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math/0012073 | We adapt the methods used in CITE to our setting. Let us prove REF . We first show that for all MATH, MATH . Write MATH so that MATH. Let MATH. Using REF, we have that MATH that is MATH. Evaluate both sides of this equality with MATH, reverse the order of the tensorands and multiply them to obtain MATH . Now, by REF , the left-hand side is equal to MATH and, by REF , the right-hand side is equal to MATH . Thus REF is proven. Let us show that MATH is invertible. Set MATH . By REF , MATH. Write MATH. Then MATH and MATH. Now MATH . It can be shown similarly that MATH. Thus MATH is invertible, MATH, and so MATH for any MATH. REF is a direct consequence of REF follows from REF, and REF. Let us show REF . For any MATH, MATH . In particular, for MATH, one gets that MATH. For the proof of the first equality of REF , set MATH. Using REF , we have also that MATH. We first show that for all MATH, MATH . By REF, MATH. Evaluate both sides of this equality with the algebra homomorphism MATH and multiply them on the right by MATH to obtain MATH . Then, using REF, one gets equality REF. Set now MATH . We have to show that MATH. Write MATH, MATH, and MATH. Then MATH. We have that MATH . Therefore, using REF for MATH and then REF , MATH . Now MATH is a right MATH-module under the action MATH . Then MATH . For any MATH-spaces MATH and any MATH, we set MATH, MATH, etc. Therefore, by REF , MATH . Write MATH and MATH. Then MATH and so MATH . If we write MATH, then MATH . Therefore MATH . Write now MATH. Then MATH . Hence MATH. Finally, write MATH. Then MATH. This completes the proof of the first equality of REF . Let us show the second one. Using the first equality of REF and then REF , we have MATH and so, by REF , MATH . It remains to show REF . We have MATH . Now MATH since MATH is invertible (by REF ) and MATH (by REF ). Hence MATH. This finishes the proof of the lemma. |
math/0012073 | Let us show REF . Denote by MATH the NAME elements of the mirror NAME MATH-coalgebra MATH to MATH (see REF). Since MATH, REF applied to MATH gives that, for any MATH, MATH . Now, by REF , MATH . Thus we obtain that MATH. Moreover MATH by REF . Hence MATH. To show REF , let MATH and MATH. Applying REF to MATH and then to MATH gives that MATH . This completes the proof of the corollary. |
math/0012073 | REF is a direct consequence of REF, and REF . Let us show REF . We have MATH . Now MATH since it is invertible and MATH (by REF ). Hence MATH. To show REF , let MATH of finite order MATH. For any MATH, using REF , we have that MATH and so MATH. Hence MATH is central in MATH. Finally, let us show REF . Using REF , we have that MATH, and so MATH. |
math/0012073 | Let us show REF . Firstly MATH by REF . Secondly, for any MATH, using REF , MATH . Thus MATH. REF follows directly from REF , and REF from the fact that MATH is a MATH-grouplike element. By REF , MATH and so REF is established. Let us show REF . By REF , MATH. Therefore MATH. Finally, to show REF , let MATH. Then, using REF , MATH. This completes the proof of the lemma. |
math/0012073 | We adapt the technique used in the proof of CITE. Let us first show REF . For any MATH, using REF, the multiplicativity of MATH, and REF , we have that MATH . Moreover, using REF , MATH. Thus MATH. To show REF , let MATH and MATH be a non-zero left integral for MATH. We first show that, for any MATH, MATH and MATH . Indeed MATH and so, since MATH is a left integral for MATH, MATH . Similarly, MATH and so, since MATH is a left integral for MATH, MATH . Write MATH. Recall that MATH. By REF , MATH. Thus MATH and so, using REF , MATH. Then MATH . Now MATH by REF and MATH by REF . Therefore MATH . Let MATH be left MATH-integral for MATH such that MATH (see the proof of REF ). Applying MATH on both sides of the last equality, we get MATH and so, since MATH, MATH . Write MATH so that MATH. Since, by REF , MATH for all MATH, we have that MATH . Therefore MATH. Finally, comparing with REF, we get MATH. Hence MATH, since MATH. This finishes the proof of the theorem. |
math/0012073 | For any MATH, MATH by REF . |
