Search Results

The question seem to be nearly as hard as finding Voronoi diagram. In other words, you have a collection of functions of the form $$f_i(t)=t^2+a_i{\cdot}t+b_i$$ and you need to find the maximal int …

The answer is "no". Consider the function $$f(x,y)=\sqrt{x^2+y^2}+|y|$$ Note that $v_0=(1,0)$ is a subgradient at $(0,0)$. The gradient of $f$ is defined if $y\ne0$ and at all these points its first …

A counterexample to both questions can be found among the surfaces of revolution for long ovals which are symmetric with respect to the axis of rotation and the curvature bit more than 1 around the en …

(This is a new answer; my original answer was completely wrong.) Assume $\mathop{\rm vol}E>\tfrac12$. Then it contains two opposite points say $x$ and $x'=x+(\tfrac12,\tfrac12,\dots,\tfrac12)$. WLOG …

Sometimes it is not possible. Let me describe a metric on the 2-disk $\mathbb{D}$ that is Riemannian everywhere except the center of $\mathbb{D}$. Make a sequence of holes in $\mathbb{D}$ that conver …

No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions. For the updated question, the answer is still "NO". Again tak …

Consider the sphere with equator 4. Divide it into spherical cubes, the central projections from an inscribed cube. Note that the exponential map from tangent plane to the sphere is short. Note also …

Consider the convex hull $L_\varepsilon$ of the points $(\pm 1,0)$ and the $\varepsilon$-disc centered at the origin for small $\varepsilon>0$. If there is a $(\varepsilon,1)$-bi-Lipschitz map from …

It is easy to construct a volume preserving map between any two shapes of equal volume with triangular Jacobian matrix. (Gromov used such map from the given domain to ball to prove isoperemetric inequ …

The answe is "yes". Assume $d_H(C_1,C_2)<\varepsilon$. Let $\lambda_1$ and $\lambda_2$ be the length-measures of $C_1$ and $C_2$. We can assume $C_1$ is surrounded by $C_2$; the general case can be …

Your estimates are not scale invariant, so I am trying to guess what you want from the picture. A closed geodesic cuts your surface into two discs. Both have geodesic as a boundary, positive curvature …

What about $$C=\{\,(x,y)\in\mathbb{R}^2\mid x>0,\quad x\cdot y\ge 1\,\}$$ and $u=(0,-1)$? If instead you assume that $u$ belongs to the interior of the asymptotic cone of $-C$, then (I suspect that) t …

The function $q$ might be unbounded even in 1-neighborhood of $X^*$; here is an example of such function $f$ on the $(x,y)$-plane. Let $\phi(t)=|t|-t$. Choose a sequence $x_n\to \infty$. For each $n$ …

It follows from the fact that the closest-point projection is short (=distance nonexpanding). Check the solution of 9.47 in our book, altho it is a bit of an overkill.

The proof does not require smoothness. Assume $B(x,r_x), B(y,r_y)\subset \Omega$. It is sufficient to show that $$B(\tfrac{x+y}2,\tfrac{r_x+r_y}2)\subset \Omega.$$ The latter follows from the convexit …