User andrey rekalo - MathOverflow

Answer by Andrey Rekalo for What is the difference between hard and soft analysis?

2010-11-20T02:33:41Z

A historical note.

Hermann Weyl mentioned in his talk "Felix Kleins Stellung in der mathematischen Gegenwart" that the dichotomy of "hard vs. soft analysis" had been suggested by Hardy. According to Hardy, there is the function theory of the "hard, sharp, narrow" kind (due to Bohr, Landau or Littlewood) as opposed to the "soft, large, vague" kind (due to Birkhoff or Koebe).

Edit. Apparently, Hardy's musings are contained in his paper "Prolegomena To a Chapter on Inequalities" (unfortunately, I don't have access to it at the moment).

Edit 2. Indeed, here's the quotation from Hardy's paper.

A thorough mastery of elementary inequalities is to-day one of the first necessary qualifications for research in the theory of functions; at any rate, in function theory of the "hard, sharp, narrow" kind as opposed to the "soft, large, vague" kind (I do not use any of these adjectives as words either of praise or blame), the function-theory of Bohr, Landau, or Littlewood, as opposed to the function-theory of Birkhoff or Koebe. It is essential to anyone working in this field to be master both of the main results and of the tricks of the trade.

Answer by Andrey Rekalo for Why are currents named currents?

2013-04-10T19:59:51Z

The classical electric current density can be modelled as a 2-form $$J=J_{ij}\wedge dx^{ij}$$ which is assumed to be locally integrable over a 3-manifold (3-dimensional domain) $X$. By integrating $J$ over a 2-submanifold (a surface) $S\subset X$, one gets the electric current through S, i.e. $$I(S)=\int_{S}J,$$ that is a simple example of a 2-current in the sense of de Rham. Extending this notion to the $n$-dimensional case, one can naturally model a density on the $n$-manifold as a twisted $(n − 1)$-form $J\in\Omega^{n-1}(X,L)$, and to treat the corresponding integral $I(S)$ as a generalized electric current through the (n-1)-dimensional submanifold $S\subset X$.

Whether this was a real motivation behind the notion of de Rham's currents, I don't know.

Answer by Andrey Rekalo for Who discovered the winding number?

2013-01-06T21:04:50Z

Grünbaum and Shephard suggest that the winding numbers (for closed polygons) have been discussed in the literature at least since 1769.

See A.L.F. Meister, Generalia de genesi figurarum planarum et inde pendentibus earum ajfectionibus, Novi Comm. Soc. Reg. Scient. Gotting. 1 (1769/70), pp. 144-180.

On linear independence of exponentials

2010-07-11T20:32:22Z

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

Answer by Andrey Rekalo for Can we cover the unit square by these rectangles?

2010-08-01T20:25:57Z

This problem actually goes back to Leo Moser.

The best result that I'm aware of is due to D. Jennings, who proved that all the rectangles of size $k^{-1} × (k + 1)^{-1}$, $k = 1, 2, 3 ...$, can be packed into a square of size $(133/132)^2$ (link).

Edit 1. A web search via Google Scholar gave a reference to this article by V. Bálint, which claims that the rectangles can be packed into a square of size $(501/500)^2$.

Edit 2. The state of art of this and related packing problems due to Leo Moser is discussed in Chapter 3 of "Research Problems in Discrete Geometry" by P.Brass, W. O. J. Moser and J. Pach. The problem was still unsettled as of 2005.

Answer by Andrey Rekalo for Does this formula have a rigorous meaning, or is it merely formal.

2011-02-03T12:54:05Z

Back in the 19th century, when people had been experimenting with determinants a lot, they might have interpreted the above definition of $B\times C$ in terms of quaternions. If $i$, $j$, and $k$ denote basis elements of $\mathbb H$ and $${\mathbf x}=x_1i+x_2j+x_3k,$$ $${\mathbf y}=y_1i+y_2j+y_3k\quad$$ are pure imaginary elements of $\mathbb H$, then the vector part $\Im(\mathbf{xy})$ of the Hamilton product $\mathbf{xy}$ is equal to the determinant

$$\Im(\mathbf{xy})=\Im(\mathbf{x})\times \Im(\mathbf{y})=\det \begin{vmatrix} i & j & k \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\\
\end{vmatrix}.$$

