User dorian - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:41:41Z http://mathoverflow.net/feeds/user/8755 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38751/a-holder-continuous-function-which-does-not-belong-to-any-sobolev-space A Hölder continuous function which does not belong to any Sobolev space Dorian 2010-09-15T00:24:19Z 2012-12-05T21:47:04Z <p>I'm seeking a function which is Hölder continuous but does not belong to any Sobolev space.</p> <p><strong>Question:</strong> More precisely, I'm searching for a function $u$ which is in $C^{0,\gamma}(\Omega)$ for $\gamma \in (0,1)$ and $\Omega$ a bounded set such that $u \notin W_{loc}^{1,p}(\Omega)$ for any $1 \leq p \leq \infty$. Take $\Omega$ to be bounded, open.</p> <p>My first guess is to do a construction with a Weierstrass function. I know this is differentiable 'nowhere' but that doesn't convince me it isn't weakly differentiable in some bizarre way. Hopefully someone knows of an explicit example.</p> http://mathoverflow.net/questions/38825/what-kind-of-lagrangians-can-we-have What kind of Lagrangians can we have? Dorian 2010-09-15T14:42:54Z 2012-11-21T09:51:59Z <p>In any physics book I've read the Lagrangian is introuced as as a functional whose critical points govern the dynamics of the system. It is then usually shown that a finite collection of non-interacting particles has a Lagrangian $\frac{1}{2}(m_1\dot{x}_1^2 + \cdots + m_n \dot{x}_n^2)$. It is then generally argued that $L=T-U$. I feel like something is missing here. </p> <p>What exactly are the physical hypotheses that go into this? Can we have other forms of the Lagrangian? How do we know those are "right"? Do we always have to compare them to the form of the equations we derived previously? For example, the Lagrangian formalism seems to be justified usually in so far as it 'works' for a finite collection of particles. Then you can solve any dynamics problem involving a collection of particles.</p> <p>I have been vague so let me try to be more precise in my question. Is the principle of least action an experimental hypothesis? Is it always true that $L=T-U$? When we don't know what the Lagrangian is, do we have to just guess and hope it is compatible with the dynamical equations we had already? Or can we perhaps start with the ansatz of a Lagrangian in some cases?</p> <p>I hope this is sufficiently precise.</p> http://mathoverflow.net/questions/100941/bonnesens-inequality-for-non-simple-curves Bonnesen's inequality for non-simple curves Dorian 2012-06-29T15:10:54Z 2012-06-30T13:43:59Z <p>Given a closed curve in the plane $\mathbb{R}^2$, it is well known that $L^2 \geq 4\pi A$ where $L$ is the length of the curve and $A$ is the area of the interior of the curve.</p> <p>For a <em>simple</em> closed curve $\gamma$, the stronger inequality due to Bonnesen holds: $L^2 - 4\pi A \geq \pi^2 (R_{out}- R_{in})^2$, where, setting $\Omega =$ Int $(\gamma)$, $R_{in}$ and $R_{out}$ denote the inner and outer radii of the sets:</p> <p>$R_{in} = \sup_{B_r \subset \Omega} r$</p> <p>$R_{out} = \inf_{\Omega \subset B_R} R$</p> <p><strong>Question:</strong> Does this inequality continue to hold <strong>without</strong> the assumption that the curve is <strong>simple</strong>? In particular, does it hold for any <em>connected, rectifiable set</em>?</p> <p>If one adds components to the interior of a simple curve, it is clear that this increases the isoperimetric defecit. However while $R_{out}$ will remain unchanged, $R_{in}$ will necessarily decrease, making it not immediately clear that the inequality would continue to hold. </p> <p><strong>Update:</strong> I just realized this is <em>false</em>. Take a countable union of points in the interior of the unit ball. Around the kth point, make a circle of radius $\epsilon/2^k$. Distribute the points well enough and $R_{in}$ can be made arbitrarily small. Then if you make $\epsilon$ small enough, you don't change the length or area too much and so the desired inequality will be violated. </p> http://mathoverflow.net/questions/98991/generalization-of-an-inequality-due-to-gage-for-curve-shortening-part-ii Generalization of an inequality due to Gage for curve shortening Part II Dorian 2012-06-06T21:20:45Z 2012-06-07T20:17:08Z <p>I asked a question recently about generalizing an inequality due to Gage. This inequality asserts that given a convex domain $\Omega$ in $\mathbb{R}^2$ with support function $p(X) = \langle X, \nu \rangle$ where $X$ is the position vector with respect to the origin and $\nu$ is the normal on $\partial \Omega$ that </p> <p>$\int_{\partial \Omega} p^2 dS \leq \frac{LA}{\pi}$, holds for some choice of origin where $L$ is the length of the curve and $A$ the area.</p> <p>I have been trying to prove the following generalization for <strong>star shaped sets</strong>:</p> <p><code>$\int_{\partial \Omega} p^2 dS \leq \int_{\partial \Omega^*} (p^*)^2 dS^* + C(L,A)(A^*-A)$</code>,</p> <p>where here <code>$\Omega^*$</code> is the convex hull of $\Omega$, <code>$p^*$</code> is the support function of the convex hull, <code>$A$</code> is the area of the set and <code>$A^*$</code> the area of the convex hull. Using Green's theorem I have been able to deduce the following: <code>$\int_0^{2\pi} p(\theta)^2 \frac{dS}{d\theta}{d\theta} = \int_0^{2\pi} p^*(\theta)^2 \frac{dS^*}{d\theta} dS^* + \int_0^{2\pi} p^*(\theta)p(\theta)( \frac{dS^*}{d\theta} -\frac{dS}{d\theta}) d\theta + \frac{1}{2} \int_{0}^{2\pi} [p(\theta)^*-p^(\theta)][r^*(\theta)^2 - r(\theta)^2] d\theta$</code>,</p> <p>where <code>$r^*$</code> and <code>$r$</code> are the lengths to <code>$\partial \Omega^*$</code> and <code>$\partial \Omega$</code> respectively and we can paramateterize by <code>$\theta$</code> since the set is star shaped. I thus need only control the term </p> <p><code>$\int_0^{2\pi} p^*(\theta) p(\theta)(\frac{dS^*}{d\theta} - \frac{dS}{d\theta}) d\theta$</code></p> <p>but this is not clear at all to me. My intuition suggests it should be negative.