Timeline for For what kind of $C^*$ algebras does the inequality $\frac{(ab+ba)}{2}\leq\frac{ a^p}{p} +\frac{b^q }{q}$ hold for $a,b>0$?
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14 events
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| Dec 14, 2022 at 7:29 | vote | accept | Ali Taghavi | ||
| Dec 5, 2022 at 12:04 | comment | added | Ali Taghavi | Thank you very much and my +1 for your very interesting answer and also the reference for tracial states. | |
| Dec 4, 2022 at 15:43 | history | edited | Jamie Gabe | CC BY-SA 4.0 |
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| Dec 4, 2022 at 13:28 | history | edited | Jamie Gabe | CC BY-SA 4.0 |
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| Dec 4, 2022 at 13:03 | comment | added | Zerox | On the other hand, there does exist a Young-like inequality for complex matrices, see this wiki page or [T. Ando (1995). "Matrix Young Inequalities". In Huijsmans, C. B.; Kaashoek, M. A.; Luxemburg, W. A. J.; et al. (eds.). Operator Theory in Function Spaces and Banach Lattices. Springer. pp. 33–38.] | |
| Dec 4, 2022 at 12:51 | comment | added | Jamie Gabe | @IosifPinelis: If $x$ is nilpotent of order $n$, then $x^{n-1}$ is nilpotent of order 2, so if your algebra contains a nilpotent element, it contains a nilpotent element of order 2. But let me add a (somewhat) detailed proof in my answer above. There's a limit to how many elementary details I can give without making the argument drown in minor details, so a lot of the steps will be easy exercises that anyone who knows elementary $C^*$-theory should be able to fill out. | |
| Dec 4, 2022 at 12:43 | comment | added | Jamie Gabe | @Zerox: First, I would say that $a>0$ means "positive and non-zero", so my examples are still valid. Second, you can simply replace my examples by $a+\delta 1$ and $b+\delta 1$ for sufficiently small $\delta$ if you want the spectrums contained in $(0,\infty)$ (since the set of positive semidefinite matrices is closed). | |
| Dec 4, 2022 at 12:40 | comment | added | Iosif Pinelis | @JamieGabe : Thank you for your response. However, as I have very little knowledge of $C^\ast$-algebras, it is hard for me to follow the logic in your comment. In particular, I don't understand why "the $C^\ast$-algebra generated by a nilpotent element of order 2 is a quotient of [...]" and why "it follows that it has a quotient isomorphic to [...]". Also, is there a gap between "a non-zero nilpotent element" and "a nilpotent element of order 2"? Since this crucial result does not seem to be readily available in the existing sources, do you mind adding a detailed proof of this fact? | |
| Dec 4, 2022 at 10:23 | history | edited | Jamie Gabe | CC BY-SA 4.0 |
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| Dec 4, 2022 at 10:12 | comment | added | Jamie Gabe | The closest thing I could find is 2.12.21 in Dixmier's $C^*$-book (which he attributes to unpublished by Kadison). Here it says that a $C^*$-algebra contains a non-zero nilpotent element if it is non-commutative (the converse is trivial). As the $C^*$-algebra generated by a nilpotent element of order 2 is a quotient of $C_0((0,1], M_2(\mathbb C))$ (by polar decomposition of the nilpotent element), it follows that it has a quotient isomorphic to $M_2(\mathbb C)$. | |
| Dec 3, 2022 at 22:33 | comment | added | Iosif Pinelis | The crucial fact here is of course that any "non-commutative [...] [$C^\ast$-algebra] $A$ contains a $C^\ast$-subalgebra which surjects onto $M_2(\mathbb C)$". Can you give a reference to this fact? | |
| Dec 3, 2022 at 12:10 | history | edited | Jamie Gabe | CC BY-SA 4.0 |
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| Dec 3, 2022 at 10:46 | history | edited | Jamie Gabe | CC BY-SA 4.0 |
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| Dec 3, 2022 at 10:15 | history | answered | Jamie Gabe | CC BY-SA 4.0 |