There is a note by Sir Arthur Cayley where he introduces the notion of a quaternion determinant. He mentions several identities of the form

$$ \det \begin{vmatrix} {\mathbf x} & {\mathbf x} \\ {\mathbf y} & {\mathbf y}
\\
\end{vmatrix} = -2\det \begin{vmatrix} i & j & k \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\\
\end{vmatrix} $$ and $$ \det \begin{vmatrix} {\mathbf x } & {\mathbf x } & {\mathbf x } \\ {\mathbf y } & {\mathbf y } & {\mathbf y } \\ {\mathbf z } & {\mathbf z } & {\mathbf z } \\
\end{vmatrix} = -2\det \begin{vmatrix} {3} & i & j & k \\ x_0 & x_1 & x_2 & x_3 \\ y_0 & y_1 & y_2 & y_3\\
z_0 & z_1 & z_2 & z_3\\
\end{vmatrix} $$ where $\mathbf x$, $\mathbf y$, $\mathbf z$ are arbitrary quaternions $${\mathbf x}=x_0+x_1i+x_2j+x_3k, \mbox{ etc.}$$

Minimal surface which divides a convex body into two regions of equal volume

2011-02-07T12:33:55Z

Question. Given a convex body $\Omega$, what is the shape of a surface $\Gamma$ of minimal area which divides $\Omega$ into two regions of equal volume?

Background/motivation.

A 2D version of the question was posed by Michael Goldberg in Monthly: find the shortest curve which divides a convex quadrilateral into two equal areas. In the latter case the solution is given by a circular arc perpendicular to two sides of the quadrilateral (or just a segment of a straight line in degenerate situations). This follows from the observation that a circular sector is the shortest curve which, together with two sides of an angle, encloses a fixed area.

Goldberg himself offered a physically intuitive solution to the problem:

The curve may be considered as a restraining member under tension produced by internal fluid pressure in the restricted area. The ends of the curve are free to slide along the sides. Hence, the curve must be normal to two sides of the quadrilateral. Furthermore, since the fluid pressure is uniform, the curve must take the form of a circular arc.

The same approach suggests that in the general multidimensional case the solution is given by a spherical "cap" that intersects $\partial \Omega$ orthogonally. Now, assuming that the intuition is correct, is there a simple formal proof of this conjecture?

Edit. As Sergei Ivanov points out the minimal surface in question need not be a spherical cap for dimensions $>2$.

A modified question: is there always at least one solution to the problem? Should one impose any smoothness conditions on $\partial\Omega$ to guarantee existence?

Volumes of n-balls: what is so special about n=5?

2011-01-24T20:27:02Z

I am reposting this question from math.stackexchange where it has not yet generated an answer I had been looking for.

The volume of an $n$-dimensional ball of radius $R$ is given by the classical formula $$V_n(R)=\frac{\pi^{n/2}R^n}{\Gamma(n/2+1)}.$$ It is not difficult to see that the "dimensionless" ratio $V_n(R)/R^n$ attains its maximal value when $n=5$.
The "dimensionless" ratio $S_n(R)/R^n$ where $S_n(R)$ is the $n$-dimensional volume of an $n$-sphere attains its maximum when $n=7$.

Question. Is there a purely geometric explanation of why the maximal values in each case are attained at these particular values of the dimension?

[EDIT. Thanks to all for the answers and comments.]

How to define a differential form on a fractal?

2010-06-13T14:08:58Z

It is well known how to construct a Laplacian on a fractal using the Dirichlet forms (see e.g. the survey article by Strichartz). This implies, in particular, that a fractal can be "heated", i.e. one can write (and solve) the heat equation on the fractal.

The question is, can one run a fluid flow through a fractal set? In other words, is there a proper way to write the Navier-Stokes equations on a fractal? In order to do this, it seems that we need a "correct" notion of divergence at least.

More generally, is there a "correct" way to define a differential form on a fractal?

