29

Krein-Milman theorem can be used to prove the following: suppose we associate to each point of the integer lattice in the plane a real number in $[0, 1]$, such that the number corresponding to each point is the average of the numbers corresponding to the nearest four points of the lattice. Then all numbers must necessarily be equal.
Although the related ...

27

That isn't a conjecture but a routine exercise assigned after the students learn about Bang's solution of the Tarski plank problem. The proof goes in 2 steps:
1) Consider all sums $\sum_j \varepsilon_i u_i$ with $\varepsilon_i=\pm 1$ and choose the longest one. Replacing some $u_j$ with $-u_j$ if necessary, we can assume WLOG that it is $y=\sum_i u_i$. ...

22

A counterexample is given in the following paper:
Roberts, James W. "A compact convex set with no extreme points."
Studia Mathematica 60.3 (1977): 255-266.
I did not see the original paper, but an exposition can be found in Section 5.6 of the book "Metric linear spaces" by Rolewicz.

answered Sep 26 '18 at 7:42

Michael Greinecker

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21

Fix $n$, and let $V(x)$ be the polynomial from the question, for which we want to show that it is
positive on the open interval $(0,1)$. Set
\begin{align*}
a(n) &= 1024n^2 - 4096/3n + 320\\
b(n) &= 12288n^3 - 14336n^2 + 4864/3n + 256\\
W(x) &= x^4\left(V(x)-(n+1)^2(2n+1)^2x^{24n-1}(1-x)^4(a(n)x + b(n)(1-x))\right).
\end{align*}
For all $n\ge12$ ...

21

The following proposition answers OP's question regarding the upper bound of $$\tau(C) \Doteq f(C)/\lambda^2(C).$$
Let $B_n$ be the closed Euclidean unit ball of $\mathbb{R^n}$ centred at $0$, that is $$B_n = \{ (x_1,\dots, x_n) \in \mathbb{R^n} \, \vert \,\,
x_1^2 + \cdots + x_n^2 \le 1\},$$
and let $\tau_n = \tau(B_n)$.
Proposition.
The Euclidean ...

17

This won't be a complete answer to Q2, but something of a starting point, at least for all two-dimensional convex $P$. Assuming for simplicity that the area of $P$ is 1, $P$ contains a parallelogram of area at least $1/2$ (triangular $P$-s are extreme in this respect). This is quite easy to prove, and for polygonal $P$, an algorithm can be produced to find ...

16

No, the Loewner ellipsoid is not monotone w.r.t. inclusion. Let $K$ be a square, whose Loewner ellipsoid is its circumcircle. Let $L$ be any other ellipse through the four vertices of $K$. The Loewner ellipsoid of $L$ is $L$ itself but it does not contain the circle. (I assume that the Loewner ellipsoid is the minimal circumscibed one. If you meant the ...

16

The second derivative of a convex function, in the distributional sense, is a non-negative bounded measure. And conversely. If this measure $\mu$ contains a sum $\sum_na_n\delta_{x=x_n}$, where $(x_n)_n$ is dense and $a_n>0$, $\sum_na_n<\infty$, then $\mu|_I\not\in L^1(I)$ for every non-void open interval $I$. The corresponding $g$ is an example for ...

answered Mar 22 '17 at 12:14

Denis Serre

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15

The standard sub-Riemannian metric on the 3-dimensional real Heisenberg group $\mathbb{H}^3$ is a counterexample. Topologically, it is homemorphic to $\mathbb{R}^3$, so in particular it is locally contractible. It is geodesic, but no ball is convex.
Thanks to the homogeneity and dilation structure, it suffices to know that, say, the unit ball centered at ...

14

I propose the axiom that any isometry between two such sets must take the center of one onto the center of the other. This axiom by itself is consistent with the existence of a center for every weakly compact convex set by the Ryll-Nardzewski fixed point theorem (we need the group of isometries of $S$ onto itself to always have a fixed point), and it alone ...

14

The statement for $C^1$ regularity is true, but with "dimension-2 projections" instead of "codimension-1 projections''. This is even stronger, if $d\ge 3$. On the other hand, for $d=2$ the statement with "hyperplane projections" fails, since $1$ dimensional projections are just closed intervals, whose boundary is certainly smooth. Also, an analogous ...

14

No. Let $X$ be a tripod (three segments with one common endpoint), $d=1$, $r=2$ and $E$ the set of the 3 leaf points.

14

Using the operator norm, as you have defined it, the fraction of the unit ball in real symmetric $n$-by-$n$ matrices that consists of positive definite matrices is $2^{-n(n+1)/2}$. Thus, this fraction shrinks very quickly as $n$ increases.
Added comment 1: Actually, I just realized that you don't need to do the calculation below; the ratio is obvious ...

answered May 2 '14 at 11:44

Robert Bryant

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14

No, as a counterexample take a closed "stadium" in $\mathbf{R}^2$ (the convex hull of two half-circles, I hope the word stadium makes it clear) and remove from it one of the endpoints of the half-circles. This is somehow related to the difference between extreme points and exposed points.

13

There has been recently a flurry of new results on provable nonconvex methods which can be guaranteed to converge to the global optimum. In other cases, the non-convex problem itself is shown to have no spurious local optima. The classical case is the singular value decomposition (SVD) which is non-convex but yet solvable. This is because only the top ...

