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seen Jul 19 at 8:00

Jul
14
comment How to evaluate the wiener measure of sets?
Thank you very much for your help! I have to think some time about your approach and probably will tell you whether it helps.
Jul
14
comment How to evaluate the wiener measure of sets?
@Nate Eldredge: I want to know the measure of the set G (which I wrote down above), which is quite explicit. Don"t you think so?
Jul
14
comment How to evaluate the wiener measure of sets?
@Kjos-Hanssen: Why do you mean the set can't be a Borel set?
Jul
14
comment How to evaluate the wiener measure of sets?
@Martin Hairer: Thank you very much for your comment. I will consider the construction via Kolmogorov soon. But I'm not a probabilist.
Jul
13
asked How to evaluate the wiener measure of sets?
May
8
accepted The first eigenvalue of the Schrödinger operator is simple.
May
8
accepted What is the right initial domain for the Dirichlet-Laplacian on a bounded domain?
Jan
4
comment What is the right initial domain for the Dirichlet-Laplacian on a bounded domain?
Ok thank you. So I assume that the Friedrichs extension of $-\Delta:C^{\infty}_0 (\Omega)\subset L^2(\Omega)\rightarrow L^2(\Omega)$ is equal to the other one. Is that right?
Jan
4
asked What is the right initial domain for the Dirichlet-Laplacian on a bounded domain?
Jul
27
awarded  Disciplined
Jul
25
awarded  Yearling
Jul
24
awarded  Commentator
Jun
26
accepted Eigenfunctions of Schrödinger Operators on the boundary
Jun
22
comment Eigenfunctions of Schrödinger Operators on the boundary
What do you mean by "test functions". Smooth functions with compact support (in our case this would mean all smooth functions)?
Jun
22
comment Eigenfunctions of Schrödinger Operators on the boundary
Thank you for the comment. Actually I wasn't looking for a physical reason (there is no ambient space in my situation).
Jun
20
asked Eigenfunctions of Schrödinger Operators on the boundary
May
20
comment The first eigenvalue of the Schrödinger operator is simple.
You are using the elliptic regularity. What happens if the potenital V is not smooth but just bounded? I think the eigenfunctions will not be smooth anymore. Does the statement remeins true for this case or are there counterexamples?
May
20
revised The first eigenvalue of the Schrödinger operator is simple.
added 426 characters in body
May
20
asked The first eigenvalue of the Schrödinger operator is simple.
May
20
accepted regularity of eigenfunctions of Schrödinger Operator