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# Operator Theory - Science topic

Explore the latest questions and answers in Operator Theory, and find Operator Theory experts.
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One of the main problems of semigroup theory for linear operators is to decide whether a concrete operator is the generator of a semigroup and how this semigroup is represented.
One idea is to write complicated operators, as a sum of simple operators. For this reason, perturbation theory has become one of the most important topics in semigroup theory. My question is about the multi-perturbed semigroups or multiple perturbation of semigroups in a Banach space. I need a recurrent formula for a semigroup perturbed by multiple (several, i.e. more than two) bounded (in general unbounded) linear operators. I have searched for it, but only found a simple case, called the Dyson-Phillips series for a semigroup generated by A0+A1. How can we find the generalisation of this formula for a semigroup generated by A0+A1+...+An for a fixed natural n? Many thanks in advance. I am looking forward to your suggestions and recommendations on this topic.
Hi professors. I hope you are doing well. This is a mathematical article that proves a new recurrence relation that is fundamental for mathematics. The article proves also that four infinite series are equivalent. Hence, this article opens new opportunities to demonstrate and develop new mathematical findings and observations. This is the link: https://www.researchgate.net/publication/364651911_A_useful_new_equation_of_four_infinite_series_and_sums_by_using_a_new_demonstrated_recurrence_relation
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Let $A$ be a densely defined symmetric (unclosed) operator and let $B\in B(H)$ be positive.
I know that if $\overline{A}B$ is normal, then $AB$ need not be normal.
My question is: If $AB$ is normal, is it necessary that  $\overline{A}B$ remains normal? (notice that, thanks to the normality of $AB$, I have shown that $B\overline{A}\subset \overline{A}B$).
Many thanks,
Hichem
Retern
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Is $TT^*$ (or $T^*T$) densely defined if $T$ is a densely defined and symmetric linear operator?
I feel this is untrue, but do you have a counterexample?
Thanks,
Hichem
I expect not. For a counter example I would consider a differential operator where the coefficients have differing differentiability conditions so the formal adjoint might not exist in a normal sense such as out of the definition of the adjoint associated with a ordinary operator where the coefficients have no smoothness and the formal adjoint is a "quasi differential operator" in the verbiage of Naimark. Naimark covered this topic in detail in his books Linear Differential Operators.
That Naimark's work in differential equations was not better known is probably a victim of the Cold War. My wife at the time was a Russian linguist and I didn't have to depend on translations and not all of his better works were - but those two volumes above were.
In the thesis of G. C. Rota, a student of Jacob Schwartz, on the extension theory of ordinary differential operators he develops a theory which might point the way to the counter example you seek. His results were published in the Communications on Pure and Applied Mathematics in 1958. Another place to look is in the classic Dunford and Schwartz "Linear Operators", vol II where there is an detailed development of the extension theory of ordinary operators on Banach spaces (L^p).
My gut tells me without other conditions the conjecture is false and the counter example lies in differential operators.
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Let $B\in B(H)$ be self-adjoint and let $A$ be a densely defined symmetric (and closed if needed) operator such that $A^2$ is densely defined. If $BA^2\subset A^2B$ say, is there a result which gives $BA\subset AB$? We may add positivity to $A$ to avoid trivialities.
Notice that I already have a counterexample when $A^2$ is not densely defined.
Cheers,
Hichem
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A quantum mechanical operator acting on the abstract state space of Dirac kets maps vectors of the space to other vectors of the space.
Assuming for simplicity that the examined quantum system is one dimensional, in position space, the same operator is a function of momentum, which is a differential operator, and position, which is a variable.
If the expression, or representation, of the operator in position space has singular points; i.e. values of the position where the operator is not defined for some reason; for instance due to a pole, is there a property of the same operator in state space that distinguishes it from operators that do not have singularities in position space?
Τhe position space is realized by using the position eigenstates which do not belong to the state space since they are not normalizable, but since the state and position spaces are physically (and mathematically) equivalent, shouldn’t a property exist that distinguishes operators with singularities from operators without singularities in position space?
