User syang chen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:33:18Z http://mathoverflow.net/feeds/user/13776 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88259/nonvanishing-of-jacobians-implies-global-injectivity Nonvanishing of Jacobians implies global injectivity? Syang Chen 2012-02-12T04:05:24Z 2012-08-17T16:22:27Z <p>I am interested in obtaining injectivity of a $C^1$ map from the nonvanishing minors of its Jacobian matrix. Here is a brief history of the topic.</p> <p>In 1953, <a href="http://en.wikipedia.org/wiki/Paul_Samuelson" rel="nofollow">Samuelson</a> asked the following:</p> <blockquote> <p>If the upper left-hand principal minors of the Jacobian matrix of a map $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ do not vanish, is it true that $F$ must be injective?</p> </blockquote> <p>In 1965, <a href="http://en.wikipedia.org/wiki/David_Gale" rel="nofollow">Gale</a> and Nikaido gave a counterexample in $\mathbb{R}^2$. In <a href="http://www.springerlink.com/content/w007438g2kw60qjt/" rel="nofollow">their paper</a> the following is proved</p> <blockquote> <p><strong>Gale-Nikaido theorem:</strong> If all the principal minors of the Jacobian matrix of $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ are <strong>positive</strong>, then $F$ is injective. </p> </blockquote> <p>Since then, some effort has been made to weaken the assumption in Gale-Nikaido theorem since the assumption seems to be too restrictive in application. A comprehensive dicussion can be found in T. Parthasarathy, <em>On Global Univalence Theorems</em>, Lecture Notes in Mathematics, Vol. 977, 1983. In the case of polynomial map, this is related to the real version of <a href="http://en.wikipedia.org/wiki/Jacobian_conjecture" rel="nofollow">Jacobian conjecture</a>.</p> <p>A possible generalization I'm interested in is the following, which seems to be open.</p> <blockquote> <p><strong>Question:</strong> If all the principal minors of the Jacobian matrix of $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ <strong>do not vanish</strong>, is $F$ necessarily injective?</p> </blockquote> <p>In Gale and Nikaido's paper, the case of $\mathbb{R}^2$ was answered in affirmative, the case of $\mathbb{R}^3$ was claimed in affirmative (yet no complete proof seems to be known).</p> <p>My motivation comes from trying to make a change of variables to globally rectify a curved coordinate system so that Plancherel theorem can be applied. Any information would be appreciated : )</p> http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures Fourier decay rate of Cantor measures Syang Chen 2012-05-30T18:13:49Z 2012-06-01T08:36:32Z <p>For $0&lt;\theta&lt;\frac{1}{2}$, denote by $C_\theta$ the Cantor set with dissection ratio $\theta$, i.e. the Cantor set obtained from dissection parttern $(\theta, 1-2\theta,\theta)$. It is known that $C_\theta$ carries a uniform measure $\mu_\theta$ which is usually called Cantor measure. And it is not hard to show that the <a href="http://en.wikipedia.org/wiki/Fourier_transform#Fourier.E2.80.93Stieltjes_transform" rel="nofollow">Fourier–Stieltjes transform</a> of $\mu_\theta$ is (up to scaling and constant multiple)</p> <p><code>$$\hat{\mu}_\theta(\xi)=\prod^{\infty}_{k=1} \cos(\theta^k\xi)$$</code> </p> <p>But unlike integrable function we do not have Riemann-Lebesgue lemma for these Cantor measures. In fact a theorem of Erdős and Salem says <code>$\hat{\mu}_\theta (\xi)=o(1)$</code> as $|\xi|\rightarrow\infty$ if and only if $\theta^{-1}$ is a not <a href="http://en.wikipedia.org/wiki/PV_number" rel="nofollow">PV number</a>. On the other hand, it is known that for some $\theta^{-1}$ not a PV number, <code>$\hat{\mu}_\theta (\xi)$</code> does not decay at any positive rate, even though $\hat{\mu}_\theta (\xi)=o(1)$.