User willie wong - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:45:09Z http://mathoverflow.net/feeds/user/3948 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128460/is-geodesic-plane-field-a-killing-field/128467#128467 Answer by Willie Wong for Is geodesic plane field a Killing field? Willie Wong 2013-04-23T11:23:53Z 2013-04-23T11:23:53Z <p>(I assume by a plane field you mean a <a href="http://en.wikipedia.org/wiki/Distribution_%28differential_geometry%29" rel="nofollow">distribution</a>.)</p> <p>No, take $\mathbb{T}^3$ parametrized as $(x,y,z) \in [0,2\pi)^3$. The field $v = \partial_x + \sin(z) \partial_y$ is geodesic. But for any $\phi(x,y,z)$ the deformation tensor $\mathcal{L}_{\phi v} g$ can be computed to be $$\begin{pmatrix} 2 \phi_x &amp; \phi_y + \phi_x \sin(z) &amp; \phi_z \newline \phi_y + \phi_x \sin(z) &amp; 2\phi_y \sin(z) &amp; \phi_z \sin(z) + \phi \cos(z) \newline \phi_z &amp; \phi_z \sin(z) + \phi \cos(z) &amp; 0 \end{pmatrix}$$ which can only vanish if $\phi_x = \phi_y = \phi_z = 0$ (from the (1,1), (2,2) and (1,3) components) and $\phi \equiv 0$ (from the (2,3) component). </p> <p>For distributions of higher dimension, you can observe that for any complete Riemannian manifold $TM$ as a top-dimensional distribution is geodesic. So it suffices to find a Riemannian manifold of the appropriate dimension that does <em>not</em> admit any Killing vector field to get a counterexample. Alternatively you can also modify the above one dimensional construction in the obvious way. </p> http://mathoverflow.net/questions/105951/finite-dimensional-mountain-pass-lemma Finite dimensional "Mountain Pass Lemma" Willie Wong 2012-08-30T15:06:31Z 2013-04-19T02:33:37Z <p><strong>Question</strong> Does anyone know of a good reference which I can cite for the <em>finite</em> dimensional version of Mountain Pass Lemma?</p> <p><strong>Motivation</strong> I am writing a paper and found myself using the following result:</p> <blockquote> <p>Let $f$ be a proper smooth real-valued function on $\mathbf{R}^3$ such that $f(0) = 0$, $f|_{B_1(0)} \geq 0$, $f|_{\partial B_1(0)} \geq 1$ and $\exists p \in {\partial B_2(0)}$ such that $f(p) = 0$. Then $\exists q\in \mathbf{R}^3 \setminus B_1(0)$ such that $f'(q) = 0$ and $f(q) \geq 1$. </p> </blockquote> <p>For the time being I referred to Ambrosetti and Rabinowitz's JFA article for the mountain pass lemma, but citing a Banach space version for a finite-dimensional Euclidean space application gives me a funny feeling. (Also, if feels like such a result could in principle be found in not-so-advanced undergraduate textbooks...)</p> http://mathoverflow.net/questions/124942/finding-an-optimal-p-such-that-u-in-lp/125022#125022 Answer by Willie Wong for Finding an optimal $p$ such that $u \in L^p$ Willie Wong 2013-03-19T21:20:03Z 2013-03-19T21:20:03Z <p>The power of 10 from <a href="http://mathoverflow.net/questions/124942/finding-an-optimal-p-such-that-u-in-lp/124989#124989" rel="nofollow">Michael Renardy's answer</a> is in fact optimal, which follows from the fact that for $y\approx 0$, the vector fields $\partial_y$ and $x\partial_y - y\partial_x$ are parallel. </p> <p>We can also get at it using a scaling argument. Let $\phi(x,y) \in C^\infty_0(\mathbb{R}^2)$. Let $\phi_{\alpha\beta}^\lambda(x,y) = \lambda \phi(\lambda^\alpha x, \lambda^\beta y)$. The usual scaling analysis shows that for $\alpha = 6$ and $\beta = 4$, we have that</p> <p><code>$$\|\phi^{\lambda}_{\alpha\beta} \|_{H^{2/3}} \leq \|\phi\|_{H^{2/3}}$$</code></p> <p>and</p> <p><code>$$\|\partial_y\phi^{\lambda}_{\alpha\beta} \|_{L^2} \leq \|\partial_y \phi\|_{L^2}$$</code></p> <p>and</p> <p><code>$$\|y\partial_x \phi^{\lambda}_{\alpha\beta} \|_{L^2} \searrow 0$$</code> </p> <p>as $$\lambda\nearrow \infty$$</p> <p>On the other hand, for $\gamma > 10$, we have that</p> <p><code>$$\| \phi^{\lambda}_{\alpha\beta} \|_{L^\gamma}^{\gamma} = \lambda^{\gamma - \alpha - \beta} \|\phi\|_{L^\gamma}^\gamma \nearrow \infty$$</code></p> <p>Now you can do the usual trick of summing a bunch of these guys with disjoint support to show that the embedding in to $\gamma > 10$ is not possible. </p> http://mathoverflow.net/questions/122147/why-is-it-called-the-diamagnetic-inequality/122165#122165 Answer by Willie Wong for why is it called the diamagnetic inequality? Willie Wong 2013-02-18T13:04:24Z 2013-02-18T13:04:24Z <blockquote> <p>Why is it called this name?</p> </blockquote> <p>Diamagnetism is a quantum effect in which material under an applied external magnetic field generate their own "opposite" magnetic field. The most prominent example happens in <a href="http://en.wikipedia.org/wiki/Meissner_effect" rel="nofollow">superconductors</a>. </p> <p>A consequence of this is that matter under an applied external magnetic field "gain" energy (or rather, removing magnetic fields decreases kinetic energy). See, for example, <a href="http://www.ams.org/mathscinet-getitem?mr=548176" rel="nofollow">this paper</a>. Now, the quantum mechanical description of the momentum operator is just $i\nabla$, and the associated (free) Hamiltonian is $-\triangle = (i\nabla)^2$. The mathematical description of a quantum mechanical system under an external magnetic field uses the gauge potential $A$ to capture the magnetic field, and postulates that the Hamiltonian is $H = (i\nabla + A)^2$. And hence results which bound the lower energy $i\nabla$ by the higher energy $i\nabla + A$ became known as "diamagnetic inequalities". See also <a href="http://www.ams.org/mathscinet-getitem?mr=1383577" rel="nofollow">this mathscinet review where this is briefly discussed</a>. </p> <p>Note that originally inequalities of the type are (among many) called Kato's inequalities. Now they are called diamagnetic inequalities because it conveniently reminds us of the physics. </p> <p>For how to derive the inequality and some more historical and practical notes, see sections 7.19 - 7.22 in Lieb and Loss, <em>Analysis</em>. </p> http://mathoverflow.net/questions/49384/tools-for-long-distance-collaboration Tools for long-distance collaboration Willie Wong 2010-12-14T13:11:45Z 2013-02-08T12:43:06Z <h2>Background</h2> <p>In general, I am aware of four and a half methods of long-distance collaboration:</p> <ol> <li>Telephone (including voice-chat, VOIP, etc.; anything that is voice based) </li> <li>Text chat (chat room, IM, gchat, things like that)</li> <li>E-mail (or other asynchronous messaging system)</li> <li>Online whiteboards, real-time collaborative text editors, desktop-sharing (or other software, graphical system)</li> <li>(The half) Adding a webcam to any of the above and call it Video-blah.