User benoît kloeckner - MathOverflow

Special coordinates for periodic metrics

2013-03-21T10:46:23Z

This question is a follow-up to that one.

Given a $\mathbb{Z}^n$-periodic metric $g$ on $\mathbb{R}^n$ (with $n>2$), is it possible to find a periodic diffeomorphism $\varphi$ such that $\varphi^*g$ makes the voronoi cell of $\mathbb{Z}^n$ convex? Or more generally, what kind of good compatibility between the metric and the affine structure of $\mathbb{R}^n$ can one expect by choosing good coordinates on the quotient torus?

Edit. Misha's comment shows that the precise part of the question is very naive. To make the remaining part more precise, one "compatibility" that would do for my need would be the following.

Call $g$ "$k$-balanced" if for all $v$ in the Voronoi cell of $0$, we have $$\sup_{p\in\mathbb{R}^n} d(p,p+v) \le k \ \mathrm{diam}(g)$$ Is it true that there is a $k=k(n)$ such that for all periodic riemmannian metric $g$, there is a periodic diffeomorphism $\varphi$ such that $\varphi^*g$ is $k$-balanced?

Answer by Benoît Kloeckner for chromatic number of the hyperbolic plane

2013-05-13T10:08:55Z

This does not really answer your questions, but I recently got a few results on the chromatic number of the hyperbolic planes. They are formulated by fixing the curvature and letting the distance vary, and I use the notation $\chi(\mathbb{H}^2,{d})$ for the chromatic number of the distance-$d$ graph on the hyperbolic plane with curvature $-1$.

for small $d$, $\chi(\mathbb{H}^2,{d})\leq 12$ (this can probably be improved, but maybe not easily to $7$),
for large $d$, $\chi(\mathbb{H}^2,{d})\leq \frac{4}{\ln 3} d + O(1)$.

The proofs can be found here: http://www-fourier.ujf-grenoble.fr/~bkloeckn/posts/2013-05-13-DistanceGraphs.html and the paper will appear on the arXiv soon. All this is not difficult, and the paper raises more questions than it answers.

My impression is that the monotony of the chromatic number with $d$ seems reasonable, but is in fact a subtle issue; and I would rather bet on a negative answer to question (2) but not too high. All in all, these questions are probably incredibly difficult, because we have only very cumbersome tools to relate the geometry with the distance graph.

For the story: about one year after I read, liked and bookmarked your question, I had forgotten about it but read "Ramsey Theory, Today, and Tomorrow", and realized I could answer some questions asked in it by Johnson and Szlam. In the course of writing a paper from these answers, I investigated the case of the hyperbolic plane. After writing a first version I happened to look at my MO favorites, and saw your question again -- which is therefore cited in the paper (will there soon be a @mathoverflow in standard bibTeX definitions?)

Algebraic topology in low regularity

2013-05-09T08:35:35Z

This question is triggered by a talk by Pierre Bousquet, who considered related questions (but not quite what I ask below).

Take a classical algebraic topological result, like the inexistence of retraction map $f:D^2\to \partial D^2$. Can we lower the regularity hypothesis (i.e., replace continuity with something weaker, or at least something not implying continuity) and still get a result?

Let me be more precise:

For which values of $p$ Does it exist a map $f:D^2\to \partial D^2$ in $W^{1,p}$ such that the trace of $f$ on the boundary is the identity?

In the same spirit:

For which values of $s,p$ must each map $f:D^2 \to D^2$ in $W^{s,p}$ have an almost fixed point in some sense (e.g. a sequence $x_n\to x$ such that $f(x_n)\to x$).

Is there a simple proof that a group of linear growth is quasi-isometric to Z?

2010-04-16T14:06:53Z

I proposed to a master's student to work, from the exercise in Ghys-de la Harpe's book, on the proof that a finitely generated group $G$ that is quasi-isometric to $\mathbb{Z}$ is virtually $\mathbb{Z}$. However I initially had in mind the result that gives the same conclusion from the hypothesis that $G$ has linear growth.

