Timeline for Dual Norm For Sum of 2-Norms
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 14, 2012 at 20:38 | vote | accept | AnonSubmitter85 | ||
| Jul 4, 2012 at 7:02 | answer | added | Mikael de la Salle | timeline score: 2 | |
| Jul 4, 2012 at 1:33 | history | edited | AnonSubmitter85 | CC BY-SA 3.0 |
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| Jul 3, 2012 at 17:34 | answer | added | Deane Yang | timeline score: 2 | |
| Jul 3, 2012 at 16:18 | history | edited | AnonSubmitter85 | CC BY-SA 3.0 |
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| Jul 3, 2012 at 16:17 | comment | added | AnonSubmitter85 | @Brain: I edited the question to use what you say is the standard definition, though I do not think it affects the answer. | |
| Jul 3, 2012 at 2:10 | comment | added | Brian Borchers | The definition of "dual norm" used in this question isn't at all standard. The standard definition is that $\| y \|_{*}=\sup \left\{ y^{T}x | \| x \| \leq 1 \right\} $ | |
| Jul 3, 2012 at 2:06 | comment | added | Will Sawin | If you expand Deane Yang's equation you end up finding the highest $C$ such that $y^T= C\sum_{i=0}^k (1/||A_i \cdot x||_2) x^T A_i^T A_i$, but I'm not sure how to solve that. | |
| Jul 3, 2012 at 1:45 | comment | added | Will Sawin | @Yemon: How is that accurate? Let $||x||$ be the standard $L_2$-norm, then $n||x||+m||x||=(n+m)||x||$, but their duals are $(1/n)||x||$, $(1/m)||x||$, and $(1/n+m)||x||$ respectively. That operation is not the maximum. | |
| Jul 2, 2012 at 23:24 | comment | added | Yemon Choi | A rule of thumb that may be useful: the dual of a norm defined in terms of sums of other norms is usually given by taking the maximum of the duals of those constituent norms | |
| Jul 2, 2012 at 23:23 | comment | added | Yemon Choi | You need some conditions on your A matrices, presumably, to ensure one has a norm and not just a seminorm | |
| Jul 2, 2012 at 19:22 | comment | added | Deane Yang | Note that $\|y||_* = \max_{x} \{ |y^T\cdot x|/\|x\|\}$. So just solve for the critical points of $x \mapsto |y^T\cdot x|/\|x\|$ and substitute the answer back into the formula. | |
| Jul 2, 2012 at 18:22 | history | asked | AnonSubmitter85 | CC BY-SA 3.0 |