</p> <p>I would appreciate any ideas or suggestions for alternative approaches to this.</p> http://mathoverflow.net/questions/99036/an-isoperimetric-type-maximization-problem-with-a-barrier An isoperimetric type maximization problem with a barrier. Dorian 2012-06-07T13:55:31Z 2012-06-07T20:10:18Z <p>I'm trying to minmize a particular functional which depends on a curve with fixed endpoints which lies below a fixed line in $\mathbb{R}^2$. Here are the details:</p> <p>Let $(r(\theta), \theta)$ be a smooth segment of a curve in polar coordinates satisfying the following for $0 &lt; \theta_1, \theta_2 &lt; \pi$. </p> <p>1) $r(\theta_i) \sin (\theta_i) = \lambda$ for $i=1,2$ where $\lambda$ is a fixed constant.</p> <p>2) $r(\theta) \sin(\theta) \leq \lambda$ for all $\theta \in [\theta_1, \theta_2]$. </p> <p>Thus $(r(\theta),\theta)$ is a curve which touches the line $y=\lambda =$ constant at its two endpoints and lies beneath this line for all other value of $\theta$. Let $\mathcal{A}$ denote the class of curves $(r(\theta),\theta)$ satisfying conditions $1)$ and $2)$. Then consider the maximization problem:</p> <p><code>$\max_{ (r,\theta) \in \mathcal{A} } \int_{\theta_1}^{\theta_2} \frac{1}{\sin^2(\theta) \sqrt{1+(\dot r/r)^2} } d\theta$</code>.</p> <p>Geometrically speaking, the above can be written as <code>$\int_{\theta_1}^{\theta_2} \frac{d\theta}{dS} \frac{r^2}{\sin^2(\theta)} d\theta$</code>, which makes it appear as a sort of inverse perimeter problem. Clearly the curve would like to follow the trajectory of a circle, but due to the constraint it cannot do this everywhere. My suspicion is that the solution should be attained by $r(\theta) = \frac{\lambda}{\sin \theta}$ but it's not clear that it would not be favorable for the curve to descend below this line if it can follow the path of a circle for some rangle of angle. </p> <p>If this type of problem is known then references would also be appreciated. </p> http://mathoverflow.net/questions/98728/generalization-of-an-inequality-due-to-gage-for-curve-shortening Generalization of an inequality due to Gage for curve shortening Dorian 2012-06-03T17:22:22Z 2012-06-04T13:21:55Z <p>There is a well known inequality due to Gage which asserts the following. Let $\Omega$ be a smooth, convex set in $\mathbb{R}^2$ and let $p = \langle X, \nu \rangle$ be the support function of $\Omega$, where $X = \langle x, y \rangle$ with respect to some origin $O$ and $\nu$ is the normal to the boundary.</p> <p>Denoting $A$ and $L$ and the area and length of the curve, then it holds that $\int_{\partial \Omega} p^2 dS \leq \frac{AL}{\pi}$ for some particular choice of origin $O \in \Omega$.</p> <p><strong>Question/Conjecture:</strong> Given an arbitrary simply connected, smooth set $\Omega$, does it hold that <code>$\int_{\partial \Omega} p^2 dS \leq \frac{LA^*}{\pi}$</code> where $A^*$ denotes the area of the convex hull of $\Omega$ and $L$ is the length of the original boundary $\partial \Omega$. </p> <p><strong>Update June 04/2012:</strong> There has been an answer to my original question so I would like to ask if a related although weaker assertion is true. Let $\partial \Omega$ be paramaterizable by the angle $\theta$ in polar coordinates, so that the curve is represented by $(r(\theta),\theta)$. Then <code>$p = p(\theta)$</code> is obviously single valued. This means precisely that the domain $\Omega$ is <em>star shaped</em>. Let <code>$p^*$</code> be the support function of the convex hull $\Omega^*$. Does it hold that <code>$\int_{\partial \Omega} p^2 \leq \int_{\partial \Omega^*} (p^*)^2 dS$</code>?</p> <p>Any direction to references on related questions would also be greatly appreciated.</p> http://mathoverflow.net/questions/93885/different-forms-of-bonnesens-strong-isoperimetric-inequality-in-the-plane Different forms of Bonnesen's strong isoperimetric inequality in the plane. Dorian 2012-04-12T17:41:07Z 2012-05-30T16:27:10Z <p>I'm sure that many readers are already familiar with the well known Bonnesen inequality in the plane for a smooth, connected curve: </p> <p>$(R_{out} - R_{in})^2 \leq \pi^2 (L^2 - 4\pi A),$</p> <p>where $R_{out}$ and $R_{in}$ are the outer and inner radii respectively, $L$ is the length of the curve and $A$ is it's area. </p> <p>This form turns out to often be quite inconvenient for me, and I was wondering if there was a version which read as follows:</p> <p>$R_{out}^2 - R_{in}^2 \leq C (L^2 - 4\pi A)$,</p> <p>where now $C$ is a constant which may depend on the support of the curve or other quantities possibly. The main difference is that when $R_{in} - R_{out}$ is small, the second inequality I've written is <em>strictly stronger</em> if one assumes the curve is constrainted to a bounded domain. The sacrifice I make though is that I do not care what the constant is, or if it depends on the support of the curve.</p> <p>This inequality seems quite reasonable but I have thus far been unable to prove it. Has anyone encountered such a version of this inequality?</p> http://mathoverflow.net/questions/25470/when-is-sobolev-space-a-subset-of-the-continuous-functions/37775#37775 Answer by Dorian for When is Sobolev space a subset of the continuous functions? Dorian 2010-09-05T02:48:36Z 2012-05-29T20:06:08Z <p>I'll give you a hint for the first one $d=1$. Consider first the case that your function $f \in H^1([0,1])$ was smooth. Then we could say $f(x) - f(y) = \int_{x}^y f'(s)ds$. Apply Cauchy Schwartz now and you'll be able to see immediately that $f$ is $1/2$ Hölder continuous.</p> <p>For higher dimensions you actually proceed similarly but you need to use the co-area formula.