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math/0010096 | Suppose such an action exists. Recall that the MATH - fold join of MATH is MATH. Hence taking joins we can obtain a free action of MATH on MATH, which contradicts a result due to CITE. |
math/0010096 | The irreducible representations of MATH are described as follows: if MATH let MATH . Then MATH. Evaluating on MATH, MATH respectively yields MATH whence we readily infer that MATH and MATH. Hence we deduce that MATH if and only if MATH; but noting that MATH we conclude that MATH if and only if MATH. |
math/0010096 | Note that MATH, MATH, MATH, hence for odd primes MATH they have cyclic MATH-Sylow subgroup and we need only consider the prime MATH. By our lemma and the existence of the characters mentioned above, the free actions on MATH can be constructed using REF . |
math/0010096 | We first consider MATH, note that it has order MATH and that MATH. Now MATH is an extra-special MATH-group MATH with MATH central. Denote by MATH the conjugacy class of this central element of order REF. From the Atlas CITE we see that MATH has one conjugacy class of involutions MATH and two conjugacy classes of elemen... |
math/0010096 | Let MATH denote a group with normal MATH - NAME subgroup MATH. Consider an effective class MATH for MATH arising from a MATH - dimensional representation of MATH as in REF . Now induce this to a representation for MATH; by the double coset formula, the resulting NAME class will restrict in MATH to MATH, which is eviden... |
math/0010096 | The first part can be found in CITE, page REF. For the second part we use an argument due to NAME CITE. Note that MATH is the minimal degree of a complex irreducible character for MATH. This is also the minimal degree for a faithful irreducible of the NAME normalizer, which is a semidirect product MATH, where MATH is e... |
math/0010096 | Clearly an NAME class is effective if and only if its MATH-localization at each prime for which MATH is effective. Moreover if MATH is an NAME class then each of its localizations MATH is a MATH-local NAME class. Now assume that MATH and that for each prime MATH for which MATH one has a fibration MATH with homotopy fib... |
math/0010102 | The first part of the result follows immediately from REF. For the second part we observe that if a point MATH in MATH is covered by the ball MATH, then MATH; so the number MATH of the balls MATH, that cover MATH, is estimated by the greatest number MATH of points MATH in MATH with MATH and we observe that, by MATH, MA... |
math/0010102 | We have to prove that there is a subsequence of MATH convergent in MATH. Taking into account assumption MATH, the ball MATH can be covered by a finite number of balls MATH, MATH, MATH where MATH depends on MATH, such that every point of MATH belongs at most to MATH balls, where MATH does not depend on MATH. Let MATH an... |
math/0010102 | Let MATH. After extraction of a subsequence, we have that MATH is weakly convergent in MATH to MATH. We suppose, without loss of generality that MATH and prove MATH when MATH. Let MATH, MATH such that MATH when MATH. Since MATH when MATH, then MATH such that for MATH . Then for MATH . |
math/0010102 | Let MATH be a sequence in MATH such that MATH in MATH as MATH, where MATH denotes the dual space of MATH. From MATH we obtain that there exists MATH such that for MATH . Then MATH . Since MATH is bounded in MATH and from REF , we know that there exists a subsequence strongly convergent in MATH and weakly to MATH . We a... |
math/0010102 | First we prove that for MATH small enough MATH for MATH. Consider the case MATH. As in REF of Ref. CITE we obtain that for every MATH there exists a constant MATH such that MATH where MATH if MATH or MATH if MATH. There exists MATH such that MATH . Choose MATH; then MATH and MATH and the result follows from the last in... |
math/0010115 | The equivalence of REF has been shown for countably generated NAME MATH-modules in REF. The implication MATH can be seen to hold setting MATH for the existing unitary operator MATH and for MATH. The demonstration of the inverse implication requires slightly more work. For the given normalized tight frames MATH and MATH... |
math/0010115 | We have already pointed out that the set of algebraic generators MATH of MATH is a frame with respect to any MATH-valued inner product on MATH which turns MATH into a NAME MATH-module. That means the inequality MATH is satisfied for two finite positive real constants MATH and any MATH, see REF. What is more, for any fr... |
math/0010115 | Since both the modular NAME bases are sets of modular generators of MATH we obtain two MATH-valued (rectangular, without loss of generality) matrices MATH and MATH with MATH and MATH such that MATH . Combining these two sets of equalities in both the possible ways we obtain MATH for MATH, MATH. Now, since we deal with ... |
