User 002 - MathOverflow

Rolle's theorem in n dimensions

2010-01-01T21:51:09Z

This looks like a statement from a calculus textbook, which perhaps it should be.

"Rolle's theorem". Let $F\colon [a,b]\to\mathbb R^n$ be a continuous function such that F(a)=F(b) and F'(t) exists for all $a<t<b$ . Then there exist numbers $a < t_1 < t_2 < \dots < t_n < b$ such that the vectors $F'(t_1),\dots,F'(t_{n})$ are linearly dependent.

We are all familiar with the case n=1. The case n=2 is not hard either: pick any $a<t^\ast<b$ and find, using the mean value theorem, numbers $a<t_1<t^\ast$ and $t^*<t_2<b$ such that $F'(t_j)$ is collinear with $F(t^\ast)-F(a)$ . Note that we avoided using the parameter value $t^\ast$ , which will be important in a moment. When n=3, we pick $a<t^\ast<b$ and project F onto the orthogonal complement of $F'(t^\ast)$ , then apply the case n=2 to the projection ($t^\ast$ will become the third chosen parameter value). So far so good.

But I get stuck at n=4. If the above process is followed, then after F is projected down to two dimensions, we must avoid two particular parameter values. Which is not possible in general: if in two dimensions F parametrizes a triangle and F(a) is a vertex, then one of points $F(t_j)$ must be one of two other vertices. Presumably this problem can avoided by a generic choice of points of projection, but how?

Determination of a symmetric convex region by parallel sections

2010-01-10T02:36:02Z

This question is partly inspired by a problem in Stewart's Calculus: "Find the area of the region enclosed by $y=x^2$ and $x=y^2$."

Suppose $f\colon [0,1]\to [0,1]$ is a convex increasing function that fixes 0 and 1. The graphs $y=f(x)$ and $x=f(y)$ enclose a convex region. To find its area, students will likely integrate the length of vertical sections: $f^{-1}(x)-f(x)$. They are unlikely to ask if two different functions can yield the same integrand. But I'll ask:

Does $f^{-1}-f$ determine $f$? (Among all convex increasing functions that fix 0 and 1).

Level set of a harmonic function

2010-01-01T03:29:50Z

Let $u$ be a nonconstant real-valued harmonic function defined in the open unit disk $D$. Suppose that $\Gamma\subset D$ is a smooth connected curve such that $u=0$ on $\Gamma$. Is there a universal upper bound for the length of $\Gamma$?

Remark: by the Hayman-Wu theorem, the answer is yes if $u$ is the real part of an injective holomorphic function; in fact, in this case there is a universal upper bound for the length of the entire level set in $D$. For general harmonic functions, level sets can have arbitrarily large length, e.g. $\Re z^n$.

How to best distribute points on two concentric circles?

2010-04-14T03:56:17Z

An N-subset $\{x_1,\dots,x_N\}$ of a compact set $X\subset \mathbb R^d$ is called a set of Fekete points (named after Michael Fekete) if it maximizes the product $$\prod_{1\le k<j\le N}|x_k-x_j|\qquad (1)$$ among all such $N$-tuples. When $X\subset \mathbb C$, one can express this product in terms of the Vandermonde determinant. In this case Fekete points are of particular interest in approximation theory (as interpolation nodes). Generally, they have no explicit form and must be found numerically. But there are exceptions.

When $X$ is a circle, Fekete points are equally spaced. This is well-known and can be proved like this: we may assume $x_k=\exp(i\theta_k)$ with $0=\theta_1<\theta_2<\dots <\theta_N<2\pi$ . For a fixed integer $m$, $1\le m\le \lfloor N/2 \rfloor$, consider the product $\Pi(m)=\prod_{k=1}^N|x_{k+m}-x_{k}|$ with indices taken mod $N$. Since each point $x_k$ appears in just two terms of this product, it is not hard to see that $\Pi(m)$ is maximal when $x_k$ are equally spaced. (Indeed, $\log \sin$ is a concave function.)

