Timeline for inequality of norms
Current License: CC BY-SA 2.5
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2011 at 16:30 | comment | added | Matthew Daws | I think, by the Closed Graph Theorem, your condition on sequences is equivalent to $P_X, P_Y$ being bounded. | |
| Mar 24, 2011 at 14:04 | comment | added | Michael Renardy | It suffices to make the following assumption: If $x_n\to x$ in X, $y_n\to y$ in Y and $(x_n,y_n)\to z$ in Z, then z=(x,y). If we assume this, then Z becomes a Banach space under the norm $\|(x,y)\|=\|(x,y)\|_Z+\|x\|_X+\|y\|_Y$. Since this norm is stronger than either $\|.\|_Z$ or $\|.\|_1$, it must be equivalent to both by the open mapping theorem. | |
| Mar 24, 2011 at 12:27 | history | answered | Matthew Daws | CC BY-SA 2.5 |