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Answer by coudy for Characterizing convex polynomials

2010-06-21T20:04:41Z

p is convex iff p'' is non-negative. And a polynomial is non-negative iff it is the modulus of the square of a polynomial with complex coefficients. So p must be of even degree (or of degree 0 or 1).

If we write $p''=|q|^2$, then we see that p is convex iff there exist complex coefficients $q_0...q_l$, $l \leq n/2$, such that, for all $k$, $$k(k-1)\ a_k=\ \ \Sigma_{i,j \ \ s.t.\atop i+j=k-2} \quad q_i \bar{q}_j$$

This is not very explicit, but at least this may be used to generate convex polynomials.

Answer by coudy for Evil Fourier Coefficients

2010-11-05T21:29:28Z

The Fourier transform of the derivative $\mu$ of the Devil staircase is explicitely stated on the wikipedia page of the Cantor distribution, in the table at the right, under the heading "cf" (characteristic function). Its value is

$$ \int_0^1 e^{itx} d\mu(x) = e^{it/2}\ \ \prod_{k=1}^\infty \cos(t/3^k)$$

Just multiply by $-1/it$, add $1/it$, and you get the Fourier transform of the Devil staircase.

A word on the proof. The Cantor distribution is the weak limit of the functions obtained by summing the indicator functions of the 2^n intervals generating the Cantor set at the nth step (after renormalization). The Fourier transform of these sums can be computed explicitely. Then let n goes to infinity.

What is the oldest open problem in mathematics ?

2010-06-04T18:30:16Z

What is the oldest open problem in mathematics ? By old, I am refering to the date the problem was stated.

Browsing wikipedia list of open problems, it seems that the Goldbach conjecture (1742, every even integer greater than 2 is the sum of two primes) is a good candidate.

The Kepler conjecture about sphere packing is from 1611 but I think this is finally solved (anybody confirms ?). There may still be some open problem stated at that time on the same subject, that is not solved. Also there are problems about cuboids that Euler may have stated and are not yet solved, but I am not sure about that.

A related question: can we say that we have solved all problems handed down by the mathematicians from Antiquity ?

Examples of $G_\delta$ sets

2011-03-04T11:23:49Z

Recall that a subset A of a metric space X is a $G_\delta$ subset if it can be written as a countable intersection of open sets. This notion is related to the Baire category theorem. Here are three examples of interesting sets that are $G_\delta$ sets.

The set of continuity points of a function $f:X\rightarrow R$.
The set of positively recurrent points of a continuous transformation $T: X \rightarrow X$. Recall that a point is recurrent if there exists some sequence $n_i\rightarrow \infty$ such that $T^{n_i}x \rightarrow x$.
The set of transitive points of a continuous transformation $T: X \rightarrow X$. Recall that a point is transitive if its orbit $\{T^n(x)\}_{n\in Z}$ is dense in $X$.

Question : Can you provide more examples of interesting $G_\delta$ sets ?

Answer by coudy for Terminology: "cocompact"

2011-02-17T19:21:06Z

1) from wikipedia "In mathematics, an action of a group G on a topological space X is cocompact if the quotient space X/G is a compact space."

2) Let $\bar{M}$ be a noncompact manifold, $\Gamma$ its fundamental group, and $M$ its universal cover. Given a riemannian metric on $\bar{M}$, we can lift the metric to $M$ and the group $\Gamma$ acts as a group of isometry on $M$. For a generic metric on $\bar{M}$, there is no other isometries on $\bar{M}$ than the one provided by $\Gamma$. In which case $G/\Gamma$ is trivial (hence compact) whereas $M/\Gamma$ is isometric to $\bar{M}$, hence noncompact. As an example, take an euclidean cylinder, give a good (asymetrical) kick in it, unroll it and you are done.