</p> http://mathoverflow.net/questions/97853/what-are-the-most-general-types-of-curves-in-mathbbr2-for-which-gauss-bonne What are the most general types of curves in $\mathbb{R}^2$ for which Gauss-Bonnet holds? Dorian 2012-05-24T17:23:38Z 2012-05-29T11:52:44Z <p>I would like to know what is the most general form of the Gauss-Bonnet theorem in the plane for curves. It is well known for that for any piecewise $C^2$ simply connected curve with corners, one has</p> <p>$\int_{\partial \Omega} \kappa(y) dS(y) = 2\pi$,</p> <p>where $\kappa$ denotes the curvature of the boundary $\partial \Omega$. This formula continues to hold for any curve for which $\hat n \cdot (1,0)$ (for instance) defines a $BV$ function.</p> <p><strong>Question:</strong> I would like to know to what extent this generalizes. For example, if $\Omega$ is a simply connected, compact set in $\mathbb{R}^2$ whose boundary has finite length and has <em>generalized</em> curvature $\kappa \in L^{\infty}(\partial \Omega)$, does it hold that $\int_{\partial \Omega} \kappa(y) dS(y) = 2\pi$? If not, are there counter examples in the plane where this fails? </p> http://mathoverflow.net/questions/94727/is-there-a-characterization-of-generalized-constant-mean-curvature-surfaces Is there a characterization of generalized constant mean curvature surfaces? Dorian 2012-04-21T10:48:39Z 2012-04-21T22:12:54Z <p>It is a well known result of Alexandrov that the <em>only compact, connected, constant mean curvature surface is the <strong>ball</em></strong>. There is a generalized notion of curvature known as <em>generalized mean curvature</em> which makes sense on rectifiable varifolds. In particular if $V$ is a recifiable varifold than the generalized mean curvature is defined as follows:</p> <p>Let $\|\delta V\|$ denote the first variation of $V$. Then if this measure is absolutely continuous with respect to $\|V\|$ then by the Riesz representation theorem:</p> <p>$\delta V(g) = - \int g \cdot H d\|V\|$, for $H$ defined $\|V\|$ a.e. </p> <ul> <li>H as defined here is the <em>generalized mean curvature</em>. </li> </ul> <p><strong>Question:</strong> Working in the plane, $\mathbb{R}^2$, assume that $\Omega$ is a set of bounded variation so $|\partial \Omega| &lt; C &lt; +\infty$ and define $H$ as above.</p> <ul> <li><p>If $H=$ constant, what does this imply about $\Omega$? Must $\Omega$ be a ball? Are there examples of singular $BV$ sets which satisfy this condition but are not balls?</p></li> <li><p>What is the generalized mean curvature of a square? </p></li> </ul> http://mathoverflow.net/questions/89983/what-is-the-constant-in-the-rate-of-exponential-convergence-for-mean-curvature-fl What is the constant in the rate of exponential convergence for mean curvature flow? Dorian 2012-03-01T18:25:45Z 2012-04-18T16:22:00Z <p>Given a domain $\Omega \subset \mathbb{R}^d$ which is convex and smooth and $| \Omega|=1$, it is well known that the metric converges exponentially fast to that of the sphere under volume preserving MCF. I would like to know the following:</p> <p><em>Question:</em> Does the rate of convergence depend on the domain $\Omega$? If so, in what way? If I know in particular that <code>$\frac{d}{dt} \|\kappa - \bar \kappa\|_{L^2(\partial \Omega)}^2 \leq -C \|\kappa - \bar \kappa\|_{L^2(\partial \Omega)}^2$</code>, can I say that $C$ is independent of $\Omega$? If not, can I see in which explicit way it does depend on $\Omega$? I suppose this is equivalent to asking if there is a particular $\Omega$ so that the convergence is <em>slowest</em>. </p> <p>Thanks.</p> http://mathoverflow.net/questions/55404/stronger-version-of-the-isoperimetric-inequality Stronger version of the isoperimetric inequality Dorian 2011-02-14T12:44:06Z 2012-04-12T15:33:46Z <p>I have been searching for a version of the isoperimetric inequality which is something like:</p> <p>$P(\Omega) - 2\sqrt{\pi} Vol(\Omega)^{1/2} \geq \pi (r_{out}^2 - r_{in}^2)$ where $r_{out}$ and $r_{in}$ are the inner and outer radius of a given set. There are of course details which I am missing such as what kind of sets this applies to (clearly connected and possibly simply connected). I was hoping somebody may recognize this inequality and be able to direct me to a source for it along with a proof.</p> <p><strong>Update:</strong> I'm curious if anyone can direct me to a some papers which relate the isoperimetric deficit to Hausdorff distance. Such as: $P(\Omega)^2 - 4\pi Vol(\Omega) \geq C d_H(\Omega,B)^2$ whre $B$ is a sphere in $\mathbb{R}^2$ which may be the inner or outer circle.</p> <p><strong>Update April 12:</strong> I would like to know if the first Bonnesen inequality written below is strictly stronger than the one in higher dimensions? In particular, if one considers the Fraenkel assymetry $\alpha(\Omega) = \min_B |\Omega \Delta B|$ where $|B|=|\Omega|$, does it hold on a bounded domain that</p> <p>$r_{out}^2 - r_{in}^2 \leq C \alpha(\Omega)$,</p> <p>for some constant $C>0$? This seems like it should be true but I can't seem to find a concise proof of it.</p> http://mathoverflow.net/questions/89980/what-does-convergence-in-the-l2-sense-to-a-constant-mean-curvature-surface-imp What does convergence in the $L^2$ sense to a constant mean curvature surface imply? Dorian 2012-03-01T18:09:10Z 2012-04-04T11:33:12Z <p>I have been thinking about the following question and have been unable to find any literature on the subject. </p> <p><em>Question:</em> Assume I have a sequence of <strong>smooth, simply connected, compact</strong> domains $\Omega_s \subset \mathbb{R}^d$ such that $|\Omega_s|=1$ and</p> <p>$\int_{\partial \Omega_s} (\kappa - \bar \kappa)^2 dS(y) \to 0$ as $s \to +\infty$,</p> <p>where here $\kappa$ is the mean curvature of the surface $\partial \Omega_s$ and $\bar \kappa$ denotes the average mean curvature over $\partial \Omega_s$. I can prove that the limit is in fact a ball in the following <strong>two</strong> cases:</p> <ol> <li>All of the sets $\Omega_s$ are convex. or</li> <li>I assume the uniform bound $\limsup_{s \to +\infty} |\partial \Omega_s| + \int_{\partial \Omega_s} \kappa^2 dS &lt; +\infty$. </li> </ol> <p>I would however like to remove these restrictions since they seem quite artificial. I have been able to rule out the standard "pinching" counter examples of a long rod with capped ends, but am not sure if there could exist other pathologies. Any direction to results in this direction would be appreciated.