Do you know of any simple (and elementary, in particular without assuming Gromov's theorem on polynomial growth groups) proof that a group of linear growth is quasi-isometric to $\mathbb{Z}$?

Answer by Benoît Kloeckner for Random graphs nonisomorphic to unit distance graphs

2013-05-02T20:47:22Z

The almost sure asymptotic chromatic number of $G$ goes to $\infty$ with $c$, see for example the precise result by Achlioptas and Naor in Annals of Math. 2005.

The chromatic number of a unit-distance graph (and in fact of the whole plane) is bounded above by $7$, see e.g. the math coloring book by Soifer (this is simple: one colors an hexagonal tiling of carefully chosen side length).

These two facts end the proof of your problem.

Answer by Benoît Kloeckner for Positively curved manifold with a codimension 1 totally geodesic submanifold.

2013-04-30T17:11:28Z

$\mathbb{CP}^n$ (with $n>1$) does indeed not have any codimension $1$ totally geodesic manifold; neither does $\mathbb{CH}^n$. You can probably find a proof in Goldman's book on complex hyperbolic geometry.

(Added later: this is true even locally: there are no open codimension $1$ totally geodesic manifold in $\mathbb{CP}^n$ nor in $\mathbb{CH}^n$.)

Note that this is an important geometrical fact, as (as far as I know) all proofs of the isoperimetric inequality that work in the real hyperbolic space use reflexions with respect to a totally geodesic codimension one manifold. This explains why we still don't know if balls are optimal for the isoperimetric problem in $\mathbb{CH}^n$ and $\mathbb{CP}^n$ (small balls in the latter case, as for large volumes balls are known not to be optimal).

Also note that it is a source of great difficulty in the study of subgroups of isometries of $\mathbb{CH}^n$: for many groups $\Gamma$ acting isometrically on $\mathbb{CH}^n$, we do not know whether they are discrete; one cannot construct a fundamental domain with geodesic faces that could be used to prove discreteness, as it is done in real hyperbolic geometry. We are therefore mainly left with arithmetic methods, and to find non-arithmetic lattices of $\mathrm{SU}(1,n)$ is an important problem, see notably the work of Martin Deraux.

Answer by Benoît Kloeckner for Random rings linked into one component?

2013-04-28T07:34:45Z

I think I can complement the answer of Ori Gurel-Gurevich to prove that indeed, when we deal with connected open sets (no need for convexity) the answer is positive.

1. There is a finite configuration of circles $C_1, \dots, C_N$ whose centers are in the domain $D$, such that any circle $C$ with center in the domain $D$ must be linked to at least one of the $C_i$.

One way to do this is to first take a very tight lattice $\Lambda$, and put horizontal circles $C_1,\dots, C_K$ with centers on $\Lambda\cap D$. Now, a circle with center in $D$ that is not linked to any of these $C_i$ ($i\leq K$) must be roughly horizontal.

Then, add circles $C_{K+1},\dots, C_N$ with center on the lattice but oriented along a given vertical plane. A circle not linked to any $C_i$ must be roughly horizontal and roughly vertical, thus don't exist.

2. The above construction is stable under small perturbation. This means that there are small open sets $U_1,\dots, U_N$ of the parameter space such that for all set of circles $C_1,\dots, C_N, C$ such that $C_i\in U_i$, $C$ must be linked to one of the $C_i$.

3. By adding more circles that link together the $C_i$ of 1., we could have assumed that the $C_i$ make a linked component (this is where we use the connectedness assumption on $D$, which in fact could be weakened). As in 2., this is stable under small perturbation, so in fact there are small open sets $U_1,\dots, U_N$ of the parameter space such that for all set of circles $C_1,\dots, C_N, C$ such that $C_i\in U_i$, the $N+1$ circles must be linked together.