math/0012073 | We first show that, for all MATH and MATH, MATH and MATH . Indeed, let MATH be the distinguished grouplike element of MATH. Since MATH (MATH is unimodular), REF gives that MATH. Now, by REF , MATH. Thus MATH and REF is proven. Moreover, REF gives that MATH, where MATH is the distinguished MATH-grouplike element of MATH and MATH. Since MATH and by REF , MATH. Thus MATH. Now MATH by REF . Hence MATH and REF is proven. Let us suppose that there exists MATH verifying REF. For any MATH and MATH, MATH and MATH . Hence MATH is a MATH-trace. Conversely, suppose that MATH is a MATH-trace. Recall that MATH is a right MATH-module for the action defined, for all MATH and MATH, by MATH . By REF , MATH is free, its rank is MATH (respectively,MATH) if MATH (respectively,MATH), and MATH is a basis vector for MATH. Thus, for any MATH, there exists MATH such that MATH. Set MATH. Let us verify that the family MATH verify REF. By the definition of MATH, REF is clearly verified. Let MATH and MATH. For any MATH, MATH . Therefore MATH. Hence MATH (since MATH is a basis vector for MATH) and so MATH. REF is then verified. Let MATH. For any MATH, MATH . We conclude as above that MATH, and so REF is satisfied. Finally, let MATH. For any MATH, MATH . Thus MATH and so MATH. Hence REF is verified and the lemma is proven. |
math/0012073 | Let MATH. If MATH is semisimple, then MATH is semisimple and thus there exists a left integral MATH for MATH such that MATH (by CITE). Now MATH by REF . Therefore, using REF, MATH. Suppose now that MATH is cosemisimple. By REF , there exists a right MATH-integral MATH for MATH such that MATH. Then MATH. Suppose finally that MATH. Let MATH be a non-zero right MATH-integral for MATH. Then MATH and thus MATH (since MATH by REF ). |
math/0012073 | Let MATH. By REF , MATH. Thus, since MATH, MATH. |
math/0012074 | This is analogous to REF, (compare also REF): under our assumptions a length MATH chain must have MATH, MATH and MATH; thus setting MATH, MATH and MATH one obtains a holomorphic triple MATH. One then proves that the stability conditions coincide. The NAME index is obviously zero from REF. |
math/0012074 | It is clear that MATH and the divisors MATH and MATH determine the bundles MATH, MATH and MATH and the sections MATH and MATH up to scalar multiplication. It is easy to check that any two NAME bundles obtained in this way are isomorphic and hence the map of the statement of the proposition is an isomorphism. To calculate the NAME index, one simply applies REF, noting that MATH and so MATH and MATH. |
math/0012077 | For all MATH close enough to zero and for any MATH equalities MATH and MATH holds. Consequently, MATH on the boundary MATH. |
math/0012078 | The orthogonality of the trigonometric functions and REF yield MATH . Now use the reflection REF to obtain REF . The calculation of REF is similar. |
math/0012078 | Multiply REF by MATH, integrate over MATH, and apply REF to give REF . Observe that the integral of MATH over MATH vanishes, so there is no contribution from MATH. The second result follows from the fact that the NAME expansion of MATH differs from that of MATH given in REF only in the sign of the last term. |
math/0012078 | The expansion REF shows that the coefficients of MATH are given by MATH . REF then yields REF . Now use NAME 's relation REF for the MATH-function to obtain REF . The proofs of REF are similar. |
math/0012078 | Let MATH in REF . |
math/0012078 | For MATH let MATH in REF . The case MATH is direct. |
math/0012078 | Let MATH in REF and use MATH . |
math/0012078 | Let MATH in REF to produce MATH . The result then follows from MATH. |
math/0012078 | Let MATH in REF to obtain MATH using REF . The vanishing for MATH odd is clear, and for MATH even the result follows from REF . |
math/0012078 | We prove REF by induction. The case MATH follows from REF and the vanishing of the integral of MATH. For MATH, integration by parts yields MATH where we have used REF for power MATH. The final form is obtained from the identity MATH. |
math/0012078 | Apply REF to write MATH using REF to go from the second to the third line. The final form follows from the identity MATH . |
math/0012078 | The generating function for the NAME polynomials REF yields MATH so that MATH . Since MATH and MATH integrates to MATH, the above sum effectively starts at MATH. Thus REF gives MATH which can be written as MATH using REF . Now replace MATH by MATH and use the evaluation MATH to yield the final result. |
math/0012078 | We discuss the case MATH; the case of MATH even is similar. Let MATH in REF . Then MATH where we have employed the identity MATH that appears in CITE:REF:REF. |