</p> http://mathoverflow.net/questions/90573/classification-of-limits-under-volume-preserving-mean-curvature-flow Classification of limits under volume preserving mean curvature flow? Dorian 2012-03-08T12:45:17Z 2012-03-09T13:33:42Z <p>It is well known that if you start with a domain $\Omega \subset \mathbb{R}^d$ which is <em>uniformly convex</em>, then it converges exponentially fast to the <em>ball</em> when evolved under volume preserving mean curvature flow (VPMCF).</p> <p>In $\mathbb{R}^2$, if one starts with a <em>smooth simply connected</em> domain, then after a finite amount of time it will become convex, and converge to the <em>ball</em>.</p> <p><em>Question:</em> If $\Omega \subset \mathbb{R}^d$ is smooth and connected, does it converge (perhaps after surgery) to a constant mean curvature (CMC) surface? Is it possible to classify the type of CMC surface it converges to based on the intial domain $\Omega$? I am hoping mostly for direction to the appropriate literature in this case, as mathscinet is flooded with papers on the subject.</p> <p><strong>Update:</strong> I would like to know in particular, on the torus $\mathbb{T}^2$, if I start with a set $\Omega \subset \mathbb{T}^2$ which satisfies $|\Omega|=1/2$, and I evolve it under VPMCF, does it eventually converge to the stripe pattern?</p> http://mathoverflow.net/questions/87020/how-to-minimize-the-length-of-a-graph-connecting-n-points-in-mathbbr3 How to minimize the length of a graph connecting n points in $\mathbb{R}^3$ Dorian 2012-01-30T11:23:06Z 2012-02-01T15:40:32Z <p>More precisely I would like to consider the following problem:</p> <p>Let <code>$\{a_i\}_{i=1}^n$</code> be n points in $\mathbb{R}^3$ and assume I have the constraint $\min_{i \neq j} |a_i-a_j| = r_{min}>0$. How can I place $n$ points in $\mathbb{R}^3$ as to minimize <code>$\sum_{i \neq j} |a_i-a_j|.$</code></p> <p>When $n=2$ this is <em>trivial</em>. When $n=3$ the minimizer will clearly be an <em>equilateral triangle</em> with a point at each vertex. When $n=4$, the solution would be a <em>tetrahedron</em>. For higher $n$ the answer is not clear to me. I would also be interested in the patterns in the asymptotic limit as $n \to +\infty$.</p> <p>I presume that if this is known, it is a well established result in graph theory, and I would appreciate any references.</p> <p>Update: Thanks for the helpful comment and the answer. This answers my question.</p> http://mathoverflow.net/questions/80357/a-variational-problem-involving-a-negative-fractional-soboblev-norm A variational problem involving a negative fractional Soboblev norm. Dorian 2011-11-08T04:06:27Z 2011-12-05T21:53:18Z <p>I've run into the problem of trying to evaluate the following:</p> <p>$\max_{ \xi} \iint_{\partial B \times \partial B} \xi(x) \Phi(|x-y|) \xi(y) dS(x)dS(y)$ </p> <p>subject to $\int_{\partial B} \xi(y)dS(y) = 0$ and $\int_{\partial B}\xi^2(y)dS(y)=1$ where $B \subset \mathbb{R}^3$ is a ball of radius $1$ and $\Phi(|x-y|)=\frac{1}{|x-y|}$ is the Newtonian potential.</p> <p>This seems to resemble an inverse fractional Soblev norm such as $H^{-1}$ and moreover appears to be related to the problem of finding an optimal Poincare constant.</p> <p>My guess is that the maximum is obtained for $\xi=+1$ on the upper half and $\xi=-1$ on the lower half. Given this however, I still cannot do an explicit calculation to determine this quantity. Is there a standard reference for such problems arising in Potential Theory perhaps which will allow one to evaluate (even approximately) such expressions?</p> <p>For instance I know I can rewrite the above as: $\int_{\partial B} |\nabla w|^2$ where $-\Delta w = \mu$ and $\mu(x) = \xi(x)dS(x)$ but I'm not sure how this can help me to evaluate such an expression.</p> <p>To summarize, I would like to try to evaluate the above double integral for the particular function $\xi = +1$ on the upper half of the ball and $\xi=-1$ on the lower half. Being able to solve explicitly the above maximization problem would be a bonus.</p> http://mathoverflow.net/questions/37924/elliptic-regularity-for-the-neumann-problem Elliptic regularity for the Neumann problem Dorian 2010-09-06T20:06:38Z 2011-11-03T14:44:21Z <p>I'm trying to understand how to establish regularity for elliptic equations on bounded domains with Neumann data.</p> <p>For simplicity, let's presume we are focusing on $-\Delta u = f$ in $\Omega$ and $\frac{\partial u}{\partial \nu} = 0$ on $\partial \Omega$. Interior regularity works the same as always.</p> <p>When proving boundary regularity, for the dirichlet boundary case we first consider some ball $B(0,1) \cap \mathbb{R}_+^n$ and let $\xi = 1$ on $B(0,1/2)$, $\xi = 0$ on $\mathbb{R}^n - B(0,1)$ and then estimate all derivatives $\frac{\partial^2 u}{\partial x_i \partial x_j}$ except $\partial^2 u/\partial x_n^2$. Two main points are needed</p> <p>1) $\xi$ vanishes on the curved part of $B(0,1) \cap \mathbb{R}_+^n$\</p> <p>2) $u=0$ on ${x_n=0}$.</p> <p>This allows us to let $-\partial_{x_i} (\xi \partial_{x_j} u)$ (with derivatvies replaced by difference quotients) be an admissible test function for our weak definitoin of a solution.