If I undertsand your question, you consider two different bases: One being the base consisting of delta(x) and the other let's say the hydrogen bound wavefunctions plus free (scattering ) states. You can transform, of course between these base states via the coefficients <x|a> where x is a position eigenstate and a a hydrogen bound or scattering state. You can also take matrix elements and expectation values of any operator in either base and transform. If you compute the matrix element in one basis and it diverges e.g. <x=0|V|x=0>, you should also get the same result by summing over a1,a2 <x=0|a1><a1|V|a2><a2|x=0>. Else your basis is not complete. If you are using complete sets of states, the properties of an operator do not depend on the choice of basis.
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Dear colleagues. The text of the question is available in the attached image or pdf-file.
Sincerely, Yaroslav Grushka.
Dear Professor,
You may want to consult the following manuscript:
Regards
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As we know, the vastness of the subject is realized by the variety of interdisciplinary subjects that belong to several mathematical domains such as classical analysis, differential and integral equations, functional analysis, operator theory, topology and algebraic topology; as well as other subjects such as Economics, Commerce etc.
Fixed point theory solved many nonlinear problems. In the future, it will play more crucial role in many real world problems.
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Recently, some new operators having the keyword "fractional" are proposed to define the non-integer order derivative.
For integer order derivatives, it is well known that the locality and validity of Libeniz rule are the main properties of integer order derivatives, but what is the characteristic property of a fractional derivative? When we can call an operator is "fractional" and what features should an operator have to call it a "derivative" operator?
Dear colleagues
Interesting question and I want to add a reply, about the Leibnitz Rule. You can build local derivatives that do not satisfy it! Conclusion: in my opinion there is no clear classification in this regard. I prefer a broader sentence: fractional derivatives are not derivatives, local generalized derivatives are not fractional.
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If B(H) is the algebra of bounded linear operators acting on an infinite dimensional complex Hilbert space, then which elements of B(H) that can't be written as a linear combination of orthogonal projections ?
I found something more interesting here https://arxiv.org/pdf/1608.04445.pdf
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My question concersns operators on Hilbert spaces.
It is well-known that any family of mutually commuting compact self-adjoint operators can be simultaneously diagonalized. Is the same true for any family of mutually commuting bounded self-adjoint operators?
I doubt that the answer can be positive even in the case of one operator.
Namely, consider the operator $T: L_2[0, 1] \to L_2[0, 1]$ that acts by the rule $(Tf)(t) = t f(t)$. This is a self-adjoint operator, its spectrum is $[0, 1]$ and its projection-valued measure $m_T$ is easy to calculate: for every Borel subset $D \subset [0, 1]$ $m_T(D)$ is the orthoprojector onto the subspace $L_2(D)$ of those $f \in L_2[0, 1]$ whose support lies (up to a subset of Lebesgue measure 0) in D.
So, for this $T$ you want to have an orthonormal basis $E= (e_k)_{k \in \mathbb N}$ such that for every Borel subset $D \subset [0, 1]$ of positive measure the intersection of $E$ and $L_2(D)$ forms a basis in $L_2(D)$. But this is impossible, because for the sequence $A_n$ of supports of $e_n$ one can find a Borel set $D$ of positive measure such that for each $n$ the set $[0, 1] \setminus D$ intersects $A_n$ by a set of positive measure, which means that for this special $D$ the intersection of $E$ and $L_2(D)$ is empty.
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Let X be a Banach space and let T be a bounded linear operator on X. We know that if X is reflexive and T is compact then there exists x in the unit sphere of X such that T attains its norm at x. Can we impose any condition on X or on T such that x is the unique such point?
In general, it is a difficult problem to identify the points at which an operator attains norm, if it attains norm at all! Having said that, there are certain estimations available in l_p^2 spaces over the real field, where p \neq 1, 2, \infty. It says that if an operator attains norm at more than 2(8p-5) number of points, then it must be a scalar multiple of an isometry.
For p=2, norm attaining point is unique (upto multiplication by \pm 1) if and only if the concerned operator is a smooth point in the operator space. In general, for a SMOOTH operator between Banach spaces, norm attaining point is unique (upto multiplication by \pm 1) if the operator attains norm.
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A most interesting project! There is still a marked need for work to identify fundamental first principles in biology!
