Timeline for Why this equality holds?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2017 at 15:31 | vote | accept | Schüler | ||
| S Nov 26, 2017 at 13:51 | history | suggested | Schüler | CC BY-SA 3.0 |
improved formating
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| Nov 26, 2017 at 6:07 | review | Suggested edits | |||
| S Nov 26, 2017 at 13:51 | |||||
| Nov 26, 2017 at 5:47 | vote | accept | Schüler | ||
| Nov 26, 2017 at 8:30 | |||||
| Nov 24, 2017 at 19:12 | comment | added | Kashayar | Thanks for your clarification. I see where the confusion is. In my original answer, The sumation is from $k=1$ to $d^n$. What I meant to say was to consider all the $d^n$ possible cases which appear in $\boldsymbol{S}^n$, and the summation in my answer, is NOT on different combinations of $\alpha$ . For example, if $d=2, n=2$, there are three choices for $\alpha$: $(2,0) , (0,2) , (1,1)$, and hence three terms in your summation on the left. However, in the summation on your right, there are four terms, $S_1 S_1 ,S_1 S_2 , S_2 S_1 , S_2 S_2$. | |
| Nov 24, 2017 at 18:37 | comment | added | Schüler | Thank you very much but perhaps you don't understand me very well: I tested your formula for $n=2$ and $d=2$ and I don't find that it is the same as the left hand side. | |
| Nov 24, 2017 at 18:36 | comment | added | Kashayar | I'm sorry, I had a few typos, so I deleted the comment and added a new one. In my comment above, take $f(x) = (\sum _{k=1}^{d^n} \Vert S_1^{\alpha_1} \ldots S_d^{\alpha_d} \Vert ^2)^{1/2}$. Then you can see if $A=\{ x\in E, \Vert x \Vert =1\}$, we get $\Vert \boldsymbol{S}^n \Vert = sup_{x\in A}\{f(x)\}$. | |
| Nov 24, 2017 at 18:32 | comment | added | Kashayar | This is a standard "take $\epsilon >0$ and prove each side is less than the other" argument. Let $F= sup _{x\in A}\{f(x)\}$. Then for each $n\in \mathbb{N}$, there exists $x_n \in A$ such that $f(x_n) > F - 1/n$. Hence assuming that $f(x)$ is non-negative for each $x\in A$ (which is the case in your example.), $sup _{x\in A}\{ f(x)^2\} \geq f(x_n)^2 > (F-1/n)^2$ for each $n$, hence $sup _{x\in A}\{ f(x)^2\} \geq F^2$. Also, for each $x\in A$, $F \geq f(x)$, therefore $F^2 \geq f(x)^2$. Now by taking supremum on both sides, we get $F^2 \geq sup _{x\in A}\{ f(x)^2\}$. This finishes the proof. | |
| Nov 24, 2017 at 17:42 | comment | added | Schüler | I think this formula is not true: $$ \Vert \boldsymbol{S}^n \Vert^2 = sup \Big\{ \sum _{k=1}^{d^n} \Vert S_1 ^{\alpha _1} S_2^{\alpha_2} \ldots S_d^{\alpha_d} \Vert^2, x\in E, \Vert x \Vert =1 \Big\} .$$ | |
| Nov 23, 2017 at 12:33 | vote | accept | Schüler | ||
| Nov 23, 2017 at 14:58 | |||||
| Nov 23, 2017 at 12:33 | vote | accept | Schüler | ||
| Nov 23, 2017 at 12:33 | |||||
| Nov 22, 2017 at 5:08 | vote | accept | Schüler | ||
| Nov 22, 2017 at 20:29 | |||||
| Nov 21, 2017 at 21:29 | history | answered | Kashayar | CC BY-SA 3.0 |