paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0004047 | By REF we can find a sequence MATH of complete lattices such that: For all MATH, every monotone MATH is represented by a polynomial with coefficients in MATH. Moreover, MATH is an end extension of MATH. Let MATH. It is easy to see that MATH is complete. [If MATH, let MATH. Since the sequence MATH is weakly decreasing, ... |
math/0004047 | We only give a sketch of the main ideas of the proof. The details will appear elsewhere. Start with a lattice MATH. Let MATH that is, MATH is the ``horizontal sum" of the disjoint ortholattices MATH and MATH, where we of course identify the two top elements of the two lattices, and also the two bottom elements. The fun... |
math/0004049 | The implications REF are trivial. To show REF assume that MATH and fix a base MATH of zero neighborhoods. If MATH is nb-bounded then MATH is bounded for some MATH. Note that MATH is another base of zero neighborhoods. For each MATH in MATH we have MATH. But MATH is bounded and so MATH for some positive MATH, that is, M... |
math/0004049 | The first equality in REF follows immediately from the definition of MATH. We obviously have MATH . In order to prove the opposite inequality, notice that if MATH, then MATH. Thus, it is left to show that MATH implies MATH. Pick any MATH with MATH, then MATH so that MATH for MATH. Further, since MATH converges to MATH ... |
math/0004049 | Fix a bounded set MATH. Since MATH converges to zero uniformly on bounded sets then for every base zero neighborhood MATH there exists an index MATH such that MATH whenever MATH. This yields MATH since MATH is bounded. Thus, MATH is bounded for every bounded set MATH, so that MATH is bb-bounded. |
math/0004049 | Fix a zero neighborhood MATH, there exist zero neighborhoods MATH and MATH and an index MATH such that MATH and MATH whenever MATH. Fix MATH. The continuity of MATH guarantees that there exists a zero neighborhood MATH such that MATH. Since MATH, we get MATH, which shows that MATH is continuous. |
math/0004049 | Suppose MATH. If MATH, then MATH for some MATH and some subsequence MATH, so that MATH as MATH goes to infinity. It follows that MATH. On the other hand, if MATH is finite and MATH then MATH for some MATH and for all sufficiently large MATH. Then MATH. |
math/0004049 | Let MATH be a linear operator on a topological vector space MATH. Since every singleton is bounded then MATH. Next, assume MATH, fix MATH such that MATH, then the sequence MATH converges to zero equicontinuously. Take a bounded set MATH and a zero neighborhood MATH. There exists a zero neighborhood MATH and a positive ... |
math/0004049 | To prove REF let MATH . Since every convergent sequence is bounded, we certainly have MATH. Conversely, suppose MATH, and take any positive scalar MATH such that MATH. Then for every MATH the sequence MATH is bounded, and it follows that the sequence MATH converges to zero, so that MATH and, therefore MATH. To prove RE... |
math/0004049 | It follows from the definition of MATH and REF that MATH . Similarly, since the balanced convex hull of a bounded set is bounded, MATH . Let MATH for every MATH. Then, rephrasing the definition of MATH and applying REF , we have MATH . Similarly, MATH . Finally, MATH . |
math/0004049 | To prove the first equality it suffices to show that MATH if and only if MATH whenever MATH is an ultimately bounded net. Suppose that MATH, and let MATH be a zero neighborhood. One can find a zero neighborhood MATH such that for every MATH there exists MATH such that MATH for each MATH. Let MATH be an ultimately bound... |
math/0004049 | Suppose MATH and MATH and let MATH be an ultimately bounded net in MATH. Then the net MATH is ultimately bounded by REF . By applying REF again we conclude that MATH converges to zero. In particular, MATH, and applying REF one more time we get MATH. |
math/0004049 | Assume without loss of generality that both MATH and MATH are finite. Suppose that MATH and take MATH and MATH such that MATH. Let MATH be an ultimately bounded net in MATH. By REF it suffices to show that MATH. Notice that the net MATH is ultimately bounded. This implies that the net MATH converges to zero. Fix a semi... |
math/0004049 | Suppose MATH is a fast null sequence in a topological vector space and MATH. Let MATH, the sequence MATH converges to zero, hence is ultimately bounded, then by REF we have MATH . |