</p> <p>My question is whether there exists $\theta$ such that <code>$\hat{\mu}_\theta (\xi)=O(|\xi|^{-\alpha})$</code> for some $\alpha>0$? How much is known about the precise decay rate of $\hat{\mu}_\theta (\xi)$? </p> <p>Thanks in advance.</p> <p><strong>Edit:</strong> To make the second question precise, I was actually wondering if there exists Salem set (as pointed out by Pablo) among these Cantor sets. So the rate of decay I was expecting is (or arbitrarily close to) half of the Hausdorff dimension $\frac{1}{2}\dim_H(C_\theta)=\frac{1}{2}\frac{\log(1/2)}{\log(\theta)}$.</p> http://mathoverflow.net/questions/90233/mean-value-property-with-fixed-radius Mean value property with fixed radius Syang Chen 2012-03-04T23:49:03Z 2012-03-05T01:30:57Z <p>Let $f$ be a continuous function defined on $\mathbb{R^n}$. It is well known that both the spherical mean value property (MVP) of $f$, i.e. $$f(x)=\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}f,\ \forall x\in\mathbb{R^n}, r>0$$ and the ball MVP, i.e. $$f(x)=\frac{1}{|B(x,r)|}\int_{B(x,r)}f,\ \forall x\in\mathbb{R^n},r>0$$ imply that $f$ is harmonic.</p> <p>Note that in the definitions we require the redius $r$ to run over all the positive numbers. Out of curiosity I tried to find non-harmonic functions which satysfy the MVPs <em>only</em> for $r=1$. I did some search and found a remarkable fact called <a href="http://www.encyclopediaofmath.org/index.php/Mean-value_characterization" rel="nofollow">Delsarte's two-radius theorem</a> saying that the spherical MVP with <em>two</em> fixed radii is enough to imply harmonicity of $f$. But for the $1$-radius MVP I haven't found any statement.</p> <p>In the case $n=1$ examples have been found nicely in this <a href="http://math.stackexchange.com/questions/116236/mean-value-property-with-fixed-radius/116486" rel="nofollow">M.SE post</a>. But it is still unclear to me how to construct similar examples in higher dimensions. Any comments would be appreciated!</p> http://mathoverflow.net/questions/88875/antiderivative-of-a-darboux-function/89151#89151 Answer by Syang Chen for antiderivative of a darboux function Syang Chen 2012-02-22T00:58:55Z 2012-02-22T00:58:55Z <p>Let $\phi(x)=x\chi_{[-1,1]}(x)+\text{sgn}(x)\chi_{[-1,1]^c}(x)$, $f$ be the <a href="http://en.wikipedia.org/wiki/Conway_base_13_function" rel="nofollow">Conway base 13 function</a>, then $\phi\circ f$ is a Darboux function and is nowhere continuous. Moreover, $\phi\circ f$ is Borel measurable and bounded (in particular, it is $L^1$). </p> <p>But $\phi\circ f$ is not a derivative because any derivative is a limit of continuous functions, hence its discontinuity points form a set of first category (in particular cannot be the whole interval). See <a href="http://mathoverflow.net/questions/88271/real-valued-function-whose-derivative-is-nowhere-continuous" rel="nofollow">this post</a>.</p> <p>In sum, $\phi\circ f$ is a bounded Borel measurable Darboux function but is not a derivative.</p> http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian/88911#88911 Answer by Syang Chen for $L^1$ norm of the Fourier transform of a truncated Gaussian Syang Chen 2012-02-19T09:03:53Z 2012-02-22T00:06:10Z <p>If I didn't miss anything, the dominated convergence argument turns out to work well. The following shows that $\widehat{G_k}$ can be dominated by a rapidly decaying function.</p> <p>Denote $\psi=\widehat{\phi}, \psi_k=\widehat{\phi_k}$, then $\psi_k(y)=2^k\psi(2^ky), \int \psi_k=0$ and $\int |\psi_k|=\int |\psi|$, hence</p> <p>$$|\widehat{G_k}(-x)| =|\widehat{\phi_k}\ast G(-x)| =|\int \psi_k(y)G(y+x)dy| =|\int \psi_k(y)[G(y+x)-G(x)]dy|$$ $$=|\int_{|y|\le |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|+|\int_{|y|\ge |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|$$ $$\le \frac{|x|}{2}\sup_{B(x,|x|/2)}|G'|\cdot \int |\psi|+C\int_{|y|\ge |x|/2}|\psi|dy$$</p> <p>The last line is a function in $x$ decaying at any rate. </p> <p>In view of the above argument, the function $\phi$ and $G$ can be replaced by any Schwartz functions where in addition $\phi$ vanishes at $0$, and we will still have $\|\widehat{G_k}\|_1 \rightarrow 0$. An estimate of the rate of decay follows in the specified case.