Answer by Andrey Rekalo for Surface equivalent of catenary curve

2011-07-09T00:05:00Z

A model equation for an inextensible, ﬂexible, heavy surface in a gravitational ﬁeld was deduced by ~~Poisson~~ Lagrange and later the problem was also studied by Poisson (see the references in the linked papers below). The equilibrium condition for a hanging heavy surface of constant mass density reads $$\sqrt{1+|\nabla u|^2}\ \nabla\cdot{}\frac{\nabla u}{\sqrt{1+|\nabla u|^2}}=\frac{1}{u+\lambda},\qquad x\in\Omega\subset\mathbb R^2,\qquad\qquad(1)$$ where $u=u(x)$ is the vertical displacement and $\lambda\in\mathbb R$ is an arbitrary constant (a Lagrange multiplier). (1) is the Euler equation of the variational integral $$I(u)=\int_{\Omega}u\sqrt{1+|\nabla u|^2}dx,$$ which can be interpreted as the vertical coordinate of the center of gravity of the surface $$\mbox{graph}(u)=\{(x,u(x)):\ x\in\Omega\}\subset\mathbb R^2\times\mathbb R.$$

Equation (1) is to be supplemented with the requirement that the surface has a prescribed area $A$ $$\qquad\qquad\qquad\qquad\qquad\int_{\Omega}\sqrt{1+|\nabla u|^2}dx=A,\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad(2) $$ and the Dirichlet boundary condition describing the curve from which the surface is being suspended $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\left.u\right|_{\partial \Omega}=g.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad(3) $$ One can check formally that a solution to (1)-(3) provides a graph of a heavy surface of prescribed area and boundary with the lowest center of gravity, so this is a precise 2D analogue of the classical catenary problem.

It is known that problem (1)-(3) has no classical solutions for the values of area $A$ outside of some bounded interval $[A_{\min},A_{\max}]$. Moreover, the corresponding variational problem has no global solutions for all $A\in\mathbb R$. A short survey of some old and relatively new results concerning well-posedness of (1)-(3) and its multidimensional analogues can be found in the paper by Dierkes and Huisken, "The N-dimensional analogue of the catenary: Prescribed area", in J. Jost (ed) Calculus of Variations and Geometric Analysis, Int. Press (1996), pp. 1-13.

Addendum. Here is a more recent survey by Dierkes: "Singular Minimal Surfaces" (in Geometric Analysis and Nonlinear Partial Differential Equations, Springer (2003), pp. 177-194).

Answer by Andrey Rekalo for Distributions on product spaces

2011-06-30T11:49:02Z

According to the Schwartz Kernel Theorem and its variants, there are the canonical isomorphisms $$\mathcal{D}^{\prime} \left(X\right)\tilde\otimes \mathcal{D}^{\prime} \left(Y\right)\simeq\mathcal{D}^{\prime}\left(X\times Y\right),$$ $$\mathcal{E}^{\prime} \left(X\right)\tilde\otimes \mathcal{E}^{\prime} \left(Y\right)\simeq\mathcal{E}^{\prime} \left(X\times Y\right),$$ $$\mathcal{S}^{\prime} \left(\mathbb R^n\right)\tilde\otimes \mathcal{S}^{\prime} \left(\mathbb R^m\right)\simeq\mathcal{S}^{\prime} \left(\mathbb R^{n+m}\right),$$ where $E\tilde\otimes F$ is the completion of the space $E\otimes F$.

Roughly speaking, this follows from the fact that the corresponding spaces of test functions $\mathcal{D}$, $\mathcal{C}^{\infty}$, and $\mathcal{S}$ are nuclear Fréchet spaces, and one has the canonical isomorphisms $$E^{\prime}\tilde\otimes F^{\prime}\simeq \left(E\tilde\otimes F\right)^{\prime}\simeq L(E; F'),$$ provided that $E$ and $F$ are nuclear Fréchet spaces. (Here the duals carry the strong dual topology and the space $L(E;F ')$ of continuous linear mappings is endowed with the topology of bounded convergence.)

As Johannes mentioned in his comment, a detailed presentation of the Schwartz Kernel Theorem and its versions for various spaces of distributions can be found in Topological Vector Spaces, Distributions and Kernels by Trèves. (More specifically, take a look at Chapt. 51, "Examples of Nuclear Spaces. The Kernels Theorem".)

Answer by Andrey Rekalo for Counterexamples in PDE

2011-06-24T07:41:28Z

Scheffer has shown that there is a nontrivial weak solution $u(x,t)\in L^2(\mathbb R^2\times\mathbb R)$ to the incompressible Euler equations in 2D

$$\left\{\begin{eqnarray} \frac{\partial u}{\partial t}+\nabla\cdot(u\otimes u) +\nabla p=0 \\ \nabla\cdot u=0\qquad\qquad\qquad\qquad\qquad\ \end{eqnarray}\right.$$ such that $u(x,t)\equiv 0$ for $|x|^2+|t|^2>1$. In other words, the solution is identically zero for $t<-1$, then "something happens" and the solution becomes non-zero, and for all $t>1$ the solution vanishes again. In the real world, this would look like if the water suddenly started to move in a cup that stands firmly on a table.