math/0004049 | For any MATH such that MATH one can find MATH such that MATH and MATH. Consider a point MATH and a base zero neighborhood MATH. Since by the definition of MATH the sequence MATH converges to zero, there exist a positive integer MATH, such that MATH whenever MATH. Therefore, MATH because MATH is balanced. Thus, if MATH,... |
math/0004049 | Suppose that MATH, then the sum MATH of the NAME series exists by REF . As in the proof of REF we denote the partial sums of the NAME series by MATH. Fix MATH such that MATH and MATH, and consider a bounded set MATH and a closed base zero neighborhood MATH. Since MATH converges to zero uniformly on MATH, there exits MA... |
math/0004049 | Let MATH. It follows from REF that the NAME series converges to MATH. Again, we denote the partial sums of the NAME series by MATH. Let MATH be such that MATH and MATH. For a fixed closed zero neighborhood MATH there exists a zero neighborhood MATH such that MATH for every MATH. Let MATH, then MATH for some MATH. Then ... |
math/0004049 | Let MATH. By REF the NAME series MATH converges to MATH. Again, we denote the partial sums of the NAME series by MATH. Fix some MATH such that MATH and MATH. There exists a base MATH of closed convex zero neighborhoods such that for every MATH there is a scalar MATH such that MATH for all MATH. Fix MATH, then for each ... |
math/0004049 | Let MATH. By REF the NAME series MATH converges to MATH. Since MATH then MATH is bb-bounded by REF . But then MATH. Notice that MATH is nb-bounded as a product of a bb-bounded and an nb-bounded operators (see REF). Suppose that MATH. Fix MATH such that MATH and MATH, then the sequence MATH converges to zero uniformly o... |
math/0004049 | If MATH is locally bounded then the identity map is an nb-bounded bijection. Suppose that MATH is an nb-bounded bijection on MATH. Then there exists a closed base zero neighborhood MATH in MATH such that MATH is bounded. Let MATH, then MATH is convex, bounded, balanced, and absorbing. It follows that the space MATH is ... |
math/0004049 | If every zero neighborhood of MATH contains a non-trivial linear subspace, then MATH cannot be one-to-one by REF. Suppose now that MATH and MATH are NAME and assume that MATH is a bijection. Let MATH be the linear inverse of MATH. The Open Mapping Theorem implies that MATH is continuous and hence bb-bounded. It follows... |
math/0004049 | Define a linear map MATH from MATH to MATH via MATH. Then the dimension of the range MATH is at most MATH. Define also a linear map MATH from MATH to MATH via MATH. It can be easily verified that MATH is well-defined. Then the range of MATH coincides with the range MATH, which is of dimension at most MATH. |
math/0004049 | Suppose MATH maps some weak base zero neighborhood MATH (MATH), to a weakly bounded set. Since the weak topology is NAME, it follows from REF that MATH. In particular, MATH whenever MATH for every MATH. Then REF implies that MATH is a finite rank operator. |
math/0004049 | If MATH is locally bounded then the result is trivial by REF. Suppose that MATH is not locally bounded, then, in view of REF, it suffices to show that MATH. Let MATH, then MATH is bb-bounded. If MATH, then it follows from MATH that MATH. Thus, MATH is a sum of an nb-bounded operator and a multiple of the identity opera... |
math/0004049 | By REF it suffices to show that MATH. Since MATH is nb-bounded, then MATH is a bounded set for some zero neighborhood MATH. Let MATH and fix a zero neighborhood MATH. Then MATH is again a zero neighborhood. In particular, since the sequence MATH converges to zero uniformly on bounded sets, we have MATH for all sufficie... |
math/0004049 | Suppose MATH is bounded for some base zero neighborhood MATH. It follows from REF , and REF, and REF that it suffices to show that MATH. We are going to show that MATH induces a continuous operator MATH on some NAME space such that MATH while MATH, and then appeal to the fact that the spectral radius of a continuous op... |
math/0004049 | Assume that MATH. Without loss of generality (by scaling MATH) we can assume that MATH. Since MATH is compact, there is a closed base zero neighborhood MATH such that MATH is compact. In particular MATH is bounded, so that MATH for some MATH. We can assume without loss of generality that MATH. We define the following s... |