</p> <p>I presume <strong>the main difficulty in neumann boundary data is making your test function be admissible</strong>. In other words, we would need $\int v = 0$ since our existence was established on $H^1(\Omega)$ restricted to mean value zero functions. </p> <p>So in order to proceed, can we just subtract off a constant from our original $-\partial_{x_i}(\xi \partial_{x_j}u)$? Is there some more natural way to establish regularity in this case? I do not want to take advantage of the fact that we have a green's function in this case however as I only chose the Laplace equation for simplicity.</p> http://mathoverflow.net/questions/77875/best-poincare-constants-on-the-surface-of-a-ball Best Poincare constants on the surface of a ball Dorian 2011-10-11T22:55:27Z 2011-10-12T05:26:30Z <p>I'm considering specifically functions $\xi:\partial B(0,1) \to \partial B(0,1)$ in $\mathbb{R}^2$ and $\mathbb{R}^3$ satisfying $\int_{\partial B(0,1)} \xi(y) dS(y) = 0$. I would like to know first of all if there is a Poincare inequality of the form:</p> <p>$\int_{\partial B(0,1)} |\nabla^{-1}\xi(y)|^2dS(y) \leq C_{B} \int_{\partial B(0,1)}|\xi|^2 dS(y).$</p> <p>Secondly (and more importantly) I'd like to know if there are any resources I can be directed to regarding the optimal constant in such a case. </p> http://mathoverflow.net/questions/72431/functionals-continuous-with-respect-to-weak-convergence/75096#75096 Answer by Dorian for Functionals continuous with respect to weak convergence Dorian 2011-09-10T13:31:45Z 2011-09-13T01:41:36Z <p>I am not going to try to find the most general conditions under which lower semi-continuity holds but for that I suggest the standard reference for all of this is "Direct methods in the Calculus of Variations" by Bernard Dacorogna which covers all of this in full detail. I will give a brief outline of the answer however.</p> <p>In order for $\int f(x,u,du)dx$ to be lower semi-continuous with respect to weak convergence one does <em>not</em> in general need any sort of <em>convexity</em> in the $u$ variable and what is more important generally is </p> <p>1) Coercivity (see below) of the functional $f$ in the $du$ variable.</p> <p>2) The right embedding along with <em>continuity</em> of $f$ in the $u$ variable with respect to the topology of strong convergence in this space.</p> <p><em>Example:</em> A good example is $E(u) = \int_0^1 |\nabla u |^2 + (1-u^2)^2$ with $u = 0$ on $\partial \Omega$ in $\mathbb{R}^d$. Observe that when $d \leq 4$ one has $H^1(\Omega) \subset \subset L^4(\Omega)$ and consequently the non-convex term depending on $u$ is lower order. Therefore one can pass to the limit in any minimizing sequence even though $(1-u^2)^2$ is very non convex (but it is continuous with respect to strong convergence). Notice however that for the energy $\bar E(u) = \int_0^1 |\nabla u|^2 + u^4$ in $\mathbb{R}^5$ with $u$ prescribed on $\partial \Omega$, the second term is <em>not</em> lower order but here we may use convexity to conclude lower semi-continuity of the second term.</p> <p>The function $p \mapsto f(x,z,p)$ is <em>coercive</em> if there exists some constant $C > 0$ so that $f(x,z,p) \geq C|p|^q$ for some range of $q$. It then depends on what spaces one is working in but the goal is to use an embedding theorem such as $L^q \subset\subset W^{1,p}$ for $1 \leq q &lt; p^*$ and to conclude that for a minimizing sequence $u_n$, there is in fact a <em>strongly</em> convergent subsequence in $L^q$ for some $q$. Then one will generally expect some sort of continuity of $f$ in the $u$ variable. </p> <p>There are many more interesting examples but in almost all cases in practice the goal is to show that one has <em>strong</em> convergence of the $u_n$s in your minimizing sequence in some $L^p$ space. The book by Braides focuses mostly on asymptotics of functionals which depend on some large (or small) parameter and I'm not sure how much he talks about the assumptions needed in the direct method.</p> http://mathoverflow.net/questions/74884/how-does-electric-potential-relate-to-mean-curvature How does electric potential relate to mean curvature? Dorian 2011-09-08T14:49:01Z 2011-09-08T15:25:37Z <p>Consider a compact, convex domain $\Omega \subset \mathbb{R}^3$ with $|\Omega|=1$ with smooth boundary $\partial \Omega$. </p> <p>Now consider the electric potential generated by this uniform mass distribution: $\phi = \int_{\Omega} \frac{1}{|x-y|} dx$. <strong>Question:</strong> I would like to know if there is a relationship between mean curvature $H$ and $\phi$. What I have in mind is an inequality of the form <code>$\|\phi - \bar \phi\|_{L^p(\partial \Omega)} \leq C \|H - \bar H\|_{L^p(\partial \Omega)}$</code> where $\bar \phi$ and $\bar H$ denote the averages over $\partial \Omega$ of $\phi$ and $H$ respectively and $p$ can be anything for the time being. </p> <p>Although one term is much more 'non-local' than the other, for a convex body it seems reasonable that the closer to being an surface of constant mean curvature, the closer one is to being an equipotential surface. This is related to a previous question of mine which ended up going unanswered regarding whether the only convex, compact equipotential surface in $\mathbb{R}^3$ was a sphere or not.</p> http://mathoverflow.net/questions/38309/stuck-on-a-convergence-argument-in-h-01-omega Stuck on a convergence argument in $H_0^1(\Omega)$. Dorian 2010-09-10T14:04:41Z 2011-04-29T04:14:05Z <p>I'm trying to verify that a functional I have satisfies the Palais Smale condition for appliction of the Mountain Pass lemma.</p> <p>However I've encountered this step along the way which seems clear to me but I'm second guessing whether or not it is true.</p> <p><strong>Question:</strong> If $u_k \to u$ in $L^{p+1}$ for $p + 1 &lt; 2^*=\frac{2n}{n-2}$ then I would like to see that $\Delta^{-1}(|u_k|^{p-1}u_k) \to \Delta^{-1}(|u|^{p-1}u)$ in $H_0^1(\Omega)$. This is of course equivalent to showing that $|u_k|^{p-1}u_k \to |u|^{p-1}u$ in $H^{-1}(\Omega)$.