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math/0006158 | Consider the category whose objects are pairs MATH where MATH is an algebraic MATH-group which is a negatively weighted extension of MATH and where MATH is a continuous, NAME dense representation which lifts MATH. A morphism MATH consists of a homomorphism MATH of MATH-groups that is compatible with the projections to ... |
math/0006158 | We need only prove the last statement. Since MATH has weights MATH, MATH has weights MATH, and its continuous dual MATH (a direct limit of finite dimensional MATH-modules) has weights MATH. It follows that the space MATH of degree MATH continuous cochains on MATH has weights MATH. Since the bracket is a morphism, the N... |
math/0006158 | The theorem above implies that MATH. Since MATH is pronilpotent, MATH is trivial. |
math/0006158 | The condition on MATH implies, by REF , that MATH. The last assertion in REF implies that MATH. The assumption about MATH and the last assertion of REF imply that MATH when MATH, so that MATH. Since MATH is an exact functor, it commutes with homology. So it follows that MATH vanishes. The result now follows from the fo... |
math/0006158 | By REF , there is a free graded NAME algebra MATH and a graded NAME algebra homomorphism MATH which is surjective and induces an isomorphism on MATH. Denote the kernel of this by MATH. Note that MATH is free if and only if MATH. Since MATH is an ideal in a negatively graded NAME algebra, MATH if and only if MATH. There... |
math/0006158 | Since MATH is NAME dense, MATH is surjective. Denote the image of MATH in MATH by MATH. Since the diagram MATH commutes, the image of MATH in MATH is trivial. It follows that MATH is contained in the kernel of MATH and that the composite MATH is trivial. It remains to show that MATH contains the kernel of MATH. Note th... |
math/0006158 | REF implies that MATH. The result follows. |
math/0006158 | We will prove the sufficiency of the condition; necessity is left as an exercise. Since MATH is reductive and acts on the finite dimensional complex MATH of NAME algebra cochains, MATH . In particular, each MATH can be represented by a continuous MATH-invariant cocycle MATH. The NAME algebra MATH can be constructed as ... |
math/0006158 | This follows directly from REF as the image of MATH is open. |
math/0006158 | This follows directly from REF , and REF . |
math/0006158 | Fix MATH. By replacing MATH by MATH for some MATH, we may assume that MATH is unipotent. By replacing MATH by MATH, we may assume that MATH is an inclusion. We will prove the result by induction on the dimension of MATH. First recall that if MATH is a unipotent group over a field MATH of characteristic zero, and MATH i... |
math/0006158 | We will use REF and its notation. Set MATH. Since MATH is a finitely generated pro-MATH group, MATH and MATH is a finite MATH-group. Also, MATH for all MATH if and only if MATH is torsion free for all MATH. Using universal mapping properties, one can show that the inclusion MATH is the MATH-adic unipotent completion of... |
math/0006158 | The first assertion follows from the assumption using the natural isomorphism MATH. Since the bracket MATH is surjective and MATH equivariant, it follows that MATH acts on MATH via the MATH-th power of the cyclotomic character. This implies the second and third assertions as it implies that every derivation of MATH tha... |
math/0006158 | The surjection MATH restricts to a surjection MATH, which implies that MATH has NAME dense image. By strictness, it induces a surjection MATH. From REF , it follows that MATH is the inverse image of MATH under the natural mapping MATH. There is thus an injection MATH . REF imply that this induces isomorphisms MATH when... |
math/0006158 | Since MATH is isomorphic to the free NAME algebra generated by MATH, it follows that MATH. By REF we have MATH . The non-degeneracy of the NAME bracket MATH implies that the mapping MATH, given by taking inner derivations, is an isomorphism. Thus MATH . |
math/0006158 | This follows directly from CITE - see REF . |
math/0006158 | Because of the previous lemma, we only need compute MATH. It is proved in CITE that there is a NAME type spectral sequence MATH . Since MATH is reductive and MATH, it follows that MATH . Since MATH is negatively weighted, the right hand side is zero when MATH. Since MATH is free and MATH is exact, MATH when MATH and MA... |
math/0006158 | The first statement follows because MATH classifies extensions of MATH by MATH with property MATH. Since MATH is negative, each extension of MATH by MATH is a weighted MATH-module and hence a MATH-module. It therefore determines an element of MATH. The second assertion follows using the NAME Spectral Sequence CITE for ... |