See V. Scheffer, "An inviscid flow with compact support in space-time", Journal of Geometric Analysis, vol. 3 (1993), pp. 343-401.

Answer by Andrey Rekalo for The classical Krein-Rutman theorem

2011-06-15T19:14:18Z

"Topological Vector Spaces" by Helmut Schaefer contains a thorough treatment of the classical Krein-Rutman theorem for compact positive operators in an ordered Banach space along with several generalizations to the case of a locally convex space with a cone. See Section 2 of the Appendix, Pringsheim's Theorem and Its Consequences. .

Answer by Andrey Rekalo for Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?

2011-06-09T23:28:45Z

Jonathan M. Borwein, David H. Bailey and Roland Girgensohn discuss this and related formulae for $\pi$ in their book "Experimentation in Mathematics" (see Section 1.1, p. 3). They claim that

The integral was apparently shown by Kurt Mahler to his students in the mid-1960s, and it had appeared in a mathematical examination at the University of Sydney in November, 1960.

They mention also a paper by Beuker who further developed the method of integral representations to obtain the irrationality estimate $$\left|\pi-\frac{p}{q}\right|\geq\frac{1}{q^{21.04...}},$$
which holds true for all integers $p$, $q$ with sufficiently large $q$. The exponent $21.04...$ is rather far from being optimal.

A draft of the book is freely available on J.M. Borwein's home page.

Answer by Andrey Rekalo for Sets of divergence of Fourier series

2011-06-07T10:27:28Z

I believe that the problem of characterizing the sets of divergence for classical Fourier series is more or less open for all interesting classes ($C$, $L^\infty$, $L^p$ with $p>1$).

The strongest result that I'm aware of is due to Buzdalin who showed that any null-set $E\in F_\sigma\cap G_\delta$ is a set of divergence for the Fourier series of some continuous complex-valued function ("Trigonometric Fourier series of continuous functions diverging on a given set", Math. USSR Sbornik, 24 (1974)).

The characterization problem is mostly solved however for several other orthogonal systems, including the Haar and Franklin systems. There is also a very recent paper by Karagulyan where it is proved, in particular, that

A necessary and suﬃcient condition for a set $E \subset [0, 1]$ to be a set of divergence for the sequence of $(C, \alpha)$-means ($\alpha>0$) of the Fourier series of some function $f \in L^\infty[0, 1]$ is that $E$ is a $G_{\delta\sigma}$-set of measure $0$.

(See G.A. Karagulyan, "Characterization of the sets of divergence for sequences of operators with the localization property", Sbornik: Mathematics, 202 (2011), pp. 9–33.)

To complicate things further, people tend to distinguish between the sets of divergence and unbounded divergence. A set $E \subset [0, 1]$ is said to be a set of divergence (resp. unbounded divergence) for a series of functions $$\sum_{n=1}^{\infty}f_n(x),\qquad x\in[0,1],$$ if the series diverges for $x ∈ E$ and converges for $x \in [0, 1] \backslash E$ (resp. diverges unboundedly for $x ∈ E$).

One may think of the two optimistic working conjectures.

Every $G_{\delta\sigma}$-set $E $ of measure $0$ is a set of divergence for the Fourier series of some function $f \in C[0, 1]$.

Every $G_{\delta}$-set $E$ of measure $0$ is a set of unbounded divergence for the Fourier series of some function $f \in C[0, 1]$.

Conjecture 2 was explicitly formulated by P.L. Ul'yanov in the late 1960s. Both conjectures seem to be open.

Answer by Andrey Rekalo for Logarithm of a matrix

2011-06-05T20:23:35Z

You may want to take a look at "Functions of Matrices: Theory and Computation" by Higham. Chapter 11 is specifically devoted to the matrix logarithm. In particular, the chapter contains a thorough comparison of four different numerical algorithms.

Answer by Andrey Rekalo for Ramanujan's eccentric Integral formula

2011-06-03T20:31:43Z

This is one of those precious cases when Ramanujan himself provided (a sketch of) a proof. The identity was published in his paper "Some definite integrals" (Mess. Math. 44 (1915), pp. 10-18) together with several related formulae.