math/0010115 | Let MATH be the set of all linear bounded operators on the NAME space MATH, and recall that MATH as a direct sum of copies of the NAME space MATH itself. Consequently, there are projections MATH such that MATH for any MATH, MATH, and MATH via partial isometries MATH with MATH and MATH. Define MATH, that is, the set of ... |
math/0010115 | Denote the frame transform of the tight frame MATH by MATH for MATH. Because the frame transforms MATH and MATH all possess the same coisometries in their respective polar decompositions we obtain the equality MATH. Since the inequality MATH is valid and both MATH and MATH are diagonalizable with a common set of eigenv... |
math/0010117 | It is enough to show that the anti-commutators form a MATH-Gröbner basis. To do this, we observe that all obstructions resolve. |
math/0010117 | There are only finitely many square-free monomials on MATH, and only the trivial such monomial if MATH. |
math/0010117 | Let MATH be a stable (square-free monomial) ideal, and let MATH be a generator, MATH . Let furthermore MATH . Then the maximal index MATH such that MATH is MATH. Hence stability implies that MATH, and since MATH is an ideal, MATH. This shows that MATH. |
math/0010117 | We start by noting that if MATH then MATH does not divide MATH. This follows since MATH has no elements of degree REF, and since quadratic monomials are lifted to ``strictly ordered" quadratic monomials. Hence MATH is part of a minimal NAME basis, and MATH is part of a minimal generating system of MATH. Mimicking the p... |
math/0010117 | Follows from REF . We can also see this directly: since the exterior algebra has finite vector space dimension, so has MATH, hence so has MATH. Hence, for some MATH, MATH contains all monomials of total degree MATH. Hence, MATH is generated in degrees MATH and is therefore finitely generated. |
math/0010117 | It is proved in CITE that (regardless of the characteristic of MATH) the generic initial ideal of MATH is strongly stable. Hence, for a generic MATH, MATH is strongly stable, hence stable, hence squeezed. By REF , MATH has a NAME basis which is the ``lift" of a NAME basis of MATH. Since the anti-commutator ideal is inv... |
math/0010117 | The proof is almost word-for-word identical with the proof in CITE of the corresponding assertion for the symmetrical algebra. To stress the similarity, we temporarily denote MATH by MATH. We start by fixing a total degree MATH and a basis MATH for the MATH-vector space MATH. Then, we let MATH, noting that there is a R... |
math/0010117 | NAME series are preserved by non-singular linear changes of coordinates, and by passing to the initial ideal. |
math/0010117 | There exists a finitely presented algebra with non-rational NAME series CITE. By REF the gin of the defining ideal must have non-rational NAME series, too. Since finitely generated monomial ideals have rational NAME series CITE, the result follows. |
math/0010121 | CASE: This follows directly from the commutative diagram MATH in which the vertical maps are isomorphisms induced by the two projections MATH and MATH and the top row is the homomorphism induced by MATH. CASE: Let MATH denote the standard comultiplication. Write MATH as the composition MATH and MATH as the composition ... |
math/0010121 | It is immediate from REF that MATH induces the identity on homology groups. Since MATH is a REF-connected NAME, MATH is a homotopy equivalence. Hence MATH. This establishes REF , and REF follows similarly. |
math/0010121 | Let MATH. Since MATH for MATH, the hypothesis gives that MATH for MATH. Hence, by REF , MATH. Now suppose MATH is any element in MATH. Since MATH and MATH, we have that MATH. Thus, by the properties of the action listed in the introduction, we have MATH . Therefore MATH is a homomorphism. Now suppose MATH is any elemen... |
math/0010121 | This can be argued using the long exact homotopy and homology sequences, together with the relative NAME theorem, for the pair MATH. Alternatively, NAME minimal models can be used. We omit the details. |
math/0010121 | We proceed as in the proof of REF. First, we claim that MATH, for all MATH. Since MATH is odd, we have MATH for MATH. Hence we must only check that MATH in degree MATH. By REF, MATH is surjective. Given MATH, write MATH, for some MATH. Then MATH since MATH, and the claim follows. Now a simple modification of the proof ... |
math/0010121 | Suppose MATH is not the identity element in MATH. Since this latter is a MATH-local group, it contains no non-trivial elements of finite order, and hence MATH, for all MATH. By CITE, we have MATH, for each MATH with MATH. From this it follows that for each MATH there is some positive integer MATH and some element MATH ... |
