Timeline for The completion of $F/\text{Ker}(M)$ is isomorphic to the closure of the range of $M$
Current License: CC BY-SA 4.0
35 events
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| Mar 22, 2019 at 11:23 | history | edited | Jochen Wengenroth |
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| Mar 22, 2019 at 10:12 | history | edited | Schüler | CC BY-SA 4.0 |
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| S Mar 22, 2019 at 9:58 | history | bounty ended | Schüler | ||
| S Mar 22, 2019 at 9:58 | history | notice removed | Schüler | ||
| Mar 22, 2019 at 9:58 | vote | accept | Schüler | ||
| Mar 21, 2019 at 10:48 | answer | added | Matthew Daws | timeline score: 4 | |
| S Mar 21, 2019 at 9:02 | history | bounty started | Schüler | ||
| S Mar 21, 2019 at 9:02 | history | notice added | Schüler | Current answers are outdated | |
| Mar 21, 2019 at 8:05 | history | edited | Schüler | CC BY-SA 4.0 |
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| Mar 18, 2019 at 7:18 | vote | accept | Schüler | ||
| Mar 20, 2019 at 16:06 | |||||
| Mar 17, 2019 at 10:42 | history | edited | YCor | CC BY-SA 4.0 |
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| Mar 17, 2019 at 9:18 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 26, 2018 at 15:39 | vote | accept | Schüler | ||
| Mar 17, 2019 at 8:54 | |||||
| May 20, 2018 at 8:09 | answer | added | MSMalekan | timeline score: 1 | |
| May 19, 2018 at 20:02 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 19, 2018 at 19:23 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 19, 2018 at 13:48 | comment | added | Jochen Glueck | @Schüler: Alright, I see. I missunderstood the definition of the scalar product on the range of $A^{1/2}$. What the authors actually do is to define a scalar product $(\cdot,\cdot)_{R(A^{1/2})}$ on $R(A^{1/2})$ by means of the formula $(x,y)_{R(A^{1/2})} = (PA^{-1/2}x, PA^{-1/2}y)$ for all $x,y \in R(A^{1/2})$. This resolves my concerns; I simply missinterpreted the notation. | |
| May 19, 2018 at 12:24 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 18, 2018 at 21:11 | comment | added | Jochen Glueck | Considering your first question, I now have severe doubts that this is true. Let $H =\ell^2$, let $M$ be the multiplication with the sequence $(1/n)$. Then $P$ is the identity, but the image of $M^{1/2}$ is a dense and proper subspace of $\ell^2$, and thus, it is not complete with respect to the standard scalar product on $\ell^2$, right? | |
| May 18, 2018 at 14:30 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 18, 2018 at 9:16 | comment | added | Schüler | @JochenGlueck Please see my edit. | |
| May 18, 2018 at 9:15 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 18, 2018 at 7:44 | comment | added | Jochen Glueck | Probably I am missing something essential, but why is the norm induced by the inner product $(x,y) \mapsto \langle Px, Py \rangle$ on $\operatorname{Im}(M^{1/2})$ complete? | |
| May 18, 2018 at 7:27 | comment | added | Jochen Glueck | @Hannes: Just to add a few details to Schüler's answer to your comment: A bounded linear operator $A$ on a complex Hilbert space is self-adjoint if and only if $\langle Ax, x\rangle \in \mathbb{R}$ for all vectors $x$; this follows from the polarisation identity. | |
| May 17, 2018 at 15:38 | comment | added | Schüler | Yes M is selfadjoint since it is positive. | |
| May 17, 2018 at 14:41 | comment | added | Hannes | Also, it is implicitly assumed that $M$ is selfadjoint, no? (Since $\langle \cdot,\cdot\rangle_M$ is supposed to be a (semi-) inner product.) | |
| May 17, 2018 at 14:29 | comment | added | Hannes | Completion of $F/\operatorname{Ker}(M)$ with respect to the $\langle\cdot,\cdot\rangle_M$ norm? | |
| May 17, 2018 at 14:11 | answer | added | solway | timeline score: 5 | |
| May 17, 2018 at 13:36 | comment | added | YCor | Yes, a Hilbert space is determined up to linear isometry by its dimension (in the Hilbert sense: cardinal of an orthogonal basis). This is one of the very first things to know about a Hilbert space. | |
| May 17, 2018 at 13:00 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 17, 2018 at 13:00 | comment | added | Schüler | @YCor But we work in infinite dimensionel Hilbert spaces | |
| May 17, 2018 at 12:56 | comment | added | YCor | Oh, actually these are two Hilbert spaces, so the whole point is to check that they have the same dimension... | |
| May 17, 2018 at 12:25 | history | edited | Schüler | CC BY-SA 4.0 |
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| May 17, 2018 at 12:11 | comment | added | YCor | Why do you expect this to be true? | |
| May 17, 2018 at 12:09 | history | asked | Schüler | CC BY-SA 4.0 |