</p> http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59988#59988 Answer by Syang Chen for Additive Subgroups of the Reals. Syang Chen 2011-03-29T17:00:52Z 2011-03-29T17:28:25Z <p>If you would like to classify the subgroups in the sense of Lebesugue measure, you may find the following facts helpful.</p> <p>(1) Any measurable proper subgroup of the real line is of measure $0$.</p> <p>(2) Any non-measurable subgroup $G$ of the real line charges fully everywhere, i.e., for any interval $I$, $m^{\ast}(G \cap I)=|I|$, where $m^{\ast}(\cdot)$ denotes the outer Lebesgue measure.</p> <p>(3) Non-measurable subgroup of the real line exists.</p> http://mathoverflow.net/questions/59748/what-structure-is-needed-to-define-a-gaussian-distribution-on-a-given-space/59835#59835 Answer by Syang Chen for What structure is needed to define a Gaussian distribution on a given space? Syang Chen 2011-03-28T13:51:37Z 2011-03-29T16:38:17Z <p>Let me assume that you seek for the generalization of Gaussian distribution in order to generalize the Brownian motion.</p> <p>As far as I know, regarding the heat kernel as the generalization of the Gaussian distribution has long been adopted in many literatures. It comes from the following observation.</p> <p>In $\mathbb {R}^1$, the following notions coincide:</p> <p>(1) Gaussian distribution $N(x,t)\sim f(t,x,y)=\frac {1}{\sqrt{2\pi t}}e^{\frac{-(y-x)^2}{2t}}$,</p> <p>(2) transition function $p(t,x,y)$ of the Brownian motion $B_t$,</p> <p>(3) (heat kernel) fundamental solution $k_t(x,y)$ of the heat equation $\partial_t k=\Delta_y k$, with initial data $\delta_x$.</p> <p>Thus, on manifolds, one way to define the Brownian motion is to construct a Markov process on the manifold whose transition function is exactly the heat kernel (let's identify the heat kernel with the Gaussian distribution in this setting). Since we always have the Laplacian-Beltrami $\Delta$ on a manifold, it is justifiable to talk about the heat equation and thus the heat kernel, and the Brownian motion in this sense is known to exist for a large class of manifolds. </p> <p>But on metric spaces, we no longer have the Laplacian-Beltrami. So, in order to talk about heat kernel/Gaussian distribution, we need to generalize the notion of Laplacian-Beltrami. The key concept on this line the so-called <a href="http://en.wikipedia.org/wiki/Dirichlet_form" rel="nofollow">Dirichlet form</a>. A Dirichlet form on metric measure space $(X,d,\mu)$ a closed symmetric form $(\cdot,\cdot)$ defined on $L^2(X,\mu)$. It should further satisfy a couple of conditions so that it behaves like its prototype $(f,g)=\int_{M} {\nabla f\cdot \nabla g dx}$ on a manifold $M$. Notice that $(f,g)=(-\Delta f,g)_{L^2(M)}$ on manifolds, in the general case, one obtains the desired "Laplacian" by the same formula. Therefore, every Dirichlet form corresponds to a "Laplacian" and thus a Gaussian distribution (and thus a Brownian motion). What's more, a reasonable Dirichlet form always exists provided the space is suitably good.</p> <p>In sum, if the space you are considering have both metric and measure structures, then the theory of Dirichlet form may provide you some satisfactory results regarding construction and properties of the Guassian distribution (and thus the Brownian motion). Roughly speaking, if we don't have a presumed measure, we may not be able to construct a reasonable probability space; if we don't have a metric, it would be hard to measure the regularity and decay of the Gaussian distribution. So metric measure structure might be the minimal structure for reasonable construction of Gaussian distribution.