Functional Analysis

Functional Analysis

  • George Stoica added an answer:
    Which empirical measures, associated to infinite dimensional stochastic processes, satisfy the moderate deviation principle?

    In 1998 we proved that the Antoniadis-Carmona processes satisfy the above requirements, in connection with the tunneling effect. Am interested in the current situation.

    George Stoica · Mathematics and Statistics

    Alternatively, the infinite dimensional Ornstein-Uhlenbeck processes with unbounded diffusion coefficient also satisfy the requirements, see: http://arxiv.org/pdf/1503.05322.pdf

  • Miodrag Mateljević added an answer:
    Are there characterization of harmonic gradient mapping from the unit ball onto itself in 3-space?

    In communication between V. Zorich and the author, the
    question was asked to find examples  of harmonic gradient mapping from the unit ball onto itself in 3-space. For example, if $u=x^2 +y^2 - 2 z^2$, then $f=
    \nabla u=(2x,2y,-4z)$ is injective harmonic gradient mapping from
    $\mathbb{B}^3$ onto the ellipsoid.

    If $u$ is real-valued function such that $f= \nabla
    u=(x,y,z)$, then $u= x^2/2 +y^2/2 +z^2/2 +c$.

    In particular, $Id$ is not harmonic gradient mapping.

    In complex plane, if $u$ is real-valued harmonic function, then $u_z=\frac{1}{2}(u'_x -u'_y)$ is analytic function and therefore
    $ \nabla u=\overline{F}$, where $F= 2 u_z$ is analytic function.

    Miodrag Mateljević · University of Belgrade

    Znam da je pretpostavka jaka. 

  • Qingping Zeng asked a question:
    Any advice on isolated points of the approximate point spectrum of a bounded operator?

    Let $X$ be a complex Banach space and $T$ be a bounded operator acting on $X$. Let $\sigma(T)$ and $\sigma_{ap}(T)$ denote the spectrum and approximate point spectrum of $T$, respectively. Let $\lambda \in \sigma(T)$. It is classical that, with the aid of the spectral projection, $\lambda$ is isolated in $\sigma(T)$ if and only if $T - \lambda$ can be decomposed as a direct sum of a quasinilpotent operator and a invertible one. Now my nature question is as follows: Let $\lambda \in \sigma_{ap}(T)$. Is it true that $\lambda$ is isolated in $\sigma_{ap}(T)$ if and only if $T - \lambda$ can be decomposed as a direct sum of a quasinilpotent operator and a bounded belowness one? Here we say that an operator is bounded below if it is injective and its range is closed. It is also nature to find the answer for other spectra (eg. left spectrum, surjective spectrum, right spectrum, essential spectrum,...). Thank you!

  • Can anyone help with a membership function question?

    Can I use different membership functions in Fuzzification layer for ANFIS type model? e.g. U1 = Bell-shape and U2 = Triangular, so that the output of this layer will be wi = U1 x U2. In most of the papers they used the same type of membership function.

    Gerardo Maximiliano Mendez · Instituto Tecnologico de Nuevo Leon

    Yes you can do that, but there is not a good reason to do that. On the other side, you can do anything using C or C++.

  • Daoud Bshouty Bshouty added an answer:
    Is there any appropriate version of the Rad\'{o}-Kneser-Choquet theorem (RKC-Theorem) in space?

    Let $\gamma$ be a closed Jordan curve and $f_0 : S^1 \overset{\text{onto}}{\longrightarrow}
    \gamma$. The basic question that they address in this paper is
    under which conditions on $f_0$ we have that $F=P[f_0]$ is a
    homeomorphism of $\mathbb{B}$ onto $D$, where $D$ denotes the
    bounded open, simply connected set for which $\partial D =
    \gamma$. The fundamental benchmark for this issue is a classical
    theorem, first conjectured by T. Rad\'{o} in 1926 , which was
    proved immediately after by H. Kneser [12], and subsequently
    rediscovered, with a different proof, by G. Choquet. Let us recall the result.

    Theorem 1.1 (T. Rad\'{o} H. Kneser G. Choquet ) If $D$ is convex, then $F$ is a
    homeomorphism of $\mathbb{B}$
    onto $D$.