4. Now, when $n\to \infty$ the probability that circles have been drawn in each of the $U_i$ increase to $1$ exponentially fast, so at some point our random configuration contains with high probability a set of circles that links all admissible circles, including all the other randomly drawn ones.

Reference for ultrametric spaces

2013-04-08T15:13:48Z

I have a research project involving ultrametric spaces, and there are some facts that I use but have a hard time finding explicitely in the literature, although I know that some of them are folklore (for example, an ultrametric space can be described as the set of leaves of a tree, endowed with the induced metric).

I would like to know whether there is a book or comprehensive survey paper on the geometry and structure of ultrametric spaces.

An important point: I am interested in purely metric spaces, without algebraic structure (I did find books on analysis in non-Archimedean fields, which are too focused on this case). I can restrict to compact spaces, but not to finite ones.

Answer by Benoît Kloeckner for geometric interpretation of Lie bracket

2013-04-17T12:27:18Z

Let me attempt to reconcile the two views on the Lie bracket.

First, one has to wonder what it should mean that a vector field $Y$ is ``constant'' along $X$. This is ambiguous, as noticed by katz. One point is that it is not a property that depends solely on the values of $Y$ along $X$, contrary to its Riemannian counterpart: it should really depends on the (local) field $Y$. Another confusion not to make is that it cannot be simply defined in charts by looking whether $Y$ is constant in the Euclidean sense: this would certainly not be chart-independant (even if we ask the chart to be a flow box for $X$).

Since the model is when $X=\frac\partial{\partial x}$ and $Y=\frac\partial{\partial y}$ in the plane, the one thing we could ask to a ``constant along $X$'' field $Y$ would be that if one follows during a given time $h$ an integral curve of $Y$ starting from any point in a integral curve $\gamma$ of $X$, then one should end up in a given integral curve $\gamma'$ of $X$ that does not depend on the starting point (but only on $t$ and $\gamma$). In fact, one should even ask that the parametrization of $\gamma$ is respected. This is what you get if you can find some chart that is a flow box for both $X$ and $Y$, that is if are part of a coordinate system (up to minor cheating on colinearity).

But this is exactly the definition of Lie bracket given in Spivak, up to a little twist: one asks if following $X$ for some time $h$ then $Y$ for time $h$ gives you the same point than following $Y$ for time $h$ then $X$ for time $Y$.

Answer by Benoît Kloeckner for Finding a good ordering of $\mathbb{Q}$

2013-04-17T12:06:12Z

The answer is no.

First, the ordering and density hypothesis are irrelevant (you do not use the ordering, and the density can be managed independently of the measure assumption we are trying to satisfy).

The Lebesgue measure of the set of $x\in(-1,1)$ such that $x\in B(x_,;r_n)$ for at least one $n>N$ is at most $2\sum_{n>N} r_n$. Your set is the intersection of these sets over all $N\in\mathbb{N}$, so that it must have measure $0$ as soon as $\sum r_n<\infty$.

Answer by Benoît Kloeckner for when constant scalar curvature implies Einstein?

2013-04-17T11:56:56Z

There is no reason for this, and the answer is indeed no.

The simplest example I can think of is the product of two $\mathbb{S}^2$, each endowed with the round metric. This manifold is homogeneous and thus has constant scalar curvature, its sectional curvature is non-negative so its Ricci tensor also is (and is in fact even positive), but the Ricci curvature in a direction $u$ depends on the angle between $u$ and the tangent spaces to the fibers of the projection on each factor (i.e., on whether $u$ is close to be horizontal or vertical or not).

Answer by Benoît Kloeckner for Is there a lower bound for variance in terms of curvature?

2013-04-15T13:38:04Z

As far as I understand the question, the answer is no: for any domain $\Omega$ and any $\delta>0$, denoting by $K_f$ the curvature function of the metric $g=f^2g_{eucl}$, we have

$$\inf_{K_f\geq\delta} \mathrm{Var}(f) = \inf_{K_f\leq-\delta} \mathrm{Var}(f) = 0$$

Indeed, given any $f$ such that $K_f\geq\delta$ and any $\lambda\in(0,1)$, the function $u=\lambda f$ has $\mathrm{Var}(u)=\lambda^2 \mathrm{Var}(f)$ and $K_u=K_f/\lambda^2\geq\delta/\lambda^2>\delta$.