math/0012078 | The NAME coefficients of MATH are MATH and MATH . These appear in CITE Thus REF follows from REF . |
math/0012078 | Let MATH in REF giving MATH . Now use REF to obtain the result. |
math/0012078 | Using REF we have MATH . The result now follows by REF and the classical value MATH . An elementary evaluation of REF appears in CITE. |
math/0012078 | The NAME coefficients of MATH appear in CITE MATH and MATH as MATH . Thus MATH where MATH is defined in REF . The evaluations MATH yield REF . |
math/0012078 | Replace in REF the variable MATH by MATH and MATH respectively. Then use REF in the first case and REF in the second. |
math/0012078 | Use REF to write MATH . The value MATH is then obtained from REF . The result now follows from REF . |
math/0012078 | Direct differentiation of REF . |
math/0012078 | In CITE MATH we find MATH . A straightforward application of REF completes the proof. |
math/0012078 | Put MATH in REF to obtain MATH . The change of variable MATH then produces MATH since the second integral equals MATH. |
math/0012078 | The value MATH in REF yields REF . To prove REF it is enough to establish the identity MATH . The internal sum in REF can be written as MATH . Logarithmic differentiation of the reflection REF for the gamma function yields MATH so that, evaluating at MATH, MATH . Thus MATH and REF is established. |
math/0012078 | The usual technique yields MATH which is REF . The proof of REF is similar. |
math/0012078 | The NAME coefficients of MATH are MATH . The main theorem then yields MATH . The last series is identified in CITE, page REF, as MATH which, for MATH, implies REF after using NAME 's relation REF . |
math/0012078 | The NAME expansion MATH yields, according to REF , MATH . But the vanishing of the integral of MATH over MATH can be written as MATH so REF is proved. REF follows from REF and NAME 's functional equation for the MATH-function. Alternatively, REF can be derived directly from the indefinite integral REF . |
math/0012078 | For MATH let MATH in REF and use REF . REF can also be checked to hold for the case MATH, using the value MATH. |
math/0012078 | Integrate by parts the identity MATH and use MATH to simplify the boundary terms. |
math/0012078 | Let MATH in REF . |
math/0012078 | Simply use REF and NAME 's formula MATH . |
math/0012078 | From REF it is seen that near MATH the NAME polynomials behave as MATH . Thus, the integrand MATH behaves as MATH or MATH, according if MATH or not. The result now follows from the fact that the singularity MATH is integrable for MATH. |
math/0012078 | This follows directly from REF and a reasoning similar to the proof of the previous example. |
math/0012078 | Integrate by parts to produce MATH which establishes the result for MATH. Repeated integration by parts yields REF . |
math/0012078 | Take MATH in REF . We obtain MATH so that MATH . Hence MATH . Now choose MATH with MATH. Then MATH in virtue of the identity MATH. Moreover, MATH and MATH. Thus the integral on the right hand side of REF is just the negative of the initial integral. The result REF follows, since MATH . |
math/0012078 | Take MATH in REF . Then MATH . Thus MATH . Using REF we have MATH . The result now follows after using the identity MATH . |
math/0012078 | We compute the limit as MATH in REF . Substitute MATH in REF and let MATH. We encounter two types of singularities as MATH: one corresponding to the pole of MATH at MATH, for MATH, and the other corresponding to the vanishing of the NAME symbol MATH, for MATH. To derive REF consider the NAME expansion of REF about MATH up to order MATH. The following expansions are employed: MATH . A direct calculation yields MATH . The coefficient of the singular term MATH is MATH which vanishes in view of the identity MATH . The rest of the terms can be collected to yield REF , after multiplying by the overall factor MATH in REF . |
math/0012078 | REF yield MATH and REF follows by l'Hopital's rule. |
math/0012079 | Let MATH for MATH. Then the sequence MATH, MATH fails to equal MATH for some MATH only if MATH has either two repeated indices REF , or if REF , or else if REF . In each of these cases MATH, and so the function MATH defined for MATH satisfies the recursion REF for MATH. Since the index of a minimal quantum NAME variety (which is a point) is MATH, and MATH, the function MATH also satisfies the initial condition for MATH. |