13

As it was noticed by Abhinav Kumar, you only can hope for equality up to $\mathrm{GL}_n(\mathbb Z)$ transformations.
The following picture shows two $\mathrm{GL}_n(\mathbb Z)$-distinct plane figures which have the same number of integer points after any rescaling.
It still might be true that such bodies are $\mathrm{GL}_n(\mathbb Z)$-equidecomposable.
P.S....

answered Nov 20 '13 at 13:01

Anton Petrunin

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13

The natural generalization for your inequality is the setting of distributive lattices. The inequality is then known as the Fortuin–Kasteleyn–Ginibre (FKG) inequality, and has a long history. See for example Graham's article "Applications of the FKG inequality and its relatives". A generalization of the FKG inequality is the Ahlswede-Daykin inequality. For ...

answered Dec 8 '13 at 0:59

Gjergji Zaimi

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12

You can parametrize the zero locus of $\Delta$ (other than the origin) in the first quadrant by
$$
(x,y) = \left(\frac{(3{-}t)\sqrt{(3{+}t)(1{-}t)}}{8},
\frac{(3{+}t)\sqrt{(3{-}t)(1{+}t)}}{8}\right)
\qquad\qquad -1\le t\le 1
$$
and then compute that the curvature is
$$
\frac{(x'y''-y'x'')}{((x')^2+(y')^2)^{3/2}}
= \frac{2^{3/2}(3{+}t^2)}{\...

answered Aug 1 '14 at 17:33

Robert Bryant

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11

This problem has been considered before:
The notes to chapter 4 of
Rahman/Schmeißer: Analytic Theory of Polynomials, Oxford University Press, 2002
mention this problem, state that conv(P) is a proper subset of hull(P) in general, and give two references:
1) J. L. Walsh: The location of Critical Points of Analytic and Harmonic Functions,
AMS Colloquium ...

11

It may be a little uncommon to post an answer to one's own question, but
for a good record-keeping I think it is worth mentioning that a complete
and very explicit answer is now known. The function in question is given
by
$$ F(x) = \min \{ k\|x\|^{1-1/k}\colon k\ge 1 \}, $$
where $k$ runs over all positive integers, and $\|x\|=\min\{x,1-x\}$. The
graph of ...

11

This proof is mainly based on some of the suggestions in the early version of this answer. Also, we are building here on the comment by Peter Mueller concerning the polynomial $a$ defined below. It turns out that similar techniques can be used for all other nontrivial polynomials arising in the proof, which allowed us to replace multiple uses of Mathematica ...

11

No, there is no ellipsoid satisfying equivariance and monotonicity.
Suppose we have such an ellipsoid. Consider the square whose corners are $(\pm 1, \pm 1)$. Since this is invariant under a quarter-rotation, so is its ellipsoid, and the ellipsoid must be a circle of radius $r$.
By equivariance, we get the following table of shapes:
\begin{matrix}
\...

10

No; here's a counterexample: let $f = 0$ and consider the minimizer $y = 0.$ Then you can construct convex functions which converge to $0$ pointwise but whose minima are always moving away from $y =0,$ e.g. $f_n(x) = (x - n)^2/n^n.$

10

You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:
If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no ...

10

Let me answer at least some of your questions. I will only talk about your first definition of the cells, since these are somewhat nicer, as Igor Rivin pointed out.
You consider the function $f(w_1)=w_1\cdot\text{Area}(w_1)$ and asked whether it is convex. I assume you mean "Is the set $\{(x,f(x)):f(x)\geq 0\}\subset\mathbb{R}^2$ convex?", or in other words:...

10

Let's call the functions defined by Ali Taghavi to be sliced functions: a continuous function $\ f:\mathbb R^2\rightarrow\mathbb R\ $ is called sliced $\ \Leftarrow:\Rightarrow\ \ \forall_{c\in\mathbb R}\ f^{-1}(c)\ $ is convex.
NOTATION:
$$\ [x;y]\ :=\ \{(1\!-\!t)\cdot x\ +\ t\cdot y\ :\ 0\le t\le 1\}\ $$
for arbitrary $\ x\ y\in \mathbb R^n\ $ and $...

answered Jan 17 '15 at 21:06

Włodzimierz Holsztyński

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10

Let me elaborate on Darij's comment: Convex functions on a closed interval can be uniformly approximated by piecewise linear convex functions, and it is a theorem of Popoviciu that piecewise linear convex functions are of the form $$f(x)=ax+b+\sum c_i\left|x-r_i\right|,$$
where the $c_i$ are non-negative.
With this in mind, inequalities like Popoviciu's ...

answered Jun 28 '15 at 19:51

Gjergji Zaimi

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10

Time to start typing. As I said, this is going to be long and boring and I have only limited amount of free time nowadays so I'll type in chunks. If you see a flaw somewhere, comment immediately, but if everything looks like a chain of correct stupid computations leading nowhere, just keep patient and wait until it is finished.
Before passing to the proof, ...

10

Negative part
The answer is no without further assumptions, here is a counterexample (a bit nasty, it is not locally simply connected) :
In the plane consider first the two positive semi axis. ($\mathbb{R}_+\times\{0\}$ and $\{0\}\times\mathbb{R}_+$).
for each $n$ add the segment from $(0,2^{-n})$ to $(2^{-n},0)$.
put on it the length metric induced by the ...

10

I sketch a method that should show $F_n$ is convex for $n$ sufficiently large.
A Taylor expansion gives that $F_n(x) = x^{2n} (1-x)^2 + O(x^{4n-1})$, and so $$F_n''(x) = 2x^{2n-2}[2n^2(1-x)^2 + x^2 - n(1-x)(1+3x)] + O(x^{4n-3}).$$
This should show that $F_n''(x) > 0$ for $n$ large, provided $x < 1-\frac{C}{n}$ for some fixed $C>0$.
On the other ...

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