math/0004049 | It can be easily verified that MATH is a bijection from MATH onto MATH and, therefore, the restriction MATH is a bijection. Notice that since MATH for each MATH and MATH for each MATH, then MATH so that MATH and MATH commute on MATH. It also follows that for each MATH we have MATH . It follows that MATH for each MATH, ... |
math/0004049 | Suppose that MATH and consider the resolvent operator MATH on MATH. Then MATH hence MATH is nn-bounded and MATH. Suppose now that MATH is dense in MATH, MATH is the smallest closed extension of MATH, and MATH. Then there exists an nn-bounded operator MATH such that MATH for each MATH. Then there is a constant MATH such... |
math/0004051 | Given a functor MATH, we define MATH and MATH. Since MATH is a left adjoint, it preserves colimits. The structure maps of the colimit are then the composites MATH . Although MATH does not preserve limits, there is still a natural map MATH for any functor MATH. Then the structure maps of the limit are the composites MAT... |
math/0004051 | Given MATH, we define MATH, with structure maps MATH. Given a map MATH, we define MATH by MATH. |
math/0004051 | To prolong MATH, take MATH to be the identity in REF . To prolong its right adjoint MATH, take MATH to be the composite MATH of the counit and unit of the adjunction. It is a somewhat involved exercise in adjoint functors to verify that the resulting prolongations MATH and MATH are still adjoint to each other. Like all... |
math/0004051 | The main point is that MATH commutes with colimits. We leave the remainder of the proof to the reader. |
math/0004051 | By adjunction, a map MATH has the left lifting property with respect to MATH if and only if MATH has the left lifting property with respect to MATH. Since a map is a cofibration (respectively, trivial cofibration) in MATH if and only if it has the left lifting property with respect to all trivial fibrations (respective... |
math/0004051 | Since MATH is a left NAME functor, every map in MATH is a level cofibration. By REF , this means that MATH for all MATH and all trivial fibrations MATH. Since a map in MATH has the left lifting property with respect to every map in MATH, in particular it has the left lifting property with respect to MATH. Another appli... |
math/0004051 | The retract and two out of three axioms are immediate, as is the lifting axiom for a projective cofibration and a level trivial fibration. By adjointness, a map is a level trivial fibration if and only if it is in MATH. Hence a map is a projective cofibration if and only if it is in MATH. The small object argument CITE... |
math/0004051 | We only prove the cofibration case, leaving the similar trivial cofibration case to the reader. First suppose MATH is a projective cofibration. We have already seen in REF that MATH is a cofibration. Suppose MATH is a trivial fibration in MATH, and suppose we have a commutative diagram MATH . We must construct a lift i... |
math/0004051 | The functor MATH obviously takes level fibrations to fibrations and level trivial fibrations to trivial fibrations. Hence MATH is a right NAME functor, and so its left adjoint MATH is a left NAME functor. Similarly, the prolongation of MATH to a functor MATH preserves level fibrations and level trivial fibrations, so i... |
math/0004051 | Suppose first that MATH does induce a NAME equivalence on the localizations, and suppose that MATH is MATH-local. Then MATH is also MATH-local. Let MATH denote a fibrant replacement functor in MATH. Then, because MATH is a NAME equivalence on the localizations, the map MATH is a weak equivalence in MATH (see CITE). But... |
math/0004051 | The only if half is clear. Since every cofibrant object is a retract of a cell complex (that is, an object MATH such that the map MATH is a transfinite composition of pushouts of maps of MATH), it suffices to show that MATH, or, equivalently, MATH, is a weak equivalence for all cell complexes MATH. Given a cell complex... |
math/0004051 | By definition, MATH is MATH-local if and only if MATH is level fibrant and MATH is a weak equivalence for all MATH and all domains and codomains MATH of maps of MATH. By the comments preceding REF , this is equivalent to requiring that MATH be level fibrant and that the map MATH be a weak equivalence for all MATH and a... |
math/0004051 | We know already that the cofibrations are the same in the stable model structure on MATH and the stable model structure of CITE. We will show that the weak equivalences are the same. In any model category at all, a map MATH is a weak equivalence if and only if MATH is a weak equivalence of simplicial sets for all fibra... |
