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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| Dec 14, 2022 at 7:29 | vote | accept | Ali Taghavi | ||
| Dec 3, 2022 at 10:15 | answer | added | Jamie Gabe | timeline score: 13 | |
| Dec 2, 2022 at 19:19 | comment | added | Jamie Gabe | One can improve on the argument I did above in $M_2(\mathbb C)$ (general case is still open, though). In fact, if one uses the projections $a=e_{1,1}$ and $b= \left( \begin{array}{cc} 3/4 & \sqrt{3}/4 \\ \sqrt{3}/4 & 1/4 \end{array} \right)$, and again compute the eigenvalues of $(1-t)a + tb - \tfrac{1}{2} (ab+ba)$, then there are negative eigenvalues when $\min\{ t, 1-t\} < 1/4$. So the inequality from the question fails in all non-commutative $C^\ast$-algebras whenever $\min\{1/p, 1/q\} < 1/4$. | |
| Dec 2, 2022 at 17:14 | comment | added | Joseph Van Name | My computer calculations from testing random matrices suggest that for all $p\in(1,\infty)\setminus\{2\}$, the first inequality fails for $M_2(\mathbb{R})$. | |
| Dec 2, 2022 at 16:26 | comment | added | Jamie Gabe | Just to expand on the failure when $\min\{1/p, 1/q\}$ is small (I'm hoping it will inspire someone else to come up with a proof), any non-commutative $C^*$-algebra has a $C^*$-subalgebra which contains $M_2(\mathbb C)$ as a quotient, so it suffices to show the failure in $M_2(\mathbb C)$. Here I took $a = e_{1,1}$ and $b = \tfrac{1}{2} \sum_{i,j} e_{i,j}$ and asked Maple to compute the eigenvalues of $(1-t)a + tb - \tfrac{1}{2} (ab+ba)$ and observed that there is a negative eigenvalue for $\min \{ t, 1-t \} < (\sqrt{2}-1)\sqrt{2}/4$. | |
| Dec 2, 2022 at 16:15 | comment | added | Jamie Gabe | The inequality holds for $p=q=2$ since $0 \leq (a-b)^2 = a^2 + b^2 - ab - ba$, and it holds in all commutative $C^*$-algebras by Young's inequality. When $\min\{1/p, 1/q\} < (\sqrt{2}-1)\sqrt{2}/4 \approx 0.146$ I can show (quite ad hoc) that it fails for all non-commutative $C^*$-algebras, so I kind of suspect that when $p\neq q$ then it fails for all non-commutative $C^*$-algebras. The inequality for tracial states holds by [Farenick, Douglas R., Manjegani, S. Mahmoud, Young's inequality in operator algebras. J. Ramanujan Math. Soc. 20 (2005), no. 2, 107–124]. | |
| Dec 1, 2022 at 19:40 | history | edited | LSpice | CC BY-SA 4.0 |
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| Dec 1, 2022 at 18:53 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Dec 1, 2022 at 17:21 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Dec 1, 2022 at 16:40 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Dec 1, 2022 at 16:36 | comment | added | Ali Taghavi | @InfiniteLooper Thank you I was mistaken I revise it | |
| Dec 1, 2022 at 15:49 | comment | added | InfiniteLooper | Are there other integers $p, q$ with $\frac{1}{p} + \frac{1}{q} = 1$ except $p = q = 2$ ? | |
| Dec 1, 2022 at 14:38 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Dec 1, 2022 at 14:24 | history | asked | Ali Taghavi | CC BY-SA 4.0 |