From a modelling perspective you possibly work from objects (and their structure) towards interactions between objects (and their equations).
If so, the recently developed Operator Theory may offer guidance by offering a set of standard objects, and by unraveling natural organization/complexity along three dimensions: operators of different complexity, organization inside an operator, and organizations consisting of many operators. (Classical approaches to hierarchy almost all suggest a single linear dimension).
Look forward hearing more from you,
Gerard Jagers op Akkerhuis
Dear Javoxir, thank you for your answer. On the website indicated above I find a broad range of important mathematical approaches. Am I right that the focus is dynamic? In that case, the foundation for any object-based or agent-based dynamic model can be sought -in my view- in the 'objects' or 'agents', which in biology are organisms, ranging from bacteria (cells) to animals with brains. For this reason the identification of the organisms (at their specific levels of complexity), and of all the other (abiotic) 'building blocks' in nature represents a necessary and fundamental basis for agent-based theoretical biology. Because of the relevancy of identifying the 'building blocks' my work in the past 25 years has focused on this aspect, with quite a few discoveries that link closely to major themes in theoretical biology (e.g. the closure of interaction networks). The resulting theory is called the Operator Theory. Potentially its approach to defining the building blocks in nature may offer a contribution to your project. It would be most insightful to get your opinion on this.
Kind regards,
dr. dr. Gerard Jagers op Akkerhuis
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Under what conditions, a linear operator can have closed extensions?
Let T be a closable linear operator. Are there any others closed extensions of T except the natural extension $\overline{T}$ ? if "yes" are there any formal constructions of these extensions?
The answer is yes, and an example can be found in Kato's book, Perturbation Theory of Linear operators, V.3.6.
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let A be an algebra and A1 the unitization of A, then multiplication algebra which is denoted by M(A) is algebra generated by identity operator and left multiplication and right multiplication also for the algebra A1, we have M(A1)
it is clearly that M(A) is subset of M(A1) when we have M(A)=M(A1)?
Please have a look at the paper : Wang J. K.: Multipliers of commutative Banach algebras. Pac. J. Math. 11, 1131–1149 (1961).
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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!
Dear Professor,
Kindly go through it.
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I learn that in a Banach algebra over F (R or C), an element a is positive if and only if a=b*b, and equivalently a is positive if a = a* and spectrum of a is a subset of R of non negative elements ([0, +\infny). My questions are:
1) If F=R, that is real Banach algebra, what can we say about it's positive elements?
2) Do we still need involution on the real Banach algebra to have a positive elements?
3) Could you please give an example to justify the answer to Q2.
Consider the example: Take any nilpotent operator J, i.e. J^2 = 0. Then
{c J : c element of F} is a Banach algebra over F with norm || cJ || := |c| and involution (cJ)^* := c* J, where c* is a conjugate of c in the field F. This algebra does not contain any positive element, apart from 0 (which is
"positive" in any Banach algebra).
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(paper and project "Weighted Anisotropic Morrey Spaces Estimates for Anisotropic Maximal Operators" and "weighted anisotropic Morrey spaces...." by Ferit Gürbüz= I pronounce my self on its/their subject and not on the novelty that its author brings to it). the rg-profile of the author has 2 (slightly different) papers with same title, but one of them has a "full text" with the title ending with: "...AND 0 -ORDER ANISOTROPIC PSEUDO-DIFFERENTIAL OPERATORS WITH SMOOTH SYMBOLS" with an additional math paragraph.
Dear Prof. Serfatti,
You have raised a very importanrt issue which affects our newly discovered Dbranes String FUNCTOR ALGEBRA CALCULUS using Mathematical Physics and Quantitative Finance experiments and is based on the Analysis provided in proving the P vs. NP problem of Millenium Maths problems (Mallick, Hamburger, Mallick (2016)).  However, so far we have been able to formulate only the Fundamental Theorem in plain language which is stated on our www.econometricsociety.org/Soumitra K. Mallick website. The point is that we have not studied convergence properties of the FAC Integrals so I cannot tell you FAC has particular group homology properties over FAC algebras. This is a long drawn process of development of the Mathematical Fields which if you are interested in developing some feel free to communicate. My son who is in this with me will be specialising in Pharmaceutical Engineering Science (now undergraduate) where he studies Analysis as part of his course. So he may take more part in expanding this Maths also. Thanks for your question.