math/0006158 | As in the proof of REF , the NAME spectral sequence implies that MATH. It thus suffices to prove that the natural mapping MATH is injective. Let MATH be a class in the left hand side corresponding to a two-step extension MATH . Since all MATH-modules are locally finite, we may assume MATH to be finite dimensional. By e... |
math/0006158 | By REF we have an injection. MATH . Since the right hand side vanishes, MATH is free by REF . The computation of MATH follows from REF and the facts (compare REF) that MATH when MATH and that there is an inclusion MATH corresponding to MATH via NAME characters. Thus, MATH where MATH is defined as in REF . |
math/0006158 | First recall the elementary fact that the set of torsion elements of a nilpotent group MATH forms a characteristic subgroup MATH and that MATH is torsion free. This implies that the set MATH of elements of MATH that are torsion modulo MATH is a characteristic subgroup, that MATH is a torsion free nilpotent group, and t... |
math/0006158 | This follows by a proof that is essentially the same as that of REF , taking the continuity of MATH into consideration in the current setting. The following lemma is a key ingredient. |
math/0006158 | The proof is similar to the proof of REF. Since MATH is finitely generated, there is a central filtration of MATH by closed normal subgroups, each of whose graded quotients is either MATH or MATH. We prove the lemma by induction on the length of this filtration. There is an exact sequence MATH where MATH is MATH or MAT... |
math/0006158 | We may assume that MATH is the upper triangular unipotent subgroup of MATH for some MATH. We denote by MATH the group of matrices whose MATH-th entry lies in MATH when MATH, is MATH when MATH, and MATH when MATH. Since MATH is finitely generated, the image of MATH is contained inside MATH for some MATH. The filtration ... |
math/0006158 | It follows from the lemma that MATH is continuous and induces continuous homomorphisms MATH . By the universal mapping property of MATH-adic unipotent completion, these induce homomorphisms MATH . But the homomorphism MATH induces a homomorphism MATH whose composite with REF is the canonical isomorphism of REF . This c... |
math/0006158 | Suppose that MATH is an automorphism of MATH. Since MATH, and since the bracket mapping MATH is surjective and commutes with MATH, we see that MATH acts trivially on MATH if and only if it acts trivially on MATH. It follows that the kernel MATH of MATH is a prounipotent group. The last statement follows from standard N... |
math/0006158 | This follows from REF as there is a natural action of MATH on MATH. |
math/0006158 | The short exact sequence MATH gives a long exact sequence CITE MATH . Since MATH is compact, the proof of CITE shows that MATH is torsion. So, after tensoring with MATH, we have an isomorphism MATH . When MATH, MATH is finite. This is obvious when MATH. When MATH it follows from class field theory. Indeed, reduce to th... |
math/0006158 | Let MATH and MATH. By CITE there is a long exact sequence MATH where MATH denotes étale cohomology with support on MATH. Thus it is enough to show that MATH vanishes for all MATH unless MATH or MATH. Let MATH denote the henselization of MATH at MATH. Then by CITE we have MATH so it suffices to establish the vanishing o... |
math/0006159 | The sketch of the proof is as follows: basically, the claim follows from REF , which implies that the denominator of any MATH in the standard basis of MATH is uniformly bounded, whence the period of the MATH-expansion of MATH is bounded as well. |
math/0006159 | It suffices to show that if MATH is weakly finitary, then MATH in question does exist. Let MATH be weakly finitary; then for any MATH there exists MATH such that MATH. Let MATH has the MATH-expansion MATH and MATH . Without loss of generality we may regard MATH to be greater than the preperiod MATH the period of the se... |
math/0006159 | Let MATH denote the distance to the closest integer, MATH be the MATH-coordinate of MATH and MATH denote the linear transformation of MATH defined by the matrix MATH. Let MATH be the mapping acting from MATH into MATH by the formula MATH . Then by REF , MATH where MATH. Therefore, it suffices to show that the diameters... |
math/0006159 | By the well-known result, for any NAME MATH and MATH (where MATH denotes the trace of an element MATH of the extension MATH, that is, the sum of all its NAME conjugates) - see, for example, CITE. Since MATH is a unit, MATH implies MATH, whence MATH . Thus, if we regard MATH as a lattice over MATH, then by REF , MATH is... |