</p> <p>Some reference books could be found in the above link. <a href="http://projecteuclid.org/euclid.aop/1022855410" rel="nofollow">This paper by Sturm</a> may allow you to have a glance at the whole picture. I am not an expert in this field. I apologize in advance for any mistake and naivety.</p> http://mathoverflow.net/questions/59115/a-set-for-which-it-is-hard-to-determine-whether-or-not-it-is-countable/59138#59138 Answer by Syang Chen for A set for which it is hard to determine whether or not it is countable. Syang Chen 2011-03-22T05:17:23Z 2011-03-27T08:24:48Z <p>An elementary example which any grad can think about. Hope it will require 10 mins for most of them to figure out (or recall) the proof:</p> <p>The set of discontinuous points of a non-decreasing function.</p> <p>Or, making it more geometric, (under suitable assumptions) the set of the radius $r$ such that a given Borel measure charges the $r$-sphere.</p> <p>Or, based on the first result, the set of non-differentiable points of a convex function on the real line.</p> http://mathoverflow.net/questions/58955/question-on-eigenvalue-square-root-subadditivity/58968#58968 Answer by Syang Chen for Question on eigenvalue square root subadditivity Syang Chen 2011-03-20T12:56:51Z 2011-03-23T09:54:03Z <p>As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).</p> <p>Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.</p> <p>Could you explain how you got the inequality? Hope we will get some clue from Seva's work.</p> <hr> <p>EDIT: Salle's answer is very instructive to me. I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.</p> <p>Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)+Tr(A)Tr(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ and then get the inequality.</p> <p>For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.</p> <p>I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.</p> <hr> <p>EDIT II: I guess you can change the title into "A generalized Hoffman-Wielandt inequality" and add the tag "Numerical Analysis". </p> <p>The Hoffman-Wielandt inequality states the following:</p> <blockquote> <p>Let $A$ and $B$ be real symmetric matrices, $a_i$, $b_i$, $c_i$ the eigenvalues of $A$, $B$, $A+B$ respectively with $a_i\leq a_{i+1}$, etc. Then we have $(\sum_i |c_i-b_i|^2)^{1/2} \leq (\sum_i |a_i|^2)^{1/2}$.</p> </blockquote> <p>A proof in spirit similar to Mikael's can be found in "The Algebraic Eigenvalue Problem" by Wilkinson. The $L^p$ variant can be found in <a href="http://www.springerlink.com/content/g526018322656280/" rel="nofollow">a paper by Rajendra Bhatia and Ludwig</a>. It seems here your taking the square root inside and $L^1$ norm outside somewhat make things tougher.</p> http://mathoverflow.net/questions/58960/a-question-about-regular-signed-or-complex-borel-measure-under-lrn-decomposition/58962#58962 Answer by Syang Chen for A question about regular signed or complex Borel measure under LRN decomposition Syang Chen 2011-03-20T10:49:19Z 2011-03-20T10:49:19Z <p>If $\nu\bot\lambda$, then $|\nu+\lambda|=|\nu|+|\lambda|$</p> http://mathoverflow.net/questions/4625/regularity-of-sparse-fourier-transforms/4626#4626 Comment by Syang Chen Syang Chen 2012-07-15T08:25:21Z 2012-07-15T08:25:21Z @Yemon, could you tell me on which page I can find the proof of the statement you stated at the beginning? I found the statement that Lip$_\alpha$ implies Fourier coefficient decays like $O(1/n^{\alpha})$, but I couldn't find the converse. Thank you. http://mathoverflow.net/questions/44192/fourier-dimension-of-the-sum-of-sets/47639#47639 Comment by Syang Chen Syang Chen 2012-06-13T19:16:32Z 2012-06-13T19:16:32Z @Pablo: Where can I find the proof of the inequality $\dim_H(A+B)\le\dim_H(A)+\dim_B(B)$? http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures/98478#98478 