math/0004051 | In view of NAME 's localization REF , we must show that MATH is a stable equivalence for all MATH. Since the domains and codomains of the maps of MATH are already cofibrant, it is equivalent to show that MATH is a stable equivalence for all MATH. Since MATH, we have MATH. In view of REF , this map is a weak equivalence... |
math/0004051 | There is a a natural map MATH which is a weak equivalence when MATH is a stably fibrant object of MATH. This means that the total right derived functor MATH is naturally isomorphic to the identity functor on MATH (where we use the stable model structure). On the other hand, MATH is naturally isomorphic to MATH and also... |
math/0004051 | The map MATH is the adjoint of the structure map MATH of MATH, where MATH denotes the counit of the adjunction, MATH denotes the unit, and MATH denotes the structure map of MATH. Thus MATH is the composite MATH . Since MATH is the identity, it follows that MATH. |
math/0004051 | The map MATH is the colimit of the vertical maps in the diagram below. MATH . Since the vertical and horizontal maps coincide, the result follows. For the second statement, we note that if MATH is level fibrant, each MATH is level fibrant since MATH is a right NAME functor (with respect to the projective model structur... |
math/0004051 | By assumption, the map MATH is a level equivalence between level fibrant objects. Since MATH is a right NAME functor, MATH is a level equivalence as well. Then the method of CITE completes the proof. Recall that this method is to use factorization to construct a sequence of projective trivial cofibrations MATH with MAT... |
math/0004051 | By definition, MATH is a stable equivalence if and only if MATH is a weak equivalence for all MATH-spectra MATH. But we have a level equivalence MATH by REF , and so it suffices to know that MATH is a weak equivalence for all MATH-spectra MATH. But, by REF , MATH is an isomorphism. |
math/0004051 | Suppose MATH is a MATH-spectrum such that the map MATH is an isomorphism. We will show that MATH as a retract of MATH; this will obviously complete the proof. We first note that there is a natural map MATH obtained by precomposition with MATH and postcomposition with MATH. Here MATH is the inverse of the map MATH, whic... |
math/0004051 | One can easily check that MATH is an isomorphism, using REF . |
math/0004051 | The first statement follows immediately from REF . By the first statement, if MATH is a stable equivalence, so is MATH. Since MATH is a map between MATH-spectra, it is a stable equivalence if and only if it is a level equivalence. The converse follows from REF . |
math/0004051 | We have MATH, by REF , where MATH denotes the left homotopy relation. We can use the cylinder object MATH as the source for our left homotopies. Then adjointness implies that MATH. Since MATH and MATH are finitely presented, we get the required result. |
math/0004051 | We wil actually show that, if MATH is a level fibration and MATH is a stable equivalence, the pullback MATH is a stable equivalence. The first step is to use the right properness of the projective model structure on MATH to reduce to the case where MATH and MATH are level fibrant. Indeed, let MATH, MATH, and MATH. Then... |
math/0004051 | We first point out that the right adjoint MATH reflects weak equivalences between fibrant objects. Indeed, suppose MATH and MATH are MATH-spectra, and MATH is a map such that MATH is a weak equivalence. Then, because MATH and MATH are MATH-spectra, MATH is a weak equivalence for all MATH. Since MATH is a NAME equivalen... |
math/0004051 | Suppose MATH has right adjoint MATH, MATH has right adjoint MATH, and MATH has right adjoint MATH. The natural transformation MATH induces a dual natural transformation MATH. Define MATH by MATH, with structure maps adjoint to the composite MATH where MATH is adjoint to the structure map of MATH. The functor MATH is an... |
math/0004051 | By REF there is a map of pairs MATH induced by MATH. By REF , MATH is a NAME equivalence. Define MATH to be the composite MATH. |
math/0004051 | We will first show that MATH is a NAME equivalence on the projective model structures. Use the same notation as in the proof of REF , so that MATH denotes the right adjoint of MATH. Then, since MATH is a NAME equivalence, MATH reflects weak equivalences between fibrant objects, by CITE. It follows that MATH reflects le... |