</p> <p><em>My idea:</em> Since I have convergence in $L^{p+1}(\Omega)$ it follows that I have convergence in all $L^q(\Omega)$ for $p+1 \geq q \geq 1$. By Sobolev embeddings I believe that it's true that <code>$||w||_{H^{-1}} \leq ||w||_{L^q}$</code> for any $q$ with $1/q + 1/r = 1$ for $1 \leq r \leq 2^*$. So this should imply the needed $H^{-1}(\Omega)$ convergence <em>if</em> I knew that $|u_k|^{p-1}u_k \to |u|^{p-1}u$ in some $L^q$ within this range. The best however I can say is that I have convergence in $L^{\frac{p+1}{p}}$ since $u_k \to u$ in $L^{p+1}$. But then $1 + 1/p > 1 + \frac{n-2}{n+2} = \frac{2n}{n+2}$ which is the conjugate exponent to $2^*$. </p> <p>This appears to work but is quite technical and messy and all of the 'proofs' I've seen hint at some "simply energy argument". This doesn't appear simple at all! Therefore I would appreciate any suggestions about a better approach or if someone could point out something wrong with how I've thought about it. I hope this fits within the paramaters of the website.</p> http://mathoverflow.net/questions/38266/spectral-galerkin-method-for-a-semi-linear-parabolic-pde Spectral Galerkin method for a semi-linear parabolic PDE Dorian 2010-09-10T04:19:37Z 2011-03-29T14:10:31Z <p>I'm trying to understand how to apply the Galerkin method to $u_t - \Delta u = u^3$. I understand how to obtain all of the a-priori estimates using Sobolev embeddings and such but my question concerns the actual discretization procedure where we project onto the finite dimensional subspace spanned by the eigenfunctions of $-\Delta$. </p> <p>In the linear case we may simply set $u_N = \sum \phi_n c_n$ and plug this in to obtain a set of $N$ O.D.Es which we then show satisfy the same energy bounds. In the non-linear case though we may not just substitute directly because of the $u^3$ term. How can this be dealt with? Is there perhaps a better way to do the approximation?</p> <p><strong>Addition:</strong> In this example if we let $u_N = \sum \phi_n c_n^N(t)$ then when we plug this into the weak form of our PDE we obtain and then choose our test function to be one of the basis elements $w_k$ we obtain $d/dt c_k^N(t) + \sum_{i=1}^n e^{ki}(t) c_k^N(t) = \int (\sum_{n=1}^N c_n^N(t) \phi_n)^3w_k$, for some coefficients $e^{ki}(t)$. My question is, how do I deal with the <em>integral on the right</em>? I would like to be able to solve this ODE and then say I have a solution.</p> http://mathoverflow.net/questions/59708/existence-of-rearrangements-of-functions-in-lp0-1-when-given-a-measure-pre Existence of rearrangements of functions in $L^p([0,1])$ when given a measure preserving map. Dorian 2011-03-27T10:27:47Z 2011-03-28T05:55:25Z <p>Given a funtion $f \in L^p([0,1])$ (take $p=\infty$ if you'd like), and also a measure preserving map $s:[0,1] \to [0,1]$ (meaning $s$ pushes Lebesgue measure forward to itself) I would like to know if there exists some $f^* \in L^p([0,1])$ such that $f^*\circ s = f$. If $s$ is invertible this is of course obvious but measure preserving maps need not be invertible (although must be onto).</p> <p>Recall that given $f$ there exists a monotone rearrangement of $f$, and a measure preserving map $t:[0,1] \to [0,1]$ which yields this rearrangement.</p> <p>However my question is in some sense the reverse question (with no monotonicity added).</p> <p>Somehow it seems intuitive that such a map should exist but I'm not able to prove it directly. It seems like something which may be well known however.</p> http://mathoverflow.net/questions/37551/a-counter-example-to-hahn-banach-separation-theorem-of-convex-sets A counter example to Hahn-Banach separation theorem of convex sets. Dorian 2010-09-02T22:32:37Z 2011-03-09T22:00:05Z <p>I'm trying to understand the necessity for the assumption in the Hahn-Banach theorem for one of the convex sets to have an interior point. The other way I've seen the theorem stated, one set is closed and the other one compact. My goal is to find a counter example when these hypotheses are not satisfied but the sets are still convex and disjoint. So here is my question:</p> <p><strong>Question:</strong> I would like a counter example to the Hahn-Banach separation theorem for convex sets when the two convex sets are <em>disjoint</em> but neither has an interior point. It is trivial to find a counter example for the <em>strict</em> separation but this is not what I want. I would like an example (in finite or infinite dimensions) such that we <em>fail</em> to have any separation of the two convex sets at all. </p> <p>In other words, we have $K_1$ and $K_2$ with $K_1 \cap K_2 = \emptyset$ with both $K_1$ and $K_2$ convex belonging to some normed linear space $X$. I would like an explicit example where there is <strong>no</strong> linear functional $l \in X^*$ such that $\sup_{x \in K_1} l(x) \leq \inf_{z \in K_2} l(z)$.</p> <p>I'm quite sure that a counter example cannot arise in finite dimensions since I think you can get rid of these hypotheses in $\mathbb{R}^n$. I'm not positive though.</p> http://mathoverflow.net/questions/38430/failure-of-regularity-up-to-the-boundary-for-a-linear-elliptic-pde Failure of regularity up to the boundary for a linear elliptic PDE Dorian 2010-09-11T23:11:29Z 2010-10-27T10:03:40Z <p>I asked a question before where I wanted a simple example where regularity up to the boundary fails for a linear elliptic PDE. I was presented an example with $\Omega = B(0,1) \backslash {0}$ (ball minus a point) which is nice but I would like something less pathalogical. I would like an example where my domain is at least Lipschitz (so a rectangle is an example). In the case where we subtract off a point, a boundary value problem doesn't really even make sense in a weak formulation to begin with.</p> <p>More precisely,</p> <p><strong>Question:</strong> I would like an example where an elliptic PDE with smooth coefficients satisfies $Lu = f$ in $\Omega$ for smooth $f$ (or zero) and $u = 0$ on $\partial \Omega$ (which is of Lipschitz class) but where $u$ somehow fails to be "regular" up to the boundary (I'm being vague on purpose here as any failure of regularity will do for the most part).