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seen Aug 26 at 8:54

Jan
7
accepted Use of Jensen's inequality on a Riemann surface
Jan
3
asked Use of Jensen's inequality on a Riemann surface
Dec
6
comment Spectral measure and Stone's theorem
@MateuszWasilewski: $\lambda $ is any real number.
Dec
6
comment Spectral measure and Stone's theorem
Thanks for your answer. However, I would like to get rid of $E$ completely and express the RHS in my equation (1) only in terms of $U$. Also, $U$ depends on $\lambda $ and $UTU^{-1}=M_\lambda $.
Dec
5
comment Spectral measure and Stone's theorem
$U_\lambda $ does not commute with $\lambda $. Compare with $\mathcal{F}(-\Delta )\mathcal{F}^{-1}=\xi ^2$ where $\mathcal{F}$ is the Fourier transform.
Dec
5
asked Spectral measure and Stone's theorem
Oct
29
awarded  Informed
Oct
29
awarded  Tumbleweed
Oct
9
awarded  Caucus
Sep
6
comment Trace, eigenvalues and functional calculus
Well, what I actually want to know is if the formula for the trace is true (and at first I did not know if the eigenspaces were finite dimensional in my case).
Sep
6
comment Trace, eigenvalues and functional calculus
@András Bátkai: Yes, $\Pi _p$ is the projection onto the point spectrum of $T$.
Sep
6
comment Trace, eigenvalues and functional calculus
In this case we can assume all eigenspaces to be finite dimensional. The assumption I have on $f$ is that it should belong to $\mathcal{S}(\mathbb{R})$ (Schwartz space of rapidly decreasing functions).
Sep
5
asked Trace, eigenvalues and functional calculus
Aug
29
accepted Why equality of singular supports?
Aug
29
comment Why equality of singular supports?
Sorry, I have made an edit. $\nu $ should be a distribution on $\mathbb{R}$. I'm trying to phrase the question in slightly simpler terms than the original.
Aug
29
revised Why equality of singular supports?
deleted 2 characters in body
Aug
29
asked Why equality of singular supports?
Aug
22
accepted Support of a distribution
Aug
21
asked Support of a distribution
Jul
24
comment Fourier transform of tempered distribution
But it really doesn't matter as I'm interested in the transform only for $\omega \ne 0$. I was just curious.