</li> </ol> <p><strong>What this question is not about</strong></p> <p>I am not asking about <a href="http://mathoverflow.net/questions/3044/tools-for-collaborative-paper-writing" rel="nofollow">tools for collaborative paper-writing</a> which has already been addressed here last year. So in particular, to limit the scope, this question is not about the part in a collaboration when all the ideas are set-out, all the heuristics checked, and all that's left is to flesh out the argument and write it up. </p> <p>I am also <em>not</em> asking for just a list of services. I am fairly confident my Google-fu is at least as good as yours. </p> <p><strong>What this question is about</strong></p> <p>I am interested in tools that help collaboration in the earlier stage when we are still brainstorming, setting the scope of the project; or the stage where we are troubleshooting to fix a flawed argument. In other words, I am interested in the scenarios where the ideal thing to do would be for a face-to-face chat while writing on a black board or a piece of paper, but when it is difficult to do so (both of you have to teach, and you are on different continents). </p> <p>In other words, I am asking about situations where real-time, instantaneous interactions are preferred (and so option 3, e-mail, should be reserved as a last resort). In this sense, voice interaction is preferred: it is a lot easier to interrupt the other party when talking then when typing, and be able to force a change of direction in the conversation. On the other hand, e-mail and a lot of the chat software has the advantage that your discussions are automatically documented and saved for future review. The main downside to a pure voice communication, however, is that (for me at least) mathematics is visual. It helps a lot when there is a black board or a piece of paper with equations on it on which I can focus my attention. So I'm especially interested in ways that I can share mathematics visually (rendered LaTeX, diagrams, things like that). </p> <h2>The Question</h2> <p>There are two questions:</p> <ul> <li>Personal testimonials: of the above solutions, which, and in what combinations, have you used and feel strongly about. I would especially appreciate it if you can say a few words about the strengths and potential weaknesses of the setup. </li> <li>Thinking outside the box: are there other solutions that I have overlooked in my list above?</li> </ul> http://mathoverflow.net/questions/109996/estimates-for-fourier-transform/110000#110000 Answer by Willie Wong for Estimates for Fourier transform Willie Wong 2012-10-18T10:03:20Z 2012-10-18T10:03:20Z <p>As stated, no. Let $f \equiv 0$, and $\mu$ with $\int \mu = 0$. You then have $|\tilde{\mu}(\xi)| = 0$ for any $\xi$, and you have no control over the non-zero frequencies $\hat{\mu}$. </p> http://mathoverflow.net/questions/106947/reflexive-besov-spaces/107006#107006 Answer by Willie Wong for Reflexive Besov spaces Willie Wong 2012-09-12T12:55:41Z 2012-09-12T12:55:41Z <p>Like Paul Garrett says, there may be difference in definitions. But if I guess correctly at your definitions, then the answer is <strong>no</strong>, the Besov space $B^1_{1,1}(\mathbb{T})$ is not reflexive. </p> <p>Reference: Theorem 2, item (iv) of Flett, <a href="http://www.ams.org/mathscinet-getitem?mr=313779" rel="nofollow">"Lipschitz spaces of functions on the circle and the disc"</a>, J. Math. Anal. Appl. 39 (1972), 125–158. (Remark: the definition/characterisation of the space $\Lambda(\alpha;p,q)$ of the paper can be found on page 134.) </p> http://mathoverflow.net/questions/106594/does-this-variable-have-an-upper-bound/106599#106599 Answer by Willie Wong for Does this variable have an upper bound? Willie Wong 2012-09-07T12:13:49Z 2012-09-07T12:13:49Z <p>Integrate a few times:</p> <p>Using that $x > 0$ we have that $|\dot{x}(t)| \leq \exp (0) = 1$ which implies that $x(t) \leq M + t$ for some $M > 0$. Plug it back in we have</p> <p>$$|\dot{x}(t)| \leq \exp \left( - \int_0^t \frac{1}{M+\tau} \mathrm{d}\tau \right) = \exp \left( - \ln (M+t) + \ln M\right) = \frac{M}{M+t}$$</p> <p>this implies that </p> <p>$$x(t) \leq M + M \left[\ln (M+t) - \ln M\right]$$</p> <p>for which we can very, very roughly estimate by </p> <p>$$x(t) \leq 2\sqrt{M' + t}$$</p> <p>for a sufficiently large $M'$. Plugging this back in we have that</p> <p>$$|\dot{x}(t)| \leq \exp \left( - \int_0^t \frac{1}{2\sqrt{M'+\tau}} \mathrm{d}\tau\right) = \exp \left( \sqrt{M'} - \sqrt{M' + t}\right)$$</p> <p>Using that $e^{-\sqrt{t}}$ is integrable from $0$ to $\infty$ to some constant, we have that $x$ must be bounded. </p> http://mathoverflow.net/questions/106387/frobenius-theorem-with-lesser-regularity/106412#106412 Answer by Willie Wong for Frobenius theorem with lesser regularity Willie Wong 2012-09-05T10:22:58Z 2012-09-05T10:22:58Z <p>Not an answer, but a pointer to the difficulties:</p> <p>If you restrict yourself to just one dimensional distributions, you immediately see a problem. </p> <p>Let $V$ be a continuous (non-vanishing) vector field on $\mathbb{R}^n$, $n > 1$, it clearly generates a $C^0$ sub-bundle of the tangent bundle. By Peano's existence theorem we have that $V$ has <em>at least one</em> integral curve passing through any initial point. On the other hand, contrasted against Picard's existence theorem which requires Lipschitz regularity of the vector field (or in your case, $C^1$ suffices), this integral curve is not guaranteed to be unique. </p> <p>Why are curves important? Frobenius in $C^1$ can be interpreted in the "integral form" as the following: To be sure that the distribution integrates to a submanifold, if you travel infinitesimally first in the $V_1$ direction, then in the $V_2$ direction, and compare with traveling first in the $V_2$ direction, then in the $V_1$ direction, the difference should be in a direction that is inside the subbundle. (In other words, you cannot move off the manifold by moving inside the manifold.) </p> <p>So to have any hope of having a Frobenius like statement you need to overcome this problem of having way too many integral curves. This, in fact, also causes problems for the existence of a foliation!