Commenting on Scatt Carnahan answer, the example of the hyperbolic upper half plane is misleading, because its isometry group is transitive. This is exceptional in the world of riemannian geometry. Just take some non compact quotient of the upper half plane, say $H/\Gamma(2)$, make a small bump on the quotient, lift the resulting metric, et voila you have killed most isometries on $H$ with respect to this new metric.

Answer by coudy for Superfluous definitions

2011-02-13T21:10:38Z

A $\sigma$ algebra of subsets of a set X is defined as a collection of subsets of X which is invariant by taking complements and denumerable unions. And which contains the empty set.

But this last condition is (almost) superfluous. If there exists an element, say $A$, in the $\sigma$-algebra, then it must contain $(A\cup A^c)^c$, which is the empty set. Hence the sole purpose of requiring the empty set to be in the $\sigma$-algebra, is to deny the empty set the right to be a $\sigma$-algebra itself.

I don't really know why the empty set should not be a $\sigma$-algebra, and I don't see any result that would suddenly fail badly if we give the empty set this promotion.

EDIT: Bourbaki, Topology, ch5, section 6 no 3 (TG IX.60) does not require the empty set to be in the $\sigma$-algebra. So I think this is really superfluous.

Answer by coudy for Can a non-Borel set be a standard Borel space?

2010-11-07T15:32:38Z

Strictly speaking, a standard Borel space can be finite or countable. Keeping in mind this minor point, A subset of $[0,1]$ (endowed with the restriction of the Borel $\sigma$-algebra) is a standard Borel space if and only if it is a Borel subset of $[0,1]$.

This comes from the fact that the image of a Borel subset by an injective Borel map between two standard Borel spaces is a Borel set. This result is proven for example in Dudley "Real analysis and Probability" or Cohn "measure theory".

Now for the proof. Let $\phi$ be the isomorphism between $([0,1],{\cal B})$ and $(E,{\cal M})$. The space E is a subset of $[0,1]$, so we get a Borel injection from $[0,1]$ to $[0,1]$ whose image is precisely $E$. Hence $E$ is a Borel subset of $[0,1]$.

Answer by coudy for Decomposition of a dynamical system into ergodic componenents

2010-11-04T11:09:43Z

Answer to the quick version. Yes it is true as soon as $(X,\mu)$ is a Lebesgue space. Beware that the transformation on the product $A_i\times B_i$ is not necessarily a true product, but instead it is a skew-product of the form $(x,y)\mapsto (T_x(y),y)$. This follows from the ergodic decomposition theorem, together with the classification of measurable partitions.

Recall that if T is an invertible measurable transformation acting on a Lebesgue space X, then there is a measurable partition $C_i$ (which may have uncountably many elements) and probability measures $\mu_i$ on $C_i$ such that all $C_i$ are invariant by T, T is ergodic w.r.t $\mu_i$ and $\mu$ is obtained by integrating the $\mu_i$. $$\mu(A) = \int_X \mu_i(A) d\mu$$ There are two kinds of ergodic components $C_i$. The one of positive measure, there are at most countably many such components. Let us remove these components from $X$, together with the periodic points, which are easily dealt with. Rohlin structure theorem on measurable partitions (1947) now says that there is a isomorphism between $([0,1]\times [0,1], \lambda)$ and $(X,\mu)$ such that the pullback of the measurable partition $(C_i)$ is the decomposition into horizontal lines $([0,1]\times \{i\})_{i\in [0,1]}$. A reference is the book of Parry, "entropy generators in ergodic theory".

Here is how the ergodic decomposition is often used. If it happens that a result is true for an ergodic transform, then it is true for an arbitrary transform in restriction to its ergodic components, and you (may) recover the result on the whole space $X$ just by using the integral formula given above. A reference for the ergodic decomposition for countable groups action is Glasner, "ergodic theory via joinings" th. 3.22.

Finally the result you are alluding in your last question is a section theorem. Given a measure preserving transform between two Lebesgue spaces X and Y, there is a section from Y to X, up to null sets, and some warning is in order here because this is not true in the Borel category. I think this is again due to Rohlin, and it can be deduced from its structure theorem for measurable partitions. Have a look at the book of Parry, but really this is overkill.