math/0012079 | Let MATH be the function defined by the sum. First observe that if MATH, then there is only the trivial summand (all MATH) and so MATH. Also, since MATH is alternating, MATH is an alternating function. We show that the function MATH satisfies the conditions of REF , when MATH is a weakly increasing sequence of non-negative integers with MATH. First, MATH satisfies the recursion of REF because the function MATH satisfies the recursion. Second, MATH, giving the initial condition. Next, since MATH is alternating, it satisfies REF . Suppose MATH. Then every summand indexed by MATH with MATH vanishes as MATH is alternating, and every summand with MATH is paired with another summand indexed by MATH, which has the same absolute value, but opposite sign, as MATH is alternating. Thus REF holds for MATH. Finally, if MATH, then either MATH and so MATH or else MATH and so MATH, giving REF and proving the theorem. |
math/0012079 | By REF , there exist MATH so that the intersection MATH is transverse and consists of exactly MATH points, and when MATH, these points of intersection are real. Furthermore, we may choose these numbers MATH so that their MATH-th powers are distinct. To prove REF , we show that these points all lie in MATH. Thus each point in REF represents a map MATH of degree MATH satisfying MATH for MATH. Let MATH be the map REF whose image is the complement of MATH in MATH. Then MATH where MATH is the linear form for MATH analogous to MATH. Let MATH be the hyperplane given by the linear form MATH. Any point in the intersection REF but not in MATH is the image of a point MATH in MATH satisfying MATH for each MATH. As the MATH-th powers of the MATH are distinct, such a point can satisfy MATH for at most one MATH. Thus MATH lies in at least MATH of the hyperplanes MATH. Since MATH exceeds the dimension MATH of MATH, there are no such points MATH, by REF applied to maps of degree MATH. |
math/0012079 | For any MATH, the intersection REF has dimension at least MATH. We show it has at most this dimension, if MATH are distinct. Suppose MATH and let MATH. Then MATH if MATH and so the form MATH defining MATH evaluated at MATH is MATH . This is a non-zero polynomial in MATH of degree at most MATH and thus it vanishes for at most MATH distinct values of MATH. It follows that REF is empty for MATH. If MATH and MATH are distinct, but REF has dimension exceeding MATH, then we may complete MATH to a set of distinct numbers MATH which give a non-empty intersection in REF , a contradiction. |
math/0012079 | We prove both parts of the theorem simultaneously, making note of the differences when MATH. We construct the sequence MATH inductively. The unique element of rank REF in MATH is MATH, where MATH is the sequence MATH. The quantum NAME variety MATH is a line in NAME space. Indeed, it is isomorphic to the set of MATH-planes containing a fixed MATH-plane and lying in a fixed MATH-plane. By REF or direct observation, MATH is then a single point, for any non-zero MATH. When MATH, this point is real. Let MATH be arbitrary. Suppose MATH are distinct points with the property that for any MATH with MATH, MATH is transverse. When MATH, we suppose further that all points of intersection are real. Let MATH be an index with MATH and consider REF-parameter family MATH of schemes defined by MATH, for MATH. If we restrict the form MATH to MATH, then we obtain MATH a polynomial in MATH with leading term MATH. Since the NAME coordinate MATH is non-zero, MATH is defined by MATH, and so MATH equals MATH by REF . Claim: The cycle MATH is free of multiplicities. If not, then there are two components MATH and MATH of MATH such that MATH is non-empty. But this contradicts REF , as MATH, where MATH is the greatest lower bound of MATH and MATH in MATH, and so MATH. Because the intersection of MATH, the fibre of MATH at infinity, with the cycle MATH is zero dimensional and free of multiplicities, it is transverse, and so the general fibre of MATH meets MATH transversally. Thus there is a non-empty NAME open subset MATH of MATH consisting of points MATH for which MATH is transverse. Choose MATH to be any point common to all MATH for MATH. When MATH, the claim implies there is a real number MATH such that if MATH, then MATH is transverse with all points of intersection real. Set MATH and let MATH be any real number satisfying MATH. |