Prof. Dr. S.K.Mallick ForMemEPS, ForMemReS, MES, MAICTE, QC
for S.K.Mallick, N. Hamburger, S.Mallick
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Does there exists a function $g \in L^1(R)$ such that it satisfy the following property
(1)$L^1$ norm of g \leq 1
(2) $\hat{g}(s) =1$ for $s\in [-\delta, \delta]$ and
(3) \hat{g} has compact support ?
More precisely
Denote $R=\mathbb{R}$, $S(R)= Schwarz Class function on R$
I know that $C_{c}^{\infty}(R) \subset S(R)\subset L^1(R)$ and Fourier transform is bijection on S(R).  So one can start with a function with property $(2)$ and $(3)$ above and take $g$ as a inverse Fourier transform of the function. But I don't know that norm of g can be bounded by 1 ?
Dear Debabrata De,
The answer is negative by the following reason. The L1 norm of the function g is, by definition, the integral over R of |g|, and the \hat{g}(0) is the integral of g over R. So, the conditions \hat{g}(0) = 1 and ||g||1 \le 1 can happen only if g is real and non-negative almost everywhere. By the same reason, if ||g||1 \le 1 and \hat{g}(t) = 1 for some non-zero t, then g(x)eitx must be real and non-negative for almost all values of x, which is impossible for a real function g.
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positive definite = positive and invertible.
operators are defined on a separable complex Hilbert space.
Check this paper:
I hope you will find it useful..
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Since,  classically we define force as rate of change of momentum and by weyl rule we can map classical functions to quantum operators. Then, why is it that we don't often hear the word 'force operator' in quantum mechanics?
Well, yes we can have a force operator in the mathematical sense as outlined by Rafik Karaman and in his referred stack-exchange discussion by Lubos Mottl. The only thing is that its not much useful or even 'appropriate' to use it in the context of quantum mechanics. This is because as an operator, it cannot give observable eigen values which can be experimentally detected with accuracy.The reason behind this is the uncertainty principle. If one elevates force from the classical definition to quantum, then in the practical sense, one is actually  making a measurement of the momentum at some initial time and subsequently at a later time which is infinitesimally close to the initial time.But whenever the measurement of momentum is performed on a state,it  would yield an eigen value corresponding to its measurement(by collapse if it is not a momentum eigenstate initially).Since fundamentally completely isolated states are not realistic(fluctuations of quantum fields in the ground state can also affect other states,as happens in atomic transitions often) the state will evolve in time and any subsequent measurement of momentum cannot give an accurate value since there would be uncertainty in momentum measurements.Therefore force as an operator will be not much of a help from a realistic point of view.Even one can argue on the grounds of relativity that position is not a 'good' operator in quantum mechanics as well(multiparticle states being created at length scales < compton wavelength). But yes, as seen from the viewpoint of expectation values(via Ehrenfest's theorem), the mathematical way of stating the 'force' as <-dv/dx> is fine.But that is just the expectation value, not pertaining to a single measurement of a system.
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Consider the normed linear space X=R^2 endowed with the usual l_p norm, p>1 and p other than 2, infinity. Consider the family of linear operators T_k (x,y)=(y, (x+y)/k), k>0. Is it true that each T_k attains their respective norms at only one pair of points of S_X? (The points would depend on k, of course!) In particular, I would like to especially know the answers for k= 1,2.
Let us restrict to the case of operator T of norm one. Then the picture is easier to understand if one applies to both sets (the unit ball U of  two-dimensional  lp and to its image V=T(U)) the map M which maps every vector (x,y) to the vector with the same signs of coordinates, but with moduli changed to |x|p and |y|p respectively. M(U) is the unit ball of two-dimensional l1, and the boundary S of M(V) can be written explicitly. Then the problem reduces to the problem of finding out whether  the straight line x+y = 1 (or x - y = 1) is a tangent line to curve S, and in how many points. The corresponding equations are well-known from Calculus. So the problem reduces to some explicitly written equations. Of course, I don't say that these equations are easy to deal with ...