math/0010121 | Let MATH and MATH be a complex with MATH and dim-MATH. It suffices to show that MATH for every MATH. Consider the last cofibre sequence in a length-MATH restricted spherical cone decomposition of MATH, MATH where MATH is a wedge of spheres and MATH. Now, since MATH, it follows that MATH. Thus, from the properties of th... |
math/0010121 | This follows by REF. |
math/0010121 | Any map MATH induces a corresponding map MATH of MATH'th NAME sections. This gives us a homomorphism MATH which is one-one. Therefore, MATH. But if MATH is the collection of all cyclic groups, then it is a consequence of REF that MATH. |
math/0010121 | If MATH, then the universal coefficient theorem (compare CITE) gives a commutative diagram with exact rows MATH where the middle homomorphism MATH can be either MATH or the identity REF. Thus there is a homomorphism MATH such that MATH. By hypothesis, MATH and so MATH. Therefore, MATH. |
math/0010121 | We shall prove that MATH acts nilpotently on MATH. Then it follows from CITE that MATH is a nilpotent group. Let MATH and consider the diagram with exact rows obtained from the universal coefficient theorem MATH . Then both MATH and MATH are identity maps. Write MATH as the sum of its free and torsion parts and let MAT... |
math/0010123 | Let MATH be the diagonal submonoid of MATH. Since MATH is finitely generated, it is rational. It follows that MATH is a product of rational sets and so is itself rational. |
math/0010123 | Pick an automaton accepting MATH. Reverse the orientation of each edge and invert the edge label. Make every terminal state an initial state and every initial state a terminal state. |
math/0010123 | Suppose MATH is accepted by a finite automaton MATH over MATH. Construct a context - free grammar with one nonterminal MATH for each vertex MATH of MATH. The start symbol is the nonterminal corresponding to the initial vertex. For each edge MATH there is a production MATH, and for each terminal vertex MATH there is ano... |
math/0010123 | If MATH is asynchronously automatic, then by CITE MATH is a rational transduction. The converse is CITE except that the automata used there are more restricted than ours. In terms of our notation the vertices of those automata are partitioned into two sets. All edges leaving the first set have labels from MATH, and all... |
math/0010123 | Let MATH be a point on MATH. If MATH, then the distance from MATH to MATH is at most MATH, the maximum length of the finitely many combing paths for elements MATH with MATH. Otherwise estimate the distance by constructing a sequence of triangles as in REF . Let MATH be a MATH - triangle whose base is MATH and whose thi... |
math/0010123 | By the Pumping Lemma for context - free languages there exists a constant MATH such that for all MATH each MATH of length MATH contains a subword MATH of length at most MATH which can be replaced by a shorter word MATH to obtain another element of MATH. Note that MATH as all words in MATH represent the same element of ... |
math/0010128 | To see that REF implies REF , suppose (as we may) that MATH is MATH-equivalent to MATH. Take MATH and let MATH be a MATH-dominated perturbation of MATH. Then MATH is norm bounded; and so, since MATH is equivalent to MATH, we may define a bounded linear operator MATH by setting MATH . Moreover, for MATH, we have, settin... |
math/0010128 | First note that if MATH is a semi-normalized sequence in MATH such that MATH for some MATH, then we necessarily have MATH for each MATH. Indeed, this is obvious for MATH, while for MATH and for MATH we have: MATH . Thus, MATH is semi-normalized if both MATH and MATH are, whence the result follows. In particular, REF ho... |
math/0010128 | We need to show that MATH for any sequence of scalars MATH. The right hand inequality follows from the triangle inequality. Towards proving the left hand inequality, let MATH be given and write MATH where MATH denotes the standard unit vector basis of MATH. Then for each MATH, we have MATH; whence, MATH as desired. |
math/0010128 | To construct the sequence heralded by REF , it suffices to find, for each integer MATH, a basis MATH of MATH such that both of the following statements hold: CASE: MATH is MATH-equivalent to the standard unit vector basis of MATH. CASE: MATH. Towards this end, let MATH and let MATH denote the standard unit vector basis... |
math/0010139 | It follows from CITE that the length is at least REF. Let now MATH denote a bounded unital homomorphism of MATH into MATH then by CITE there exists a *-representation MATH of MATH on MATH and an invertible MATH in MATH such that MATH. Let MATH then it follows that MATH also is a *-representation and we can - and will -... |