math/0006159 | We have MATH, where MATH. As was mentioned above, the dimension of the unstable foliation MATH is REF, whence MATH, and since MATH, we have MATH, that is, MATH. Now the claim of the lemma follows from REF . |
math/0006159 | Suppose MATH is fundamental. Then the homoclinic point MATH whose MATH-coordinate is MATH can be represented as a finite linear integral combination of the powers MATH, that is, MATH whence MATH. Therefore, MATH is invertible in the ring MATH. Conversely, if MATH, then using the same method, we show that the claim of t... |
math/0006159 | By REF , MATH is finite and since it is shift-invariant, it must contain purely periodic sequences only. Let MATH. Then by REF , MATH as MATH, whence from REF , therefore, MATH, and MATH, because MATH. |
math/0006159 | Let MATH denote the set of all partial limits of the collection of sequences MATH, where MATH is the sequence MATH whose ``value" is MATH. It suffices to show that MATH. Let MATH; by definition, there exists a sequence of positive integers MATH such that MATH . Then MATH, and we are done. |
math/0006159 | Let the mapping MATH be given by REF and MATH. Let MATH and MATH . The NAME Separation Lemma CITE says that there exists a constant MATH such that if MATH and MATH are two sequences in MATH and MATH, then MATH. Hence MATH where MATH denotes the NAME measure on MATH. Since for any MATH is equivalent to MATH and the corr... |
math/0006159 | The claim follows from the definition of MATH (see Introduction) and the fact that the positive root MATH of the equation MATH is the smallest NAME number CITE. Indeed, MATH and MATH is a subshift of finite type, namely, MATH . Now the desired claim follows from CITE asserting that if MATH, then MATH. |
math/0006159 | We have MATH . Since MATH we have MATH . Now by REF , MATH being weakly finitary (see REF ) and the deifnition of MATH we have MATH for any MATH. Hence MATH and from REF we finally obtain the estimate MATH whence one can take MATH, and REF is proven. |
math/0006159 | We showed that MATH, whence MATH. |
math/0006159 | We already know that any bijective arithmetic coding is parametrised by a fundamental homoclinic point. Let MATH be such a point for MATH; then any other fundamental homoclinic point MATH satsifies MATH, where MATH and MATH are the corresponding MATH-coordinates and MATH - the proof is essentially the same as in REF . ... |
math/0006159 | Let MATH denote the bijective arithmetic coding for MATH parametrised by MATH and MATH. If MATH, then one can consider the mapping MATH; it will be well defined on the dense set and we may extend it to the whole torus. By the linearity of both maps, MATH is a toral endomorphism. Thus, we have MATH . Let MATH is given b... |
math/0006159 | Let MATH be written column-wise as follows: MATH. Then by REF and the definition of MATH, MATH whence by the fact that MATH, we have MATH for MATH. |
math/0006159 | Let MATH, where MATH is a certain bijective arithmetic coding for MATH. Then MATH is a linear mapping from MATH onto itself defined a.e.; let the same letter denote the corresponding toral endomorphism. Then MATH. Therefore the matrix MATH of the endomorphism MATH satisfies REF , whence by REF , MATH for some MATH. Hen... |
math/0006159 | REF : see REF ; REF : see REF ; REF : also follows from REF ; REF : it is obvious that MATH satisfies this property (take MATH). Hence so does any MATH which is conjugate to MATH. |
math/0006159 | By REF , the definition of MATH and the NAME Theorem, MATH . |
math/0006159 | We have MATH. The case MATH thus leads to REF yields REF . |
math/0006159 | Since MATH and MATH are conjugate, they have one and the same characteristic polynomial. By the definition of MATH we have MATH which is equivalent to REF . |
math/0006161 | Consider a monad MATH in MATH. Let MATH be the (object part of) the coinserter of the pair MATH: MATH . Notice that REF-cell MATH sets up a lax cocone over MATH as follows: MATH . But this lax cocone need not satisfy the equations involving MATH and MATH, so we must enforce them. For the unit, pasting MATH and MATH, we... |
math/0006161 | CASE: Recall that for a given object MATH we have the string of adjunctions MATH . For MATH, cartesianness implies that MATH is the comma object MATH. Thus since MATH is left adjoint to MATH, also MATH is left adjoint to the projection MATH; and so MATH is a split fibration. The same argument in MATH yields the split c... |
math/0006161 | CASE: MATH: Given a bimodule MATH define the invertible REF-cell MATH as the pasting composite MATH where the left isomorphism is that of REF and the right one is given by the homomorphism MATH applied to the naturality square for MATH. CASE: MATH: Use the same argument in MATH. MATH . |
