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Jan
4
comment Does there exist a diagonal dominance concept for integral kernels?
On every inner product space every self-adjoint operator $P$is positive if for all non-zero $u$ we have $\langle Pu, u \rangle > 0$.
Jan
4
comment Does there exist a diagonal dominance concept for integral kernels?
@SimenK. But, I do not see understand why my decomposition would not be the way to proceed. Self-adjointness amounts to real eigenvalues and so on.
Jan
3
answered Does there exist a diagonal dominance concept for integral kernels?
Jan
3
comment Does there exist a diagonal dominance concept for integral kernels?
So, is the question about the positivity of Hilbert-Schmidt operators? Hit it with the spectral theorem to decouple your kernel in that case.
Jan
3
comment Growth $\beta X\setminus X$ of a Banach space $X$
Yes, thank you for the link. I had something like that in mind - Topologically dualize a test function space and sqeeuze the Banach space in between.
Jan
3
answered Fundamental solutions of PDEs
Jan
2
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.
Jan
2
revised Growth $\beta X\setminus X$ of a Banach space $X$
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Jan
2
revised Growth $\beta X\setminus X$ of a Banach space $X$
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Jan
2
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.
Jan
2
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...
Jan
2
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.
Jan
2
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.
Jan
2
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!
Jan
2
comment Initial segments order-isomorphic
It is quite peculiar someone went through the trouble of downvoting this like today.
Jan
2
revised Growth $\beta X\setminus X$ of a Banach space $X$
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Jan
2
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...
Jan
2
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!
Jan
2
revised Growth $\beta X\setminus X$ of a Banach space $X$
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Jan
2
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?