It might be instructive to look first at the simpler identity (i.e. the limiting case when $b\to\infty$; the identity mentioned in the original question can be obtained by a similar approach): $$\int\limits_{0}^{\infty} \prod_{k=0}^{\infty}\frac{1}{ 1 + x^{2}/(a+k)^{2}}dx = \frac{\sqrt{\pi}}{2} \frac{ \Gamma(a+\frac{1}{2})}{\Gamma(a)},\quad a>0.\qquad\qquad\qquad(1)$$ Ramanujan derives (1) by using a partial fraction decomposition of the product $\prod_{k=0}^{n}\frac{1}{ 1 + x^{2}/(a+k)^{2}}$, integrating term-wise, and passing to the limit $n\to\infty$. He also indicates that alternatively (1) is implied by the factorization $$\prod_{k=0}^{\infty}\left[1+\frac{x^2}{(a+k)^2}\right] = \frac{ [\Gamma(a)]^2}{\Gamma(a+ix)\Gamma(a-ix)},$$ which follows readily from Euler's product formula for the gamma function. Thus (1) is equivalent to the formula $$\int\limits_{0}^{\infty}\Gamma(a+ix)\Gamma(a-ix)dx=\frac{\sqrt{\pi}}{2} \Gamma(a)\Gamma\left(a+\frac{1}{2}\right).$$

There is a nice paper "Wallis-Ramanujan-Schur-Feynman" by Amdeberhan et al (American Mathematical Monthly 117 (2010), pp. 618-632) that discusses interesting combinatorial aspects of formula (1) and its generalizations.

Answer by Andrey Rekalo for Determinant of sum of positive definite matrices

2011-05-19T12:54:05Z

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO thread).

Answer by Andrey Rekalo for The Ramanujan Problems.

2011-05-17T13:35:29Z

There is a survey article by Berndt, Choi, and Kang devoted to the set of 58 Ramanujan's problems. They indicate that the questions had originally appeared in the problems section of the Journal and apparently the editors published readers' solutions in subsequent issues.

Concerning your question 1, let me just quote from the Introduction to the survey:

Several of the problems are elementary and can be attacked with a background of only high school mathematics. For others, significant amounts of hard analysis are necessary to effect solutions, and a few problems have not been completely solved.

An elementary solution to the specific geometric problem you've mentioned can be found in Ramanujan's Notebooks, Part III by Berndt (Springer, 1991, pp. 244-246). The problem stems from Ramanujan's work on modular equations of degree 3...

Answer by Andrey Rekalo for A question regarding a claim of V. I. Arnold

2011-05-09T16:33:45Z

The limit $$ \lim_{x\to 0}\frac { f(x) - g(x) } { f^{-1}(x) - g^{-1}(x)} = -\left(f'(0)\right)^6 $$ appears in the Problems section of Mathematics Magazine, where it is calculated under the assumption that $f$ and $g$ are analytic in a neighborhood of $0$ odd functions such that $f'(0)=g'(0)\neq 0$, $f^{(3)}(0)= g^{(3)}(0)$, and $f^{(5)}(0)\neq g^{(5)}(0)$ (Problem 1672, vol. 77, No. 3, June 2004, pp. 234-235) .

In Arnold's example, we have that $$\sin(\tan x)= x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55x^7}{1008}+O(x^9) $$ $$\tan(\sin x)= x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040}+O(x^9)$$ $$\arcsin(\arctan x)=x-\frac{x^3}{6}+\frac{13x^5}{120}-\frac{341x^7}{5040}+O(x^9)$$ $$\arctan(\arcsin x)=x-\frac{x^3}{6}+\frac{13x^5}{120}-\frac{173x^7}{5040}+O(x^9)$$ Now, $$ \lim_{x\to 0} \frac { \sin(\tan x) - \tan(\sin x) } { \arcsin(\arctan x) - \arctan(\arcsin x) } = \lim_{x\to 0} \frac { -\frac{55x^7}{1008}+\frac{107x^7}{5040} +O(x^9)} { -\frac{341x^7}{1008}+\frac{173x^7}{5040}+O(x^9)} $$ $$=\lim_{x\to 0} \frac{-\frac{168x^7}{5040}+O(x^9)}{-\frac{168x^7}{5040}+O(x^9)}=1.$$

Given that one has to use the Taylor series expansion of $f$ and $g$ up to the seventh order, I find it somewhat difficult to see the result just by inspecting the graph.

Edit. And for the sake of completeness, here's the original argument from Arnold's book (Birkhauser Verlag 1990, P. 108).

