Timeline for Inequality regarding $\ell_p$ norms, $p<1$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 9, 2012 at 14:40 | comment | added | newuser | I just realized that Mikael's example solves this problem too. Thanks a lot! | |
| Nov 9, 2012 at 13:11 | comment | added | newuser | Sorry, I didn't realize which $1$'s you are reffering to. Ok, thanks, that definitely convinces me. Can you also prove (un)boundedness when the marginals are fixed? | |
| Nov 9, 2012 at 3:10 | comment | added | fedja | Really, think a bit and stop just adding new restrictions. If I say that $\sum x_{ij}$ is completely irrelevant because it scales in a totally wrong way, it won't convince you, so let's do everything explicitly. Take an $n$ by $n$ matrix of $n^{-1-\frac 1p}$'s as before, which gives $\sum x_{ij}=n^{1-1/p}\to 0$. Add $x_{n+1,n+1}=1$. Then the first two sums are at $2$ exactly and the third tends to $1$. I hope you'll not ask me to prove that there may be no bound discontinuous at $(2,2,1)$... | |
| Nov 8, 2012 at 15:03 | comment | added | newuser | Fedja, I certainly realize that. The bound, if it exists, would depend on $\sum_{i,j} x_{i,j}$ in general. I only gave probability distributions as an example and simplification. | |
| Nov 8, 2012 at 8:46 | comment | added | Mikael de la Salle | Take $p_j = q_j = C^{-1} 1/j^\alpha$ with $1/p<\alpha<(2-p)/p$ and $C=\sum_j 1/j^\alpha$, so that $\sum_j q_j^p<\infty$. For any integer $n$, define $x_{i,j}=1/n^{\alpha+1}$ if $i,j \leq n$ and complete this family arbitrarily to have the correct marginals. Then $\sum_{i,j} x_{i,j}^p \geq n^{2 - p \alpha -p}$ which is not bounded as $n \to \infty$. | |
| Nov 8, 2012 at 0:32 | comment | added | fedja | @mladjoni Look, any function of $1$ (or, if you prefer, of two ones) is a constant, no matter how complicated it is. | |
| Nov 7, 2012 at 10:52 | history | edited | newuser | CC BY-SA 3.0 |
added 180 characters in body; edited tags; added 76 characters in body
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| Nov 4, 2012 at 11:25 | comment | added | newuser | Tom, I had infinite sequences in mind. Sorry for not being clear. Fedja, upper bound can involve any function of $ \sum_{i} (\sum_{j} x_{i,j})^p $ and $ \sum_{j} (\sum_{j} x_{i,j})^p $, their product for example, etc. Your example does not work in that case. | |
| Nov 4, 2012 at 2:14 | comment | added | fedja | Tom, the bound $n^{1-p}$ times the second quantity, say, is equally trivial if you think of the question for more than one millisecond and apply Holder to the inner sums. What finalized my decision to vote to close is that OP can think that the term of different order of homogeneity can be used in a bound like this. Otherwise, I would, probably, let it float for a while... | |
| Nov 4, 2012 at 1:23 | comment | added | Tom Leinster | fedja: if $(x_{i,j})$ is an $n \times n$ matrix and the upper bound is allowed to mention $n$, then your comment doesn't show that the question is trivial. (On the evidence of your example, the first quantity could be $\leq$ $n^{1-p}$ times the max of the second and third.) It's true that this is being slightly generous to the original question, because "double sequence" suggests double infinite sequence, in which case your example does indeed answer it with a rather trivial "no". But it doesn't trivialize it for the finite case. | |
| Nov 3, 2012 at 23:52 | comment | added | fedja | Take $n\times n$ matrix with all entries equal to $a$. You want to bound $n^2a^p$ in terms of $n^{1+p}a^p$. It $a=n^{-1-\frac 1p}$, the second quantity is $1$ but the first is $n^{1-p}$. Voting to close. | |
| Nov 3, 2012 at 23:51 | comment | added | Tom Leinster | Is the upper bound allowed to use knowledge of how many $i$s and $j$s there are? | |
| Nov 3, 2012 at 23:44 | history | asked | newuser | CC BY-SA 3.0 |