Timeline for About an argument in the paper "Commutators on $\ell_\infty$" by Dosev and Johnson
Current License: CC BY-SA 4.0
4 events
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| Jun 25, 2019 at 6:53 | comment | added | A beginner mathmatician | @ Bill. Thank you for the hint. I am writing the details just to complete the argument. Following Bill, we have $X=X_n\oplus \tilde{X}_n$, $Y=Y_n\oplus \tilde{Y}_n$ (As dim $\tilde{X}_n$ & $\tilde{Y}_n$ finite, this is ok.) Let $\|\alpha_k-\beta_k\|\to 0,$ $\alpha_k\in X,\beta_k\in Y$ and $\alpha_k=t_k+s_k$ and $\beta_k=r_k+l_k$ chosen according to the decomposition. Since $\tilde{X}_n+\tilde{Y}_n$ is complemented in $X+Y$ we clearly have that $\|s_k-l_k\|\to 0.$ This establishes Bill's claim. Now choosing block basis is easy since we have freedom to cut off the series in $x_n$'s and $y_n$'s. | |
| Jun 23, 2019 at 14:34 | comment | added | Bill Johnson | Hint: Check that for each $n$, $d(X_n,Y_n) =0$, where $X_n$ is the linear span of $x_n, x_{n+1}, \dots$ and similarly for $Y_n$. | |
| Jun 23, 2019 at 14:33 | history | edited | Yemon Choi |
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| Jun 23, 2019 at 11:04 | history | asked | A beginner mathmatician | CC BY-SA 4.0 |