math/0006161 | CASE: The square is a pullback and the lower arrow is a split monic (MATH), hence the top arrow is monic. CASE: Given `global generic' elements MATH, we must stablish a REF correspondence MATH . REF-cells into MATH are classified by REF into MATH and those into MATH are classified by REF into MATH. In the following dia... |
math/0006161 | We must first verify that MATH is indeed a monad. This follows by a routine calculation using the unit and associativity axioms for a lax algebra REF . The normality condition is straightforward. It is furthermore clear that starting with a normal monad MATH we obtain the data for a normal lax algebra by taking adjoint... |
math/0006161 | CASE: MATH CASE: MATH . The monad axioms follow from those of MATH and pseudo-naturality of MATH. MATH . |
math/0006161 | CASE: Given MATH, define MATH as follows: MATH CASE: Given a MATH-algebra MATH, we have a lax cocone MATH where MATH is the adjoint transpose of MATH (associativity for MATH). Thus we have an induced morphism MATH from the NAME object MATH. Once again, associativity for MATH and universality imply that this induced mor... |
math/0006161 | Since the counit of the required adjunction is the identity, we must simply define the unit. Notice that MATH and MATH is the morphism of algebras MATH uniquely determined in MATH by the lax cocone MATH given by the adjoint transpose of MATH, while the underlying morphism of MATH is MATH. To give a REF-cell MATH amount... |
math/0006161 | CASE: Let MATH have an adjoint-pseudo-algebra structure MATH, with MATH and MATH the unit and (invertible) counit of the adjunction MATH. Recall that the underlying morphism of MATH is MATH. Define MATH. We now intend to show that MATH represents MATH. Define MATH as the pasting MATH . We claim MATH is an isomorphism: ... |
math/0006161 | We must verify the compatibility with units and multiplications: CASE: MATH follows immediately from the definition of MATH that MATH (for MATH). CASE: MATH we have by definition of MATH that the right pasting instantiated at MATH is the upper morphism in the following commuting diagram, MATH while the bottom map is th... |
math/0006161 | The details are essentially contained in the proof of REF . Given a (pseudo-)MATH-algebra MATH (with structural isomorphisms MATH and MATH) we have MATH by definition of MATH. In the other direction, MATH is an isomorphism of (pseudo-)MATH-algebras. MATH . |
math/0006161 | Let MATH be an endomorphism in MATH with MATH a monad. Let MATH be the NAME object of MATH in MATH. Since MATH preserves it, there is a uniquely determined morphism MATH mediating between lax cocones in the following diagram: MATH . Uniqueness of mediating morphisms between lax cocones means that the algebra axioms for... |
math/0006161 | Given a monad MATH in MATH, we transform it into a monad MATH in MATH. We claim that the resulting representable NAME object MATH in MATH yields one for MATH in MATH. Indeed, we obtain a lax cocone for MATH as follows MATH the latter correspondence by MATH being locally fully faithful. The universal property of such a ... |
math/0006161 | Notice first that for a given pseudo-MATH-algebra MATH, the (morphism part of the) monad MATH corresponds to a morphism in MATH. Now, our REF allows to apply REF . The corresponding NAME object yields the required left biadjoint by virtue of the analysis preceeding REF , while this latter guarantees that the unit of th... |
math/0006161 | For an internal category MATH, the corresponding `hom' bimodule is MATH and therefore MATH. MATH . |
math/0006161 | We define an explicit inverse MATH to the given REF-functor. Given a multicategory MATH we obtain a category MATH by pulling back along MATH: MATH . CITE. We then construe MATH itself as a bimodule MATH, whose action is obtained by `restriction' of the composition operation of MATH. Furthermore, the monad structure of ... |
math/0006161 | Given a monad MATH, we obtain via the lax functor MATH a monad MATH in MATH which is therefore an internal category. This category MATH is the (vertex of the lax cocone of the) NAME object of MATH. To define a lax cocone MATH REF-cell MATH yields an identity-on-objects functor MATH: MATH the description of MATH in the ... |
math/0006165 | The decomposition MATH determines a projection MATH such that MATH and MATH is also a projection, MATH. The same for MATH and MATH. The inclusion MATH gives MATH, that is, MATH; similarly, MATH. So, our projections commute with each other. Introducing MATH we have MATH; that is, MATH is also a projection, and MATH. It ... |