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Please can anyone explain how the Bernstein durrmeyer operator becomes a contraction?
See  Theorem 1 in the paper by J. A. Adell, J. de la Cal in Computers Math. Applic., V. 30, No 3 - 6, pp. 1 - 14 (1995).
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Hi, I have a linear bounded operator A between tow Hilbert spaces with a dense domain and dense range, is it surjective? If the answer is no, are there more conditions on A to be surjective?
Dear Hafdallah,
A can be not surjective. For instance, let A: l_2 -> l_2 be as follows
A(x_1,..., x_n, ...)=(x_1/2,..., x_n/2^n, ...)    (x_n is divided by 2^n).
It is clear that the set F of all finite sequences Y=(y_1,...,y_m,0,0,0,...) is contained in   Im A (m depends on Y and can be arbitrary). Clearly, F is dense  in l_2.
But there is no pre-image for, say, Y=(1,1/2,...,1/n,...) \in l_2.
You should be more specific about A.
Best regards, Yuri
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Before, I did some work about the queuing model by using functional analysing method(operator theory), it was related to abstract Cauchy problem. Now ,can we discuss the abstract Cauchy problem by changing it to convex programming? is it possible?
Dear Ehmet Ablet,
Can you please be more specific and write down the abstract Cauchy Problem and the equivalent Convex Program?!
Thank you!
Milen
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Are there any unitary operators on the Bergman space $A^2(D)$?
Define $U_{t}: A^{2}(D)\rightarrow A^{2}(D)$ by the rule $(U_{t}f)(z)=f(e^{-it}z)$. This operator is unitary, and it  is very important in the description of the  radial functions.
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Let M be a convex and compact subset of m by n matrices (m is strictly less than n). If for each A in M, A has  full rank, then there is an d in Rn such that for each A in M, the matrix [AT dT] has full rank.
I was wondering if someone proves this conjecture or provides a counterexample.
Moslem,
I am sorry
There is  an obvious mistake in my proposed argument for $m>1$, however I think that  my argument for $m=1$ is correct.
My mistake for $m>1$ is that the complement S of full rank matrices is not convex, but in Hahn banach theorem we need convexity of both compact set K and closed set S.
BTW, you called your question as a "conjecture". Where is  a  reference for this problem. Is it a well known conjecture?
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in functional analysis
If you mean that the spectrum is finite, a necessary and sufficient condition is that the operator is algebraic  over  C i.e. satisfies a polynomial  \prod_i  (X - \lambda_i).
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Any suggestion by using the finite element method?
If you send me the problem i can propose to you an idea
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There are unitary, self adjoint and normal operators in operator theory. Do each collection posses nice property?
The class of normal operators on a Hilbert space have two nice subsets:  the self-adjoint operators and the unitary operators. These two have properties the resembles the real line and the unit circle in the complex plane. [Youngson, Linear Functional Analysis, pg.134]
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For what type of separable  C* algebras, there is no any integer n and any nontrivial morphism  from A to the Cuntz algebra O_{n}? In the other word  for  what type of  C* algebras, Hom(A, O_{n}) is trivial for all n?
The motivation:  I was thinking to consider a NC analogy for singular homology as follows:
A singular object is  a continuous map from Delta^{n}\to X. So We have a natural morphism from C(X)\to  C(\Delta^{n}).
With some (or  a lot of) abuse of notations  and equations, the equation \sum t_{i}=1,  which defines the standard simplex  \Delta^{n},  could be considered as a commutative picture of universal property of Cuntz algebra  O_{n}. This is  a motivation to construct   a  complex  as  followos:   We  put  C_{n}(A)= Free abelian group generated by Hom(A,, O_{n}). Now we can define  a  complex
.......  C_{n}(A) \to  C_{n+1}(A) \to....
using n+1 embeddings O_{n} to O_{n+1}
Does this idea leads to triviality?For what type of  C* algebras, this  construction is useful?
A separable C*-algebra embeds in the Cuntz C*-algebra \mathcal{O}_2 if and only if it is an exact C*-algebra (a result by E. Kirchberg)
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All sets are in the set of complex numbers.