If the graphs of non-coincident analytic functions $f$ and $g$ touch the line $y = x$ at the origin (Fig. 37), then the ratios $|AB|/|BC|$ and $|BC|/|ED|$ tend to one as $A$ tends to the origin. Therefore the required limit of the ratio $|AB|/|D'E'|$ is equal to one.

The Invariant Subspace Problem: examples

2010-12-19T17:48:11Z

Question. Is there a concrete example of a bounded linear operator on a Hilbert space for which it is not known if it has a non-trivial closed invariant subspace?

[Added 24.01.2011: According to Bernard Beauzamy (Introduction to Operator Theory and Invariant Subspaces, Elsevier (1988), p. 345),

the operator which is "closest" to a counter-example is the one built by the present author: it has one hypercyclic point $x_0$, and for every polynomial $p$ with complex coefficients, $p(T)x_0$ is also hypercyclic. Therefore, the operator has a vector space of hypercyclic points (thus solving a question raised by P. Halmos), but it may still have points which are not cyclic at all, thus having Invariant Subspaces.

Beauzamy refers to his manuscript "The orbits of a linear operator". I have not been able to find an electronic version of this manuscript (or paper) online. Does anyone know where one may find a description of the example? Is it presently known whether the operator in Beauzamy's example has an invariant subspace?]

Answer by Andrey Rekalo for Which Fréchet spaces have a dual that is a Fréchet space?

2011-04-29T07:34:09Z

For any locally convex and metrizable space $E$, its strong dual is metrizable if and only if $E$ is normable.

This and related properties of (F)-spaces are discussed in detail in Topological Vector Spaces I by Köthe (see §29.1, pp. 393-394 in the English edition).

Answer by Andrey Rekalo for Elementary+Short+Useful

2011-04-03T17:44:18Z

The Banach fixed point theorem.

Answer by Andrey Rekalo for Example of noncomplete quotient of complete lcs mod closed subspace

2011-03-07T13:15:18Z

A counterexample for both the first and third questions can be found in Counterexamples in Topological Vector Spaces by Khaleelulla (p. 108).

Let $W$ denote the space of all $\mathbb C$-valued sequences $(x_n)$ and $\Phi$ the space of finite sequences. Let $E=E_1\oplus E_2$ where $E_1$ is the topological countable direct sum of copies of $W$ and $E_2$ is the topological countable product of copies of $\Phi$. $E$ is a complete locally convex space but the quotient $E/M$ where $M=\{(u,u):\ u\in E_1\cap E_2 \}$ is not even sequentially complete.

Answer by Andrey Rekalo for Constant in Poincare Inequality

2011-03-05T01:02:05Z

This is a fairly standard stuff. Suppose that the Stokes operator $A=-\Delta$ is defined on smooth divergence-free vector fields $u$ which satisfy the standard no-slip boundary condition $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.u\right|_{\partial\Omega}=0.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(1)$$ Let $H$ be the closure in $L^2$-norm of the space of smooth divergence-free vector fields with compact support: $$H=\overline{\{u=(u_1,\dots,u_d)\in (C_0^\infty(\Omega))^d:\ \mbox{div}\ u=0\}}^{L^2}$$

It is well known that $A$ gives rise to a self-adjoint operator with compact inverse on $H$ provided that $\Omega\subset \mathbb R^d$ is a bounded domain with Lipschitz boundary (see, e.g., Chapter 1 of Navier-Stokes Equations by Temam). Moreover, one can show that $D(A^{1/2})\subset (H^1(\Omega))^d$ and that, for any $u\in D(A^{1/2})$, $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\|A^{1/2}u\|_{L^2}^2=\|\nabla u\|_{L^2}^2.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad(2)$$

Now, let $\{e_k\}_{k\in\mathbb N}$ be the orthonormal basis in $H$ which consists of eigenvectors of $A$. The required estimate is implied by (2) and the trivial inequality $$\|(I-\Pi_N)v\|_{L^2}^2=\sum\limits_{k\geq N} |(v,e_k)|^2\leq \frac{1}{\lambda_N}\sum\limits_{k\geq N}\lambda_k |(v,e_k)|^2=\frac{1}{\lambda_N}\|A^{1/2}v\|^2_{L^2}\qquad\qquad$$ which holds true for any $v\in D(A^{1/2})$ such that $\Pi_N v =0$.

The similar estimates remain valid if (1) is replaced with the Navier or periodic boundary conditions.