Now let $X$ be the union of two concentric circles of radii $r$ and $R$. Assume that $N=2n$ is even and that exactly $n$ of the points lie on each circle. With this additional assumption we are not quite in the Fekete problem anymore, but we have an obvious candidate for the maximum -

Under the assumptions of the preceding paragraph, is the product (1) maximized by equally spaced, interlaced points? E.g., by $$x_k=\begin{cases} r\exp(2\pi i k/n), \quad k=1,\dots,n \\ R\exp(\pi i (2k-1)/n), \quad k=n+1,\dots,2n \end{cases} \qquad (2)$$

Presumably, a solution would consist of two steps.

Show that the maximum is attained by interlaced points, i.e., when ordered by the argument, the points alternate between the two circles.
Show that for interlaced points, maximum of (1) is attained when they are equally spaced.

One can hope that step 2 is doable with standard calculus tools, since there should be no other local maxima for interlaced points. But Step 1 seems to require a more imaginative approach.

Why do I find this question interesting? There is a natural way to distribute N points on a circle (i.e., at equal distances), and this configuration is known to give the solution for many extremal problems, such as the Fekete problem, isoperimetric N-gon problem, etc. There is also a natural way to distribute two sets of N points -- namely, by formula (2), but I don't know of any nontrivial problem for which the optimality of such configuration has been established.

Answer by 002 for Approximately holomorphic functions

2010-04-01T16:44:49Z

Men'shov proved in 1936 that if $f\colon D\to\mathbb C$ is continuous and approximately differentiable outside of a countable set, then it is holomorphic in $D$ 1 (Russian original with French summary). In the same paper he gives an example, attributed to Lusin, which shows that continuity cannot be dropped entirely even if the function is approximately differentiable at every point. Let $\varphi\colon\mathbb C\to\mathbb C$ be an entire function that tends to $0$ as $z\to\infty$ within any sector $\{|\arg z|<\pi-\epsilon\}$ (approximation theory can be used to create such examples). Then $f(z)=z\varphi(1/z)$ has an approximate derivative everywhere in $\mathbb C$, but of course it is not differentiable or even continuous at $0$.

Men'shov asked if continuity can be replaced with boundedness. This was answered affirmatively by Telyakovskii 50 years later 2. In fact, boundedness can be weakened to logarithmic integrability.

Brodovich 3 proved that an injective function with an approximate derivative at every point is holomorphic. (MR review omits the injectivity assumption). A survey of results in this area was written by Dolzhenko 4, but it predates the work of Brodovich.

Answer by 002 for Solutions to a Monge-Ampère equation on the simplex

2010-02-23T22:46:50Z

Any set of values of $u$ at the vertices of $\Delta_k$ can be attained just by adding an affine function to $u$, which does not change $M[u]$. To see that the solution of your problem is not unique, consider $u(x,y)=ax^2+a^{-1}y^2+\mathrm{(affine\ terms)}$ with $a>0$. Clearly $M[u]=4$ for any $a$.

On the other hand, Theorem 1.6.2 in the book The Monge-Ampère equation by C. Gutierréz states that there is a unique convex solution of $M[u]=\mu$ (with $M[u]$ properly understood) with prescribed continuous boundary values in a strictly convex domain $\Omega$. Without strict convexity we can't allow arbitrary continuous boundary data; it must be at least consistent with some convex function in $\Omega$. (But I don't know if that's enough). The uniqueness part holds without strict convexity; see Corollary 1.4.7 in the same book.

Answer by 002 for Riemann surface disconnected at infinity

2010-02-13T16:51:44Z

This is an extended version of my comment. Suppose we stay on the surface $z^2+w^2=1$ but away from the origin. The identity $|z^2+w^2|^2=|z|^4+|w|^4+2Re((z \bar w)^2)$ tells us that the square of $z \bar w$ has negative real part. The set of complex numbers $\zeta$ such that $Re(\zeta^2)<0$ has two connected components: it's the disjoint union of two open sectors. Finally, note that switch from (z,w) to (-z, w) involves going from one component to the other.

Answer by 002 for What m minimizes E(|m-X|^3) for a random variable X?

2010-02-11T02:17:57Z

The minimizer $m$ is the nearest point projection of $X$ onto the subspace of $L^p$ formed by the constant functions ($p=3$ in your case). This $m$ is sometimes called the $p$-prediction or $p$-predictor of $X$. Apparently, this terminology began with Andô and Amemiya. Some of later papers are Landers and Rogge (who wrote a few other papers, e.g. this one), and Cuesta and Matrán. The term "generalized (conditional) expectation" also appeared.