</p> <p>An example: </p> <p>Consider the vector field $V$ on $\mathbb{R}^2$ given by $$V(x,y) = (1, \sqrt{|y|})$$ Integral curves through $(0,0)$ include $$\gamma_{0}(t) = (t,0)\qquad \gamma_{+}(t) = (t,t^2 / 4) \qquad \gamma_{-}(t) = (t,-t^2/4)$$ for positive time $t$, and only $$\gamma(t) = (t,0)$$ is admissible for negative time $t$. (In fact, one can also transition from $\gamma_0$ to a suitably translated $\gamma_{\pm}$ at any $t \geq 0$.) This shows that every integral curve of $V$ must become tangent to the $x$ axis in finite negative time. Hence it is impossible to obtain a foliation to this non-vanishing, continuous vector field. </p> <p>Thus you see that the issue of regularity occurs when just one dimension is considered, much before the statement of Frobenius theorem, which is a compatibility condition on multiple dimensions, comes into play. </p> http://mathoverflow.net/questions/105951/finite-dimensional-mountain-pass-lemma/106224#106224 Answer by Willie Wong for Finite dimensional "Mountain Pass Lemma" Willie Wong 2012-09-03T07:27:20Z 2012-09-03T07:27:20Z <p>For historical interest: A friend pointed me to the book </p> <ul> <li>Youssef Jabri, <a href="http://books.google.ch/books/about/The_Mountain_Pass_Theorem.html?id=uGWaffeFbroC&amp;redir_esc=y" rel="nofollow"><em>The Mountain Pass Theorem: Variants, Generalizations and Some Applications</em></a>, CUP</li> </ul> <p>which asserts that one of the earliest known published version of the finite dimensional mountain pass theorem was due to </p> <ul> <li>Richard Courant, <em>Dirichlet's Principle, Conformal Mapping, and Minimal Surfaces</em>, Interscience</li> </ul> <p>published originally in 1950. The version stated and proven by Courant does not, technically speaking, imply the result I stated in the question text (the points $0$ and $p$ are assumed to be local minima of the function $f$). But a simple modification of the deformation lemma (for example, as in Liviu's book that he mentioned) would do. </p> http://mathoverflow.net/questions/105883/can-we-actually-find-any-fixed-points-with-brouwers-theorem/105909#105909 Answer by Willie Wong for Can we actually find any fixed points with Brouwer's theorem? Willie Wong 2012-08-30T07:31:48Z 2012-08-30T07:31:48Z <p>(This answer is in a similar direction to that of Johannes Hahn's and to Will Sawin's)</p> <p>I think (I am not 100% sure) that one may get away without doing the triangulation and simplicial approximation (of the "usual" Sperner's Lemma proof) if you take the approach using van Maaren's version of Sperner's Lemma (there is an outline of the proof in Schechter's <em>Handbook of Analysis and its Foundations</em> with some typos). </p> <p>One first obtains the van Maaren's version of Sperner's Lemma, which is purely combinatorial/order theoretical and is a constructive statement on finite sets (the proof just gives the algorithm). </p> <p>Using that one gets an approximate fixed point statement (roughly speaking for every $\epsilon$ you find a point that is $\epsilon$ away from being a fixed point). To get the approximate fixed point at size $1/k, k\in\mathbb{N}$ you only need to consider <em>the</em> finite subset formed by the lattice of spacing $1/(3kn)$ where $n$ is the dimension. In this step the convexity comes into play (but not continuity; compactness only enters via Heine-Borel as boundedness). </p> <p>Note that this does not require being able to approximate the function $f$: it just requires being able to evaluate $f(x)$ for given $x$ to arbitrary accuracy (in particular you need to know whether the $i$-th coordinate of $f(x)$ is less than or equal to the $i$-th coordinate of $x$). </p> <p>Then you take limit as $k\to \infty$ (and here continuity and compactness are used, by convexity is no longer relevant) (the practicality of this last step, of course, is questionable; and as Noam Elkies and Michael Greinecker alluded to, this method gives no rate of convergence, so cutting off the computation at a finite $k$ doesn't guarantee that you are near a bona fide fixed point at all). </p> http://mathoverflow.net/questions/103205/work-on-an-einstein-hilbert-type-action-but-with-the-absolute-value-of-scalar-c/103220#103220 Answer by Willie Wong for Work on an Einstein-Hilbert type action but with the *absolute value* of scalar curvature? Willie Wong 2012-07-26T17:21:36Z 2012-07-26T17:21:36Z <p>In the vacuum case this is not greatly different from the Einstein-Hilbert action. </p> <p>Let $(M,g)$ be a classical solution to the variational problem as you posed. Suppose $p\in M$ is such that $R(p) \neq 0$, then by continuity in a small neighborhood of $p$, the scalar curvature $R$ is signed, and hence locally in that neighborhood it is also a critical point to the Einstein-Hilbert action. But then it must be Ricci flat, contradicting the assumption that $R \neq 0$ at $p$. </p> <p>Conversely, if $(M,g)$ is a classical solution to the Einstein-Hilbert variational problem, then it is Ricci flat and hence scalar flat. And hence you have that all Einstein-vacuum solutions are also solutions to the critical point problem you posed. </p> <p>Going back forwards again, note that by definition any scalar flat 4 manifold will be a minimizer of the action. Hence you have that for the vacuum problem of your proposed action:</p> <blockquote> <p>There are no critical points which do not minimise the action; the action minimisers are precisely the scalar flat Lorentzian 4 manifolds. </p> </blockquote> <hr> <p>In any case, if you really are interested in this action, for literature searches the relevant keyword is <a href="http://en.wikipedia.org/wiki/F%28R%29_gravity" rel="nofollow">f(R) gravity theories</a>. </p> http://mathoverflow.net/questions/103215/dissipative-operator/103219#103219 Answer by Willie Wong for Dissipative operator Willie Wong 2012-07-26T17:07:01Z 2012-07-26T17:07:01Z <p>No. </p> <p>Let $H = \mathbb{R}^2$ and set </p> <p>$$A = \begin{pmatrix} -1 &amp; 0 \newline 0 &amp; -100 \end{pmatrix}$$</p> <p>which is clearly dissipative. Now take </p> <p>$$y = \begin{pmatrix} 0 \newline 1\end{pmatrix} \qquad z = \begin{pmatrix} 1 \newline 0 \end{pmatrix}$$</p> <p>and you have a counter example. </p> http://mathoverflow.net/questions/102731/reference-daprato-and-grisvard-parabolic-pdes/102735#102735 Answer by Willie Wong for Reference: DaPrato and Grisvard parabolic PDEs. Willie Wong 2012-07-20T12:10:33Z 2012-07-20T12:10:33Z <blockquote> <p>Has anyone read the paper?</p> </blockquote> <p>I haven't; <a href="http://www.ams.org/mathscinet-getitem?mr=551075" rel="nofollow">but someone did</a>. <code>:-)</code></p> <blockquote> <p>Is it all in French?</p> </blockquote> <p>Yes</p> <blockquote> <p>Are there articles/papers that cover the same material...? </p> </blockquote> <p>It depends on which results you are looking for. It maybe that the result that you need is already contained in some previous work (some of which are in English). The following is translated/excerpted from the first page of the article. </p> <p>For existence of weak solutions (using monotone operators), the method in the paper is similar to those of references 15, 14, 8, 7, 4. </p> <p>(All references refer to the references in the paper, which you can see <a href="http://rd.springer.com/article/10.1007/BF02411952" rel="nofollow">on this page</a>.)</p> <p>For classical solutions, the basic technique is that of references 21, 14, 12, 20, and 19. The main difference is that in the DaPrato-Grisvard article, the equation is not assumed to be a perturbation of a linear equation, which they achieve by developing some new interpolation spaces. </p> http://mathoverflow.net/questions/99404/wavefront-set-of-a-product/99414#99414 Answer by Willie Wong for Wavefront set of a product Willie Wong 2012-06-13T08:55:51Z 2012-06-14T09:34:40Z <p>In general you can only say that $$WF(uv) \subseteq \Bigg[ WF(u) \cup WF(v) \cup \Big[WF(u)+WF(v)\Big]\Bigg]$$ where the set $$WF(u) + WF(v) = \left\lbrace (x,\xi +\eta)| (x,\xi)\in WF(u), (x,\eta)\in WF(v) \right\rbrace$$ (note that a sufficient criterion for the product to be well-defined is precisely when the above set contains no points of the form $(x,0)$). </p> <p>See, e.g. chapter 11 of Friedlander and Joshi <em>Introduction to The Theory of Distributions</em>.