    Laugesen (see Duren's book}, p. 54-56) constructed a homeomorphisam
    of the the unit sphere in $\mathbb{R}^3$ onto itself, whose
    Poisson extension to a vector-valued harmonic function fails to
    be univalent in the ball.

    Daoud Bshouty Bshouty · Technion - Israel Institute of Technology

    None that I am aware of.

  • Gul E Hina Aslam added an answer:
    Minkowski type inequality in Banach algebras
    Under which circumstances it is true that ||(A+B)ⁿ||¹/ⁿ ≤ ||Aⁿ||¹/ⁿ + ||Bⁿ||¹/ⁿ for elements A and B in a Banach algebra and a natural number n?
    Gul E Hina Aslam · National University of Science and Technology

    How can you even talk about the relation $\leq$ in a Banach algebra? Unless you mention it as an ordered Banach algebra or Banach lattice algebra. 

  • A.J. Muñoz-Vazquez added an answer:
    Does a Hölder continuous function have a bounded fractional derivative?
    Some nowhere differentiable functions are fractional differentiable and comply with the Hölder condition.
    A.J. Muñoz-Vazquez · Center for Research and Advanced Studies of the National Polytechnic Institute

    By example, boundedness of the Caputo derivative of f at every interval, implies Hölder continuity of f in the same order

  • Christopher Jason Larsen added an answer:
    How can one characterize the boundary of a convex set?

    I am working on a part of a paper related to topological properties of boundary points. It is important for me to realize the topological and algebraic behavior the boundary points of the convex sets. I would be grateful if someone could help me around this issues by giving some ideas or references related to it.

    The general question provided in following;

    Let B be a closed set in n−dimensional Euclidean space. What other properties B should have in order to be guaranteed that there exist the closed convex set A such that ∂A=B. How about infinite dimensional spaces?

    Christopher Jason Larsen · Worcester Polytechnic Institute

    I suppose B should be n-1 dimensional, connected, without boundary, with all principal curvatures of the same sign. 

  • Hayder Dibs added an answer:
    Can someone help with the interpolation between the Bloch and the Moebius invariant H^1 space?

    The space $H'$ consists of functions $f$ analytic in the unit disc such that $f'\in H^1$. The Bloch $B$ space is defined by the requirement 

    $$\sup (1-|z|^2)|f'(z)| <\infty, $$

    Endowed with the obvious semi-norms, both $H'$ and $B$ are M-invariant in the strong sense: $\|f\circ \m\|=\|f\|,$ for every Moebius transformation $m$ of the disc.

    The question is: What is the real or complex interpolation space $(H',B)_...$ ?   

  • Harish Kumar Kotapally added an answer:
    Can anyone prove how Legendre wavelet forms an orthonormal basis for L^2(R)?

    Please explain......

    Thanks in advance.............

    Harish Kumar Kotapally · Indian Institute of Technology Indore

    Sir, It may help you(I proved it with the help of information given in the links)
    http://planetmath.org/orthogonalityoflegendrepolynomials
    http://physicspages.com/2011/03/18/legendre-polynomials-orthogonality/

  • Cenap Özel added an answer:
    If A is a subspace of a normed space X and \{x_n\} is a sequence in A such that \{x_n\} converges to z, does z \in A?
    Please note that A is a subspace.
    Cenap Özel · Dokuz Eylul University

    If A is finte dimensional space A is closed then it contains limit

  • James F Peters added an answer:
    Any suggestions on the classification of Moebius invariant Besov spaces?

    The Besov space $ B^{p,q}_s $ of analytic functions on the unit disc D consists of those f for which 

    $$ \int_0^1 M_p^q(r,D^n f)(1-r)^{(n-s)q-1} dr < \infty,\quad 0<p,q\le\infty $$

    where $D^nf$ is the $n$-th derivative of $f$, $s$ is an arbitrary real number,and $n>s$. The definition is independent of a particular choice of $n$. It is known that the space $B^{p,p}_{1/p}$ is M-invariant for all $p$. 

    What is about the general case?

