Timeline for Equivalence classes of norms on $R^n$ under symmetries
Current License: CC BY-SA 3.0
18 events
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| Sep 21, 2017 at 10:16 | comment | added | Surb | @JonasAdler An absolute norm is a norm that satisfies $\|\, |x|\, \| = \|x\|$ for all $x$ where the absolute value is taken component wise. In particular, it can be shown that such norms satisfy $\|x\|\leq \|y\|$ for all $x,y$ such that $0\leq x_i\leq y_i$ (it is in fact an equivalence). So if $\|\cdot\|_{\alpha}$ is absolute, for the triangle inequality we get $$\| \|x+y\|_1,\ldots,\|x+y\|_n\|_{\alpha}\leq\| \|x\|_1+\|y\|_1,\ldots,\|x\|_n+\|y\|_n\|_{\alpha}$$ $$\leq\| \|x\|_1,\ldots,\|x\|_n\|_{\alpha}+\| \|y\|_1,\ldots,\|y\|_n\|_{\alpha}$$ as desired. | |
| Sep 21, 2017 at 8:59 | comment | added | Jonas Adler | What would you mean by an absolute norm in this setting? | |
| Sep 21, 2017 at 6:02 | comment | added | Surb | @gsa you are right! I forgot to add that $\|\cdot\|_{\alpha}$ should be an absolute norm. | |
| Sep 20, 2017 at 21:03 | comment | added | gsa | @Surb If you try to prove the triangle inequality, it doesn't work. For a concrete counterexample consider the matrix $A = [4,-4;1,1]$ and the norm $\lVert x\rVert := \lvert Ax\rvert_2 = (16(x_1-x_2)^2 + (x_1+x_2)^2)^{1/2}$. Then $N(x) = \lVert \lvert x\rvert_1, \lvert x\rvert_\infty\rVert$ is not a norm in $\mathbb{R}^2$. For $x=(1,0), y=(0,1)$ you get $N(x+y) = 5$ but $N(x) = N(y) = 2$. | |
| Sep 20, 2017 at 16:16 | comment | added | Surb | @gsa why not? (note that everything is finite in my expression) | |
| Sep 20, 2017 at 15:59 | comment | added | gsa | @Surb Maybe I misunderstood what you are saying, but in general the expression $\lVert \lVert\cdot\rVert_1, \ldots\rVert_{\alpha}$ is not a norm. | |
| Sep 20, 2017 at 12:56 | history | edited | Jonas Adler | CC BY-SA 3.0 |
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| Sep 20, 2017 at 11:52 | answer | added | Denis Serre | timeline score: 6 | |
| Sep 20, 2017 at 11:45 | history | edited | Denis Serre | CC BY-SA 3.0 |
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| Sep 20, 2017 at 11:35 | answer | added | Peter Michor | timeline score: 4 | |
| Sep 20, 2017 at 11:02 | history | edited | Jonas Adler | CC BY-SA 3.0 |
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| Sep 20, 2017 at 10:59 | comment | added | Ali Taghavi | every norm is the pull back of sup norm on $C[0,1]$ however my comment is not so relevant to the main question. | |
| Sep 20, 2017 at 9:15 | comment | added | Gro-Tsen | The question seems to be essentially to classify (balanced, bounded and absorbing) convex shapes in a linear space, which stated that way seems a bit hopeless. On the other hand, there could be much interesting to say about the topology or some other structure (and remarkable points, etc.) of the space of all norms. | |
| Sep 20, 2017 at 9:09 | comment | added | Surb | Note that the norm you mention is the special case where $\|\cdot\|_{\alpha}$ is a weighted $1$ norm on $\Bbb R^n$. Anyway, don't get me wrong, although I believe what you are asking is a very difficult question, I think it is of high interest. | |
| Sep 20, 2017 at 9:04 | comment | added | Jonas Adler | You can also do $\|x\| = \sum_i c_i\|x\|_i$ for any norms, so there certainly are a few things that can be done, but perhaps we can classify that "a norm is given by combinations of $p$ norms under these operations", or something similar. Anothing thing we may classify is some set of "fundamental" norms, that cannot be derived from any other norms. | |
| Sep 20, 2017 at 9:02 | comment | added | Surb | I'm not sure that the case $n=2$ is much easier as you can always define a norm $\|x\|=\| \|x\|_1,\ldots,\|x\|_n\|_{\alpha}$ where $\|\cdot\|_i$ are norms on $\Bbb R^2$ and $\|\cdot \|_{\alpha}$ is a norm on $\Bbb R^n$. | |
| Sep 20, 2017 at 8:50 | review | First posts | |||
| Sep 20, 2017 at 9:10 | |||||
| Sep 20, 2017 at 8:48 | history | asked | Jonas Adler | CC BY-SA 3.0 |