EDIT: following the comments of R.W., here is a counterexample to having a true product, instead of just a skew-product. On $[0,1]\times [0,1]$ take $(x,y)\mapsto (x+y\ \ mod\ \ 1,y)$, together with Lebesgue measure. The restrictions to the fibers $[0,1]\times \{y\}$ are ergodic for a.e. y, and give uncountably many different isomorphic systems, as can be shown by looking at their spectra.

Answer by coudy for Ergodic limits along subsets of $\mathbb{N}.$

2010-10-24T18:49:00Z

These are called good universal sets. Bourgain (1987) proved that sequences of the form $p(n)$, $n \in {\bf N}$, $p$ a non constant polynomial, are good.

He also proved (1988) that the set of primes is a good universal set for $L^p$ functions, $p> {(1+\sqrt{3})\over 2}$. This was later improved to $p>1$ by Wierdl, see its article for a short introduction to the problem http://www.springerlink.com/content/e027w4211n7784h1/fulltext.pdf . There is now an extensive litterature on the problem, following Bourgain's articles.

In another direction, note that a transformation T is mixing iff the ergodic theorem for T holds for all subsequences (see e.g. the book of Krengel, "ergodic theorems").

Answer by coudy for Group cannot be the union of conjugates

2010-08-05T08:30:49Z

(Let me drop the finite index requirement as in the other answers)

This remains true for discrete virtually solvable groups. Indeed the property of having no proper subgroup containing a conjugate of every element in the group is stable by extension.

I let you check that if the property holds for G_1 and G_2, and G is such that there is a sequence 0->G_1->G->G_2->0, then the property also holds for G.

Note that this property is not stable by direct limit. If K is the algebraic closure of a finite field, SL_n(K) is the direct limit of finite groups, yet the property is not verified. This gives an example of an amenable group for which the property does not hold.

It is not yet known if there are amenable groups of finite type which do not satisfy the property.

On the other hand, the property does not hold for non-abelian free groups, or, in the realm of uncountable groups, for connected semi-simple complex lie groups that are not solvable.

Answer by coudy for Let a function f have all moments zero. What conditions force f to be identically zero?

2010-07-10T14:17:56Z

The answer to the first question is no. The following example is standard in probability theory, see e.g. Billingsley "probability and measure", example 30.2.

$$f(x) = {1\over \sqrt{2\pi}\ x}\ e^{-{(\ln x)^2\over 2}}\ \sin(2\pi \ln x) \ {\bf 1}_{[0,\infty[}(x)$$

You can check that all the moments are zero using the change of variable $\ln x = s+k$.

This example is used to show that a probability measure is not always determined by its moments.

Answer by coudy for Kalinin's formulation of the Anosov closing lemma

2010-07-07T20:52:23Z

The closing lemma as stated by Kalinin can be found in many textbooks e.g. Katok-Hasselblatt "Introduction to the modern theory of dynamical systems", corollary 6.4.17.

The closing lemma really gives a periodic point close to x, with iterates also close to the iterates of x until the orbit of x returns. That's not just the fact that periodic points are close to non-wandering points.

The point y is obtained by taking the intersection of the local stable set of x with the nth pull-back of the local unstable set of $f^n(p)$. Draw a picture to understand what's going on. There are "geometric" proofs of the closing lemma that build y before p. And of course the original article of Livsic contains such a proof.

Minimal elements of minimal R^k actions

2010-06-22T08:39:02Z

C. Pugh and M. Shub showed in 1971 that, given an ergodic action of $G=\mathbb{R}^k$ on some separable finite measure space $(X,\mu)$, then all elements of $G$ , off a countable family of hyperplanes, are ergodic.

Is there an analogous statement in the topological setting, with ergodicity replaced by minimality (i.e. all orbits are dense) and $X$ assumed to be compact ?

