Timeline for If the operators $B_i'$ satisfy an inequality, prove that $B_1'+\dotsb+ B_n'$ also satisfies the same inequality
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| S Jan 31 at 21:06 | history | bounty ended | CommunityBot | ||
| S Jan 31 at 21:06 | history | notice removed | CommunityBot | ||
| S Jan 23 at 19:11 | history | bounty started | matilda | ||
| S Jan 23 at 19:11 | history | notice added | matilda | Draw attention | |
| Jan 21 at 16:50 | comment | added | matilda | @LSpice- Thank you, appreciate the edits. | |
| Jan 21 at 15:15 | comment | added | LSpice |
TeX notes: for norms, you want \lVert \rVert. Compare, for example, $||f||$ ||f|| to $\lVert f\rVert$ \lVert f\rVert. (\| \| is better but can fail, e.g., $\|-f\|$ \|-f\| vs. $\lVert-f\rVert$ \lVert-f\rVert.) Also, \dots can only work their magic with sufficient context; compare, e.g., $B_1' + \dots B_n'$ B_1' + \dots B_n' to $B_1' + \dots + B_n'$ B_1' + \dots + B_n'. You can explicitly tell the compiler you want "dots for binary operators" using \dotsb, as, for example, $B_1' + \dotsb B_n'$ B_1' + \dotsb B_n', if that is what you want. I edited both of these.
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| Jan 21 at 15:13 | history | edited | LSpice | CC BY-SA 4.0 |
TeX; deleted "thank you"
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| Jan 21 at 14:58 | history | asked | matilda | CC BY-SA 4.0 |