Jonas Teuwen
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 Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ Nothing. Does that answer your question 8-)? I would like to know what $\beta W_0^{m, p}\setminus W_0^{m, p}$ is nevertheless. Jan2 revised Growth $\beta X\setminus X$ of a Banach space $X$ added 236 characters in body Jan2 revised Growth $\beta X\setminus X$ of a Banach space $X$ added 1113 characters in body Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ At least it was stupid enough for someone to downvote it! I will attempt to phrase the question in a more meaningful way. Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ I wish to figure out how common this phenomenon is. It is clear that I cannot make $W^{m, p}$ larger in this sense as that would involve things having non-finite norm. But I can make $W_0^{m, p}$ --which is as much Banach as $W^{m, p}$-- larger. I am starting to wonder if the question would be of enough 'level' to ask on MO. I did not as I might need to be able to phrase it better... Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ @JonasMeyer Yes. As in my Sobolev space, if I take the circle and functions with compact support on the circle and take the closure in the norm $W^{m, p}$ where the functions are at least $C^m$ then I get a different space than if I start with $C^m$ functions without the compact support. Namely, the first space will have functions with trace null, whereas the second one will have much more. They only coincide whenever the complement if their domain is $(m , q)$-polar. Both norms make sense. But $W_0^{m, p}$ is a strict subset. Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ @JonasMeyer I had in mind the above Sobolev spaces on domains where one only can describe Dirichlet problems. I need to figure out how to characterize that. Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ @JonasMeyer Yes, I think I must have misunderstood the reply. I do want $X$ to at least continuously be included in $Y$. It is quite a simple example and was not really what I had in mind. Showing that Banach spaces are not so easy to grasp intuitively! :-). So you have understood the question! Jan2 comment Initial segments order-isomorphic It is quite peculiar someone went through the trouble of downvoting this like today. Jan2 revised Growth $\beta X\setminus X$ of a Banach space $X$ added 27 characters in body Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ suPset. I do know a little bit of elementary metric space theory. Maybe not that much, but at least that... Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ 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! Jan2 revised Growth $\beta X\setminus X$ of a Banach space $X$ added 383 characters in body Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ 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? Jan2 comment Growth $\beta X\setminus X$ of a Banach space $X$ But many interesting examples are not separable. Take even a countable tensor product of Hilbert spaces, a bosonic Fock space for instance. Jan1 awarded Custodian Jan1 reviewed Close Show $\sum_{n=1}^N e^{i n\theta}.$ is bounded for $0< \theta < 2 \pi$, $\forall N \in \{1,2,…\}$ Jan1 reviewed Reject What do $\pi$ and $e$ stand for in the normal distribution formula? Jan1 revised Growth $\beta X\setminus X$ of a Banach space $X$ edited body Jan1 asked Growth $\beta X\setminus X$ of a Banach space $X$