Answer by coudy for Minimal elements of minimal R^k actions

2010-07-07T19:38:45Z

A colleague pointed out the following counterexample. Let $h_t$ be the horocyclic flow on a negatively curved compact surface S. This R action is known to be minimal. Now Consider the $R^2$ action on $S\times S$ given by $(s,t)\rightarrow (h_s,h_t)$. This action is again minimal.

The action of the diagonal $\{(s,s), s\in R\}$ is not minimal since the orbit of any point (x,x) stays in the diagonal.

Let $\theta\in R$. The action of the line $\{(s,\theta s), s\in R\}$ is not minimal because it is conjugated to the diagonal action. This comes from the fact that the two actions $h_{\theta s}$ and $h_s$ are conjugated by the geodesic flow.

As a result, there are no elements in $R^2$ acting minimally, although $R^2$ itself acts minimally.

Answer by coudy for Fixed point theorems and equiangular lines

2010-07-07T18:28:45Z

The book "Fixed point theory" by Dugundji and Granas is a nice reference. The headers of the sections in the book give some kind of classification of fixed point theorems.

results based on compactness
order theoretic results
results based on convexity
Borsuk theorem and topological transitivity
homology and fixed points
Leray-Shauder degree and fixed point index

Part VI of the bibliography is really extensive and contains a finer classification of fixed point theorems.

Answer by coudy for Demystifying complex numbers

2010-07-01T12:18:54Z

You can solve the differential equation y''+y=0 using complex numbers. Just write $$(\partial^2 + 1) y = (\partial +i)(\partial -i) y$$ and you are now dealing with two order one differential equations that are easily solved $$(\partial +i) z =0,\qquad (\partial -i)y =z$$ The multivariate case is a bit harder and uses quaternions or Clifford algebras. This was done by Dirac for the Schrodinger equation ($-\Delta \psi = i\partial_t \psi$), and that led him to the prediction of the existence of antiparticles (and to the Nobel prize).

Answer by coudy for Is there a central limit theorem for bounded non identically distributed random variables ?

2010-06-30T08:58:07Z

Theorem (Billingsley, "probability and measure", example 27.4)

Let X_i a sequence of independent, uniformly bounded random variables with zero mean, such that $\sigma(S_n)$ goes to infinity with n. Then $S_n/\sigma(S_n)$ converges in law to the normalized Laplace-Gauss distribution.

This follows from the Lindeberg triangular array theorem. As pointed out in the other answers, the convergence can be slow. The Bernstein inequality may be used to bound the tail. Under the assumption of the previous theorem, for all n, we have

$$P(S_n>t) \leq exp(-t^2/(\sigma^2(S_n)+Ct/3))$$

where C is a bound for the |X_i|'s.

a.e. convergence of the powers of an operator built from rotations

2010-05-17T20:18:38Z

Consider two numbers $a,b\in R/Z$ and some integer $p\geq 1$. Let $T:L^p(R/Z)\rightarrow L^p(R/Z)$ be the operator given by $$T(f)(x)=1/2(f(x+a)+f(x+b))$$ For which values of $a,b$ do we have almost everywhere convergence of the sequence $T^nf$ for all $f \in L^p$ ?

If $a-b$ is rational, it's not difficult to show that a.e. convergence fails. If $a=-b$ and $p>1$, then a.e. convergence follows from Stein (1961), Rota (1962). Note that if $a-b$ is not rational, then $T$ is ergodic and the mean $1/n\ \Sigma_k^n\ T^kf$ converges almost everywhere to a constant. So maybe this is the right condition (together with p>1 ?).

Answer by coudy for Convergence of orthogonal polynomial expansions

2010-06-24T15:21:58Z

Let me recall a quick $L^2$ proof of the uniform convergence of the Fourier series of a $C^1$ function $f$. Let $c_n$ be its Fourier coefficients. Then

$$|f(x)| \leq |c_0|+ \Sigma|c_n|\ n \ {1\over n}\ \ \leq |c_0| + \ \ \sqrt{\Sigma \ n^2 |c_n|^2}\ \ \sqrt{\Sigma\ 1/n^2}$$

Replacing f by f minus its partial sum, and noting that $\Sigma \ n^2|c_n|^2 = ||f'||_2^2 \ $ is finite, you get uniform convergence.