</p> <p>But quite obviously the $\subseteq$ is not always an equality: just take $u,v$ two compactly supported distributions with distinct supports. Note that given $WF(u)$ and $WF(v)$ you only know the singular support of $u$ and $v$ and not their actual supports. </p> http://mathoverflow.net/questions/98580/uniform-equicontinuity-of-a-family-of-indefinite-integrals Uniform equicontinuity of a family of indefinite integrals Willie Wong 2012-06-01T15:53:26Z 2012-06-01T15:59:18Z <p>Let $f_k$ be a sequence of measurable functions on $\mathbb{R}^k$ where $k > 1$. (Let us be generous and also assume that $f_k$ is locally integrable.) Does anyone know what the phrase</p> <blockquote> <p>uniform equicontinuity of the indefinite integrals $\int f_k(x) \mathrm{d}x$ </p> </blockquote> <p>means? </p> <p>(For that matter, what is an indefinite integral on $\mathbb{R}^k$? The best I can figure is that it refers to the signed measure $Q \mapsto \int_{Q} f_k(x) \mathrm{d}x$, but I am not entirely sure how to interpret equicontinuity in this context.) </p> <p>For what it is worth, the quote comes from Shatah and Struwe, <em>Geometric Wave Equations</em>, p.67 in the proof of Segal's theorem. I'll include the full context below. </p> <hr> <p>The relevant setting: $f:\mathbb{R}\to\mathbb{R}$ is a locally Lipschitz function with $f(0) = 0$. We define $f_k:\mathbb{R}\to\mathbb{R}$ to be equal to $f$ on $[-k,k]$, and set $f_k(x) = f(k)$ if $x > k$ and similarly $f_k(x) = f(-k)$ if $x &lt; -k$ (so it is a globally Lipschitz truncation of $f$). </p> <p>Suppose we are now giving a sequence of functions $u_k \in L^2(I\times \mathbb{R}^m)$ where $I$ is a closed interval. We are given that $u_k \to u$ strongly in $L^2_{\text{loc}}$ and that $u_k \to u$ almost everywhere. We also assume that $u_k$ is uniformly bounded in $L^2(I\times\mathbb{R}^m)$. The claim is that knowing</p> <p>$$\int_I \int_{\mathbb{R}^m} |u_k f_k\circ u_k| \mathrm{d}x \leq \int_I\int_{\mathbb{R}^m} (u_k f_k\circ u_k + u_k^2) \mathrm{d}x + \int_I\int_{\mathbb{R}^m} u_k^2 \mathrm{d}x \leq C$$</p> <p>and</p> <p>$$\int_I \int_{\mathbb{R}^m} |u f\circ u| \mathrm{d}x \leq \liminf_{k\to\infty} \int_I\int_{\mathbb{R}^m} (u_k f_k\circ u_k + u_k^2) \mathrm{d}x + \int_I\int_{\mathbb{R}^m} u^2 \mathrm{d}x \leq C$$</p> <p>we can derive that the family of indefinite integrals $\int f_k\circ u_k \mathrm{d}x$ (no, no typos, it is not multiplied by $u_k$) is uniformly equicontinuous, and from this result we can get convergence of $f_k(u_k)\to f(u)$ in $L^1_{\text{loc}}$. </p> <p>Now, from the final conclusion it appears that one may want to derive the conclusion using something like <a href="http://en.wikipedia.org/wiki/Vitali_convergence_theorem" rel="nofollow">Vitali's theorem</a>, which would mean that perhaps the authors intended the condition to be <em>uniform integrability</em>. Is that a reasonable interpretation?</p> http://mathoverflow.net/questions/97976/approachable-french-masters/98502#98502 Answer by Willie Wong for Approachable French Masters Willie Wong 2012-05-31T16:11:25Z 2012-05-31T16:11:25Z <p>For the <em>really</em> analytically inclined :), while I have not finished reading it myself, I have been told by multiple of my French colleagues that Leray's <a href="http://dx.doi.org/10.1007/BF02547354" rel="nofollow">original paper on Navier-Stokes</a> has interesting mathematics and quite penetrable language. </p> http://mathoverflow.net/questions/97654/on-the-set-of-divergence-to-infinity-for-sequences-of-positive-continuous-functio/97729#97729 Answer by Willie Wong for On the set of divergence to infinity for sequences of positive continuous functions Willie Wong 2012-05-23T08:46:15Z 2012-05-23T08:46:15Z <p>In addition to Michael Renardy's reference, another proof (obtained perhaps independently?) was given by Sierpinski in 1921. </p> <p>Sierpiński, W. Sur l'ensemble des points de convergence d'une suite de fonctions continues. (French) [J] Fundamenta math. 2, 41-49 (1921). <a href="http://www.zentralblatt-math.org/zmath/search/?an=JFM%2048.0274.02" rel="nofollow">ZMath Link</a></p> <p>(It has the advantage of (a) being in French so I can actually understand it and (b) published in a Journal which now offers [I think] open access to the old articles; one can obtain a copy by <a href="http://pldml.icm.edu.pl/mathbwn/search/general.action" rel="nofollow">searching here</a>.)</p> <p>The main theorem in the paper states the following (all sets are subsets of $\mathbb{R}$):</p> <p><strong>Theorem</strong> For a set $E$ to be the set of convergence points of a sequence of continuous functions $f_1, f_2, \ldots$ (meaning that $f_n(x)$ converges if and only if $x\in E$) it is necessary and sufficient that $E$ be $F_{\sigma\delta}$</p> <p>The proof proceeds via a series of Lemmas. </p> <p><strong>Lemma 1</strong> Any $F_\sigma$ can be decomposed into the sum of an $F_\sigma$ with no interior points which we call $P$ with an at most countable collection of mutually disjoint intervals whose endpoints all appear in $P$. </p> <p><strong>Lemma 2</strong> An $F_\sigma$ which contains no interior points can be written as a sum of at most countably many disjoint closed sets. </p> <p>Together the above yields</p> <p><strong>Lemma 3</strong> An $F_\sigma$ can be written as a union of $P\cup Q$ where $P$ can be written as an at most countable union of mutually disjoint closed sets and $Q$ can be written as an at most countable union of mutually disjoint intervals whose endpoints lie in $P$. </p> <p>Given the above we have</p> <p><strong>Lemma 4</strong> For $E$ an $F_\sigma$ there exists a sequence of bounded continuous functions which converge to 0 on $E$ and does not converge otherwise. </p> <p><em>Sketch of construction</em>:</p> <p>Write $E = P \cup Q$ as above. Write $P = F_1 \cup F_2\cdots$ where the $F_i$ are mutually disjoint. Let $S_n = \cup_1^n F_i$. Let $\delta_n = \mathrm{dist}(S_n,F_{n+1}) > 0$ since we have disjoint closed sets. Let <code>$T_n = \{ x : \mathrm{dist}(S_n,x) \geq \delta_n / (3 + n \delta_n) \}$</code>. </p> <p>Define a sequence of functions $\varphi_n(x)$ such that for $n$ odd, $\varphi_n \equiv 0$. For $n = 2k$ even, let $\varphi_n(x) = 1$ on $T_k$ and $0$ on $S_k$, and linearly interpolate in between (we can do this because $T_k\cup S_k$ is closed and its complement is a union of open intervals). </p> <p>Now define $f_n(x)$ to be equal to $\varphi_n(x)$ on the complement of the interior of $Q$. The remaining portion is again