math/0010139 | Let MATH denote a bounded projection from MATH onto MATH. We will first prove that we may assume that MATH is generated - as a NAME algebra - by a countable set of operators. Suppose that the theorem has been proven for such NAME algebras. Then for a general finite factor as MATH and any NAME subalgebra say MATH of MAT... |
math/0010139 | As in the case above where the algebra is a complemented subspace of some larger C*-algebra we will like to show first that it is sufficient to prove the result for a NAME factor which is countably generated. So suppose that the result has been established in this case and let MATH denote a continuous REF-cocycle on MA... |
math/0010141 | Suppose MATH and MATH immerses in MATH. Then MATH has a thickening (an immersed in MATH regular neighborhood) which is an aspherical MATH-manifold with fundamental group MATH. But then the universal covering action violates REF . |
math/0010141 | The case of MATH was discussed above. For the general case, assume on the contrary that MATH violates the proposition. Define MATH by MATH. The following lemma then implies that MATH classifies into MATH, a contradiction. |
math/0010141 | We may assume that MATH since otherwise MATH is homeomorphic to MATH. Let MATH be a (small) open set homeomorphic to MATH. Consider the diagram MATH . Note that MATH fibers over MATH with fiber MATH; thus MATH and similarly MATH; moreover, inclusion MATH is a homotopy equivalence. Since for MATH the two spaces in the f... |
math/0010141 | Again we first consider the case MATH. Assuming the contrary, consider the homotopy MATH defined by declaring that MATH is the class of parallel rays containing the ray from MATH through MATH. Then MATH covers a classifying map and therefore has degree REF. Let MATH be a Euclidean ball centered at the origin containing... |
math/0010141 | Let MATH be the cycle for MATH. We define MATH to have twice as many elements: for every MATH put MATH and MATH into MATH. It is straightforward to check REF . To verify REF , choose a general position map MATH, and let MATH be a perturbation to a general position map. Put the cone point high above the hyperplane MATH ... |
math/0010141 | Let MATH be the cycle for MATH, MATH. We define MATH to have MATH elements: for each MATH and MATH we put the following two pairs in MATH: MATH and MATH. REF is vacuous for MATH as there are no MATH-simplices in MATH. To verify that MATH is a cycle, suppose first that MATH and MATH are disjoint simplices of MATH (with ... |
math/0010141 | The special cases when MATH are clear. Let MATH be a MATH-obstructor complex and MATH a proper, NAME, expanding map. Let MATH be a contractible manifold with a proper metric and MATH a uniformly proper NAME map satisfying the contractibility function requirement. Consider the composition MATH. Now extend MATH inductive... |
math/0010141 | The latter two statements are obvious, while the first one follows from the Join REF . The product MATH can naturally be viewed as MATH and the product map into MATH satisfies the requirements. (If one of the two groups is REF-ended, use the Cone Lemma instead.) |
math/0010141 | Let MATH be as in the definition. Since MATH is a MATH-set in MATH, there is a homotopy MATH with MATH, MATH and MATH. Restricting to MATH and reparametrizing yields an expanding map MATH. |
math/0010141 | Functions MATH are constructed by induction on MATH, with the case MATH being the definition. The inductive step consists of defining MATH on the boundary of MATH so that the last item above holds and then extending to the interior by coning off from the first vertex. More precisely, if MATH is a linear map with MATH a... |
math/0010141 | If MATH (or MATH) is finite, then MATH is quasi-isometric to MATH (or MATH) and equality holds. If MATH (or MATH) is REF-ended, we can use MATH (or MATH) in the proof below and appeal to the Cone Lemma. If both MATH and MATH are REF-ended, then MATH is virtually MATH and thus MATH, MATH, so equality again holds. Let MA... |
math/0010142 | We inductively find indices MATH, MATH, as above, open sets MATH and points MATH with MATH and MATH for some index MATH, so that MATH . This will prove the Lemma, for if MATH and MATH for some MATH then, since MATH it will follow that MATH. Since MATH, there is MATH with MATH. Let MATH. Using MATH and the form of the M... |
math/0010142 | Assume that MATH for some MATH-recurrent point MATH. We will find MATH such that MATH has nonzero spectral radius. We may scale MATH so that there exists a relatively compact open neighbourhood MATH of MATH such that MATH for all MATH. Since MATH when MATH, we may also assume that MATH. Since MATH is MATH-recurrent, th... |
