User syang chen - MathOverflow

Nonvanishing of Jacobians implies global injectivity?

2012-02-12T04:05:24Z

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.

In 1953, Samuelson asked the following:

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?

In 1965, Gale and Nikaido gave a counterexample in $\mathbb{R}^2$. In their paper the following is proved

Gale-Nikaido theorem: If all the principal minors of the Jacobian matrix of $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ are positive, then $F$ is injective.

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, On Global Univalence Theorems, Lecture Notes in Mathematics, Vol. 977, 1983. In the case of polynomial map, this is related to the real version of Jacobian conjecture.

A possible generalization I'm interested in is the following, which seems to be open.

Question: If all the principal minors of the Jacobian matrix of $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ do not vanish, is $F$ necessarily injective?

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).

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 : )

Fourier decay rate of Cantor measures

2012-05-30T18:13:49Z

For $0<\theta<\frac{1}{2}$, 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 Fourier–Stieltjes transform of $\mu_\theta$ is (up to scaling and constant multiple)

$$\hat{\mu}_\theta(\xi)=\prod^{\infty}_{k=1} \cos(\theta^k\xi)$$

But unlike integrable function we do not have Riemann-Lebesgue lemma for these Cantor measures. In fact a theorem of Erdős and Salem says $\hat{\mu}_\theta (\xi)=o(1)$ as $|\xi|\rightarrow\infty$ if and only if $\theta^{-1}$ is a not PV number. On the other hand, it is known that for some $\theta^{-1}$ not a PV number, $\hat{\mu}_\theta (\xi)$ does not decay at any positive rate, even though $\hat{\mu}_\theta (\xi)=o(1)$.

My question is whether there exists $\theta$ such that $\hat{\mu}_\theta (\xi)=O(|\xi|^{-\alpha})$ for some $\alpha>0$? How much is known about the precise decay rate of $\hat{\mu}_\theta (\xi)$?

Thanks in advance.

Edit: 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 $\frac{1}{2}\dim_H(C_\theta)=\frac{1}{2}\frac{\log(1/2)}{\log(\theta)}$.

Mean value property with fixed radius

2012-03-04T23:49:03Z

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.

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 only for $r=1$. I did some search and found a remarkable fact called Delsarte's two-radius theorem saying that the spherical MVP with two fixed radii is enough to imply harmonicity of $f$. But for the $1$-radius MVP I haven't found any statement.

In the case $n=1$ examples have been found nicely in this M.SE post. But it is still unclear to me how to construct similar examples in higher dimensions. Any comments would be appreciated!

Answer by Syang Chen for antiderivative of a darboux function

2012-02-22T00:58:55Z

Let $\phi(x)=x\chi_{[-1,1]}(x)+\text{sgn}(x)\chi_{[-1,1]^c}(x)$, $f$ be the Conway base 13 function, 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 $L^1$).

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 this post.

In sum, $\phi\circ f$ is a bounded Borel measurable Darboux function but is not a derivative.

Answer by Syang Chen for $L^1$ norm of the Fourier transform of a truncated Gaussian

2012-02-19T09:03:53Z

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.

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

$$ |\widehat{G_k}(-x)| =|\widehat{\phi_k}\ast G(-x)| =|\int \psi_k(y)G(y+x)dy| =|\int \psi_k(y)[G(y+x)-G(x)]dy|$$ $$=|\int_{|y|\le |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|+|\int_{|y|\ge |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|$$ $$\le \frac{|x|}{2}\sup_{B(x,|x|/2)}|G'|\cdot \int |\psi|+C\int_{|y|\ge |x|/2}|\psi|dy$$

The last line is a function in $x$ decaying at any rate.

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.

Answer by Syang Chen for Additive Subgroups of the Reals.

2011-03-29T17:00:52Z

If you would like to classify the subgroups in the sense of Lebesugue measure, you may find the following facts helpful.

(1) Any measurable proper subgroup of the real line is of measure $0$.

(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.

(3) Non-measurable subgroup of the real line exists.

Answer by Syang Chen for What structure is needed to define a Gaussian distribution on a given space?

2011-03-28T13:51:37Z

Let me assume that you seek for the generalization of Gaussian distribution in order to generalize the Brownian motion.

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.

In $\mathbb {R}^1$, the following notions coincide:

(1) Gaussian distribution $N(x,t)\sim f(t,x,y)=\frac {1}{\sqrt{2\pi t}}e^{\frac{-(y-x)^2}{2t}}$,

(2) transition function $p(t,x,y)$ of the Brownian motion $B_t$,

(3) (heat kernel) fundamental solution $k_t(x,y)$ of the heat equation $\partial_t k=\Delta_y k$, with initial data $\delta_x$.

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.

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 Dirichlet form. 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.

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.

Some reference books could be found in the above link. This paper by Sturm 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.

Answer by Syang Chen for A set for which it is hard to determine whether or not it is countable.

2011-03-22T05:17:23Z

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:

The set of discontinuous points of a non-decreasing function.

Or, making it more geometric, (under suitable assumptions) the set of the radius $r$ such that a given Borel measure charges the $r$-sphere.

Or, based on the first result, the set of non-differentiable points of a convex function on the real line.

Answer by Syang Chen for Question on eigenvalue square root subadditivity

2011-03-20T12:56:51Z

As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

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 $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

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.

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.

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.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.

EDIT II: I guess you can change the title into "A generalized Hoffman-Wielandt inequality" and add the tag "Numerical Analysis".

The Hoffman-Wielandt inequality states the following:

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 $(\sum_i |c_i-b_i|^2)^{1/2} \leq (\sum_i |a_i|^2)^{1/2}$.

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 paper by Rajendra Bhatia and Ludwig. It seems here your taking the square root inside and $L^1$ norm outside somewhat make things tougher.

Answer by Syang Chen for A question about regular signed or complex Borel measure under LRN decomposition

2011-03-20T10:49:19Z

If $\nu\bot\lambda$, then $|\nu+\lambda|=|\nu|+|\lambda|$

Comment by Syang Chen

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.

Comment by Syang Chen

2012-06-13T19:16:32Z

@Pablo: Where can I find the proof of the inequality $\dim_H(A+B)\le\dim_H(A)+\dim_B(B)$?

Comment by Syang Chen

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.

Comment by Syang Chen

2012-06-02T03:55:52Z

Nice. So generally the "neat" Cantor sets (with neat positions, ratios) can not be Salem?

Comment by Syang Chen

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...

Comment by Syang Chen

2012-06-02T03:28:13Z

Thanks. The argument above is really amazing...

Comment by Syang Chen

2012-06-01T06:37:01Z

Are you familiar with the paper you mentioned above? I have a question about (3.3) in that paper.

Comment by Syang Chen

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 $0$?

Comment by Syang Chen

2012-05-31T04:02:22Z

@Igor, Thanks. But the paper gives only the decay rate in the (ball) average sense.

Comment by Syang Chen

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.

Comment by Syang Chen

2012-05-30T21:20:30Z

@Will: $\hat{\mu}_\theta$ is the Fourier transform of $\mu_\theta$, reference added.

Comment by Syang Chen

2012-05-30T18:28:01Z

Thanks for editing.

Comment by Syang Chen

2012-05-12T17:33:04Z

@Misha: Thank you. Zorich's theorem and the results on en.wikipedia.org/wiki/Quasiregular_map 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.

Comment by Syang Chen

2012-03-05T18:26:26Z

@Igor Awesome! Just a minor correction: the implication "ball MVP ⟹ harmonicity" does not hold in one dimension.

Comment by Syang Chen

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.