a union of open intervals so we can interpolate linearly on it. </p> <p>It is clear that $f_n$ converges to 0 on $E$ by definition. It takes a little bit of computation to show that for $x\not\in E$, $\limsup f_n(x) = \limsup \varphi_n(x) = 1$. </p> <p><em>Sketch of construction for the Theorem</em></p> <p>Now let $E = E_1 \cap E_2 \cap \cdots$ where $E_k$ are $F_\sigma$. Let $\tilde{f}_{k,n}(x)$ denote the sequence found by applying Lemma 4 to $E_k$. Let <code>$\bar{f}_{k,n}(x) = \frac{1}{k} \tilde{f}_{k,n}$</code>. The sequence we want in the theorem can be obtained by the diagonal method: </p> <p>$$f_1 = \bar{f}_{1,1} \quad f_2 = \bar{f}_{2,1} \quad f_3 = \bar{f}_{1,2} \quad f_4 = \bar{f}_{3,1} \ldots$$</p> http://mathoverflow.net/questions/70740/image-of-the-trace-operator Image of the trace operator Willie Wong 2011-07-19T14:05:34Z 2012-05-10T08:34:19Z <p>It is well-known that we have the trace theorem for Sobolev spaces. Let $\Omega$ be an open domain with smooth boundary, we know that the map </p> <p>$$T: C^1(\bar\Omega) \to C^1(\partial\Omega) \subset L^p(\partial\Omega)$$</p> <p>by $Tu(y) = u(y)$ for $y\in\partial\Omega)$ can be extended continuously to a linear map on Sobolev spaces for $p > 1$</p> <p>$$T: W^{1,p}(\Omega) \to L^p(\partial\Omega)$$</p> <p>We also know that this map is <strong>not</strong> surjective, since the Trace Theorem (Sobolev embedding) tells us that when dropping 1 dimension, we have that the image of $T$ actually lives (<em>[Edited May 10 2012]</em> <strong>caveat</strong>: see my comment on the answer below) in a fractional Sobolev space, </p> <p>$$T: W^{1,p}(\Omega) \to W^{1-1/p, p}(\partial\Omega) \Subset L^p(\partial\Omega)$$</p> <p>On the other hand, we know that this map $T$ has dense image in $L^p$, just using the density of $C^1$. </p> <p><strong>Question</strong>: Is there a known characterisation of precisely what the image set of $T$ is? A slightly weaker question is: consider‡ $w \in W^{s,q}(\partial\Omega)$ for $1 - 1/p \leq s \leq 1$ and $q \geq p$, does there necessarily exist some function $u\in W^{1,p}(\Omega)$ such that $Tu = w$?</p> <p>For example, if we assume that $w$ is Lipschitz on $\partial\Omega$, then we can extend (almost trivially) $w$ to a Lipschitz function $C^{0,1}(\bar\Omega)\subset W^{1,p}$ for every $p$. So the case $s = 1, q = \infty$ has a <em>positive</em> answer. Whereas the Sobolev embedding theorem mentioned above tells us that it is impossible to go below $s &lt; 1-1/p$ and $q &lt; p$. </p> <p>‡ The lower cut-off here is clearly not sharp. The trace theorem combined with Sobolev embedding can be used to trade differentiability with integrability. Out of sheer laziness I will not include the numerology here. One should interpret the conditions on $s,q$ to be that $s \leq 1$, $q \geq p$ plus the requirement that $(s,q)$ is at least as good as what can be guaranteed by Sobolev embedding and the trace theorem. </p> http://mathoverflow.net/questions/96532/the-inverse-for-the-trace-theorem/96552#96552 Answer by Willie Wong for the inverse for the trace theorem Willie Wong 2012-05-10T08:09:39Z 2012-05-10T08:09:39Z <p>I think your question is answered in </p> <p><a href="http://www.ams.org/mathscinet-getitem?mr=500920" rel="nofollow">Article (JonWal1978)</a> Jonsson, A. &amp; Wallin, H. A Whitney extension theorem in $L_p$ and Besov spaces Ann. Inst. Fourier (Grenoble), 1978, 28, vi, 139-192</p> <p>and </p> <p><a href="http://www.ams.org/mathscinet-getitem?mr=884984" rel="nofollow">Article (Marsch1987)</a> Marschall, J. The trace of Sobolev-Slobodeckij spaces on Lipschitz domains Manuscripta Math., 1987, 58, 47-65</p> <p>Theorem 2 of the latter paper states that if $\Omega$ is a Lipschitz domain, with $s,p$ satisfying some inequalities (as usual), then $W^{s,p}(\Omega)$ traces to (as a surjection) some Besov space on the boundary $\partial\Omega$ and that the <strong>trace operator has bounded linear right inverse</strong> if $s-1/p$ is not an integer. </p> http://mathoverflow.net/questions/96414/represented-as-a-series-of-periodic-function/96442#96442 Answer by Willie Wong for represented as a series of periodic function Willie Wong 2012-05-09T12:42:34Z 2012-05-09T12:42:34Z <p>Every function can be written as the pointwise sum of a sequence of periodic functions. </p> <p>Given $f$. Let $f_1 = f$ on $(-2,2]$ and extend as a function of period 4. And follow the following recursive definition:</p> <p>Let $f_{k+1}(x) = f(x) - \sum_{n= 1}^k f_k(x)$ for $x\in (-2^{k+1},2^{k+1}]$ and extend as a function of period $2^{k+2}$. </p> <p>Observe that if $|x| &lt; 2^{k}$, by definition $f_j(x) = 0$ for all $j > k$. Hence the series converges for all $x$ pointwise. </p> <hr> <p>Note that the above construction also gives a counterexample to your final question: just start with any $f$ not identically zero with compact support. </p> <p>Of course, the construction given does not converge uniformly (quite far from it, usually). And the answer will likely be different if you impose an additional requirement that the periods of the various functions $f_k$ remain bounded. </p> http://mathoverflow.net/questions/93089/hodographic-transformation/93101#93101 Answer by Willie Wong for hodographic transformation Willie Wong 2012-04-04T12:04:26Z 2012-04-04T12:04:26Z <p>Forget about $x,t$. Consider a $C^1$ mapping $\phi:(\zeta,\eta)\mapsto (u,v)$. Locally if $|d\phi|\neq 0$ we can invert it. Let the inverse be $\psi: (u,v)\mapsto (\zeta,\eta)$, so $\psi\circ\phi(\zeta,\eta) = (\zeta,\eta)$ and $\phi\circ\psi(u,v) = (u,v)$. </p> <p>An elementary computation shows that the Jacobian matrices of $\phi$ and $\psi$ are inverses. That is, evaluated at a fixed $u,v,\zeta,\eta$ where $(u,v) = \phi(\zeta,\eta)$, </p> <p>$$\begin{pmatrix} \partial_1\psi^1 &amp; \partial_1\psi^2 \newline \partial_2\psi^1 &amp; \partial_2\psi^2\end{pmatrix}^{-1} = \begin{pmatrix} \partial_1\phi^1 &amp; \partial_1\phi^2 \newline \partial_2\phi^1 &amp; \partial_2 \phi^2\end{pmatrix}$$</p> <p>For disambiguation: write the function $\phi = (U,V)(\zeta,\eta)$ and the function $\psi = (Z,N)(u,v)$. The above implies</p> <p>$$\begin{pmatrix} U_\zeta &amp; V_\zeta \newline U_\eta &amp; V_\eta\end{pmatrix} = \frac{1}{Z_u N_v - Z_v N_u} \begin{pmatrix} N_v &amp; - N_u \newline -Z_v &amp; Z_u\end{pmatrix}$$</p> <p>Your system (1) gives $U_\eta = V_\zeta$ which from the matrix inequality immediately implies that $Z_v = N_u$, which is the first equation of system (2). </p> <p>Using that $Z_uN_v - Z_vN_u = |d\psi| \neq 0$, the equality of the matrix components also gives you that the second equation of (2) is obtained from the second equation of (1) by the replacement $U_\zeta \to N_v$, $V_\zeta \to -N_u$, $U_\eta \to -Z_v$ and $V_\eta \to Z_u$. </p> <p><strong>Notice that this works because (a) we have a 2-by-2 system and (b) the system (1) is <em>quasilinear</em>. If it were genuinely nonlinear, we cannot "factor" out from the equation the common factor of $|d\psi|$ to end up with a purely linear equation.