    James F Peters · University of Manitoba

    This is a good question.   I also agree with @Romesh Kumar that this question is interesting.

    Moebius invariant Besov spaces are considered in terms of boundary behaviour in

    K.T. Hahn, E.H. Youssft, Tangential boundary behaviour of M-harmonic Besov functions in the unit ball,  J. of Math. Anal. and Applications 175, 1993,  206-221:

    file://localhost/Users/jfpeters/Downloads/546275230cf2c0c6aec1b2d9.pdf

    The space of holomorphic Besov functions in the diagonal Besov space are shown to be Moebius invariant subsets of the Bloch space (see remark p. 208).   Also see Corollary 1.5, page 209: the elements of the Besov space  extend continuously to the closed unit ball.    See, also,

    F. Beatrous, J. Burbea, Holomorphic  Soblev spaces on the ball, Diss. Math., 276, 1989

    and

    K.T. Hahn, E.H. Youssfi, Moebius invariant Besov p-spaces and Hankel operators in the Bergman space on the ball in $\mathcal{C}^n$, Complex Variables 17, 1991, 89-104.

    Another good place to look for answer is

    Journal of Computational Analysis and Applications 10, 2008, no. 1:

    http://www.eudoxuspress.com/images/JOCAAA08-VOL10.pdf

  • Octav Olteanu added an answer:
    A power series summation a_n z^n such that a_n tends 0 as n goes infinity. How can we show it does not have pole on unit circle?

    A power series summation a_n zn such that a_n tends 0 as n goes infinity. How can we show it does not have pole on unit circle?

    Octav Olteanu · Polytechnic University of Bucharest

    A related nice property holds. Assume that the power series is convergent in the whole closed unit disc, and denote by "f" its sum. If the coefficients are non negative, then z_0=1 is a maximum point for  |f| on the closed unit disc.

  • Martin Ritterath added an answer:
    Does anybody know any continuous function in time domain whose frequency response looks like the attached figure?

    I want a function in time domain whose frequency response should be of the form shown in the following figure.

    Martin Ritterath · Systag

    Yes, interpolate the polynomial and do invers Fourier transform.

    Good luck

  • Haridas Kumar Das added an answer:
    What are the differences between "Rieman integrable" function and "Lebesgue integrable" function?
    .
    Haridas Kumar Das · Concordia University Montreal

    Also you can compare with the following. The main difference between the Lebesgue and Riemann integrals is that the Lebesgue method takes into account the values of the function, subdividing its range instead of just subdividing the interval on which the function is defined. This fact makes a difference when the function has big oscillations or discontinuities. However, the Lebesgue method needs to compute the measure of sets that are not intervals. 

    Reference: http://demonstrations.wolfram.com/RiemannVersusLebesgue/

  • Yury Brychkov added an answer:
    Space of rapidly decreasing test function is not invariant with respect to fractional integral. Is there any example to understand this?

    Lizorkin space is a subspace of a rapidly decreasing function. The space of a rapidly decreasing test function is not invariant with respect to fractional integral. Where as lizorkin space is invariant w.r.t. fractional integral and differentiation. 

    I am not getting an example of this failure so please suggest few examples so that I can get this clearly.

    Yury Brychkov · Dorodnicyn Computing Centre of RAS

    Look at the definition of the fractional derivative , and you will see that it will have a singularity and power behaviour for Riemann-Liouville derivative. Lizorkin space implies a special definition of derivative related to multiplication by (1+x^2)^r for Fourier image which is multiplicator in space of r.d.t.f.

  • Mohamed El Naschie added an answer:
    Convex bodies with equi-measurable radial functions. Does regularity of bodies imply the bodies are rotations of each other?

    If K and L are convex bodies whose radial functions have the same distribution function, i.e. equi-measurable, what can be said? In particular, does this mean that the two radial functions(of the sphere) are equal up to composition with some measure-preserving transformation? If so, does the convexity of the two bodies imply any regularity of this measure-preserving transformation of the sphere? If possible, how much regularity must the bodies possess to force this transformation to be a rotation or orthonormal linear transformation of the sphere?