math/0010142 | Let MATH be arbitrary and MATH. Then MATH which is zero since MATH is supported on MATH and MATH is supported on the disjoint set MATH. This shows that all NAME coefficients of MATH will vanish, and hence MATH. It follows that all products MATH vanish and hence MATH. On the other hand, choosing functions MATH equal to ... |
math/0010142 | Let MATH be a relatively compact open set. We wish to find a MATH-recurrent point in MATH. Since MATH is not MATH-wandering, there exists MATH such that MATH. Hence there is a nonempty, relatively compact, open set MATH with MATH such that MATH. Since MATH contains no MATH-wandering subsets, a similar argument shows th... |
math/0010142 | If the strongly recurrent points are dense in MATH, then by REF there are no nonzero monomials in the NAME radical of MATH. But we have already observed that an element MATH is in the NAME radical if and only if each monomial MATH is. Thus MATH is semisimple and hence semiprime. Suppose that MATH is semiprime. Then REF... |
math/0010142 | CASE: To see that MATH, note that if MATH then for every neighbourhood MATH of MATH (in MATH) the set MATH is a neighbourhood of MATH in the relative topology of MATH, so there exists MATH such that MATH. Thus MATH showing that MATH. On the other hand if MATH then for each relative neighbourhood MATH of MATH, since MAT... |
math/0010142 | As a MATH-recurrent point cannot be MATH-wandering, MATH. If MATH for some MATH, then by REF the set MATH of MATH-recurrent points of the subsystem MATH equals MATH, so MATH; but MATH, and so MATH. Finally, if MATH is a limit ordinal and we assume that MATH for all MATH then MATH. This shows that MATH and so MATH since... |
math/0010142 | If MATH is the support of MATH then MATH; in other words the compact set MATH consists of MATH-wandering points for MATH. This means that each MATH has an open neighbourhood MATH so that the (relatively open) set MATH is MATH-wandering for MATH. Each MATH has an open neighbourhood MATH such that MATH is empty (and so M... |
math/0010142 | Since MATH is dense in MATH, it suffices to prove that any MATH is contained in MATH. Suppose MATH . By REF we may write MATH as a finite sum MATH where each MATH is supported on a compact set which is MATH-wandering. Since MATH for some MATH as observed above, by REF we have MATH and so MATH. Thus MATH. Suppose the re... |
math/0010142 | Let MATH be a monomial contained in MATH and let MATH be the support of MATH. Then REF shows that MATH must vanish on MATH. On the other hand, let MATH be as in the statement of the Theorem, so that MATH vanishes on MATH (where MATH). We will show that MATH is in MATH. It is enough to suppose that the support MATH of M... |
math/0010148 | We re-write MATH. We shall find the values MATH such that MATH. Since MATH, we assume MATH. Applying NAME 's Theorem repeatedly for the base MATH representations (MATH), MATH and MATH, we obtain MATH . For MATH, we get MATH . Let MATH. Since for any MATH, MATH, if MATH, then MATH. Thus, the numbers MATH with MATH are t... |
math/0010148 | We use the identity MATH. Using REF we get that the power of MATH dividing MATH is MATH . Let MATH. Thus, REF becomes MATH since MATH. |
math/0010148 | As before MATH. A number MATH, which does not satisfy the divisibility, must satisfy (see REF ) MATH which implies MATH. Assume first that MATH. Therefore, MATH, therefore MATH must be even, say MATH, so MATH . Assume now that MATH, and MATH. It follows that MATH, MATH. Therefore, MATH and since MATH, we get that MATH.... |
math/0010148 | Straightforward. |
math/0010148 | By REF , if MATH, then MATH so MATH . If MATH, then MATH, for some MATH. Working modulo MATH implies MATH. Thus, MATH. Assume MATH . We obtain MATH. Modulo MATH, this transforms into MATH which will imply MATH, for some MATH. Since MATH, we get MATH. We obtain, for MATH, MATH, MATH with the condition MATH. |
math/0010148 | First we prove that MATH. By squaring we get MATH which is equivalent to MATH . The last inequality is equivalent to MATH, which is certainly true if MATH. Now, let MATH. We evaluate MATH . Therefore, MATH . Taking MATH, in REF, we get MATH . If MATH, then REF implies our claim that REF is true for MATH sufficiently la... |
math/0010148 | If MATH, REF , valid for MATH changes into MATH which is certainly true for MATH. REF imply that the exceptions (if they exist) are of the form MATH. Therefore, since the number of pairs MATH, giving different numbers of the above form, is less than MATH, we get the theorem. |
math/0010148 | If MATH, REF , valid for MATH, changes into MATH which is true for MATH. As in the previous proof, we get that the exceptions (if they exist) are of the form MATH. Therefore, since the number of triples MATH, giving different numbers of the above form, is less than MATH, we get the result. |