This seems to have to do with normalization or the volume form which is used, but I do not know how one could formulate an alternative problem with a positive answer.

Is displacement controled by stable norm?

2013-03-06T13:20:13Z

Let $T^n$ be the $n$-dimensional torus and $g$ be a Riemannian metric on $T^n$. Let $\tilde g$ be the induced metric on the universal covering; using suitable coordinates, $\tilde g$ is therefore a $\mathbb{Z}^n$-periodic metric on $\mathbb{R}^n$ (I shall conflate the lattice $\mathbb{Z}^n$ with the fundamental group of $T^n$ in the sequel).

Let $d$ be the distance induced by $\tilde g$ and $t:\mathbb{Z}^n\to (0,+\infty)$ be the function defined by $t(\gamma)=d(0,\gamma(0))$. Recall that the stable norm $\Vert\cdot\Vert_S$ is a norm on $\mathbb{R}^n$ defined by the property that for any $\gamma\in\mathbb{Z}^n$, $$ \Vert\gamma\Vert_S = \lim_k \frac{t(\gamma^k)}{k} $$

Question: is it true that for all $\gamma\in\mathbb{Z}^n$, we have $$t(\gamma)\le \Vert \gamma \Vert_S+2\mathrm{diam}(g)?$$

If not, does some similar control hold? The formula could depend on $n$ but not on the metric $g$.

A pointer to good literature on this kind of metric Riemannian geometry would already be much appreciated.

Important edit: if needed, I am ok with the additional, very strong assumption that the sectional curvature of $g$ is bounded above by some positive $\varepsilon=\varepsilon(n)$. The formula for a lower bound on $\Vert\cdot\Vert_S$ can depend upon $\varepsilon$ (explicitely).

As for motivation, I need this kind of control for a project of showing some constraints on Riemannian metrics on the torus $T^n$ by using quantitative versions of Milnor's argument in his paper on the growth of fundamental groups and volume of Riemannian manifolds.

Answer by Benoît Kloeckner for Research level applications of "row rank = column rank"?

2013-02-28T13:00:20Z

There is this proof of the De Bruijn-Erdös theorem: $p$ points in the plane, not all on the same line, at least $p$ lines go through at least two of the points.

The linear algebraic proof goes like this: let $A$ be the incidence matrix of points versus lines (each row is labeled by a point, each column by a line going through at least two of the points, and the $ij$ coefficient is $1$ if the given point is on the given line, $0$ otherwise). Then it is easily seen that $det(AA^T)\neq0$. In particular the rank of $A$ is $p$, and since this is its column rank the number of columns must be at least $p$.

Smoothing of piecewise Euclidean Riemannian metrics

2013-01-30T13:17:00Z

Let $M$ be a smooth closed manifold and $T$ be a triangulation of $M$. Endow each simplex of $T$ with the Euclidean metric making it a regular simplex; this gives a piecewise Euclidean metric $g_0$ on $M$, which is singular on (part of) the codimension $2$ skeleton of $T$.

Is it possible to approximate $g_0$ by a smooth Riemannian metric? The approximation should in particular change length of curves and the volume by arbitrarily small amounts.

I guess the answer is positive and well-known, but I did manage to find a reference (in particular, several works ask the smoothing to satisfy certain curvature assumptions, which I do not). Is there a reference or are there obstruction to smoothing?

Answer by Benoît Kloeckner for Reference question: Poncelet theorem

2013-02-07T19:44:04Z

There seems to be a misconception here: Poncelet Theorem (at least the great one, which I believe to remember is the one he proved while in jail) is a much deeper, and more difficult statement than what you state.