    Mohamed El Naschie · Alexandria University Alexanderia Egypt

    true. as an engineer  who learned this the hard way i say it with some regret. 

  • Mika Yasuda added an answer:
    Is functional alpha diversity equal to functional richness?

    Pool et al (2014) quantify alpha functional diversity as the volume of the convex hull filled by the fish species of each community in two-dimensional functional space using the values from the first two functional axes.

    But I wonder taxonomic alpha diversity is simply the species richness, so the alpha functional diversity can be functional richness...

    Mika Yasuda · Chinese Academy of Sciences

    Thank you very much for your answer, Thiago,

    I will read these paperes as well.

    Best wishes,

    Mika

  • George Stoica added an answer:
    Are there any Baum-Katz-type results proved for non-commutative random sequences?

    E.g., for free random sequences, in non-commutative L^p or Orlicz spaces.

    George Stoica · Mathematics and Statistics

    Also worth reading: http://tapchi.vnu.edu.vn/TL_S3_07/Quang_Tien.pdf

    (see reference [10] by R. Jajte).

  • Elnaz Shahryari asked a question:
    How can I summate absolute values of D5 coefficients extracted by discrete wavelet transform over one cycle and get fig 11 of attached paper?

    I get the figure 10 of the attached paper using discrete wavelet transform. Now I need to summate the absolute value of signal over one cycle. But I don't understand the decrease points of figure 11.

    can anyone help me doing this summation.

    Thanks

  • David G. Ebin added an answer:
    Can anyone help me find the simplest demonstration that the limit of the function with complex variables described below does not exist?

    The limit is limz->0 exp(1/z)/z (and z is a complex variable) We encouraged students to solve this by computing the residue of the corresponding series, however I would need a not so complicated solution for this limit. Thank you very much

    David G. Ebin · Stony Brook University

    Let z_n = 1/n  so the function evaluated at z_n will be n exp(n) which clearly goes to infinity.  Then take a sequence w_n = -1/n.  the function evaluated at w_n is -exp(-n) n

    which clearly goes to zero. Hence the limit does not exist.

  • Wiwat Wanicharpichat added an answer:
    Can you suggest any material, book or paper on connection of Crossed products of C*-algebras and semigroup C*-algebras?
    Can you suggest any material, book or paper on connection of Crossed products of C*-algebras and semigroup C*-algebras?
    Wiwat Wanicharpichat · Naresuan University

    1) William Arveson, An Invitation to C*-Algebras (Graduate Texts in Mathematics 39), Springer-Verlag, New York, Heidelberg, Berlin, 1976.

    2) X. Li, Semigroup C*-algebras and amenability of semigroups, J. Funct. Anal. 262 (2012) 4302–4340.

     http://wwwmath.uni-muenster.de/42/fileadmin/Einrichtungen/reine/Elke/Li-semigroup-C_v2.pdf

     Please see the attachment. 

  • Octav Olteanu added an answer:
    Is it possible to prove the following idea?

    Let ||x||_1\leq ||y||_1. Is it possible to say ||x||_p\leq||y||_p  for 1<p? In another way, the inequality in l1 is protect in lp or not?

    Octav Olteanu · Polytechnic University of Bucharest

    Dear colleagues, here is a possible answer to one of the previous questions by 

    N. Yilmaz  (in which case the property holds true?).

  • Alireza Borhani added an answer:
    Is the method from the paper below applicable to other functional or probabilistic situations?

    It worked well for level-3 large deviations, and for functional laws of the iterated logarithm in infinite dimensions.

    Alireza Borhani · Universitetet i Agder

    Hi, 

    If I understood you well, yes, why not! I cannot see anything against it!

    Dr. A. Borhani

  • Jean-Louis Honeine added an answer:
    Is it appropriate (correct) to calculate the median instantaneous frequency (scale) of a signal in time?

    I would like to quantify the changes in frequencies (scales to be more precise since I am using continous wavelet transformation) coefficient in time of signals composed of low frequencies 0.1-4 Hz. The signal goes from one steady state to another. Is it correct to calculate the median frequencies at each bin and see how the median frequency evolves in time? I obviously would loose some resolution since I would have to calculate my median frequency in bins that have the same size, so I have to accept the size of the biggest bin (of the lowest frequencies).  It would be really helpful if somebody has a better idea or can send me some references to read. Thank you.

