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Is there an analytic characterisation of the Čech-Stone compactification (in the norm topology, which is a normal space) of a Banach space $X$? The reason I ask is because I want to know what the maximal compact extension of such a space could be. For instance, does it matter if the space has the Radon-Nikodým property, as these give a more finite dimensional feel to the measures on the space?

Related is the question that if $X$ is a normed space it can certainly be "too small" to be a Banach space, but if I recall correctly a Tychonoff space always has a completion, so there does exist a Banach space with the norm on $X$. Can $X$ also be too large to be a Banach space? That is, when $X$ itself is already Banach, in what case are larger sets (as in: strict inclusion) no more Banach?

More explicit:

  1. Let $(X, \|\cdot\|_X)$ be a Banach space, that is, a complete metric space in the induced metric. Can we describe $\beta X \setminus X$ with the topology induced by 1) the norm 2) the weak topology.
  2. As in 1., let $(X, \|\cdot\|_X)$ be a Banach space. When can we find a proper supset $Y$ of $X$ (that is $Y \supsetneq X$) such that $(Y, \|\cdot\|_X)$ is still a Banach space?

Example: The Sobolev spaces $W_0^{m, p}$ and $W^{m, p}$ which are the closure in the Sobolev $(m, p)$-norm, of the compactly supported smooth and $C^m$ functions respectively.

Even more explicit: Consider the open unit disk $\Omega$ in $\mathbf R^2$. On the $L^p(\Omega)$ space we can find subspaces which will carry the Sobolev norm $$\|u\|_{W^{m, p}(\Omega)} := \sum_{|\alpha| \leqslant m} \|D^\alpha u\|_{L^p(\Omega)}.$$ The compactly supported $C^m$ function in the closure of the $W^{m, p}$ is the space $W_0^{m, p}$. If I take the $C^m$ functions which have finite $W^{m, p}$-norm then I obtain the space $W^{m, p}$ (this is a result by Meyers and Serrin [1964]). Both spaces are obviously Banach. However, $W_0^{m, p}$ will only consist out of functions which have trace zero, that is intuitively speaking, functions which are zero on the boundary of the space. $W^{m, p}$ does not posses such property.

Additionally, we have the obvious chain of embeddings: $$W_0^{m, p} \hookrightarrow W^{m, p} \hookrightarrow L^p.$$

So, in general, the question is (1) is if $(X, \|\cdot\|_X)$ is a Banach space, does there exist a set $Y$ which contains $X$ and still is normed by $\|\cdot\|_X$ that is also Banach and has $X$ embedded?

Indeed, as Old John remarked in chat: $\|\cdot\|_X$ would not be defined on $X$. But, if I am correct and also, given the answer, I can make $X$ 'larger' and still the restriction to $X$ would keep its norm, and there $Y$ could sit.

I hope this is a better question.

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All separable infinite-dimensional Banach spaces are homeomorphic, so $\beta X \setminus X$ does not depend on any Banach space properties of $X$. Can you clarify what you mean by "too large to be a Banach space"? Metrizability of a space is an evident obstruction, cardinality isn't, viz $\ell^1(S)$. – Martin Jan 1 '13 at 23:58
But many interesting examples are not separable. Take even a countable tensor product of Hilbert spaces, a bosonic Fock space for instance. – Jonas Teuwen Jan 2 '13 at 0:04
I answered your question on the RN-property which a separable Banach space may or may not have, while $\beta X \setminus X$ is always the same in this case. – Martin Jan 2 '13 at 0:12
Ah - I see! But separable spaces are usually not the interesting cases. $L^\infty$? I would be quite surprised if you now tell me all non-separable ones are homeomorphic as well! – Jonas Teuwen Jan 2 '13 at 0:14
Then maybe you should separate the two (seemingly) unrelated questions. In any case: I doubt that there is much explicit to say about $\beta X \setminus X$. After all, what is $\beta \mathbb{R}^n \setminus \mathbb{R}^n$? As I mentioned in my first comment: for separable infinite-dimensional spaces you get $\beta \mathbb{R}^\infty \setminus \mathbb{R}^\infty$ irrespective of your Banach space. – Martin Jan 3 '13 at 0:05

Concerning the second part of your question, when $X$ is a Banach space, one can always find a "larger" Banach space containing $X$ as a proper, closed subspace.

Indeed, if $Y$ is another Banach space one can endow the direct product $X \times Y$ with a norm making it a Banach space with $X \simeq X \times \{ 0 \}, Y \simeq \{ 0 \} \times Y$ as closed subspaces. Such a norm is certainly not unique, but one example is given by $|| (x, y) || = || x || _{X} + || y || _{Y}$. There exists Banach spaces $Y$ of arbitrary cardinality, since $L^{1}(\Omega)$ is a Banach space of cardinality at least $| \Omega |$ for any set $\Omega$, so one can make $X \times Y$ arbitrarily large too.

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The question is giving a Banach space $X$ with a norm $\|\cdot\|_X$, can we enlarge $X$ to make the enlargement with norm $\|\cdot\|_X$ not complete anymore? – Jonas Teuwen Jan 2 '13 at 0:05
@JonasTeuwen: I do not understand what that means. Could you please elaborate? – Jonas Meyer Jan 2 '13 at 19:48

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