math/0006165 | We introduce a Gaussian type space MATH and identify MATH with MATH. Subspaces MATH generate sub-MATH-fields MATH. Note that MATH generates MATH, the least MATH-field containing both MATH and MATH. The following two lemmas (and one definition) complete the proof. |
math/0006165 | `Only if': take a Gaussian measure MATH such that MATH makes MATH orthogonal, then MATH makes MATH independent. `NAME: some MATH makes MATH independent; however, MATH need not be Gaussian. We take any Gaussian measure MATH and introduce MATH-measurable densities MATH . The measure MATH still makes MATH independent, and... |
math/0006165 | We have MATH such that MATH and MATH are MATH-independent, while MATH and MATH are MATH-independent. Consider the MATH-measurable density MATH and the measure MATH, then MATH. The MATH-independence of MATH and MATH implies their MATH-independence. For every MATH, MATH, MATH . |
math/0006165 | We identify MATH with MATH of a Gaussian type space MATH and use the natural homeomorphism MATH between the space MATH of admissible norms and the space MATH of Gaussian measures. Subspaces MATH generate corresponding sub-MATH-fields MATH. The set MATH corresponds to MATH, where MATH consists of all measures making MAT... |
math/0006165 | We have to prove that every MATH is measurable with respect to MATH. It suffices to prove that the set MATH belongs to MATH for every MATH such that the set MATH is negligible (indeed, such MATH are dense in MATH). We take MATH such that MATH in MATH, consider sets MATH and note that MATH and MATH. |
math/0006165 | Assume that MATH is nonempty; we have MATH such that MATH and MATH are MATH-independent. Let MATH be a finite set and MATH a MATH-measurable function satisfying MATH. Take MATH-measurable functions MATH such that MATH in MATH and MATH. The MATH-independent MATH-fields MATH are also MATH-independent, where MATH (since M... |
math/0006165 | We choose an admissible norm on MATH, thus turning MATH into a NAME space. REF becomes MATH that is, MATH where the angle is defined by MATH. It is equivalent to MATH where MATH runs over finite-dimensional subspaces, and MATH. The following lemma completes the proof, provided that MATH tends to MATH slowly enough. Nam... |
math/0006165 | We equip MATH with the norm MATH, thus turning MATH into a NAME space, and consider orthogonal projections MATH onto MATH respectively. Introduce subspaces MATH, MATH, MATH (here MATH is the orthogonal complement of MATH); the subspaces are orthogonal to each other, and invariant under both MATH and MATH. Therefore MAT... |
math/0006165 | We choose an admissible norm on MATH, thus turning MATH into a NAME space. Let MATH be a finite-dimensional subspace, MATH. For any given MATH consider the pair MATH. Its geometry may be described (similarly to the proof of REF ) via angles MATH, MATH. This time, zero angles are allowed, since MATH need not be MATH. It... |
math/0006165 | Similarly to the proof of REF, we take MATH such that MATH are MATH-independent and MATH. We consider an arbitrary finite set MATH, an arbitrary MATH-measurable function MATH, and the corresponding vector MATH. We construct MATH-measurable functions MATH such that MATH in MATH, and corresponding vectors MATH. Independe... |
math/0006165 | REF gives us MATH such that MATH and MATH (both in the FHS sense), and MATH, and MATH is asymptotically orthogonal to MATH. REF gives us a representation MATH for an arbitrary unit vector MATH; here MATH are unit vectors of MATH, MATH are unit vectors of MATH, and these MATH (unlike MATH) do not depend on MATH. We have... |
math/0006165 | Let MATH, MATH, MATH, then MATH (the notation being self-explanatory), which implies MATH. For every MATH however, MATH and MATH (since MATH); we have MATH. Note that MATH, since MATH for all MATH. We have MATH so, MATH . Denote MATH. We need only one ray of vectors MATH such that MATH. For every MATH there exists such... |
math/0006165 | REF gives REF . Assume REF ; we have to prove REF . We choose a Gaussian measure MATH and apply REF to MATH: MATH which implies that MATH and MATH (when MATH) for all MATH, MATH (the dual to MATH). The latter, MATH, shows that MATH. The former, MATH, shows that MATH. |
math/0006165 | MATH, therefore MATH is a NAME subset of the standard NAME space MATH. |
math/0006165 | The projection is the limit (for MATH) of the orthogonal projection of MATH to MATH. Measurability of the latter implies that of the former. |
math/0006165 | The relation MATH means that, first, MATH are orthogonal in some admissible norm, and second, MATH is dense in MATH. The latter condition defines a measurable set of pairs MATH due to REF . The former condition may be expressed in terms of the infinite matrix MATH where MATH are as in REF. The relevant set of matrices ... |
