (at least, as is). Also, your guess is right - I was implicitly assuming the existence of an (extended) Schauder basis. Now, what's the right way to go in this case? Should I edit the OP and update it accordingly? Or would it be better to open a new topic? | |
| Oct 28, 2011 at 11:26 | comment | added | Salvo Tringali | @Pietro Majer. Sorry for the big delay in replying! I was indeed sloppy in many respects, let's take this step-by-step. First, the question about the "realisation" of a complex normed space (btw, is there a standard naming for referring to this process? We have the term complexification for the dual one, but I'm not aware of any analogue). Finally, I agree that it's enough to deal with the real case (for gloomy reasons, I thought that some extra compatibility between the original norm and the one obtained by restriction of the scalar field should have been necessary for this to work). | |
| Sep 16, 2011 at 12:03 | comment | added | Pietro Majer | Also I forgot to say that the question may be of interest in the hypothesis of homeomorphic X and Y; as you observed if the normed spaces X and Y are assumed to be isometric, by Mazur-Ulam theorem they are actually linearly isometric, and they share whatever (real)normed-space-theoretic notion of dimension you like. | |
| Sep 16, 2011 at 8:11 | comment | added | Pietro Majer | Actually, Evans&Tapia's paper does not claim that all separable Banach spaces do have "extended Schauder basis". I suspect it's not true indeed. In this case, question 2 has to be reformulated. Maybe assume that X and Y have an extended basis. I also implicitly assumed this in my answer, that however now I withdraw as less interesting. | |
| Sep 13, 2011 at 18:58 | comment | added | Pietro Majer | The norm of the complex normed space X also makes X a real normed space, when we restrict the field of scalars. | |
| Sep 13, 2011 at 13:48 | comment | added | Salvo Tringali |
@Pietro Majer. I'm not yet convinced with your 1st comment on the finite dimensional case. Let $\mathbf{X} \equiv (X, \|\cdot\|)$ be a complex normed space. Of course, we agree that $X$ can be as well regarded as a real vector space, and indeed $\dim_\mathbb{R}(X) = 2\, \dim_\mathbb{C}(X)$. But I'm not so sure that $\mathbf{X}$ can also be seen as a real normed space (essentially because $|a|+|b| \ne |a+ib|$ for arbitrary $a,b\in\mathbb{R}$). What do I miss here? Still, as you pointed out below, the finite dimensional case is settled by looking at the Hausdorff-Besicovitch dimension.
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| Sep 13, 2011 at 13:38 | comment | added | Salvo Tringali | Note. By error, I deleted my previous comment (dated 25 Aug 2011). Now, I've tried to remember what it should be and posted it again. Sorry for the inconvenience! @Samuele. Not really what I was seeking, but still useful. Thanks! | |
| Sep 13, 2011 at 13:34 | comment | added | Salvo Tringali | @Pietro. You're absolutely right with your 2nd comment! The key point is that the Hamel dimension of an $\infty$-dimensional (real or complex) Banach space cannot be less than $|\mathbb{R}|$ (even if the CH fails!) as proved in H. E. Lacey, The Hamel Dimension of any Infinite Dimensional Separable Banach Space is $\mathfrak{c}$, The AMM, Vol. 80 (1973), p. 298. I confess, this is somehow surprising for me as I really thought the answer should not depend on the completeness of the space, and I'm editing the OP accordingly to pose the question in the right (normed) setting. | |
| Aug 25, 2011 at 16:15 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Addenda
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| Aug 25, 2011 at 15:13 | comment | added | Samuele | As for question 1, there is an article by J. Bourgain, Real isomorphic complex Banach spaces need not be complex isomorphic, in Proc. AMS, vol 96, n.2 (1986) where an example is given of two complex Banach spaces which are isometric but not linearly isomorphic (over complex numbers). I don't remember the details, so I'm not sure it has really something to do with your question, but it could be a start. | |
| Aug 25, 2011 at 13:28 | comment | added | Pietro Majer | Also, isn't the Hamel dimension of a (real or complex) infinite dimensional Banach space equal to its cardinality? | |
| Aug 25, 2011 at 13:07 | comment | added | Pietro Majer | As to the first question, isn't the real case sufficient? A complex vector space is also a real vector space by restriction of the scalar field, and the real Hamel dimension (finite or not) is twice the complex Hamel dimension. | |
| Aug 25, 2011 at 12:29 | history | asked | Salvo Tringali | CC BY-SA 3.0 |