</p> <p>My guesses have been try try looking at the upper right quadrant $[{ (x,y) : x, y > 0}]$ but nothing has come from this so far. Any suggestions/ideas are welcome and appreciated.</p> http://mathoverflow.net/questions/38054/a-simple-example-where-elliptic-boundary-regularity-fails-due-to-a-kink-in-the-do A simple example where elliptic boundary regularity fails due to a kink in the domain Dorian 2010-09-08T13:44:47Z 2010-10-08T10:27:20Z <p>I'm seeking a simple example of where elliptic (preferably linear) boundary regularity fails due to a simple kink in the domain.</p> <p>So far my gueses were to look at $-\Delta u = f$ on $[0,2\pi] \times [0,2\pi]$ with $0$ Dirichlet boundary conditions and choose an $f$ which was far from $0$. This hasn't seem to produce any results (I was checking regularity directly by the method of Fourier series).</p> <p>So more precisely, I would like an example where </p> <p>1) $Lu = f$ in $\Omega \subset \mathbb{R}^n$ with $f$ smooth </p> <p>2) $L$ is elliptic and $u = 0$ on $\partial \Omega$ </p> <p>3) $\Omega$ is <em>not smooth</em> and consequently $u$ is not smooth up to the boundary.</p> http://mathoverflow.net/questions/41421/compatibility-conditions-for-parabolic-regularity Compatibility conditions for parabolic regularity Dorian 2010-10-07T14:41:40Z 2010-10-07T22:11:50Z <p>I'm trying to understand the compatibility conditions for regularity of second order parabolic equations. Let's consider the equation $u_t - Lu = f$ with $u(0)=g$ on $\Omega \times [0,T)$ with $u = 0$ on $\Gamma_T$, the parabolic boundary of <code>$\Omega \times [0,T)$</code>.</p> <p>We need the assumptions that <code>$\frac{d^{m-1}f}{d^{m-1}}(0) - Lg_{m-1} \in H_0^1(\Omega)$</code> for the $L^2$ theory to work to conclude smoothness up to the boundary (I'm presuming $f$ and $\Omega$ are smooth). As far as I can tell this is required so that we don't have the following situation:</p> <p>$u_t - u_{xx} = 0$ on $[0,1] \times [0,1]$ with $u(0,x)=1$ and $u(t,0)=u(t,1)=0$. Here we clearly can't have differentiability up to the boundary since $1 \neq 0$. However after time $t = 0$ it would appear to me that this problem goes away, ie. that the solutions are smooth up to the boundary <em>except</em> at $t=0$. Is this the case? Can people perhaps share insightful examples?</p> http://mathoverflow.net/questions/40572/fourier-series-of-non-periodic-functions Fourier series of non-periodic functions Dorian 2010-09-30T03:58:57Z 2010-09-30T04:50:06Z <p>It's well known that the Fourier series converges uniformly when a function is $C^2$ and periodic on say $[0,2\pi]$. If the function is not periodic you can have lack of uniform convergence near the endpoints due to "Gibbs" phenomenon. However I would like to understand if and when you still have uniform convergence on <em>compact</em> subsets of $[0,2\pi]$? This seems to be geometrically obvious but I can't find a good source for it. Any suggestions would be appreciated. For instance $f(x) = x$ on $[0,1]$. It would appear that the Fourier series converges uniformly on compact subsets of $[0,2\pi]$ but this doesn't seem to follow from analysis of the Fourier coefficients. Isn't it the same as if I had done an odd extension of $f(x)$ so that I have $-|x|$ on $[-2\pi,2\pi]$ and find the fourier series for this new domain? In the latter case I clearly get uniform convergence and I don't think this should depend on the fact that I reflected and chose a different domain.</p> http://mathoverflow.net/questions/40037/inconsistency-in-definition-of-characteristics-for-a-linear-pde-folland-versus-f Inconsistency in definition of characteristics for a linear PDE? Folland versus Fritz John. Dorian 2010-09-26T16:54:17Z 2010-09-26T16:54:17Z <p>There seems to be a major inconsistency (perhaps due to my lack of understanding) between what Folland calls a "characteristic" and what I had previously thought was a characteristic.</p> <p>For example, Folland says that the characteristics of the equation $\partial_x u = 0$ are <code>$\{ \xi : \xi_1 = 0\}$</code> (in $\mathbb{R}^2$). This confuses me to no end since I thought the characteristic hypersurface was the $x_1$ axis here? According to this definition it's the orthogonal compliment to this. The same issue arises with the heat operator $L=u_t-u_{xx}$ where he says the characteristics are <code>$\{\xi \neq 0 : \xi_x = 0\}$</code>. Aren't the characteristics $t=$ cosntant which are orthogonal to these?</p> <p>Sorry if this question is elementary but it's given me a real headache and I'm not sure what it is I'm missing here.</p> http://mathoverflow.net/questions/39923/can-a-self-adjoint-operator-have-a-continuous-set-of-eigenvalues/39924#39924 Answer by Dorian for Can a self-adjoint operator have a continuous set of eigenvalues? Dorian 2010-09-25T02:50:12Z 2010-09-25T03:19:44Z <p>You are confusing two notions. First of all, the point spectrum just means eigenvalues; there is no assumption that these form a discrete set. The shift operator is a simple example where the spectrum is "continuous".</p> <p>The condition for the eigenvalues to be discrete is precicsely that the operator $A:H \to H$ is <em>compact</em>. It is however possible for non-compact self adjoint operators to have a discrete spectrum. The simplest example of this is the orthogonal projection operator $P:H \to Y$ where $Y$ is a closed subspace of the Hilbert space $H$. Here the spectrum is $0$ and $1$.</p> <p>Bounded self adjoint operators have no <em>residual spectrum</em> but they do indeed have a continuous spectrum. Take any compact operator $A:H \to H$ where dim$H=+\infty$. Then $0$ belongs to the continuous spectrum because otherwise $A:H \to H$ would be invertible, implying that dim$H &lt;\infty$. Continuous = "exists a set of approximate eigenvectors".