math/0012079 | If MATH is a generator of MATH, then MATH and MATH are incomparable in the poset MATH. Thus if MATH is a saturated chain, at most one of MATH or MATH lies in the chain MATH, and so MATH lies in the ideal MATH. Suppose now that MATH is a monomial not in the initial ideal MATH. Then the variables appearing in MATH have indices which are comparable in the poset MATH. Thus we may write MATH with MATH. There is some chain MATH containing the indices MATH and so the monomial MATH does not lie in the ideal MATH. This proves the equality of the two monomial ideals. |
math/0012079 | The degree of the quantum NAME is the degree of its ideal MATH. By NAME 's Theorem CITE (see also CITE), this is the degree of the initial ideal, MATH, which is equal to the number of chains in MATH, by REF . |
math/0012079 | Let MATH be this determinant. Since MATH where MATH indicates that MATH is omitted, we seek the determinant of the matrix whose MATH-th entry is MATH. If we subtract the first column from each of the rest, we obtain a matrix in block form MATH where the entries of MATH in position MATH (note the shift from the original matrix) are MATH . Dividing the common factors of MATH from the columns of MATH gives the matrix with entries MATH, and so we have the recursive formula MATH . Since MATH, this completes the Lemma. |
math/0012080 | Let MATH such that MATH is a basis of MATH. Then we have topological direct sum splittings MATH . This implies the claim. |
math/0012080 | MATH implies MATH and, in view of the previous lemma, we may choose a sequence MATH with MATH in MATH. Then MATH and MATH in MATH. Thus MATH and MATH. |
math/0012080 | Fix MATH and consider MATH. Then we have MATH and hence MATH for almost all MATH. Moreover, the function MATH satisfies the differential equation MATH for almost all MATH. Thus, by the uniqueness theorem for first order differential equations we have MATH. Moreover, since MATH, MATH . Since MATH we infer MATH. Since MATH were arbitrary we have proved that MATH is independent of MATH. This implies the rest of the assertions. |
math/0012080 | First we prove REF. For any MATH and MATH one has MATH hence MATH . We note that MATH in MATH that is MATH for MATH. Thus one infers MATH . It follows that each MATH admits a unique decomposition MATH where MATH is the unique element in MATH such that MATH. Furthermore, MATH . Hence MATH . Since the opposite inclusion is obvious one gets MATH . In view of REF this relation implies REF . To complete the proof it remains to note that MATH . |
math/0012080 | CASE: If MATH is arbitrary then MATH and we have proved that MATH. If MATH then by REF we have MATH, thus MATH . Since MATH by definition and since MATH is continuous we conclude that MATH. We have proved MATH. Furthermore we infer MATH and MATH in view of REF. CASE: Let MATH and let MATH. We put MATH. Then MATH, that is, MATH. Consequently, there is a MATH such that MATH and hence MATH is an absolute continuous representative of MATH which satisfies MATH. CASE: Let MATH with representatives MATH. Then MATH . If MATH then by REF we have MATH and hence MATH. Moreover, MATH and thus MATH. This implies MATH a.e. and thus MATH. Conversely, let MATH and MATH. Then MATH represents the same element MATH as MATH. Moreover MATH implies MATH, hence MATH. Since MATH this argument also shows MATH. |
math/0012080 | In view of REF and the previous considerations we have MATH. Hence, from REF we infer that we may choose MATH such that MATH is a basis of MATH. This implies the assertion. |
math/0012080 | For simplicity we will give the proof for MATH. CASE: Let MATH be left - closed and let MATH. Then choose MATH and put MATH . Since MATH is left - closed we then have MATH and hence MATH. Thus MATH has dense range and consequently MATH. The same construction shows for any interval MATH that if MATH then the function MATH has compact support in MATH and thus MATH. CASE: Let MATH be any representative and put MATH. Then MATH is absolutely continuous. Using integration by parts and REF one obtains for any pair MATH . By REF we have MATH, thus REF implies MATH . Since the MATH in REF satisfy MATH we apply REF to conclude that there is a MATH such that for all MATH one has MATH . Note that by integration by parts one has MATH for all MATH, even if MATH. REF implies that MATH is an absolute continuous representative of MATH with MATH. CASE: We may assume that MATH and MATH where MATH. Then