Consider an ellipse inside another ellipse, and play inner-outer billiard with them. This means that you start from a point on the outer ellipse, choose one of the two line from this point tangent to the inner ellipse, and take the second intersection point of this line with the outer ellipse. You continue, always taking the next line tangent to the inner ellipse from the current point, and the other intersection point with the outer ellipse from the current line.

Theorem (Poncelet) $-$ If one orbit of this dynamical system is periodic, then all orbits are periodic.

This, if I remember well, is in Berger's Geometry. There might be a reference there.

Answer by Benoît Kloeckner for longest simple closed geodesic

2013-01-22T15:40:55Z

The question of the relation between the length of the shortest closed geodesic and the area of a surface is called systolic geometry. You can notably look at the work of Balacheff, Bavard, Croke Gendulphe, Katz, Parlier, Saboureau.

Answer by Benoît Kloeckner for New grand projects in contemporary math

2012-12-31T09:40:54Z

Optimal transport. Both its study (generalizations, Monge problem, regularity issues, and geometric properties to cite the part I work in) and its applications (to geometry notably with the Work of Sturm and Lott-Villani, to image processing and recognition, etc.) have developed hugely since the 90's.

Answer by Benoît Kloeckner for New grand projects in contemporary math

2012-12-31T09:38:25Z

Ricci flow. It did solve Poincaré's conjecture and the $1/4$-pinching conjecture, but has also become an object of study. More generally, it has launched a large amount of work on geometric flows (mean curvature flow and others), notably with the idea that some other problems can be solved by designing an ad-hoc flow.

Answer by Benoît Kloeckner for Convexity in $\{0,1\}^n$

2012-11-24T12:42:40Z

If you want a stable notion of convexity, you can ask for $C\subset \{0,1\}^n$ to be convex that for all $x,y\in C$, every minimal path between $x$ and $y$ is contained in $C$.

Concerning the Brunn-Minkowski inequality in the hypercube, there is a recent result of Ollivier and Villani :

"A curved Brunn-Minkowski inequality on the discrete hypercube, Or: What is the Ricci curvature of the discrete hypercube?" SIAM J. Discr. Math. 26 (2012), n°3, 983--996. (paper available e.g. on Yann Ollivier's web page).

The result is as follows: call midpoint of $x$ and $y$ any point that lies on a minimal path between them and is halfway (if $d(x,y)$ is even) or as close as halfway as possible (otherwise). For all $A,B\subset \{0,1\}^n$ , the set $M$ of midpoints of pairs $(a,b)\in A\times B$ has cardinal bounded below: $$ \ln |M| \ge \frac12 \ln |A| + \frac12 \ln|B| +\frac1{16n} d(A,B)^2.$$

Answer by Benoît Kloeckner for Proof synopsis collection

2012-10-17T07:29:10Z

I am surprised that this one did not already occurred : Perelman's proof of the Poincaré conjecture.

Endow a simply connected three-manifold with any Riemannian metric. Let the metric evolve under the Ricci flow. When singularities occur, cut them out and smoothly glue a cap in the hole, checking that the topology has not changed. After some time, you get a round metric so your manifold is a sphere.

Answer by Benoît Kloeckner for Isoperimetric inequality in complex hyperbolic space

2012-09-21T20:45:22Z

I am pretty sure it is a somewhat reputed conjecture, but I do not have a clear reference where it is stated. It might be evoked in a paper of Hsiang and Hsiang in Inventiones, where they prove that the isoperimetric domains in products of hyperbolic and euclidean spaces are invariant under the group of all isometries fixing the center of gravity. It seems a reasonable conjecture that this is true in all symmetric spaces of non-positive curvature. That conjecture might be stated in the Hsiang and Hsiang paper, and is a broad generalization of the conjecture you are interested in.

Answer by Benoît Kloeckner for examples of totally geodesic subset

2012-09-12T08:34:26Z

The obvious answer is an equatorial sphere (= intersection with a linear subspace of any codimension) in the unit sphere of $\mathbb{R}^n$.