</p> <p>If you want a continuous range of spectrum take $Af = x f(x)$ on $L^2([0,1])$. Then the range of the spectrum is just $[0,1]$. There are no eigenvalues for this operator and moreoever since the residual spectrum is empty for self adjoint operator, $[0,1]$ is the spectrum and it is equivalent to the continuous spectrum.</p> <p>So a final point, continuous just means that $R(A-\lambda I)$ is <em>not dense</em> but that $\lambda$ is <em>not an eigenvalue</em>. It has nothing to do with the actual spectrum being discrete or continuous.</p> <p>So I think you were mixing up two notions but hopefully I've provided examples for both.</p> http://mathoverflow.net/questions/100517/convert-distance-time-to-vector-in-terms-of-x-y Comment by Dorian Dorian 2012-06-24T11:04:51Z 2012-06-24T11:04:51Z This type of question is not appropriate for math overflow. Math overflow is reserved for research level math questions. Try stackexchange. http://mathoverflow.net/questions/100191/kernel-range-of-l Comment by Dorian Dorian 2012-06-20T23:25:56Z 2012-06-20T23:25:56Z This kind of question is not suitable for mathoverflow. Try stackexchange. This site is reserved for research level questions. http://mathoverflow.net/questions/98991/generalization-of-an-inequality-due-to-gage-for-curve-shortening-part-ii/99022#99022 Comment by Dorian Dorian 2012-06-07T10:18:00Z 2012-06-07T10:18:00Z By improves, is a strictly weaker inequality. Sorry. http://mathoverflow.net/questions/98991/generalization-of-an-inequality-due-to-gage-for-curve-shortening-part-ii/99022#99022 Comment by Dorian Dorian 2012-06-07T09:55:39Z 2012-06-07T09:55:39Z Thank you for this interesting example, but if $C(L,A) &gt; 0$ this <i>improves</i> the inequality no? If you see the last term that I added in my calculation, the term <code>$\int&#95;0^{2\pi} [p^&#42;(\theta)-p(\theta)][(r^&#42;)^2(\theta)-r(\theta)] d\theta$</code>, this term indeed is controlled by $C(A^*-A)$ where $C&gt;0$. Thus your example shows that what I want is certainly false for $C(L,A)=0$ but the question remains if adding such a term compensates. http://mathoverflow.net/questions/98728/generalization-of-an-inequality-due-to-gage-for-curve-shortening/98739#98739 Comment by Dorian Dorian 2012-06-05T22:53:33Z 2012-06-05T22:53:33Z Thanks for your answer. Do you think the result will hold if one assumes that the curve is star shaped and symmetric with respect to reflections across the origin? This seems quite reasonable. http://mathoverflow.net/questions/98887/wasserstein-distance-between-two-diffusion-processes Comment by Dorian Dorian 2012-06-05T19:00:27Z 2012-06-05T19:00:27Z Are you familiar with the Brenier theorem for existence of optimal transport plans for the $L^2$ Wasserstein metric? This may be useful. http://mathoverflow.net/questions/98728/generalization-of-an-inequality-due-to-gage-for-curve-shortening Comment by Dorian Dorian 2012-06-04T14:10:12Z 2012-06-04T14:10:12Z My rough idea has been the following: Take a ray of angle $\theta$ where $\theta$ is very small. By Green's theorem, for the piece of the curve below the convex hull, we will have $p^* \Delta L^* = p^2 \Delta L + A_{\theta}^* - A_{\theta} + o(\theta)$ as $\theta \to 0$ where $A$ and $A^*$ here denote the area contained in the ray intersected with the original set, and convex hull respectively. Then take the limit as $\theta \to 0$ so that we get something pointwise and go from there. Haven't been able to solidify anything though. http://mathoverflow.net/questions/98728/generalization-of-an-inequality-due-to-gage-for-curve-shortening Comment by Dorian Dorian 2012-06-04T13:51:54Z 2012-06-04T13:51:54Z Interesting, however with many wiggles, observe that p will be very close to zero very frequently since the oscillating segments will be almost parallel to the normal vector, makign $X \cdot \nu \sim 0$ on these pieces. http://mathoverflow.net/questions/98728/generalization-of-an-inequality-due-to-gage-for-curve-shortening Comment by Dorian Dorian 2012-06-04T13:22:05Z 2012-06-04T13:22:05Z Yes, corrected this. Thanks. http://mathoverflow.net/questions/98728/generalization-of-an-inequality-due-to-gage-for-curve-shortening Comment by Dorian Dorian 2012-06-04T13:06:18Z 2012-06-04T13:06:18Z Ok I have hopefully clarified the question. Yes I believe that the name for what I'm saying is star shaped. http://mathoverflow.net/questions/98728/generalization-of-an-inequality-due-to-gage-for-curve-shortening Comment by Dorian Dorian 2012-06-04T11:12:45Z 2012-06-04T11:12:45Z Hopefully clarified this. I mean that the curve is $(r(\theta),\theta)$ in polar coordinates. http://mathoverflow.net/questions/55404/stronger-version-of-the-isoperimetric-inequality/55411#55411 Comment by Dorian Dorian 2012-06-02T16:05:21Z 2012-06-02T16:05:21Z I was just thinking about this inequality. As written I think it is incorrect. Take a circle which is not centered at the origin. Then r_1-r_2 is not zero but the left side is zero.. http://mathoverflow.net/questions/97853/what-are-the-most-general-types-of-curves-in-mathbbr2-for-which-gauss-bonne Comment by Dorian Dorian 2012-05-29T10:42:13Z 2012-05-29T10:42:13Z I am looking for a class of curves, in paritcular rectifiable verifolds with generalized mean curvature in $L^{\infty}(\partial \Omega)$.. http://mathoverflow.net/questions/93885/different-forms-of-bonnesens-strong-isoperimetric-inequality-in-the-plane/93892#93892 Comment by Dorian Dorian 2012-04-13T09:54:35Z 2012-04-13T09:54:35Z Very nice counter example. Thank you Rbega. http://mathoverflow.net/questions/89983/what-is-the-constant-in-the-rate-of-exponential-convergence-for-mean-curvature-fl/93141#93141 Comment by Dorian Dorian 2012-04-07T18:18:23Z 2012-04-07T18:18:23Z This is for regular mean curvature flow, not volume preserving mean curvature flow.