choose MATH satisfying MATH and put MATH . Here MATH and MATH are the characteristic functions of the intervals MATH and MATH respectively. Then we define MATH by REF with MATH replaced by MATH . It is clear that MATH and MATH . Furthermore, for MATH one gets MATH . CASE: Let MATH. According to REF we may choose MATH with compact support such that MATH. For each MATH we have on the one hand MATH . Since MATH have compact support we may integrate by parts and thus find MATH . Thus MATH is orthogonal to MATH, that is MATH . To prove the last assertion let MATH with representatives MATH. Then MATH and hence MATH is an absolute continuous representative of MATH. Moreover, MATH and MATH. Consequently, MATH and MATH. |
math/0012080 | Consider MATH. Then MATH and hence MATH . Then the definiteness implies MATH and we are done. |
math/0012080 | Consider MATH satisfying MATH . We have to show that MATH. REF translates into MATH . REF implies that MATH a.e. Thus the set MATH has positive NAME measure. A set of positive NAME measure contains an accumulation point of itself; the reason is that a subset of the reals which does not contain an accumulation point of itself is at most countable. So let MATH be an accumulation point of MATH. Then MATH and by REF MATH. Since MATH is invertible we infer MATH and hence MATH. Since MATH is a solution of the homogeneous first order equation MATH this implies MATH. |
math/0012080 | Consider MATH. This means MATH and in view of Theorem regthm there exists MATH such that MATH. Thus MATH. This shows that the quotient map MATH is surjective. Next let MATH. This means that MATH and MATH. Thus MATH. Hence MATH consists of the solutions of MATH. This space is isomorphic to MATH (compare Subsections secREF, secREF) and hence MATH and we reach the conclusion. |
math/0012080 | This follows immediately from REF and Proposition ML-SREF |
math/0012080 | It is clear that the differential REF has MATH linear independent solutions. Hence MATH. From REF we infer that MATH is nonzero if and only if MATH. This implies MATH. |
math/0012080 | We put MATH . It is clear that MATH is a closed dissipative extension of MATH that is MATH and MATH for any MATH . In fact, we show that MATH is a maximal dissipative relation in MATH . To prove this fact it suffices to check that MATH where MATH denotes the resolvent set of MATH. For a dissipative linear relation MATH and MATH one has for MATH . Hence MATH is injective with closed range and thus it suffices to verify that MATH is dense in MATH. Let MATH be orthogonal to MATH, that is MATH . In particular, we have for MATH . Hence MATH and MATH . From the latter and REF we infer MATH . Hence MATH. Summing up, we have proved that MATH and hence MATH is maximal dissipative. On the other hand for each proper extension MATH the inclusion MATH is equivalent to the fact that MATH is transversal to MATH (see CITE). Hence MATH and MATH are transversal and this is equivalent to the direct sum decomposition REF . |
math/0012080 | Let MATH . It follows from REF , that for each MATH . |
math/0012080 | Similar to the proof of REF we put MATH . It is clear that MATH is a symmetric extension of MATH and the subspaces MATH and MATH are linearly independent since MATH . Therefore MATH. On the other hand the NAME formula for linear relations REF yields MATH . Combining these relations we obtain REF. |
math/0012080 | REF has been established in the proof of Proposition ML-SREF REF is implied by REF since MATH . REF is a special case of REF since MATH . CASE: Injectivity of the map MATH follows again from the assumption that MATH is definite. Indeed, let MATH . Then MATH satisfies the homogeneous equation MATH, that is MATH. Since MATH we have MATH and therefore MATH . Since MATH is definite the latter implies MATH. NAME has been established in REF is a consequence of REF . CASE: Without loss of generality we may assume MATH. By definiteness we then have MATH . It follows from the NAME REF with MATH that each symmetric extension MATH is given by the second NAME formula MATH where MATH is a linear subspace and MATH is an isometric operator from MATH onto MATH . The corresponding symmetric extension MATH is given by MATH . It is clear from REF that this establishes the asserted bijective correspondence. Compare also Proposition ML-SREF We know from REF that MATH . REF now imply MATH . |