</strong> </p> <p>And yes, this is an example of the hodographic transformation. In general the same procedure works for any quasilinear system of first order partial differential equations with 2 dependent and 2 independent variables. </p> http://mathoverflow.net/questions/87708/partitions-of-unity/89946#89946 Answer by Willie Wong for Partitions of Unity Willie Wong 2012-03-01T09:51:44Z 2012-03-01T09:51:44Z <p>A counterexample: Let $M$ be the unit circle. Let the two charts be the arcs $A = (0,2\pi)$ and $B = (\pi/2, 5\pi/2)$. For each $n$, consider the partition of unity subordinate to ${A,B}$ given by </p> <p>$$\psi_{A,n} = \sin^2 ( (2n+1) \theta )$$</p> <p>and</p> <p>$$\psi_{B,n} = \cos^2 ( (2n+1) \theta )$$</p> <p>Check that $\psi_{A,n} (0) = 0 = \psi_{B,n} (\pi/2)$, and clearly $\psi_{A,n} + \psi_{B,n} = 1$. </p> <p>By the scaling property it is easy to check that </p> <p>$$\| \partial^k\psi_{A,n} \| = (2n+1)^k \| \partial^k\psi_{A,1} \|$$</p> <p>And using that </p> <p>$$\sin^2(\theta) = \frac12 (1-\cos 2\theta)$$</p> <p>you see immediately that your desired uniform bound is impossible. </p> http://mathoverflow.net/questions/88741/methods-for-determining-domains-of-influence/89419#89419 Answer by Willie Wong for Methods for determining domains of influence Willie Wong 2012-02-24T16:39:27Z 2012-02-24T16:39:27Z <p>I highly doubt the result you actually asked for is true. </p> <p>Consider the <strong>linear wave equation</strong> on $(1+3)$ Minkowski space. The <em>analytic domain of influence</em> of a point $x$ as Lax defined it, which morally says that $y$ is in the analytic domain only if one can find perturbations in arbitrary small neighborhoods of $x$ that change $y$ (if I interpret your question statement correctly), actually consists of only the null cone emanating from $x$ and nothing more, since strong Huygen's principle holds. </p> <p>The same is true for the linear wave equation on $(1+(2k+1))$ Minkowski spaces. </p> <p>The opposite conclusion can be drawn on $(1+2k)$ dimensional Minkowski spaces, where the Green's function have support inside the cone. </p> <p>For more general situations, you may want to consult the classical result of Atiyah-Bott-Garding on the existence of <a href="http://en.wikipedia.org/wiki/Petrovsky_lacuna" rel="nofollow">Petrowsky lacunae</a>. For any linear hyperbolic equation that admits a lacuna, the analytic domain of influence cannot cover the entirety of the geometric one. </p> <p>But for the result that you seem to actually want, where you should replace the analytic domain of dependence by a suitable "convex" envelope of it, I don't know if such a result is proven anywhere, but my guess is that, at least for the "local" version one can approach it using some sort of geometric optics construction. </p> <p>For possible references (I haven't actually finished reading either, so they may not contain what you want), maybe you want to look at Michael Beals' book on propagation of singularities (sorry, the title escapes me at the moment) or Rauch's notes on <em>Hyperbolic PDEs and Geometric Optics</em> which I think you can find floating around on the internet. </p> http://mathoverflow.net/questions/87418/the-ground-state-is-signed-and-symmetric The ground state is signed and symmetric Willie Wong 2012-02-03T11:31:59Z 2012-02-07T16:56:57Z <p><strong>Background</strong></p> <p>In <a href="http://www.ams.org/mathscinet-getitem?mr=695535" rel="nofollow">Berestycki and Lions</a> it is asserted that (on page 316), if I am not misreading, that the "ground state", i.e. action minimizer among nontrivial solutions, corresponding to the action $$S[u] = \int_{\mathbb{R}^d} |\nabla u|^2 + G(u) dx$$ where $G$ satisfies certain conditions must be signed and spherically symmetric. In the paper they showed that under certain conditions on $G$ that such a ground state (that is, signed and spherically symmetric) exists; but they do not show that all grounds states must have these symmetries (by which I mean that they don't show the "action minimizing solutions" must be <em>both</em> single-signed and spherically symmetric). </p> <p>The only reference I can find on this claim is the paper of <a href="http://www.ams.org/mathscinet-getitem?mr=468913" rel="nofollow">Coleman, Glaser, Martin</a>; but I am not entirely convinced that they have established the necessity. Their proof uses the fact that a ground state must be a minimizer of $\int |\nabla u|^2$ under the constraint that $\int G(u)$ is fixed. They then use the Polya-Szego principle: under spherical rearrangements the latter integral is unchanged, while the former can only decrease. </p> <p>But the decrease is not necessarily strict. <a href="http://www.ams.org/mathscinet-getitem?mr=929981" rel="nofollow">Brothers and Ziemer</a> gave a counterexample in the case that the distribution function of $u$ is not absolutely continuous. </p> <p><strong>My question</strong></p> <p>I know how to complete the proof and get that ground states must be spherically symmetric, and monotonic radially, <em>provided one assume that $G$ is $C^{1,1}$</em>. This one can do by unique continuation principles for elliptic PDEs or equivalently by uniqueness theorems for ODEs. But in Berestycki-Lions or in Coleman-Glaser-Martin, $G$ is only assumed to be $C^1$, for which non-uniqueness, at least in the case of ODEs, is well known as a possibility. </p> <p>So, is the "uniqueness" statement true for $G$ merely $C^1$? Are there known counterexamples? </p> http://mathoverflow.net/questions/87418/the-ground-state-is-signed-and-symmetric/87816#87816 Answer by Willie Wong for The ground state is signed and symmetric Willie Wong 2012-02-07T16:56:57Z 2012-02-07T16:56:57Z <p>A little bit more digging turned up <a href="http://www.ams.org/mathscinet-getitem?mr=2486598" rel="nofollow">this paper of Mihai Mariş</a>. </p> <p>He shows that under two technical conditions:</p> <ul> <li>Minimizers are $C^1$ (which we have from elliptic regularity)</li> <li>If $u$ is an admissible function ($H^1$ in our case) and $v$ is a unit vector in $\mathbb{R}^d$, the function $\tilde{u}(x) = u(x)$ when $x\cdot v \geq 0$ and $\tilde{u}(x) = u(x - 2 x\cdot v v)$ otherwise is also admissible. </li> </ul> <p>The solution to a constrained minimisation problem with $k$ constraints must be radially symmetric about an axis of dimension $k-1$ if the functionals are translation and rotationally invariant. </p> <p>A very bad description of the process is follows:</p> <p>Roughly speaking, the main idea is the following: by the Borsuk-Ulam Lemma we can find inductively $\sim d-k$ mutually orthogonal hyperplanes each of which split all the constraints exactly in half. The reflection symmetry inherent in the problem implies that from each orthant we can extend to a solution that