So maybe you can use a similar computation in case of a family of orthogonal polynomials ?

Answer by coudy for Sheaf cohomology question

2010-06-24T15:01:17Z

Sheaves of continuous or smooth functions with values in a vector bundle have zero cohomology groups because of the existence of partitions of unity.

Still they provide interesting resolutions for the locally constant sheaf, e.g. by considering the bundle built from exterior algebras on the tangent space, and that's a way to prove that De Rham cohomology coincides with singular or Cech cohomology for paracompact manifolds.

Of course, we can look at the De Rham complex obtained by tensoring the exterior forms by an arbitrary bundle on the manifold. In which case we obtain an interesting cohomology with "values" in a vector bundle instead of a coefficient ring. This is similar to singular cohomology with twisted coefficient. And I think that's the kind of examples that gave rise to the concept of sheaf.

Answer by coudy for Extension of harmonic function at infinity

2010-06-22T18:02:14Z

Just take (x,y)->y. This is harmonic, but unbounded at infinity.

Yet any positive harmonic function h on the upper half plane can be obtained by integrating the Poisson kernel with respect to some finite measure on the boundary of the upper half plane (Riesz-Herglotz representation theorem). So one may say that there is an extension of h to the boundary in that case, and this extension is a finite positive measure.

Answer by coudy for Counterexamples in Algebra?

2010-06-22T08:00:29Z

Grigorchuk 1984 example of a finitely generated group with intermediate growth (there are no such linear group).

Answer by coudy for Does anyone know an intuitive proof of the Birkhoff ergodic theorem?

2010-06-22T07:55:25Z

I know of six proofs of the Birkhoff ergodic theorem.

using a maximal inequality (Birkhoff, Riesz, Wiener, Yosida, Kakutani, Garsia...)
based on martingales and upcrossing inequalities (Bishop 1966)
using non-standard analysis (Kamae 1982)
based on variational inequalities (Bourgain 1988)
using a filling scheme (Chacon)
Katznelson-Weiss proof (1982) and derivatives (Keane, Petersen)

The last one seems the most easy to remember for me. This may look like a combinatorial trick, but with some thought, it appears quite natural, and I know several results that use some similar idea.

So let $\epsilon>0$ and x in X. We can find an $n(x)$ depending on x such that $$ \overline{\lim}\ {1\over n}\ \Sigma_0^{n-1}\ f(T^k(x)) \leq \lim {1\over n(x)} \sum_{k=0}^{n_(x)-1} f(T^k(x))\ +\ \varepsilon$$ Note that n(x) is finite everywhere, hence bounded on a set R with complement of arbitrary small measure. Then cut the Birkhoff sum according to the sequence $n_{i+1}(x)=n_i(x)+n(T^{n_i}(x))$ if $T^{n_i}(x)$ is in R, $n_{i+1}(x)=n_i(x)+1$ otherwise. A picture should make clear what is going on. The rest of the proof is routine check.

Of course if you are in the business of non-standard analysis, Kamae's proof is both short and enlightening but then you need some work to get the standard statement.

Answer by coudy for Closed form of divergent infinite product?

2010-06-21T20:15:09Z

I would suggest the development of the Gamma function

$$1/\Gamma(z) = z e^{\gamma z}\ \Pi_{n=1}^\infty\ (1+{z\over n})\ e^{-{z\over n}}$$

Answer by coudy for Learning Roadmap for Software Engineer

2010-06-21T12:26:03Z

Read Claude Shannon's article "a mathematical theory of communication".

Answer by coudy for Haar Measure Existence/A problem with Borel sets and regularity.