Answer by 002 for Constraints on the Fourier transform of a constant modulus function

2010-02-09T17:11:36Z

If $g$ happens to be in $L^1$, then the amplitude of the Fourier transform of $fg$ is bounded by the $L^1$ norm of $g$, for any unimodular $f$. This is the only restriction from above since you can always choose $f$ so that $fg\ge 0$, thus bringing the (essential) supremum of $\widehat{fg}$ up to $\|g\|_{L^1}$.

Another part of the question is how small we can make $A$. I guess "arbitrarily small", but don't have a proof. (Except for special case: if $g$ is in $L^1$, then we can chop it into pieces with disjoint supports and small $L^1$ norm, and then use $f$ to move the Fourier transforms of pieces far from one another.)

Answer by 002 for cauchy product for general case

2010-02-08T21:30:24Z

It's not a problem to multiply the series: the product is $\sum_{(t,k)\in\mathbb Z^2} a_tb_k$. The question is how to sum the double series that we have.

For series with nonnegative terms summation is not a problem either: we take the supremum of all finite sums. And since any finite sum is contained in a sufficiently large square, it follows that $\sum_{(t,k)\in\mathbb Z^2} |a_tb_k|$ is finite whenever $\sum_{t\in\mathbb Z} |a_t|$ and $\sum_{k\in\mathbb Z} |b_k|$ are.

In general, $\sum_{(t,k)\in\mathbb Z^2} a_tb_k=S$ if for any $\epsilon>0$ there is a finite subset $A\subset \mathbb Z^2$ such that $|\sum_{(t,k)\in B}a_tb_k - S|<\epsilon$ whenever $B$ is finite and $B\supset A$. Now if both given series converge absolutely, then the contribution from outside of a large square is small, and it follows that $S$ is the product of two sums.

Answer by 002 for When can a function be recovered from a distribution?

2010-02-08T02:52:09Z

First of all, $T$ must have order zero, i.e., $|T(\varphi)|\le C(K)\sup|\varphi|$ for any test function $\varphi$ supported on a compact set $K$. By Riesz representation theorem, $T$ is a measure. To be a locally integrable function, it must be absolutely continuous with respect to the Lebesgue measure. One way to express this condition: $C(K)\to 0$ as the Lebesgue measure of $K$ tends to zero, which $K$ staying within a fixed compact set.

Answer by 002 for Maximizing Sparsity in l1 Minimization?

2010-02-07T04:37:53Z

The ultimate sparseness occurs when $Ax^*(\lambda)=0$, which is the case when the minimizer $x^*$ is the projection of $b$ onto $\ker A$. For this to happen, $\lambda$ must be small enough so that the restriction of $A$ to the orthogonal complement of its kernel is bounded from below by a constant greater than $2\lambda \mathrm{dist}(b,\ker A)$ . Here the lower bound for operator is understood in the $\ell^2\to\ell^1$ norm.

Answer by 002 for Bernstein inequality for multivariate polynomial

2010-01-19T16:37:20Z

Tung, S. H. Bernstein's theorem for the polydisc. Proc. Amer. Math. Soc. 85 (1982), no. 1, 73--76. MR0647901 (83h:32017)

(from MR review): Let $P(z)$ be a polynomial of degree $N$ in $z=(z_1,\cdots,z_m)$; suppose that $|P(z)|\leq 1$ for $z\in U^m$; then $\|DP(z)\|\leq N$ for $z\in U^m$ where $\|DP(z)\|^2=\sum_{i=1}^m|\partial P/\partial z_i|^2$.

Here $U^m$ is the polydisc. Same author proved Bernstein-type inequality for the ball, Tung, S. H. Extension of Bernšteĭn's theorem. Proc. Amer. Math. Soc. 83 (1981), no. 1, 103--106. MR0619992 (82k:32013)

Answer by 002 for Baire category theorem

2010-01-18T23:14:55Z

Maybe if you allow an exceptional set, as in "For every $x$ outside of some meagre set, there is a $\delta>0$ such that $B(x,\delta)$ is contained in one of the $\overline{A}_n$ "

(Indeed, the set of all $x$ where the above fails is a closed subset of $X$ with empty interior.)