Comment by Syang Chen Syang Chen 2012-06-02T16:37:08Z 2012-06-02T16:37:08Z That would be nice. Could you tell me precisely in which paper can I find the result? If I understand correctly, it contradicts Bluhm's result (Theorem 5). In his construction, the ratios are of the form $\theta^k$, only the positions are randomized. http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures/98478#98478 Comment by Syang Chen Syang Chen 2012-06-02T03:55:52Z 2012-06-02T03:55:52Z Nice. So generally the &quot;neat&quot; Cantor sets (with neat positions, ratios) can not be Salem? http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures/98478#98478 Comment by Syang Chen Syang Chen 2012-06-02T03:39:29Z 2012-06-02T03:39:29Z The paper proves that the randomized Cantor set (keeping the dissection number constant 2) is almost surely Salem. But I couldn't quite go through the proof. Maybe I was just confused... http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures/98488#98488 Comment by Syang Chen Syang Chen 2012-06-02T03:28:13Z 2012-06-02T03:28:13Z Thanks. The argument above is really amazing... http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures/98478#98478 Comment by Syang Chen Syang Chen 2012-06-01T06:37:01Z 2012-06-01T06:37:01Z Are you familiar with the paper you mentioned above? I have a question about (3.3) in that paper. http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures/98488#98488 Comment by Syang Chen Syang Chen 2012-06-01T06:21:18Z 2012-06-01T06:21:18Z Sorry, I couldn't quite follow the edit. For example, why is the cumulative correction at most $\theta+\theta^2+\cdots$? On the other hand, is there still hope to obtain the desired decay rate for $\theta$ not close to $0$? http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures/98440#98440 Comment by Syang Chen Syang Chen 2012-05-31T04:02:22Z 2012-05-31T04:02:22Z @Igor, Thanks. But the paper gives only the decay rate in the (ball) average sense. http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures Comment by Syang Chen Syang Chen 2012-05-31T03:19:04Z 2012-05-31T03:19:04Z @mike, Yes I have read that paper. I think that is where the name of the above theorem comes from. http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures Comment by Syang Chen Syang Chen 2012-05-30T21:20:30Z 2012-05-30T21:20:30Z @Will: $\hat{\mu}_\theta$ is the Fourier transform of $\mu_\theta$, reference added. http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures Comment by Syang Chen Syang Chen 2012-05-30T18:28:01Z 2012-05-30T18:28:01Z Thanks for editing. http://mathoverflow.net/questions/88259/nonvanishing-of-jacobians-implies-global-injectivity Comment by Syang Chen Syang Chen 2012-05-12T17:33:04Z 2012-05-12T17:33:04Z @Misha: Thank you. Zorich's theorem and the results on <a href="http://en.wikipedia.org/wiki/Quasiregular_map" rel="nofollow">en.wikipedia.org/wiki/Quasiregular_map</a> are very impressive. Unfortunately the maps I am concerned with are not quasiregular. Their Jocobians collapse when two coordinates coincide whereas their gradients do not. http://mathoverflow.net/questions/90233/mean-value-property-with-fixed-radius/90241#90241 Comment by Syang Chen Syang Chen 2012-03-05T18:26:26Z 2012-03-05T18:26:26Z @Igor Awesome! Just a minor correction: the implication &quot;ball MVP ⟹ harmonicity&quot; does not hold in one dimension. http://mathoverflow.net/questions/88259/nonvanishing-of-jacobians-implies-global-injectivity/88352#88352 Comment by Syang Chen Syang Chen 2012-02-15T03:27:06Z 2012-02-15T03:27:06Z It is interesting. But their results don't seem to go beyond the Gale-Nikaido theorem. Maybe I'm missing something, it seems all the sufficient conditions they obtained (in term of properties of graphs associated to the Jacobian) guarantee injectivity via G-N theorem.