is symmetric across each of the planes, which implies that it is symmetric under $x\to -x$ inside a subspace. This however implies that any linearly dependent hyperplane of the given ones will also split all the constraints in half. </p> <p>The last step is to show that using multiple reflections and a bit of geometry, one can show that the original solution much also be reflectively symmetric across each of the planes. This shows radial symmetry. </p> http://mathoverflow.net/questions/87418/the-ground-state-is-signed-and-symmetric/87660#87660 Answer by Willie Wong for The ground state is signed and symmetric Willie Wong 2012-02-06T13:03:01Z 2012-02-06T16:37:21Z <p>A few results that addresses, but not quite answers my question:</p> <ul> <li>In the seminal paper of Gidas-Ni-Nirenberg, it is mentioned that, at least in the bounded domain case, that a positive solution to $-\triangle u = g(u)$ need not be radially symmetric when $g$ is not at least Lipschitz continuous. In particular, they gave a family of solutions and equations for which $g \in C^{0,1-2/p}$ for any $p > 2$. </li> <li>Along the same veins, it is not too hard to construct a counterexample to Gidas-Ni-Nirenberg if $g$ is merely continuous. Let $u(x) = v(x-x_v) + w(x- x_w)$, where $v$ is a radially symmetric bump function supported in the unit ball and monotonic radially, and $w$ is a radially symmetric function such that $w = 1$ in the ball of radius 5 and monotonically decreasing to 0. Then one can define $g$ by considering the case $x_v = x_w = 0$, which by a bit of undergraduate analysis one can show must be continuous. But the local nature of the equation means that for $|x_v - x_w| &lt; 2$, the same equation will still be satisfied. Note that if one can construct one such example for which $u$ is actually the minimal action solution, then this will give a bona fide counterexample. </li> <li><a href="http://www.ams.org/mathscinet-getitem?mr=1378680" rel="nofollow">Franchi, Lanconelli, and Serrin</a> shows an existence theorem for radially symmetric, non-negative solutions to a large class of PDEs, which includes $-\triangle u = g(u)$ for $g$ merely continuous. The corresponding uniqueness theorem they proved in the paper, however, is only concerning "uniqueness among radially symmetric solutions" and in particular does not use the variational structure at all. In particular, this result needs to be consider together with <a href="http://www.ams.org/mathscinet-getitem?mr=1364001" rel="nofollow">that of Cortazar, Elgueta, and Felmer</a> which shows that for a particular equation of the form above with a manifestly non-Lipschitz nonlinearity, the equation admits compactly supported nonnegative solutions. This gives yet another counterexample to Gidas-Ni-Nirenberg in the low regularity case. </li> </ul> <p>Neither of the above uses the variational structure of the "ground state". One result which I found that does use the variational ground state is that of </p> <ul> <li><a href="http://www.ams.org/mathscinet-getitem?mr=1617704" rel="nofollow">Flucher and Muller</a>, which was able to show that under the assumption that, roughly speaking $g(0) = g'(0) = 0$ and that the corresponding $G$ is non-negative, all variational ground states agree <em>up to a translation outside a compact set</em>, and that if $g$ itself is non-negative, non-negative ground states are in fact positive and strictly decreasing radially, hence unique (by the result of Brothers and Ziemer mentioned in the question). So in the general case they were not able to rule out the possible counterexample given in the second bullet point above. </li> </ul> http://mathoverflow.net/questions/82442/whitney-approximation-without-second-countable Whitney approximation without second countable Willie Wong 2011-12-02T10:17:45Z 2011-12-02T10:17:45Z <p>One version of Whitney's approximation theorem states the following:</p> <blockquote> <p>Let $N$ be a smooth, Hausdorff, second-countable (or paracompact) manifold, then given any continuous function $F:N\to \mathbb{R}^k$ and any continuous function $\epsilon:N\to\mathbb{R}_+$, one can approximate $F$ by a smooth function $\tilde{F}$ such that the error is bounded by $\epsilon$. </p> </blockquote> <p>The proof I know uses the fact that (a) the statement is true for $N$ being an open subset of $\mathbb{R}^n$ (which one has Whitney's version with analytic approximation, or one can get some sort of $C^\infty$ approximation via mollifiers) (b) one can take a partition of unity on $N$. </p> <p><strong>Question</strong></p> <p>Are there analogues to Whitney's theorem when second-countable is weakened? Or is there an easy counterexample? </p> http://mathoverflow.net/questions/81257/yang-mills-gradient-heat-flow-on-4-torus/81818#81818 Answer by Willie Wong for Yang Mills gradient/heat flow on 4-torus Willie Wong 2011-11-24T16:09:23Z 2011-11-24T16:09:23Z <p>Donaldson and Kronheimer wrote their book by 1990. There were some further developments about long time behaviour of Yang-Mills flow on four manifolds by, among others, Struwe and collaborators. You may try starting with <a href="http://www.ams.org/mathscinet-getitem?mr=1443269" rel="nofollow">Schlatter's dissertation</a>. </p> <p>Crawling through MathSciNet reference links may get you "somewhere", but it is my impression that higher dimensional Yang-Mills heat flow is still somewhat of an open problem. Is there any reason why you expect $\mathbb{T}^4$ would be better behaved than, say, the unit ball?</p> http://mathoverflow.net/questions/81128/when-is-it-appropriate-to-entitle-a-paper-a-note-on-or-on-the/81159#81159 Answer by Willie Wong for When is it appropriate to entitle a paper "A note on..." or "On the ..."? Willie Wong 2011-11-17T09:48:30Z 2011-11-17T09:48:30Z <p>Some journals explicitly frown upon titles of that form. For example, the <em>AMS Bulletin</em> (strangely enough, I only found this information in the back matter of the printed/PDF version of the journal, but not directly on the website) states (emphasis original):</p> <blockquote> <p>The first page must consist of a short <em>descriptive title</em> ... The <em>descriptive title</em> should be short but informative: useless or vague phrases such as "some remarks about" or "concerning" should be avoided.