2010-06-20T20:13:08Z

The answer to your first question is no. A not very interesting counterexample is given by the interval [0,1] together with the discrete topology. All sets are open, hence the Borel $\sigma$-algebra contains all the subsets of [0,1]. On the other hand, compact sets have a finite number of elements. So the $\sigma$-algebra they generate contains only sets that are countable, or with countable complements.

If you assume that your space is metric separable, then it's ok, as explained by F.Dorais in his answer. Let me add a few words with respect to regularity. First recall that a Polish space is a topological space which is homeomorphic to a complete separable metric space. Locally compact separable metric spaces are Polish. We have the following result due to Oxtoby and Ulam.

Theorem let X be a Polish space and $\mu$ a finite Borel measure on X. Then $\mu$ is inner regular: all Borel set can be written as a countable union of compact sets together with a set of zero measure.

As a result, a finite measure on a Polish space is uniquely determined by its values on compact sets.

The book by Halmos is very good but a bit outdated in my opinion. I would suggest instead the book of Donald L. Cohn, "measure theory", who deals, among other things, with the Haar measure.

Answer by coudy for nontrivial theorems with trivial proofs

2010-06-20T15:16:27Z

Here is what Grothendieck says in Recoltes et semailles.

Dans le cas cohérent, la démonstration du théorème de bidualité est d’ailleurs triviale. Cela n’empêche que c’est ce que j’appelle sans hésitation un théorème profond”, car il donne une vision simple et profonde de choses qui ne sont pas comprises sans lui. (Voir à ce sujet l’observation de J. H. C. Whitehead sur “le snobisme des jeunes, qui croient qu’un théorème est trivial, parce que sa démonstration est triviale”, observation que je reprends et sur laquelle je brode dans la note “Le snobisme des jeunes — ou les défenseurs de la pureté”, n◦27).

May be someone who has the english version can provide a translation. This is note 947, page 763 in the french pdf version, a search for the word biduality should find it quickly.

So, in short, the biduality theorem is a profound theorem with a trivial proof in the coherent case. And the quote you are refering to is probably due to Whitehead.

Answer by coudy for Convergence of Fourier Series of $L^1$ Functions

2010-06-16T23:31:06Z

The answer to your first question is no. There is an L^1 function with Fourier series not converging in measure.

In the Kolmogorov example of an L^1 function $f$ with a.e. divergent Fourier series, there is in fact a set of positive measure E and a subsequence n_k such that for all x in E, the absolute values of the partial sums S_{n_k} of the Fourier series goes to infinity with k.

$$\forall x\in E,\ \ |S_{n_k}f(x)|\rightarrow \infty$$

This can be checked from the construction of f in the original article of Kolmogorov, in its selected works.

If $S_nf$ converges in measure, then $S_{n_k}f$ must also converges in measure. This implies that there is a subsequence $n_{k_l}$ such that $S_{n_{k_l}}f(x)$ converges a.e. x, a contradiction.

Answer by coudy for Why do we care about L^p spaces besides p = 1, p = 2, and p = infinity?

2010-06-15T05:54:14Z

Here is an example from probability theory. Let $X_i$ be a sequence of independent identically distributed random variables in $L^1$. The strong law of large numbers asserts that the mean converges to the expectation.

$$a.e. \quad {1\over n}\ \sum_{k=0}^{n-1} X_k \rightarrow E(X_0).$$

What can be said about the speed of convergence ? If we assume that the $X_i$ are in $L^p$ for some $p\in ]1,2[$, then we have:

$$a.e. \quad {1\over n}\ \sum_{k=0}^{n-1} X_k = E(X_0) + o(n^{1/p-1}).$$

Convergence of the sum of squares of averages of a sequence whose sum of squares is convergent

2010-06-14T23:09:33Z

Can we find a sequence $u_n$ of positive real numbers such that $\sum_{n=1}^\infty u_n^2$ is finite, yet $\sum_{n=1}^\infty ({u_1+u_2+...+u_n\over n})^2$ is infinite ?