The answer is yes. Consider the projection x0 of x onto A and the circle passing through the point x1 = (x + x0)/2 and whose center is x0 + t(x - x0). This circle does not contain point x if t < 3/4 but A is contained in the circle provided that | t | is large enough.
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Let f: \R^n -> \R^n be a function. What would you call the function g: x -> x' f(x) ? Furthermore, if F:\R^n -> \R^n is a linear operator and D \in \R^{n X m} a matrix, how would you refer to the operator G: D -> D' F (D) ? Does this appear in a particular physics or engineering context? Note, I use dash ' to mean transpose.
Just to add to Peter's first answer: It's technically not even a quadratic form if f is nonlinear. I don't know off-hand of any specific name for it though other than, as Peter said, it's <x,fx>.
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Can you suggest any material, book or paper on connection of Crossed products of C*-algebras and semigroup C*-algebras?
Visit:
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The same as lacI. I am curious about the when dimer tetramer structure forms in vivo, is it a inducible procedure based on the operator sequence?
Tough to say. Reading a paper it appears that it dimer before binding, and drug just changes a distance between the monomers, inducing right conformation for binding.
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In the standard proof showing that hermitian operators have real eigenvalues we exploit the  symmetry of the operators to show that they only have real eigenvalues.Does this hold for ALL hermitian operators in general?
In an infinite-dimensional Hilbert space a bounded Hermitian operator can have the empty set of eigenvalues. A typical example is the operator of multiplication by t in the space L2[0,1], i.e. the operator T that maps every function f(t) to function tf(t).
That is why in Functional Analysis one defines the spectrum of an operator T: H --> H not as the set of all eigenvalues, but in a more complicated way, namely as the set of those complex numbers a, that (T - aI) is not invertible, where the letter I'' stands for the identity operator. Now the basic facts:
1. The spectrum of a bounded linear operator is not empty and is a closed bounded set.
2. The spectrum of a bounded linear operator contains all the eigenvalues of the operator, but also can contain some other points.
3. All the points of the spectrum of a bounded Hermitian operator are real numbers.
Remark: in Functional Analysis Hermitian operators are usually called self-adjoint operators''.
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By space of measurable functions, I mean L_0(m) where m is a non-atomic sigma-finite measure space.
In my opinion, you can start with the book by Martin Vaeth "Ideal Spaces" LNM 1664. But, of course, you should think about classes of measurable functions...
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Yes.
see the spectral theorem via functional analysis.and then study the completely continuous operator via Hilbert space.
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Suppose X be a linear space. My question is: How can we say that X is not reflexive in any norm. Reflexive means X is linearly isometric to its second dual X''
If X is a Banach space with respect to one of the norms, it is no longer a Banach space if you use an inequivalent norm (either that or you have to throw some elements out of X). For any normed vector space X, X' is a Banach space, and so is X''. So for X to be reflexive, X = X'' must be a Banach space, which in turn implies that we have to stay with equivalent norms.
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Give a concrete example of linear operator whose resolvent satisfies a weak parabolic estimate in the region Re z \geq - C(1+ | Im z |)^\ alpha, where 0<\alpha<1.
Thanks
This means that the resolvent R(z,A) satisfies ||R(z,A)||\leq C(1+|z|)^-\beta for Rez|geq -c(1+|Imz|)^\alpha, with 0<\beta\leq\alpha\leq 1,
See Favini and Yagi, Degenerate differential Equations in Banach Spaces, M.Dekker, 1999, page 47.
Some exaples with \alpha<1 are described in a russian oaper already translated in english appearea in 2010-2011, which I do not succeed in finding.I received it from M.Balayev but I losed it and Balayev does not answer to my messages..
Best regards,Angelo
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We equip $L(H)$ with the natural Jordan product $a \circ b = \frac12 (a b + ba)$ and its natural involution. We consider the Jordan subalgebra of $L(H)$ generated by $e$ and $e^*$.
Yes it's true, but pheraps if $e$ is an isometry, you can consider the linear contraction $U=ee^*$ where $U\in A$ and $A$ tha Jordan alg. generated by $e$ and $e^*$.
We have ort. projection $P=[ker(I-U)]=[ker(I-ee^*)]=$