math/0004051 | We define the action of MATH on MATH levelwise. That is, given MATH and MATH, we define MATH. The structure map is given by MATH . One can easily verify that this makes MATH tensored over MATH. Similarly, define MATH, with structure maps MATH adjoint to the composite MATH where MATH is the evaluation map, adjoint to th... |
math/0004051 | Given a left proper cellular MATH-model category MATH, we have seen in REF that MATH is a MATH-model category. Just as in REF , a MATH-Quillen functor MATH induces a functor MATH, as does its right adjoint MATH. Since MATH is defined levelwise, it preserves the action of MATH. It is easy to check that MATH preserves le... |
math/0004051 | We first show that the pushout product MATH is a (trivial) cofibration when MATH is a cofibration in MATH, and MATH is a cofibration in MATH (and one of them is a level equivalence). As explained in CITE. we may as well assume that MATH and MATH belong to the sets of generating cofibrations or generating trivial cofibr... |
math/0004051 | We only prove the cofibration case, as the trivial cofibration case is analogous. If each map MATH is a cofibration, then we can show that MATH is a projective cofibration by showing MATH has the left lifting property with respect to level trivial fibrations. Indeed, we construct a lift by induction, just as in the pro... |
math/0004051 | We prove this theorem in the same way as REF . Since the cofibrations in the stable model structure are the same as the cofibrations in the projective model structure, the only thing to check is that MATH is a stable equivalence when MATH and MATH are cofibrations and one of them is a stable equivalence. We may as well... |
math/0004051 | Note that MATH is certainly enriched, tensored, and cotensored over MATH. Now use the equivalence of categories MATH to transport this structure back to MATH. |
math/0004051 | The functor MATH induces a MATH-functor MATH, which takes the symmetric sequence MATH to the symmetric sequence MATH. It follows that MATH induces a MATH-functor MATH, that takes the symmetric spectrum MATH to the symmetric spectrum MATH, with structure maps MATH . Let MATH denote the right adjoint of MATH. Then the ri... |
math/0004051 | The map MATH induces a map of commutative monoids MATH. This induces the usual induction and restriction adjunction MATH . That is, if MATH is in MATH, then MATH. Restriction obviously preserves level fibrations and level equivalences, so is a NAME functor with respect to the projective model structure. One can easily ... |
math/0004051 | We take MATH, where the functor MATH used to form MATH is defined by MATH and is part of the MATH-model structure of MATH (see the comment following REF ). This means that the structure map of MATH involves the twist map. By REF , MATH is a MATH-Quillen equivalence. On the other hand, consider MATH, where now we use th... |
math/0004051 | We let MATH be the mapping cylinder of MATH. That is, we take MATH to be the pushout in the diagram below. MATH . The map MATH is then the composite MATH, and the map MATH is the map that is MATH on MATH and MATH on MATH. It follows that MATH, as required. The map MATH is defined to be the identity on MATH and the comp... |
math/0004051 | The reader is well-advised to draw a picture in the topological or simplicial case, from which the proof should be clear. We think of MATH as the interval whose left half is MATH and whose right half is MATH. In particular, MATH is a cylinder object for MATH, where MATH and MATH. Then, because the tensor product preser... |
math/0004051 | We define MATH, MATH, MATH and a homotopy MATH from MATH to MATH, where MATH is a unit interval, inductively on MATH, using REF . To get started, we take MATH, MATH to be the identity, MATH to be MATH, and MATH to be the constant homotopy (with MATH). For the inductive step, we apply REF to the diagram MATH and the hom... |
math/0004051 | We first reduce to the case where MATH is itself symmetric. So suppose the generating cofibrations of MATH have cofibrant domains, and suppose MATH is symmetric and weakly equivalent to MATH; this means there are weak equivalences MATH, where MATH denotes a fibrant replacement functor. This means the total left derived... |