After several attempts, I think this is not possible, but I can't prove that the finiteness of the first sum implies the finiteness of the second sum.

Comment by

2012-09-22T18:53:49Z

indeed, there was a typo in my answer. This is now corrected, thanks.

Comment by

2010-11-09T21:15:50Z

Indeed, this is not stated in Dudley, sorry for pointing in the wrong direction. This is done in Cohn, and also in Kechris "classical descriptive set theory", thm 15.2 p89, online on google books, books.google.com/… Hope that helps

Comment by

2010-11-06T18:08:32Z

I edited my answer, following RW remarks. Hope this helps.

Comment by

2010-11-06T15:32:15Z

@R W. The quick question asks for a product action, without giving a precise meaning to the term "product". The technical term is "skew-product" here, the action on the ith level C_i depends on i. So the answer is yes if the term "product" is understood that way. Does it answer your question ?

Comment by

2010-11-06T13:56:46Z

Yes, I forgot to remove the periodic points, but I don't think this is a problem here. The action on the set of periodic points of period n is isomorphic to {1,..n}xB, where the action on B is just identity. So the result also holds for the set of periodic points.

Comment by

2010-11-06T13:20:01Z

May I point out that the question is about probability measure preserving actions (pmp) ?

Comment by

2010-11-04T14:42:28Z

There are a lot of books that state the ergodic decomposition theorem for countable group actions, but few actually prove it. A recent reference is Glasner, "Ergodic theory via joinings". The original articles of Rohlin ("On the fundamental ideas of measure theory", 1952 english translation) and Varadarajan (1963, jstor.org/stable/1993903) are well written.

Comment by

2010-08-07T14:18:24Z

@Pak. Can we deduce the continuous version of the Hairy ball theorem from the discrete version, by some limiting process ?

Comment by

2010-08-05T08:40:10Z

@gomaff. How about letting A be the complement of the geodesic together with all its images by the tiling group ?

Comment by

2010-07-10T15:14:46Z

@Zen. To see how your problem connects to the problem of moments in probability theory, just write f as $f_+-f_-$. Now saying that f has zero moments amounts to saying that the two measures $f_+dx$ and $f_ dx$ have the same moments. Does that imply that the two measures are equal ? The link provided by Mariano may lead you to the [[Carleman's condition][en.wikipedia.org/wiki/Carleman%27s_condition]], which traduces back into a condition on the decreasing rate of the moments of $|f|$.

Comment by

2010-07-09T07:42:45Z

Yes, the action of the horocyclic flow is ergodic and mixing of all order. This implies that the diagonal action $(h_s,h_s)$ is also ergodic on SxS. So there are points $(x,y)$ with dense orbit under the diagonal action. But of course this does not imply that all orbits are dense.

Comment by

2010-06-24T20:58:41Z

Am I missing something ? Shouldn't the condition be Hf in L^2 ? With $f_n=1/(n \log n)$, I don't see how to get the uniform convergence of $\sum f_n \psi_n$ from the finiteness of <f,Hf>.

Comment by

2010-06-24T18:07:46Z

There is a chapter (9.1 ff p368) in the book of G. Sansone "Orthogonal functions" about uniform convergence of series of Hermite polynomials. There is a preview of the book here books.google.com/….

Comment by

2010-06-23T18:17:15Z

@Helge. An invariant measure gives a unitary representation of G on $L^2$. Here is a proof of the Pugh-Shub result for $k=1$. Let f be a $g_{t_0}$ invariant function for some $t_0$. Then $F(x) = \int_{0}^{t_0}\ f(g_s(x))\ e^{-2\pi i s/ t_0}\ ds$ is an eigenvector for the flow $g_t$, associated to the eigenvalue $e^{2\pi i/t_0}$. Eigenvectors associated to different eigenvalues are orthogonal. The conclusion follows, assuming $L^2$ is separable.

Comment by

2010-06-23T08:03:27Z

@Dmitri. Yes, sorry, I forgot the compactness assumption.