math/0006165 | One may take MATH here MATH is equal to MATH if MATH and MATH otherwise. |
math/0006165 | CASE: The condition MATH may be expressed in terms of the measure MATH on MATH. (The choice of MATH does not matter.) That is the joint probability distribution of random variables MATH on MATH, and its marginal distributions are MATH. Clearly, MATH is a product measure if and only if MATH are independent with respect ... |
math/0006165 | CASE: Follows immediately from measurability of MATH in MATH. CASE: Let MATH be another probability measure on MATH, equivalent to MATH. Consider the distribution MATH we know that MATH is jointly measurable in MATH and MATH, and MATH is measurable in MATH, therefore MATH is jointly measurable in MATH and MATH. However... |
math/0006165 | We choose some MATH and replace each MATH with the corresponding MATH according to their unitary correspondence determined by MATH, MATH . The disjoint union of MATH over all MATH is naturally a standard NAME space, since it is a NAME subset of the disjoint union of all subspaces of MATH. Thus, a NAME structure appears... |
math/0006165 | Treating MATH as a subspace of MATH we choose measurable maps MATH such that every MATH is spanned by MATH. We have MATH where MATH, MATH. Applying the orthogonalization process we ensure that MATH form an orthogonal basis of MATH. Introducing NAME functions MATH for MATH we have MATH . |
math/0006165 | A sub-MATH-field MATH belongs to MATH if and only if MATH; here an element of MATH is identified with a pair MATH of elements of MATH. (The choice of a measure MATH does not matter.) For such MATH and MATH, MATH must be disjoint to the open set MATH. |
math/0006165 | Due to REF it suffices to prove measurability of the map MATH from MATH to MATH; of course, MATH means MATH. We know (see the proof of REF ) that MATH implies MATH. Therefore, for any open set MATH the set of all MATH such that MATH is closed. |
math/0006165 | We take a Gaussian type space MATH and consider MATH, so that MATH. Every subspace MATH generates a sub-MATH-field MATH. It is easy to see that MATH if and only if MATH. However, MATH is measurable in MATH by REF , and MATH is measurable by REF . |
math/0006165 | Consider some MATH treated as a sub-MATH-field of MATH. Given some MATH, we introduce on MATH a measure MATH for MATH; here, as before, an element of MATH is represented by a pair MATH where MATH. There is also a measure MATH on the MATH-field MATH such that MATH whenever MATH. The pair MATH belongs to the graph MATH t... |
math/0006165 | We know that every MATH generates MATH (see the proof of REF ), and the map MATH is measurable. The transition from MATH to the corresponding element of MATH is measurable by REF . |
math/0006165 | It is enough to consider the case MATH for an arbitrary MATH. The equality MATH, in combination with REF , reduces the problem to measurability of MATH in MATH, and MATH in MATH. However, a monotone function is always measurable. |
math/0006165 | By REF , MATH is measurable in MATH. According to REF there are MATH, measurable in MATH, such that every MATH is spanned by MATH. Therefore the graph of MATH is spanned by pairs MATH. Each pair is measurable in MATH and continuous in MATH (recall REF ), therefore, measurable in MATH (see CITE). |
math/0006165 | REF allows us to reformulate the conditions in terms of density matrices in product systems, thus making explicit their invariance under isomorphisms. |
math/0006165 | Choosing any MATH we have for large MATH where MATH, MATH. Taking into account that MATH and MATH, we get MATH . So, MATH which is the first claim of the lemma. In order to prove the second claim we note that the only properties of the function MATH used till now are the finite variation of MATH, and MATH for large MAT... |
math/0006165 | First, the function MATH on MATH is of finite variation, thus, using REF , MATH for MATH. Second, MATH, since MATH is of finite variation on MATH. Hence MATH and MATH . Third, MATH and MATH, hence MATH . It is enough to prove that MATH for every function MATH as in REF . Taking into account that MATH we transform it in... |
math/0006165 | Here is an equivalent formulation: there is an operator MATH such that MATH and MATH is an NAME, in other words, an equivalence operator in the sense of CITE. It means that MATH is one-to-one onto, has a bounded inverse, and MATH (the NAME class of operators). The latter is equivalent to MATH, see CITE. Matrix elements... |
math/0006165 | Vectors MATH (defined by REF) are orthogonal with respect tosome admissible norm on the NAME MATH; the proof is quite similar to the proof of REF . |
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