math/0004051 | For MATH, this follows immediately from the definition of MATH, REF , and REF . The proof for MATH is similar. |
math/0004051 | This is immediate, since limits in MATH and MATH are taken levelwise. |
math/0004051 | We will prove the proposition only for MATH, as the MATH case is similar. Throughout this proof we will use REF , which guarantees that subcomplexes in MATH are determined by their cells. Choose an infinite cardinal MATH such that the domains and codomains of MATH are all MATH-compact relative to MATH. When dealing wit... |
math/0004052 | If the action is MATH-filling, let MATH be nonempty open subsets of MATH. There exist elements MATH, with MATH, such that MATH, MATH. By hypothesis there exist MATH such that MATH. Then if MATH there exists MATH such that MATH. Therefore MATH, that is, MATH. Thus MATH. Conversely, suppose the stated assertion holds. Fi... |
math/0004052 | (Inspired by CITE.) Denote by MATH the canonical conditional expectation. Fix MATH. In order to prove the result it is enough to show that there exist MATH such that MATH. Put MATH. Let MATH. There exists MATH such that MATH. Write MATH, where MATH and MATH is finite. Note that MATH, and so MATH. It follows that MATH .... |
math/0004052 | There are two cases to consider. CASE: Suppose that MATH is not an isolated point of MATH. Then there exist pairwise disjoint nonempty open sets MATH contained in MATH. Let MATH be the MATH-subalgebra of MATH generated by MATH. By functional calculus, there exist MATH, MATH with MATH and MATH. CASE: Suppose that MATH i... |
math/0004052 | By REF , there exist MATH, with MATH, MATH, MATH for MATH, and MATH. Since the action is MATH-filling, there exist MATH such that MATH. Therefore MATH . Put MATH. Now MATH and so MATH. Finally, we have MATH . |
math/0004052 | Denote by MATH the canonical map from MATH onto MATH. We first show that the action of MATH on MATH is not MATH-filling. Choose a linear subspace MATH of MATH of dimension MATH. Let MATH, which is a nonempty open subset of MATH. If MATH REF then MATH. For the subspace MATH of MATH has dimension at least one, and so con... |
math/0004052 | Choose MATH with MATH and an open neighbourhood MATH of MATH such that MATH. Since the action is minimal, the family MATH forms an open covering of MATH. By compactness, there exists a finite subcovering MATH. Let MATH be nonempty open subsets of MATH. Since the action of MATH on MATH is minimal, we may choose elements... |
math/0004052 | This is an easy consequence of the definitions. |
math/0004053 | Apply NAME 's three term recurrence relation CITE with its parameters specialized as follows: MATH . This gives an expansion of the form REF for explicit coefficients MATH and MATH, which at a first glance still depend on MATH. Using the functional equation MATH for theta functions, it is easily seen that the coefficie... |
math/0004053 | Direct verification. |
math/0004053 | We substitute the definition of the dual NAME function transform MATH into the left hand side of the desired identity, and we use that MATH for MATH and MATH, see REF for the second equality. We arrive at MATH . Since MATH, all integrations are over compact subsets, so we may use NAME 's theorem to interchange the orde... |
math/0004053 | The proof of the desired identity simplifies when we slightly perturb the parameters MATH. The proof for parameters MATH can then be derived from the perturbed case using the fact that the left and right hand side of the identity depend continuously on the parameters MATH. Let us indicate one class of possible perturba... |
math/0004053 | We first observe that the asymptotic behaviour of the discrete weights MATH REF is given by MATH where MATH is the positive constant MATH . This is a direct consequence of the explicit REF for MATH (MATH). For the moment we assume MATH and MATH in order to be able to apply the MATH-function expansion for MATH (MATH). W... |
math/0004053 | As in the proof of REF , we assume for the moment that MATH and that MATH. In view of REF, these generic conditions can be removed at the end of the proof by applying the dominated convergence theorem. We fix MATH, then by REF , MATH where we used that MATH is real valued on MATH for the first equality. It remains to e... |