    Jean-Louis Honeine · University of Pavia

    Thank you. I am getting so far very good results taking into consideration the limitations.  I am actually calculating the moving median frequency of a lot of signals. Now I wish to average them which hopefully will give a better result.

  • Ali Merina Houria added an answer:
    Is the chebyshev wavelet formed a wavelet basis of the space L2(w)?

    we say Chebyshev wavelet family or basis and why ??

    Ali Merina Houria · Université Abdelhamid Ibn Badis Mostaganem

    Thank you for answering Mr Hammad Khalil,

    But in the legendre one the weight function w(x)=1 but in chebyshev one  we have four kinds of chebyshev wavelets and four weight functions wich are =~1 , my question is :when we construct the discret wavelet we will translate and dilate the wavelet mother, to form the chebyshev wavelet ,and we will translate and dilate the weight function too ,the result that we get is many spaces L2(wn) , where is the orthonormal basis here  and in which space !!!!!

  • Octav Olteanu added an answer:
    For a subset X of RxR with the property that every continuous function f:A-->R attains its maximum in R. Is X compact?

    For a subset X of RxR with the property that every continuous function f:A-->R attains its maximum in R. Is X compact? What if f is bounded but does not attain maximum in X.

    Octav Olteanu · Polytechnic University of Bucharest

    @Rogier Brussee. Of course, your assertion is true (it is a well-known result in functional analysis). An other well-known nontrivial surprising result in Banach space theory is the famous Theorem of James: "Let A be a bounded and weakly closed subset of the real Banach space X. If every continuous linear functional on X attains its supremum on A, then A is weakly compact". In a way, this is partially a (hard) converse of the following obvious assertion: "Every continuous linear functional on X attains its supremum on a weakly compact subset A". For the Theorem of James, see R. B. Holmes, "Geometric Functional Analysis and its Applications", Springer, 1975.

  • Xiangnan Guan added an answer:
    Can someone help me with in vitro functional analysis of exhausted CD8+ T cells?

    PD-1/PDL-1 blockage restores functions to exhausted CD8+ T cells. In the literature, I always see that people isolate CD8+ T cells with high-PD-1 expression, characteristic of exhaustion. These isolated cells are cultured in vitro, and have impaired functions, but have improved functions when blockade antibody (anti-PD-1 or anti-PDL-1) is added. My question is: Isolated PD-1 high CD8+ T cells, when cultured in vitro, how do they maintain the exhausted phenotypes given that no PDL-1 ligand is added to the medium? If there is no ligand, PD-1 is not activated, then why the anti-PD-1 or anti-PDL-1 even matter? Thanks a lot. 

    Xiangnan Guan · The Ohio State University

    @Julio

    Thanks for your help! I will read that review. Thanks.

  • George Stoica added an answer:
    Is there a classification of all subspaces of measurable functions for which weak-convergence implies convergence in measure?
    I ask because this seems to allow one to replace compactness with weak-sequential-compactness(and weak-compactness for Banach spaces of measurable functions) in some arguments. For example, if X is such a space of measurable functions then any weakly-sequentially-compact subset will have the property that convergence in measure is equivalent to convergence in X.
    George Stoica · Mathematics and Statistics

    Dear Geoff,

    Long shot, but one never knows: Riesz' theorem asserts that norm bounded sequences in L^p (p>1) contain a weakly convergent subsequence in the same L^p. It is also known that, norm boundedness in L^p (p>1) and a.e. convergence (or in measure) imply the weak convergence in L^p of the entire sequence. However, there are classes of random variables, such as independent & identically distributed, exchangeable, spreadable (introduced by Kallenberg), for which norm boundedness in L^p (p \geq 1) implies weak convergence in L^p of the entire sequence, without the requirement of a.e. convergence or in measure.

    Sincerely,

    George

About Functional Analysis

Geometry of Banach space, Infinitely dimensional functional theory

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