Timeline for norm inequalities
Current License: CC BY-SA 4.0
32 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2018 at 20:58 | comment | added | Paata Ivanishvili | @Iosif, see the answer below. I could not fit it in the comment. | |
| Jul 5, 2018 at 20:57 | answer | added | Paata Ivanishvili | timeline score: 1 | |
| Jul 4, 2018 at 15:20 | comment | added | Paata Ivanishvili | @Iosif, I will try to do it when I find some time since it is not a short computation. | |
| Jul 4, 2018 at 14:08 | comment | added | Iosif Pinelis | @Paata : Thank you for your latest comments. Can you also provide a proof of "the claim [in your penultimate comment] that needs to be proved"? | |
| Jun 30, 2018 at 4:46 | vote | accept | Arnold Neumaier | ||
| Jun 30, 2018 at 4:01 | comment | added | Paata Ivanishvili | @Iosif, Homogeneous Monge--Ampere, or the fact that the surfsace has the zero gaussian curvature helps a little bit, for example, it is known that such a surface consists by stright line segments, i.e., it is a developable surface, such that the gradient of the function (graph of which represents locally the surface) is constant along each such line. This helps to understand how to try to easily parametrize such surface in 2D, 3D, 4D, and in general in an arbitrary dimension. See for example projecteuclid.org/euclid.ijm/1348505534 to see how it helped to find Burkholder's function | |
| Jun 30, 2018 at 3:52 | comment | added | Paata Ivanishvili | @Iosif, if you ask why this explicit formula that I wrote on $B(x,y)$ represents the upper boundary of the convex hull of $(t,t^{2},t^{p})$, then here is the claim that needs to be proved: Let $\gamma(t)=(t,f(t),g(t))$ for $t \in [0,1]$ be a smooth 3D curve, so that tha graph $(t,f(t))$ is strictly convex in $\mathbb{R}^{2}$. Assume that the curvature of $\gamma$ never vanishes, and the torsion is nonnegative. Then the surface $\{s(c,f(c),g(c))+(1-s)(1,f(1),g(1)), \, a \in [0,1],\, c \in [0,1]\}$ represents the upper boundary of the convex hull of the curve $\gamma$. | |
| Jun 30, 2018 at 3:47 | comment | added | Paata Ivanishvili | @Iosif, if you ask why $B(p,q)=\sup_{x}\{ \|x\|_{p}^{p}, (\|x\|_{1}, \|x\|_{2}^{2})=(p,q), \, \|x\|_{\infty}\leq A\}$ is this upper boundary of the convex hull of the 3D curve $(t,t^{2},t^{p})$, then it follows, for example, from Theorem 1 in arxiv.org/pdf/1402.4690.pdf | |
| Jun 28, 2018 at 1:41 | comment | added | Iosif Pinelis | @Paata : Thank you for your responses. At this point, I have just two related questions: (i) How do you prove that the graph of $B$ is the upper boundary of the convex hull and (ii) precisely what role does the Monge--Ampere equation play in that proof? Also, I think your comments certainly deserve to be presented as a formal answer, and that may even be preferable. | |
| Jun 27, 2018 at 20:13 | comment | added | Paata Ivanishvili | The graph of $B$ represents the upper boundary of the convex hull of the 3D curve $(t,t^{2}, t^{p})$, $t \in [0,A]$. Just by definition this is exactly $B(p,q)=\sup_{x} \{ \|x\|_{p}^{p}, \; (\|x\|_{1}, \|x\|_{2}^{2})=(p,q), \|x\|_{\infty}\leq A\}$. And since by Caratheodory any point of the convex hull is the convex combination of $3$ points of its extreme points (This justifies Robert Israerl's guess), then extremizers PDF is the sum of at most 3 delta masses. But in this case $2$ is enough because $B$ happens to be linear on $\ell_{c}$, $c\in [0,A]$ which folliates $\Omega_{A}$. | |
| Jun 27, 2018 at 20:02 | comment | added | Paata Ivanishvili | @Iosif, regarding optimality, $z_{j}$ being the same that is not necessary for the equality cases. For example, $B$ is linear along a certain family of line segments, $\ell_{c}:=\{ t(c,c^{2})+(1-t)(A,A^{2}), t\in [0,1]\}$ where $c$ is any fixed number $c \in [0,A]$. So we can have equalities if and only if when the points $(z_{j}, z_{j}^{2})$ belong to $\ell_{c}$ for some fixed $c$. I would agree with you if $B$ were strictly convex, but it is not! | |
| Jun 27, 2018 at 19:54 | comment | added | Paata Ivanishvili | @Iosif, regarding typo, yes it should be $B(x,y)=\frac{(y-x^{2})A^{p}+(Ax-y)^{p}(A-x)^{2-p}}{(A-x)^{2}+y-x^{2}}$. In other words, B is just the upper bound where instead of $\|x\|_{1}=x$, $\|x\|_{2}^{2}=y$, and $A=\|x\|_{\infty}$. | |
| Jun 27, 2018 at 19:15 | comment | added | Iosif Pinelis | @Paata : Also, there seems to be a typo in your definition of $B$ -- your boundary condition does not check for me. | |
