User vkrouglov - MathOverflow

Solution to the fractional differential equation

2013-03-26T07:54:58Z

What is the solution of the fractional differential equation $$ f^{(\alpha-1)}(t) = tf(t) $$

where $(\alpha)$ denotes the fractional derivative of order $\alpha$

EDIT: Background behind this question.

I am interested in this equation in relation with the alpha-stable version of the Stein's lemma. Recall, that if $X \in N(0,1)$ then $$ E(X g(X)) = E(g'(X)) $$ for every function $g$ for which the expectations in the left and right parts exist.

The simplest way to prove this is to use the property of normal density $f'(x) = -x f(x)$ and integration by parts.

Let $\phi$ be a characteristic function of the standard symmetric stable density $S(\alpha, 1, 0)$. If I apply the usual derivative to $\phi$ and take a Fourier transform I will get the equation above (modulo signs and coefficients). A closed form solution would give an analytic presentation of the stable density (highly unlikely).

2nd question: Is it possible to obtain a differential equation for the stable density to be used in the proof of the Stein's lemma?

Thanks, Vladimir

One-sample test for the alpha-stable distribution

2013-01-20T21:03:49Z

Hi,

This is a repost of the question from http://math.stackexchange.com/questions/281920/one-sample-prametric-test-for-stable-distributions/281929#281929.

For my needs the K-S test appears to unapplicable due to a small number of observations. Maybe there are some alpha-stable specific statistics that can have better confidence interval

Thanks,

What is the maximal sparsity of a matrix?

2013-01-15T17:08:49Z

Given a $m \times 2m$ matrix G with rank(G)=m, what is the maximal number of zeros in the product $WG$ where $W$ is an $m \times m$ nondegenerate matrix?

An obvious lower bound is $m(m-1)$ and $W$ is given by a matrix that transforms the first $m$ columns of $G$ to a standard basis in $\mathbb{R}^m$. When $m=1$ this lower bound is in fact attained. Is it possible to have more zeros when m>>1?

When the adjoint of a hypoelliptic operator hypoelliptic

2012-01-13T14:00:39Z

Assume, $M$ is a smooth manifold with a measure $\mu$ and let $L^2(M, \mu)$ be a space of all square-integrable functions on $M$.

Recall that $L$ is a hypoelliptic differential operator, if for every $f \in \mathcal{D}(L)$, if $Lf$ is in $C^\infty(M)$ then $f$ is also in $C^\infty(M)$.

Could anyone give a reference to the example when $L$ is hypoelliptic but its adjoint w.r.t to $\mu$ is not hypoelliptic? Could this happen to the Hormander operators, when $L$ is defined as $$ L = \sum_i X_i^2 + X_0 $$ and ${X_i}$'s are bracket generating?

I am mostly interested in the case when $\mu$ is induced by the Riemannian metric and $M$ is complete.

Thanks,

Reference Request: Hamiltonian and quantum completeness.

2012-01-23T20:48:08Z

Let $L$ be a differential operator in $L^2(M, dvol)$ wrt to a Riemannian volume form (say). Let us call it quantum complete if it is essentially-self-adjoint. Consider $H$ - the symbol of $L$. It is a function on a cotangent bundle of $M$, i.e. a Hamiltonian. This Hamiltonian defines a Hamiltonian flow on $T^\ast M$.

I am looking for a good reference which discusses the relation between the quantum completeness of $L$ and the completeness of the Hamiltonian flow of $H$. Some information is contained in Reed and Simon but they seem to consider only the case of $M=(0, \infty)$.

Particular questions I am interested in:

1) Sometimes the Hamiltonian flow blows up in finite time but the corresponding $L$ is complete. What is the physical meaning of this?

2) Is it possible to use the completeness of the Hamiltonian flow to prove that $L$ is ess. self.adjoint? For example, is it possible to apply this logic to prove that the Laplace-Beltrami operator is ess. self-adjoint on a geodesically complete manifold (which is a Hamiltonian flow of its symbol)?

Thanks,

Invariant complement to invariant subspace.

2012-01-15T10:14:00Z

Let $G$ be a compact group and $\rho: G \to End(U)$ its linear representation in a finite dimensional vector space $U$. Fix $V \subset U$ - a subspace invariant under $\rho(G)$. Then it is well known that $V$ has some invairant complement.

What are the sufficient conditions (on $G$, $\rho$ or $V$) to ensure this complement is unique?

Stated another way, starting with any scalar product on $U$, invariant complements to $V$ can be found by averaging it w.r.t. to a Haar measure on $G$ and taking $V^\perp$. In this case my question becomes

What are the sufficient conditions that for any choice of the initial scalar product, the resulting $V^\perp$ is unique.

There are examples when the complement is unique. For example, consider $\mathbb{R}^3$ and the action of $SO(2)$ given by the rotation around the $z$-axis. Then if $V$ is a $z$-axis it has a unique complement.

Particular example I am trying to understand is the following: consider the group $U(n)$ acting on $\mathbb{R}^{2n}$ in a standard way. This action induces a representation $\rho$ in $U=Hom(\mathbb{R}^{2n} \wedge \mathbb{R}^{2n}, \mathbb{R}^{2n})$ given by $$ (g S) (u\wedge v) = g^{-1}S (gu \wedge gv) $$ Consider a map $A: Hom(\mathbb{R}^{2n}, \mathfrak{u}(n)) \to Hom(\mathbb{R}^{2n} \wedge \mathbb{R}^{2n}, \mathbb{R}^{2n})$ defined by $$ (AS)(u \wedge v) = S(u)v - S(v)u $$ and define $V = A(Hom(\mathbb{R}^{2n}, \mathfrak{u}(n)))$.

For what $n$, there is a unique invariant complement to $V$?

Such construction appears in the Cartan's method of equivalence. Each $U(n)$-invariant complement to $V$ corresponds to a certain $U(n)$-invariant linear connection.

Thanks,

Comment by vkrouglov

2012-01-16T08:19:14Z

Thanks Gil, your help is really appreciated.