math/0004053 | The proof is similar to the proof of CITE and of CITE, where the analogous statement was derived for the big and the little MATH-Jacobi function transforms, respectively. Since some care has to be taken in order to match the constants, we repeat here the proof in some detail. We prove the proposition with respect to du... |
math/0004053 | We establish the desired identity with respect to dual parameters. Let MATH and MATH. By REF we have MATH and MATH respectively. In particular, we may assume without loss of generality that MATH is continuous on MATH, and supported within MATH. For MATH we now define MATH by MATH. Observe that MATH is continuous on MAT... |
math/0004054 | We use the explicit expression of MATH to perform an estimate of MATH: MATH . We introduce the notation MATH and the change of variable MATH . The integral MATH is now given by MATH . We use the obvious inequalities MATH to infer that MATH . It is now clear that MATH is bounded independently of MATH, that is, of MATH b... |
math/0004054 | By an integration by parts, MATH . We estimate MATH: we first observe that MATH . We estimate MATH from below by MATH; we also observe that MATH is bounded, thanks to estimates REF; therefore we have the following estimate, where we have used again REF : MATH . Next step is to calculate MATH: we use REF and we find tha... |
math/0004054 | The easiest part is the estimate on MATH . We can see that the absolute value of this expression is estimated by MATH . We recognize expressions which have already been estimated in REF. Therefore, it is immediate that MATH . Next comes the slightly more complicated expression MATH . For all MATH, MATH is at most equal... |
math/0004054 | The ball of radius MATH about MATH will be invariant by the mapping MATH provided that MATH . Choose MATH small enough for the following inequality to hold MATH . Choose then MATH so large that REF holds. Let us prove now that for an adequate choice of MATH and MATH and for all MATH, the mapping MATH is a strict contra... |
math/0004054 | The first statement is an almost immediate consequence of REF: we have MATH so that on MATH and REF follows. In order to compare MATH and MATH, we write the differential equations that they satisfy: MATH . Therefore, if we subtract the second of these equations from the first, we deduce that MATH . The initial data van... |
math/0004054 | The differential equation satisfied by MATH defined by REF is deduced from REF and is given by MATH . Recall that the principal part MATH is defined by REF. Let us estimate MATH: MATH satisfies the differential equation MATH . Therefore, if we let MATH, and if we denote MATH we find that MATH . Hence, for all MATH, MAT... |
math/0004054 | The initial part of the motion is described thanks to REF. Estimate REF proves that MATH tends to MATH uniformly on compact sets of MATH; relation REF enables us to conclude. |
math/0004054 | We will show in REF that the expression MATH tends to MATH as MATH tends to infinity, uniformly on MATH. If MATH, then MATH and in consequence, MATH . Let us estimate MATH when MATH is at most equal to MATH. We can see that MATH . Therefore, for MATH large enough, MATH is a strict contraction from the ball of radius MA... |
math/0004054 | The integral MATH is analogous to the one defined in REF. We define MATH by MATH and we consider three cases: CASE: MATH: in this case MATH cannot be neglected relatively to MATH. CASE: MATH: in this interval, the dominant term in MATH will be MATH and an elementary computation shows that this expression vanishes for M... |
math/0004054 | REF implies the uniform equivalence REF, and REF is an immediate consequence of REF. Let us prove an estimate of the derivative MATH at MATH: MATH . We observe that MATH . Therefore, MATH . There exists a constant MATH such that for all MATH . We use the method which gave estimate REF: we cut the integration interval i... |
math/0004054 | Denote MATH . With these notations, REF can be written MATH . We observe that in the domain MATH, REF has exactly one critical point at MATH . This critical point is attractive, as an examination of the linearization of REF around MATH shows. Moreover, there is a NAME functional given by MATH . Therefore, given MATH wi... |
math/0004054 | We know from REF that MATH is an increasing function of MATH; if there is a time MATH for which MATH, the conclusion is clear. Assume otherwise; then, with the notations of REF, we can see that MATH and the conclusion is also clear. |
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