Without more details on your motivation, it is difficult to judge whether this answer is satisfying or not.

Answer by Benoît Kloeckner for Laplace-Beltrami operator expression

2012-08-29T21:12:57Z

The $v_i$ are vector fields, and as such are derivations. The square usually means that you apply it twice (so, e.g. in the Euclidean space one can take $v_i=\frac{\partial}{\partial x_i}$ and its square is simply $\frac{\partial^2}{\partial x_i^2}$).

Answer by Benoît Kloeckner for curvature of curves in the space of gaussians measures

2012-08-18T09:55:49Z

Your question is not specific enough about what you do not understand in the quoted paper; if you want help on this, you should at least explain what you understand and where the problem appears. Here is a little information about bibliography, that might help (or miss the point, I am not sure).

For an introduction to optimal transport and Wasserstein spaces, you can have a look at Villani's books ("Topics on ..." is more elementary, but the beginning of "... Old and New" is not as difficult to read as the size of the book might lead you to think, and I like it a lot). A more concise introduction can also be found in a nice little book by Nicola Gigli, a version of which seems to be at http://math.unice.fr/~gigli/Site_2/Publications_files/users_guide%20-%20final.pdf (but I am not sure this is exactly the text I read).

You should also now about Lott's paper "Some geometric calculations on Wasserstein space", Comm. Math. Phys. 277, p. 423-437, which computes the curvature of the Wasserstein space of a manifold.

Concerning the notion of curvature of a discretized curve, you might be interested in the concept of Menger's curvature, which applies in a very broad context.

Answer by Benoît Kloeckner for Not especially famous, long-open problems which anyone can understand

2012-07-12T13:58:50Z

In an oriented graph, is there always a vertex from which there are at least as many vertices that one can access by moving along exactly two edges, than there are vertices that one can access by moving along one edge?

This is known as Seymour's second neighborhood conjecture, and might be on the verge to being too famous (but it seems few of my colleagues know it).

Answer by Benoît Kloeckner for Number of neigbour Voronoi cells for a random set of points on S^k or cube [-1, 1]^k?

2012-07-04T10:04:44Z

When $k=2$, you can use combinatorics to avoid any relation with probability. From Euler formula, one gets that the mean degree of a graph with $N$ vertices on $S^2$ is $6-\frac{12}N$ (see e.g. Proofs from the book, section on Euler formula).

Applying this to the neighbors graph of your tesselation, you get that the average number of neighbors of a cell is bounded by $6$ independently of $N$.

For this you have to rule out cells that touch by a corner only (otherwise the graph need not be planar), but I guess this only happen with null probability.

Answer by Benoît Kloeckner for Metric Deformations from Non-Negative to Positive Curvature

2012-05-27T13:38:50Z

I think the answer is no, because if I remeber well $\mathbb{R}\mathrm{P}^2 \times \mathbb{R}\mathrm{P}^2$ does not admit a positively curved metric. My reference for this is Gallot-Hulin-Lafontaine, but I do not have the book at hand right now.

Answer by Benoît Kloeckner for basic question about rectifiability

2012-05-24T19:43:56Z

I think you are true that the definition is void when $k=0$, but this only means it is an interesting for $k>0$ only. It even seems to me that this notion makes sense mostly when $n$ is the Hausdorff dimension of the considered set; the important theorem to bear in mind is a decomposition result which I may not remember very precisely, but that roughly says that any (closed ?) set of dimension $n$ is the union of a $n$-rectifiable set and a $n$-totally unrectifiable set (which means a set that has small intersection with every $n$-rectifiable set). The classical example of totally unrectifiable set is the four-corner Cantor set, which has Hausdorff dimension $1$ but meets every Lipschitz curve along a set of null one-dimensional measure.

