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
16 events
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| Oct 29, 2019 at 16:42 | comment | added | Matthew Daws | For an example, let $H$ be an infinite dimensional Hilbert space, let $H_1=H$ and let $H_0$ be any dense subspace, and let $T$ be the inclusion map. Then $T$ is not onto, but the completion of $H_0$ is (isometrically isomorphic to) $H$, and the continuous extension of $T$ is the identity, which is unitary. | |
| Oct 29, 2019 at 16:40 | comment | added | Matthew Daws | No. In the abstract, suppose I have two inner-product spaces $H_0, H_1$ which are not complete and so not Hilbert spaces. It is quite possible to have an isometric linear map $T:H_0\rightarrow H_1$ which is not surjective (that is, $T(H_0)$ is a proper subspace of $H_1$) but that once we extend $T$ by continuity to a map $\overline{H_0} \rightarrow \overline{H_1}$ we do obtain a surjection (which is then unitary). In my answer the image of $V$ is the image of $M$ which might be a proper subspace of the image of $M^{1/2}$ | |
| Oct 29, 2019 at 15:19 | comment | added | Schüler | But we want that it will be necessary surjective in order to obtain an unitary operator. The completion is unique. Please correct me if I'm wrong. Thanks a lot. | |
| Oct 29, 2019 at 14:00 | comment | added | Matthew Daws | Because $V$ is defined between incomplete spaces. The continuous extension to the completions can be surjective without the original being surjective. | |
| Oct 29, 2019 at 12:41 | comment | added | Schüler | Dear Professor, please now I read again the proof. I'm facing confusion in the end of your proof: ''Thus $$ V:\big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im M^{1/2}, (\cdot|\cdot)_0 \big) $$ is an isometry, as you want. Notice that $V$ is not (in general) onto. Finally, we can extend $V$ to the completion of $F/\ker M$, and then we will obtain a unitary transformation onto $\overline{\im M^{1/2}}$, completion with respect to $(\cdot|\cdot)_0$.'' Since $V$ is not onto how we get an unitary operator? Thanks a lot. | |
| Mar 22, 2019 at 10:57 | history | edited | Matthew Daws | CC BY-SA 4.0 |
Add final remark about completions.
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| Mar 22, 2019 at 9:58 | history | bounty ended | Schüler | ||
| Mar 22, 2019 at 9:58 | vote | accept | Schüler | ||
| Mar 22, 2019 at 9:55 | comment | added | Matthew Daws | Yes, I think that is right. If you want, extend to the closures on both sides. | |
| Mar 21, 2019 at 21:58 | comment | added | Matthew Daws | @Schüler: Ah, I think I see. The hint is actually at the very bottom of page 4 of the paper you link to. I have yet again altered by answer. | |
| Mar 21, 2019 at 21:57 | history | edited | Matthew Daws | CC BY-SA 4.0 |
Finally think we have the theorem
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| Mar 21, 2019 at 16:30 | comment | added | Matthew Daws | @Schüler: I think I understand. I have edited my answer to correct a mistake, and to directly address your question, I hope. | |
| Mar 21, 2019 at 16:29 | history | edited | Matthew Daws | CC BY-SA 4.0 |
Correct a mistake; add comment on original question
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| Mar 21, 2019 at 12:18 | comment | added | Schüler | $\text{Im}(M^{1/2})$ is a Hilbert space with respect to the following inner product $$( M^{1/2}x,M^{1/2}y)=\langle P_{\overline{\text{Im}(M)}}x, P_{\overline{\text{Im}(M)}}y\rangle,\;\forall\, x,y \in F.$$ $P_{\overline{\text{Im}(M)}}$ is the orthogonal projection onto $\overline{\text{Im}(M)}$. | |
| Mar 21, 2019 at 12:15 | comment | added | Matthew Daws | I don't fully understand. Firstly, $\operatorname{Im}(M^{1/2})$ is not closed, so it is not a Hilbert space. Secondly, in what sense does $M^{1/2}$ map between $\overline{\operatorname{Im}(M)}$ and $\operatorname{Im}(M^{1/2})$? Do you closure here? But then $\overline{\operatorname{Im}(M)} = \overline{\operatorname{Im}(M^{1/2})}$ so $M^{1/2}$ does not map between them bijectively. Thirdly, you keep writing $\langle\cdot,\cdot\rangle$ but this is used in lots of different ways in your question, so I don't really know what inner product is meant. | |
| Mar 21, 2019 at 10:48 | history | answered | Matthew Daws | CC BY-SA 4.0 |