Answer by Andrey Rekalo for Quantitative Weierstraß approximation

2011-02-25T09:51:32Z

The answer is yes. The results which allow to infer the modulus of continuity of a function from the value of function's best approximation by polynomials are collectively known as Bernstein-type theorems.

Theorem (S.N. Bernstein). Let $E_n(f)$ be the best approximation to the function $f(.) \in C([a, b])$ by algebraic polynomials of degree at most $n$, and let $$E_n(f)=O(n^{-\alpha})$$ with some $0< \alpha \leq1$.

If $\alpha < 1$, then $f(.)$ is uniformly Hölder continuous with exponent $\alpha$ on each segment $[a', b'] \subset(a, b)$, i.e. $$\omega(f,\delta)=O(n^{-\alpha}).$$

If $\alpha=1$, then $f(.)$ is uniformly almost Lipschitz on each such segment, i.e. $$\omega(f,\delta)=O\left(\delta\ln \frac{1}{\delta}\right).$$

The theorem cannot be improved in the sense that $E_n(f)=O(n^{-1})$ does not imply that $f(.)$ is Lipschitz. This is somewhat easier to see in case of the essentially equivalent problem of the best approximation of periodic functions by trigonometric polynomials on $\mathbb R$. The modulus of continuity of the Weierstrass function $$f_0(x)=\sum_{k=1}^{\infty}\frac{\cos(3^kx)}{3^k} $$ is $M\delta\ln\frac{1}{\delta}$. At the same time, one can show that the best approximation of $f_{0}(.)$ by trigonometric polynomials of degree $\leq n$ is equal to $$E_{n}^{Т}(f_0)=\frac{3}{2}\frac{1}{n}.$$

A very good reference on Bernstein-type theorems and related results is Theory of Approximation by Akhiezer.

Answer by Andrey Rekalo for Applications of mathematics

2011-02-24T19:44:26Z

Sending a man to the Moon (and back).

Hilbert once remarked half-jokingly that catching a fly on the Moon would be the most important technological achievement. "Why? "Because the auxiliary technical problems which would have to be solved for such a result to be achieved imply the solution of almost all the material difficulties of mankind." (Quoted from Hilbert-Courant by Constance Reid, Springer, 1986, p. 92).

The task obviously required solving plenty of scientific and technological problems. But the key breakthrough that made it all possible was Richard Arenstorf's discovery of a stable 8-shaped orbit between the Earth and the Moon. This involved the development of a numerical algorithm for solving the restricted three-body problem which is just a special non-linear second order ODE (see also my answer to the previous MO question).

Another orbit, also mapped by Arenstorf, was later used in the dramatic rescue of the Apollo 13 crew.

Answer by Andrey Rekalo for Stronger version of the isoperimetric inequality

2011-02-14T13:44:08Z

There is a sharpened version of the plane isoperimetric inequality due to Benson which involves the inner and outer radii. Let $$\Gamma=\{(r,\theta):\ r=r(s),\theta=\theta(s)\}$$ be a simple closed rectifiable curve on the plane, parametrized by the arc length $s$, and let $$r_1=\sup\{r:\ (r,\theta)\in\Gamma\},\qquad r_2= \inf\{r:\ (r,\theta)\in\Gamma\}.$$ Assume that $\Gamma$ winds once arround the inner circle. Then

$$L^2-4\pi A\geq\frac{(2FA-2\pi E-\pi/(2F))^2}{1+4EF},$$

where $L$ is the perimeter of $\Gamma$, $A$ is the area of the enclosed region, and $$F=\frac{1}{r_1-r_2},\qquad E=\frac{r_1r_2(r_1+r_2)}{(r_1-r_2)^2}.$$

The reference is: D.Benson, "Sharpened Forms of the Plane Isoperimetric Inequality", The American Mathematical Monthly, Vol. 77 (1970), pp. 29-34.

Answer by Andrey Rekalo for Is this number already known to be transcendental? Is there a survey about up-to-date trascendence results?

2011-02-14T11:37:59Z

I have checked with Introduction to Algebraic Independence Theory, where it is mentioned in the preface (p. V) that

D. Bertrand and independently D. Duverney, Ke. Nishioka, Ku. Nishioka, I. Shiokawa (DNNS) deduced results on algebraic independence of the values of theta-functions at algebraic points and in particular derived the transcendence of the sums $\sum_{n=1}^\infty q^{n^2}$ for any algebraic $q$ satisfying $0 < |q| < 1$.