Answer by Benoît Kloeckner for $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$

2010-07-09T18:35:45Z

A group of french mathematicians and computer scientists are currently working on this. The project is named Hévéa, and has already produced a few images. Edit: a few images and the PNAS paper have been released, see http://math.univ-lyon1.fr/~borrelli/Hevea/Presse/index-en.html

Just a few word to explain what I understood of their method (which is by using h-principle) from the few image I saw in preview. Start with a revolution torus. The meridians are cool, because they all have the same length, as expected from those of a flat torus. But the parallels are totally uncool, because their lengths differ greatly: they witness the non-flatness of the revolution torus.

Now perturb your torus by adding waves in the direction of the meridians (like an accordion), with large amplitude on the inside and small amplitude on the outside. If you design this perturbation well, you can manage so that the parallels now all have the same length. Of course, the perturbed meridian have now varying lengths! So you do the same thing by adding small waves in another direction, getting all meridians to have the same length again. You can iterate this procedure in a way so that the embedding converges in the $C^1$ topology to a flat embedded torus. But to prove that the precise perturbation you chose in order to get a nice image does converge, and that your maps are embeddings needs work (getting an immersion is easier if I remember well).

Also, the Hévéa project plans to draw images of Nash spheres, that is $C^1$ isometric embeddings of spheres of radius $>1$ inside a ball of unit radius.

Comment by Benoît Kloeckner

2013-05-19T14:31:52Z

This is basic differential geometry, not research-level. I thought that Lev Soukhanov answer would show you where is the problem, but now the ongoing discussion does not belong here. Voting to close.

Comment by Benoît Kloeckner

2013-05-19T11:46:15Z

In other word, your "pull-back" vector field depends on both $f$ and $r$, while to properly define a $f^*X$ you would like it to depend only on $f$.

Comment by Benoît Kloeckner

2013-05-17T22:14:32Z

@Ryan Budney: the action of $\mathrm{PGL}$ does not contain the conformal group, as the former preserves the antipody relation while the latter doesn't.

Comment by Benoît Kloeckner

2013-05-17T13:01:11Z

What is the question?

Comment by Benoît Kloeckner

2013-05-14T18:03:47Z

Convexity is an assumption that may give you something, as stressed by Tom Bachmann below, connectivity is not.

Comment by Benoît Kloeckner

2013-05-14T16:07:16Z

Given the median and the number of values, you can easily compute the largest and smallest possible mean, then answer your question. This is not a suitable question for MO.

Comment by Benoît Kloeckner

2013-05-13T13:32:02Z

I took the liberty to improve your title and retag your question.

Comment by Benoît Kloeckner

2013-05-13T13:18:02Z

This is not the place to get a crash course on distributions or duality, and anyway you gave much too little information to get useful advice.

Comment by Benoît Kloeckner

2013-05-13T13:16:28Z

This looks like homework, and anyway is not suited for this site.

Comment by Benoît Kloeckner

2013-05-10T16:55:35Z

What is $T$? What do the $0$s mean? What is $\mathcal{D}$? Is this question research-level?

Comment by Benoît Kloeckner

2013-05-10T13:41:38Z

This question is not suited for this site, please read the FAQ.

Comment by Benoît Kloeckner

2013-05-10T11:58:11Z

You should explain a bit what the VRP is, or provide a link to definitions.

Comment by Benoît Kloeckner

2013-05-09T11:31:03Z

@Ricardo Andrade: you are of course right, but one can take the trace - I edited the question accordingly.

Comment by Benoît Kloeckner

2013-05-09T07:46:31Z

If one wants a crude asymptotic like the one Didier suggests, finding the dominant term for the lower bound and giving the obvious upper bound is sufficient. In any case, all ingredients are given in various comments, so either one cook up an answer that Granger will be able to accept, or we close as « off topic » (if the question is considered too simple) or « non longer relevant », but there is no need letting it popping up.

Comment by Benoît Kloeckner

2013-05-08T15:26:19Z

Your notation is a bit confusing when dealing with a functional equation, you should use indices $1$ and $2$ for the derivatives rather than variable names.