math/0012080 | Fix MATH. By NAME 's theorem on monotone convergence we have MATH and thus we may choose MATH such that MATH . Now choose MATH large enough such that MATH and put MATH . MATH has the desired properties with MATH. |
math/0012080 | CASE: According to Lemma ML-SREF let MATH be absolutely continuous with bounded derivative, MATH and MATH . For MATH we choose, according to REF , representatives MATH and put MATH . Since MATH vanishes if MATH is not invertible the function MATH is well - defined. Moreover MATH hence MATH lies in MATH and it converges to MATH in MATH. Finally, we calculate MATH . Thus MATH and MATH and the claim is proved. The proof of REF proceeds along the same lines with minor modifications. REF follows from integration by parts if MATH have compact support. To prove it in general we consider MATH and MATH. Then REF holds true for MATH and MATH. Noting that MATH is independent of MATH we obtain the result by taking the limit as MATH. |
math/0012080 | This follows immediately from the estimate REF. |
math/0012080 | MATH implies that MATH . Hence REF holds since MATH . Moreover, from MATH we infer that MATH is bounded and hence MATH. Hence Theorem ML-SREF applies. |
math/0012080 | Since MATH it is clear that MATH and MATH are invertible for all MATH. Furthermore, MATH hence MATH for all MATH. Setting MATH, one has MATH and thus MATH since MATH is bounded on MATH. Hence REF is fulfilled and we reach the conclusion. |
math/0012080 | Differentiation by MATH and integration by parts yields in view of REF MATH and by definition of MATH this is MATH (compare REF). Note that all terms in ML-GREF are real. The last statement is clear. |
math/0012080 | Assume that MATH for some MATH. Choose a self - adjoint extension, MATH, of MATH on the interval MATH. This is possible since in view of REF the deficiency indices of MATH on the finite interval MATH are given by MATH. Next let MATH be the strong solution of the wave equation for MATH. The local energy estimate above shows that for MATH small enough, MATH has compact support in MATH and hence can be extended by MATH to a strong solution of the wave equation for MATH. The uniqueness follows immediately from the local energy estimate. |
math/0012080 | If REF is fulfilled then the previous result shows that for each MATH there exists a unique strong solution MATH, of the NAME problem for the wave REF and MATH for all MATH. Hence the result follows from Theorem ML-SREF |
math/0012080 | It suffices to prove the Theorem for the linear relation MATH. As noted in Remark ML-SREF it follows from REF that MATH is definite. Therefore by REF MATH . Thus it suffices to prove the assertions for MATH . Let MATH be a solution of REF with MATH . Let MATH be any sequence converging to MATH. Then integrating by parts and taking REF into account one gets MATH . Thus MATH exists and MATH . On the other hand we find using REF MATH . We claim that there is a sequence MATH such that MATH. For if this were not the case then we had an estimate MATH for MATH. This would contradict REF and MATH. In view of REF we have MATH . Combining REF one gets MATH . By the uniqueness theorem for first order differential equations the map MATH is an embedding of MATH into MATH. Moreover, the quadratic form MATH is positive (respectively, negative) on MATH (respectively, MATH). Since MATH is just the number of positive (respectively, negative) eigenvalues of the quadratic form MATH we obtain MATH. On the other hand we have in view of REF MATH and thus equality holds. We emphasize that although we did not prove REF in full generality the relation REF was proved completely in REF . |
math/0012080 | This follows immediately from REF . |
math/0012080 | It follows from definiteness and REF that REF are equivalent. Hence it suffices to prove one of them. We put MATH and MATH . By REF we have MATH for each MATH . Moreover, REF implies that for each MATH there exists MATH with compact support such that MATH . Hence MATH . In view of Proposition ML-SREF and Remark ML-SREF the same argument applies to MATH and MATH. Hence MATH. On the other hand since MATH is a closed symmetric extension of MATH it follows from the second NAME REF with MATH and MATH replaced by MATH and MATH respectively, that MATH. Combining this formula with the obvious equalities MATH we obtain REF and thus also REF. |