</p> </blockquote> http://mathoverflow.net/questions/129924/what-is-quillens-contribution-to-index-theorem Comment by Willie Wong Willie Wong 2013-05-07T07:29:50Z 2013-05-07T07:29:50Z Hum... learning something from Quillen does not necessarily imply learning something from <i>reading one of Quillen's papers</i>. It could be that the learning took place in a different setting. Unless the book indicated otherwise, I would not assume that the quote is necessarily referring to any of Quillen's three papers that you indicated. http://mathoverflow.net/questions/129881/elliptic-equation Comment by Willie Wong Willie Wong 2013-05-07T07:26:38Z 2013-05-07T07:26:38Z And what is $M$ and $N$? This question is not very clearly written in the state it is. But I suspect that it may be outside the scope of this website. (Please see the FAQ for what kinds of questions are generally asked here.) You may have better luck with <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a> http://mathoverflow.net/questions/129373/isothermal-coordinates-on-a-surface-with-smooth-metric Comment by Willie Wong Willie Wong 2013-05-02T07:49:33Z 2013-05-02T07:49:33Z @kvshud: let me quote the Wikipedia page which you &quot;did check&quot;, &quot;Results for H&#246;lder continuous metrics were obtained by Korn (1916) and Lichtenstein (1916). Later accounts were given by Morrey (1938), Ahlfors (1955), Bers (1952) and Chern (1955). A particularly simple account using the Hodge star operator is given in DeTurck &amp; Kazdan (1981).&quot; Since you are working in two dimensions, isothermal coordinates are harmonic, the results on existence of harmonic coordinates also apply. (See e.g. p251 of <a href="http://www.numdam.org/item?id=ASENS_1981_4_14_3_249_0" rel="nofollow">numdam.org/item?id=ASENS_1981_4_14_3_249_0</a> ) http://mathoverflow.net/questions/129163/examples-of-some-linear-and-nonlinear-operators Comment by Willie Wong Willie Wong 2013-04-30T09:30:05Z 2013-04-30T09:30:05Z what do you mean $\forall x,y\in H$? In the $\limsup$ you are already quantifying over $x,y$, it seems. http://mathoverflow.net/questions/127549/increasing-regularity-for-l2-function Comment by Willie Wong Willie Wong 2013-04-29T09:33:32Z 2013-04-29T09:33:32Z You need to specify what $r$, $\partial/\partial\theta$, and $\partial_x$ mean. Partial derivatives only makes sense when you have a full coordinate system. http://mathoverflow.net/questions/128814/weighted-graph-plot Comment by Willie Wong Willie Wong 2013-04-26T10:33:48Z 2013-04-26T10:33:48Z I don't think this question is on topic for MathOverflow. It may be a better fit at <a href="http://stats.stackexchange.com/questions/tagged/data-visualization" rel="nofollow">stats.stackexchange.com/questions/tagged/&hellip;</a> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/128686#128686 Comment by Willie Wong Willie Wong 2013-04-25T08:51:04Z 2013-04-25T08:51:04Z You may be interested in <a href="http://mathoverflow.net/questions/51436/bertrand-theorem-central-forces/92689#92689" rel="nofollow" title="bertrand theorem central forces">mathoverflow.net/questions/51436/&hellip;</a> http://mathoverflow.net/questions/128410/asymptotic-decay-for-the-wave-equation/128449#128449 Comment by Willie Wong Willie Wong 2013-04-23T09:19:46Z 2013-04-23T09:19:46Z For the above analysis you need $\epsilon$ to be sufficiently small compared to the first eigenvalue of the Laplacian. Consider the case $\Omega = [0,\pi]\subset \mathbb{R}$, and $\epsilon = 4$ which is more than twice the first eigenvalue. One checks that $$y(x,t) = e^{(-2 + \sqrt{3})t} \sin(x)$$ is a solution and decays strictly slower than $e^{-2t}$ ... See <a href="http://en.wikipedia.org/wiki/Damping#Over-damping_.28.CE.B6_.3E_1.29" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a> also for the classical harmonic oscillator analogue. http://mathoverflow.net/questions/127234/measurable-projection-theorem Comment by Willie Wong Willie Wong 2013-04-11T15:38:26Z 2013-04-11T15:38:26Z About half a year ago I scribbled in my notebooks that the measurable projection theorem can be found in the fourth volume of Fremlin's books. <a href="http://www.essex.ac.uk/maths/people/fremlin/mt.htm" rel="nofollow">essex.ac.uk/maths/people/fremlin/mt.htm</a> But unfortunately I didn't write down the page number. If you just google the phrase &quot;measurable projection theorem&quot; you will also find a wealth of information. http://mathoverflow.net/questions/127072/about-sectional-curvature Comment by Willie Wong Willie Wong 2013-04-10T12:48:59Z 2013-04-10T12:48:59Z In the interest of not duplicating efforts, a partial answer has already been given on Math.SE: <a href="http://math.stackexchange.com/questions/356989/about-sectional-curvature" rel="nofollow" title="about sectional curvature">math.stackexchange.com/questions/356989/&hellip;</a> http://mathoverflow.net/questions/127072/about-sectional-curvature Comment by Willie Wong Willie Wong 2013-04-10T12:48:08Z 2013-04-10T12:48:08Z Hi, welcome to Mathoverflow. As there are large number of user overlaps between MO and Math.StackExchange, generally we discourage immediate cross posting a question between the two forums. For future reference, please just post your question in one of the two places. If you don't get a response after a reasonable amount of time (say, about a week), then maybe it would be worthwhile considering cross posting to the other forum. Thanks. http://mathoverflow.net/questions/126953/covarient-derivative-of-electromagnetic-field-tensor Comment by Willie Wong Willie Wong 2013-04-09T11:02:45Z 2013-04-09T11:02:45Z This is probably off topic for this site. (See the FAQ.) For future reference Math.StackExchange would be a better venue. //// Now to your question: you have to use the curl relation. Using that $\nabla$ commutes with $g$, you have $$\nabla^a (F_{ac}F_b{}^c) = \nabla^aF_{ac}F_{b}{}^c + F_{ac} \nabla^a F_b{}^c$$ The first term evaluates to zero. The curl relation gives $$0 = F^{ac}\nabla_a F_{bc} + F^{ac} \nabla_b F_{ca} + F^{ac} \nabla_c F_{ab}$$ Antisymmetry in $ac$ from $F$ means that the first and third terms are equal. So $$\nabla^a (F_{ac}F_b{}^c) = \frac12 F^{ac} \nabla_b F_{ac}$$ http://mathoverflow.net/questions/16312/how-helpful-is-non-standard-analysis/126832#126832 Comment by Willie Wong Willie Wong 2013-04-08T11:51:05Z 2013-04-08T11:51:05Z I make no comment on the verity of the sentiments expressed by that quote. I take issue with your statement &quot;In fact, I was unable to find such a claim there&quot;, which for better or for worse sounds like you are accusing Steve Huntsman of fabricating the quote out of thin air. http://mathoverflow.net/questions/16312/how-helpful-is-non-standard-analysis/126832#126832 Comment by Willie Wong Willie Wong 2013-04-08T11:09:51Z 2013-04-08T11:09:51Z (BTW, several of the results you mention are already discussed in the various other answers to this question below, and it would be great if you can add links to the ones which aren't [for example, a link or actual citation reference to the relevant papers of Arkeryd would be wonderful!]) http://mathoverflow.net/questions/16312/how-helpful-is-non-standard-analysis/126832#126832 Comment by Willie Wong Willie Wong 2013-04-08T11:01:14Z 2013-04-08T11:01:14Z @katz: Wikipedia evolves. See the version from which Steve Huntsman quoted when he wrote his answer three years ago: <a href="http://en.wikipedia.org/w/index.php?title=Non-standard_analysis&amp;oldid=345543442#Applications" rel="nofollow">en.wikipedia.org/w/&hellip;</a>