Sets that can be mapped onto R^n by a polynomial

2010-01-15T21:29:42Z

The question was edited several times. Most recent version, suggested by Fedja:

Does there exist an open set $U\subset \mathbb R^n$ (n>1) that contains balls of arbitrarily large radius and such that no polynomial mapping $p\colon \mathbb R^n\to\mathbb R^n$ takes $U$ onto $\mathbb R^n$? (Take $n=2$ if you prefer.)

Older version (with bounty) was answered by Fedja in the affirmative:

Does there exist a topological ball $U\subset \mathbb R^n$ of infinite volume such that no polynomial mapping $p\colon \mathbb R^n\to\mathbb R^n$ takes $U$ onto $\mathbb R^n$?

Discussion:

Motivated by a recent question, I wonder if there is a geometric characterization of open sets $U\subset \mathbb R^n$ that can be mapped onto $\mathbb R^n$ by a polynomial $p$. Let $U$ be a topological ball to simplify matters. The following tail volume condition is necessary for the existence of such $p$.

(TVC) $\int_1^{\infty} r^{m} |U\setminus B(r)|=\infty$ for some $m>0$.

Indeed, the absolute value of the Jacobian of $p$ must have infinite integral over $U$. Since the Jacobian is a polynomial, $\int_U |x|^N dx=\infty$ for some $N$. The latter can be rephrased in terms of tail volume: $\int_1^{\infty} r^{N-1} |U\setminus B(r)| =\infty$ . The example of $U=\{(x,y)\in \mathbb R^2\colon x>1, 0<y<x^{-M}\}$ shows that (TVC) is somewhat sharp. This particular $U$ can be mapped onto $\mathbb R^2$ by $(x,y)\mapsto (x,x^{M+1}y)$ followed by translation and the power map $(x+iy)\mapsto (x+iy)^8$.

I am interested in other obstructions besides small tail volume, as well as in reasonably general sufficient conditions. Initially I hoped to get a stronger necessary condition from an affirmative answer to the question below, but Bjorn Poonen answered it in the negative.

"If $f\colon U\to\mathbb R^n$ is a polynomial surjection, does there exist $\epsilon>0$ such that $p(U\cap B(r))$ contains $B(r^\epsilon)$ for large $r$?"

Answer by 002 for Asymptotics of iterated polynomials

2010-01-13T23:04:21Z

Have you tried the method of section 8.7, i.e., solving Abel's equation $\phi(P(x))-\phi(x)=1$? Here we expect $\phi(t)=t^{-1}+\sum_{n=1}^{\infty}c_n t^n$ , and you can find the coefficients of $\phi$ one by one. For example, I took $P(x)=x-x^2+x^3+x^4$ and immediately found $c_1=-2$ , $c_2=-5/2$ , $c_3=-7/2$ , $c_4=-17/4$ ... Not a general formula, but you can get as many terms as you want for a given $P$.

Answer by 002 for Taylor series of a complex function that is not holomorphic

2010-01-12T17:49:15Z

Another option is $\sum c_{mn}z^m \bar z^n$, which still keeps track of the complex structure. For instance, harmonic functions will have $c_{mn}=0$ unless $mn=0$.

Notation for "the inclusion map is a homotopy equivalence"

2010-01-12T03:07:10Z

It's sometimes convenient to have different notations for "$A$ is a subset of $B$" depending on what the inclusion map does:

If it's non-surjective, $A\subsetneq B$ or $A\subset B$, depending on your religion
If it's surjective, $A=B$ :)
If the image is a precompact set, $A\Subset B$

Does there exist notation to indicate that the inclusion $A\hookrightarrow B$ is a homotopy equivalence? I'd like to use something similar to 1-3.

Answer by 002 for Compact Convex sets and Extreme Points

2010-01-05T07:38:48Z

By definition, a non-extreme boundary point lies on an open line segment contained in the set, which happens to be an open subset of the boundary in two dimensions. Hence the set of extreme points is a closed subset of the boundary.

Answer by 002 for Lebesgue measure of boundary of Caccioppoli set

2010-01-05T02:10:35Z

The answer is no. Take countably many disjoint closed balls $B_i$ contained in the square $Q=[0,1]\times [0,1]$ and such that:
(i) Sum of areas of $B_i$ is less than 1
(ii) Sum of perimeters of $B_i$ is finite
(iii) $\bigcup B_i$ is dense in $Q$
Since the series $\sum \chi_{B_i}$ converges in BV norm, the set $E=Q\setminus \bigcup B_i$ has finite perimeter. It also has positive measure and empty interior. Any representative $F$ of the set $E$ also has empty interior and therefore $\partial F$ is not Lebesgue null.