The precise references are not given but a little googling turned up the paper by D. Bertrand, "Theta Functions and Transcendence", The Ramanujan Journal, Vol. 1 (1997), pp. 339-350, which seems to be relevant. The second reference is DNNS, "Transcendence of Jacobi's theta series", Proc. Japan Acad. Ser. A Math. Sci., Vol. 72 (1996), pp. 202-203.

Answer by Andrey Rekalo for Existence/Uniqueness of solutions to quasi-Lipschitz ODEs

2011-02-13T12:41:11Z

Yes. This follows from the classical uniqueness theorem due to Osgood (the original paper appeared in 1898).

Osgood's Criterion. Let $\omega(t,u)=\phi(t)\psi(u)$ where $\phi(t)\geq 0$ is continuous on the interval $(0,a)$ and $\psi(u)$ is continuous on $\mathbb R_{+}$, $\psi(0)=0$, $\psi(u)>0$ for $u>0$, and $$\int_{0}^{\epsilon}\phi(t)dt<\infty,\qquad \int_{0}^{\epsilon}\frac{du}{\psi (u)}=\infty$$ for some $\epsilon>0$. Suppose that the mapping $f:[0,a]\times B_R(x_0)\to \mathbb R^d$ satisfies the condition $$||f(t,x_1)-f(t,x_2)||\leq\omega (t,||x_1-x_2||)$$ for any $t\in(0,a]$ and any $x_1,x_2\in B_R(x_0)$. Then the initial value problem $$\dot{x}=f(t,x),\qquad x(0)=x_0$$ has at most one solution on the interval $[0,\delta]$ with some $\delta>0$.

Osgood's theorem allows for the mappings $f(t,x)$ which are discontinuous at $t=0$. (Actually, the condition that $\phi(t)$ is continuous on $(0,a)$ can be replaced with an assumption of mere integrability.) Of course, the existence of a local solution is implied by the Peano theorem under the additional assumption that $f$ is continuous in $(t,x)$.

Moreover, Wintner showed that Osgood's uniqueness condition implies the convergence of successive Picard iterations to a local solution on a sufficiently small interval (A. Wintner, "On the Convergence of Successive Approximations", Amer. Journal of Math. Vol. 68 (1946), pp. 13-19).

Comment by Andrey Rekalo

2013-04-28T13:10:37Z

The sister site quant.stackexchange.com might be a better place for this question.

Comment by Andrey Rekalo

2013-01-14T14:49:22Z

What is $|Du|$?

Comment by Andrey Rekalo

2012-12-03T12:26:05Z

@Denis Serre: Sorry, my reading skills are horrible today. I misread your statement (as if you wrote that $H^{-1/2}$ is contained in $L^2$). Stupid me.

Comment by Andrey Rekalo

2012-10-01T17:36:45Z

Thank you so much for the answer and reference.

Comment by Andrey Rekalo

2012-08-05T09:03:00Z

@Alexandre Eremenko: Many thanks! I stand corrected.

Comment by Andrey Rekalo

2012-07-03T10:26:11Z

@Aldanor: you may get more answers on the SE site which is devoted to Quantitative Finance quant.stackexchange.com

Comment by Andrey Rekalo

2012-04-20T21:32:18Z

A similar question has already been asked here: mathoverflow.net/questions/28147/…

Comment by Andrey Rekalo

2012-04-03T08:48:23Z

Here's a related MO question: mathoverflow.net/questions/90233/…

Comment by Andrey Rekalo

2012-03-25T11:44:19Z

A related MO question: mathoverflow.net/questions/28651/…

Comment by Andrey Rekalo

2012-02-05T21:21:26Z

The question might be suitable for the sister stackexchange site devoted to Quantitative Finance: quant.stackexchange.com

Comment by Andrey Rekalo

2011-11-06T17:51:21Z

Why is it "not really research mathematics"?

Comment by Andrey Rekalo

2011-10-16T16:01:40Z

What makes you think the sum can be expressed in closed form?

Comment by Andrey Rekalo

2011-10-09T15:08:21Z

Any specific motivation to study $\omega(A)$?

Comment by Andrey Rekalo

2011-09-13T12:43:19Z

A related MO question mathoverflow.net/questions/42929/…

Comment by Andrey Rekalo

2011-09-12T08:42:13Z

@Gjergji Zaimi: Thanks a lot for the comments.