math/0010149 | We evaluate MATH . If MATH odd, then associating MATH, we get MATH . If MATH even, then then associating MATH, except for the middle term, we get MATH . |
math/0010149 | We use REF . The first two identities are straightforward. Now, MATH since MATH and MATH. |
math/0010149 | We evaluate MATH . Assume MATH odd. Then, associating MATH, we get MATH . Assume MATH even. Then, as before, associating MATH, except for the middle term, we get MATH . |
math/0010149 | Let MATH . If MATH, then MATH, and MATH . |
math/0010149 | Straightforward using the NAME formula for MATH and MATH. |
math/0010149 | We use REF . Associating MATH, except for the middle term in MATH, we obtain MATH . We did not insert the middle term, since it is equal to MATH . Assume first that MATH is odd. Using REF into REF, and observing that MATH, we get MATH . Assume MATH even. As before, MATH . In the same way, associating MATH, except for t... |
math/0010149 | The second, third and fifth identities follow from the previous theorem. Now, using REF , with MATH, we get MATH . Furthermore, the fourth identity follows from MATH since MATH. |
math/0010149 | We use MATH in REF . Associating MATH in MATH, we obtain MATH since MATH and MATH, by REF . In the same way, associating MATH, with the middle term zero, MATH since MATH and MATH, by REF . Therefore, for MATH even, MATH and for MATH odd, MATH . |
math/0010149 | The first identity is simple application of REF . The identities for even powers are consequences of REF . Now, using REF , we get MATH since MATH. |
math/0010153 | As we mentioned the simplicial relations are already held and it remains to check the following extra relations: MATH . To prove REF , we first compute MATH : MATH . By a similar argument, we can deduce MATH . Continuing, MATH and eventually, MATH . We leave it to the reader to check the remaining equations. To prove t... |
math/0010153 | We should verify that MATH commutes with MATH . For MATH it is obvious that MATH commutes with MATH . For MATH we have MATH where we have made use of the MATH-trace property of MATH. We leave it to the reader to check MATH. Finally, we show that MATH commutes with MATH: MATH and MATH . |
math/0010153 | We must show that MATH commutes with MATH and MATH. In this proof we show MATH and leave the other cases to the reader. At first, let MATH, so MATH. Therefore MATH. Now, MATH so that MATH . Next, we have MATH . On the other hand, MATH . |
math/0010153 | It can be shown that MATH. |
math/0010153 | As always it is needed to verify the relations REF,.,REF We only check REF and leave the others to the reader. MATH . By a similar argument we get, MATH and finally we have MATH . |
math/0010153 | We leave it to the reader the first part of proof and just prove the second part. We must verify that the following diagram is commutative. MATH . We have MATH and MATH . |
math/0010153 | By the above remark we have MATH where MATH denotes cyclic double complex MATH. Since MATH is contractible and MATH is a Homotopy equivalence the double complex MATH is a resolution for MATH, where MATH is MATH . On the other hand, MATH . So, to complete the proof it suffices to compute MATH . But by finding a NAME res... |
math/0010153 | If MATH then it is obvious that MATH is a simplicial module by the following faces and degeneracies: MATH . Now, MATH can be computed by the above complex. Let MATH . We leave to the reader to show MATH is a simplicial map and in fact, MATH is a simplicial isomorphism by the following inverse map MATH . |
math/0010155 | Let MATH and MATH be two mutually independent sequences of NAME variables (that is, each is complex-valued and uniformly distributed on the unit circle). By applying the unconditionality of the NAME and the NAME inequality it is sufficient to show the existence of a constant MATH so that for any MATH we have: MATH . To... |
math/0010155 | CASE: We use the remark that it is enough to establish REF for distinct operators MATH . If MATH and MATH then MATH and hence MATH . This proves REF and indeed a rather stronger result. CASE: Let MATH where MATH and MATH is a finitely nonzero collection of complex numbers with MATH . Suppose MATH . We first note that M... |
math/0010155 | We give the proof in the NAME case, the others being similar. We first make the observation that it suffices to consider the case when MATH as one can make the transformation MATH . In this case we have the formula MATH . We write MATH and MATH so that MATH . Note that for a suitable constant MATH we have an estimate M... |
math/0010155 | We can assume MATH . For a suitable constants MATH and MATH we have MATH and the last quantity is finite. For the last part observe that for any bounded sequence MATH and MATH the series MATH must converge to MATH where MATH . |
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