By the way, any Lebesgue measurable set E has a representative F with the property

(*) $0<|F\cap B(x,r)|<|B(x,r)|$ for all $x\in\partial F$ and all $r>0$.

The proof is straightforward: add the points x for which $|E\cap B(x,r)|=|B(x,r)|$ for some r, and throw out all points x such that $|E\cap B(x,r)|=0$ for some $r$. (See Prop. 3.1 in "Minimal surfaces and functions of bounded variation" by E. Giusti.) By virtue of (*) the set $F$ has the smallest (w.r.t inclusion) topological boundary among all representatives of $E$, so if this representative doesn't help you, nothing does.

Answer by 002 for Number of uniform rvs needed to cross a threshold

2010-01-03T07:24:36Z

Another way to put it: the expected value of the sum right after it crosses x is x+1/3. If you simply conditioned the sum to be in (x,x+1) then the expectation would be about x+1/2, with the sum almost uniformly distributed. But you also condition on the overshoot being less than the last jump. The expectation of the smaller of two uniform [0,1] iids is 1/3.

Answer by 002 for The orthodrome of n-spheres.

2010-01-01T03:04:38Z

The two-dimensional formula applies (why?): the great-circle distance is $\cos^{-1}(\vec u\cdot \vec v)$ where $\vec u$ and $\vec v$ are position vectors of the points.

Comment by 002

2010-04-14T18:03:37Z

This is a question, not an answer, and should be posted as such (with a link to Anton's question).

Comment by 002

2010-04-07T17:25:57Z

And then I deleted the clarification and put the proof together.

Comment by 002

2010-04-07T03:21:21Z

@miwalin I added a clarification.

Comment by 002

2010-04-07T01:58:49Z

@timur: integration over $k$ appears in the statement of the problem.

Comment by 002

2010-03-02T02:14:38Z

I just thought it would be easier to deal with the sum than the product. You are right about the singularity; for small k one should use another estimate such as $|\hat f(k)|\le 1$.

Comment by 002

2010-02-15T20:55:26Z

-1 to bring the total of votes to 0 and return the problem to the "Unanswered" category where it belongs.

Comment by 002

2010-02-13T23:13:09Z

For independent RVs it's true because correlation is zero.

Comment by 002

2010-02-12T16:03:23Z

Another fix is to use `$\Omega=\{|z|^2<\mathrm{Re} w\}$' (Siegel half-space) which is strictly pseudoconvex and is therefore the domain of holomorphy for some function $g$. Then add square roots to $g$ as above.

Comment by 002

2010-02-08T22:52:51Z

Something like this? functions.wolfram.com/…

Comment by 002

2010-02-08T02:17:26Z

That's not my understanding either. I meant the following: if $(x_1,\dots, x_n)$ is a permutation of an arithmetic progression, then the rank of $x_i$ is a linear (or rather affine) function of $x_i$.

Comment by 002

2010-02-08T00:08:19Z

Actually, it seems that $\rho=r$ for uniform random variables. Indeed, $\rho$ is by definition $r$ for the ranks, and for uniform distributions rank is a linear function of the variable itself.

Comment by 002

2010-02-05T17:52:08Z

@Gerald: Radius of convergence greater than 1 would work (assuming f is analytic. For a sequence, uniform convergence on a disk of such radius is enough.

Comment by 002

2010-02-05T16:18:22Z

No, uniform convergence of coefficients does not imply convergence of derivatives even when the functions themselves converge uniformly. E.g. $n^{-1/2}x^n$ on [-1,1].

Comment by 002

2010-02-05T00:33:57Z

The origins of this problem may have something to do with a survey article by T.H. MacGregor (1972), where the following is proved. Let A=f(D) where D is the unit disk and f:D->A is a holomorphic function with |f'(0)|=1. Then A-A contains the disk |z|<\pi/2.

Comment by 002

2010-02-01T20:27:02Z

@ Georges: then the answer is yes. Since X is a one-dimensional manifold, it's either R or S^1. It can't be S^1 because then the image would be compact. Any locally invertible map R->R is globally invertible.