7,309 reputation
12654
bio website fa.its.tudelft.nl/~teuwen
location Delft, Netherlands
age 28
visits member for 4 years
seen Aug 5 '13 at 10:51

About me

I'm a PhD candidate in analysis (harmonic analysis). I have a Master of Science degree in mathematics (analysis).

I am interested in many topics in analysis, but in particular: harmonic analysis, functional analysis, operator theory and special functions.

My email address is jonasteuwen@gmail.com.

I would appreciate it if you would correct my spelling and grammatical errors.


Me elsewhere


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 $X$
added 27 characters in body
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 $X$
added 383 characters in body
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?
Jan
2
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.
Jan
1
awarded  Custodian
Jan
1
reviewed Close Show $\sum_{n=1}^N e^{i n\theta}.$ is bounded for $ 0< \theta < 2 \pi$, $\forall N \in \{1,2,…\}$
Jan
1
reviewed Reject suggested edit on What do $\pi$ and $e$ stand for in the normal distribution formula?
Jan
1
revised Growth $\beta X\setminus X$ of a Banach space $X$
edited body
Jan
1
asked Growth $\beta X\setminus X$ of a Banach space $X$
Jan
1
awarded  Nice Question
Dec
29
comment Definition of $L^0$ space
Yes, indeed. But, needs a bit of work as you need to mention the topology first. 8-).
Dec
29
comment Definition of $L^0$ space
@Martin: True, I'll modify that. You want to end up with convergence in probability.
Dec
28
answered Definition of $L^0$ space