| Jun 27, 2018 at 19:05 | comment | added | Iosif Pinelis | @Paata : I think your comments are interesting, in that the Monge--Ampere eq. may provide a useful method. However, I don't see how it can produce the optimal upper bound in this particular setting. Indeed, it seems that your instance of Jensen's inequality turns into the equality only if all the $z_j$'s are the same, and then necessarily $\|z\|_1=\|z\|_2=\|z\|_\infty$. On the other hand, your upper bound coincides with the one in my answer (given the assumption $\|x\|_\infty=1$), and I wonder how this can be. Also, how do you conclude that your bound is optimal? | |
| Jun 27, 2018 at 16:18 | comment | added | Paata Ivanishvili | To verify concavity of $B$ in $\Omega_{A}$ it is a little bit tricky, and it depends which tools one is allowed to use. First one can verify that $B_{xx}B_{yy}-B_{xy}^{2}=0$, i.e., the graph of $B$ has zero Gaussian curvature (and in fact this is the way $B$ was constructed initially as a solution of homogeneous Monge--Ampere equation with the boundary condition $B(t,t^{2})=t^{p}$). Now to finish proving concavity of $B$ it remains to verify only $B_{yy}\leq 0$. One can also do it in a different way as well. | |
| Jun 27, 2018 at 16:09 | comment | added | Paata Ivanishvili | @Thomas Dybdahl Ahle, for the upper bound the claim is that given $A>0$ the function $B(x,y) = \frac{(y-x^{2})A^{p}+(Ax-y)^{2}(A-x)^{2-p}}{(A-x)^{2}+y-x^{2}}$ is concave in the domain $\Omega_{A}=\{(x,y)\, :\, Ax\geq y\geq x^{2}, \, x\geq 0\}$, with the boundary condition $B(t,t^{2})=t^{p}$.If so, then take any $z=(z_{1},\ldots, z_{n})$, with $\|z\|_{\infty}\leq A$. WLOG $z_{j}\geq 0$. Then By Jensen we have $ \frac{1}{n}\sum_{j=1}^{n}z_{j}^{p} = \frac{1}{n}\sum_{j=1}^{n}B(z_{j},z_{j}^{2})\leq B(\|z\|_{1},\|z\|_{2}^{2}) $. Again here I mean normilized $\|z\|_{q}$ norms. Put $A=\|z\|_{\infty}$ | |
| Jun 27, 2018 at 15:55 | comment | added | Paata Ivanishvili | @Thomas Dybdahl Ahle, lower bound follows from Holder's inequality: $\|x\|_{2}\leq \|x\|_{1}^{\theta}\|x\|_{p}^{1-\theta}$ where $\theta$ solves the equation $\frac{\theta}{1}+\frac{1-\theta}{p}=\frac{1}{2}$, i.e., $\theta =\frac{p-2}{2(p-1)}$ | |
| Jun 27, 2018 at 15:35 | comment | added | Thomas Dybdahl Ahle | @Paata How do you prove that? | |
| Jun 26, 2018 at 22:15 | comment | added | Paata Ivanishvili | Here by $\|x\|_{q}$ I mean normalized $\ell_{q}$ norm, i.e., $\|x\|_{q} = \left( \sum_{j=1}^{n}|x_{j}|^{q}/n\right)^{1/q}$ | |
| Jun 26, 2018 at 21:46 | answer | added | Iosif Pinelis | timeline score: 10 | |
| Jun 26, 2018 at 20:01 | comment | added | Paata Ivanishvili | And the lower bound seems to be $\|x\|_{2}^{2(p-1)}\|x\|_{1}^{2-p}$. Both are independent of $n$ and sharp. These are bounds for $\|x\|_{p}^{p}$ (not for $\|x\|_{p}$). | |
| Jun 26, 2018 at 19:35 | comment | added | Paata Ivanishvili | I think the upper bound is $\frac{(\|x\|_{2}^{2}-\|x\|_{1}^{2})\|x\|_{\infty}^{p}+(\|x\|_{\infty}\|x\|_{1}-\|x\|_{2}^{2})^{p}(\|x\|_{\infty}-\|x\|_{1})^{2-p}}{(\|x\|_{\infty}-\|x\|_{1})^{2}+\|x\|_{2}^{2}-\|x\|_{1}^{2}}$. | |
| Jun 26, 2018 at 18:59 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added a conjecture
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| Jun 26, 2018 at 18:29 | answer | added | Robert Israel | timeline score: 2 | |
| Jun 26, 2018 at 16:46 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added a relaxed form of the question
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| Jun 26, 2018 at 16:01 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added 1 character in body
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| Jun 26, 2018 at 16:00 | history | edited | YCor |
edited tags
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| Jun 26, 2018 at 15:59 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added 2 characters in body
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| Jun 26, 2018 at 15:55 | comment | added | Arnold Neumaier | @MarkL.Stone: Global optimization gives bounds for specific values of $p,n$ and the norms, but for each choice a different optimization problem must be solved. I want closed formulas. | |
| Jun 26, 2018 at 15:52 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
mentioned better upper bound
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| Jun 26, 2018 at 15:50 | comment | added | Mark L. Stone | As I'm sure you know, the sharpest lower and upper bounds on $\|x\|_p$ given particular values of $\|x\|_1, \|x\|_2, \|x||_\infty$ can be obtained by numerical global optimization: minimizing (for lower bound) and maximizing (for upper bound) $\|x\|_p$ subject to $x\in R^n$ and the values of $\|x\|_1, \|x\|_2, \|x||_\infty$. I suppose you want an analytical bound? | |
| Jun 26, 2018 at 15:32 | history | asked | Arnold Neumaier | CC BY-SA 4.0 |