math/0012080 | The set MATH has positive NAME measure in view of REF. Therefore by Proposition ML-SREF the system is definite. Hence it suffices to consider the formal deficiency indices. CASE: Let MATH be a solution of REF with MATH. We show that MATH. REF reads MATH . It follows that MATH . Adding REF and integrating from MATH to MATH one gets MATH . We put MATH and recall (compare REF) that MATH . Using this and the inequality MATH we obtain MATH . For brevity we assume in the sequel that MATH that is MATH . Now combining REF and integrating by parts we have MATH . Furthermore, the assumption MATH yields MATH . Thus setting MATH we infer from REF that MATH . We rewrite the latter inequality as MATH or as MATH . We claim that MATH for MATH . Assuming the contrary one finds MATH such that MATH, hence MATH for MATH since MATH is non - decreasing. Therefore in view of REF MATH . On the other hand choosing MATH such that MATH, one derives from REF MATH . This inequality contradicts REF . Thus MATH and MATH . CASE: Next we estimate using REF MATH . By REF and NAME - NAME we know that MATH is integrable. In view of REF we infer exactly as in the proof of REF that there is a sequence MATH such that MATH. Also as in the proof of REF one now completes the proof, noting that MATH . CASE: Now assume that REF is satisfied. We reduce this case to the previous one. For this purpose it suffices to construct an absolutely continuous function MATH such that MATH for MATH and MATH satisfies both REF . Since MATH, one gets that MATH . Therefore the function MATH is absolutely continuous and monotone increasing for MATH . Denote by MATH the corresponding distribution function, MATH . Next we put MATH and observe that MATH is monotone increasing because so are MATH and MATH. Besides it is clear that MATH . Following CITE (see also CITE) one puts MATH for MATH and then extends MATH to the semi-axis MATH by linear interpolation: MATH . It is clear that MATH for MATH . Moreover MATH is globaly NAME, MATH . Finally, we put MATH and check that MATH has the desired properties. Indeed, MATH, since MATH, and therefore MATH . Further, MATH is absolutely continuous because so is MATH and MATH is NAME. Now it follows from REF that MATH and MATH which completes the proof. |
math/0012080 | As elaborated in Example SDM-SREF the system REF can be transformed into the first order system REF MATH with MATH defined in REF. Namely, putting MATH and MATH one reduces REF to the system MATH . Since the corresponding linear relations are unitary equivalent, we apply REF and reach the conclusion. |
math/0012080 | This follows immediately from Theorem ML-SREFEF and Example SDM-SREF |
math/0012080 | The condition MATH is equivalent to the fact that the constant vector MATH is in MATH. Thus MATH and MATH . Since MATH is of positive type the canonical system MATH is definite. Therefore by REF we have MATH and MATH . Now REF implies the assertion. |
math/0012080 | The gauge transformation MATH transforms the system into a canonical one with Hamiltonian MATH and MATH (see REF). A canonical system is definite if and only if the Hamiltonian is of positive type. Hence MATH is of positive type. Since a gauge transformation preserves the deficiency indices we may apply REF and reach the conclusion. |
math/0012080 | Sufficiency. REF is equivalent to REF with MATH hence by REF MATH. On other hand MATH and thus MATH. Necessity. Assume that MATH . By REF also MATH and in particular MATH admits self - adjoint extensions. Fix one of them, say MATH. It follows from REF that there exists a linear relation MATH in MATH, satisfying MATH and such that MATH. To calculate the resolvent MATH we have to find the solution MATH of the equation MATH for an arbitrary MATH, or what is the same, the solution MATH of the equation MATH with MATH satisfying some (self - adjoint) boundary conditions at zero and at infinity. It is well - known (see CITE, CITE) that MATH where MATH . Here MATH is the fundamental MATH matrix solution of REF (with MATH) satisfying the initial condition MATH and MATH is some function. It follows from REF that MATH does not depend on the representative MATH of MATH. Thus MATH is well defined on MATH and in view of REF MATH . Combining REF - REF and MATH we see that the resolvent MATH is a NAME operator for MATH . Consequently the spectrum MATH is discrete. Since MATH is finite - dimensional the existence of a self - adjoint extension of MATH with compact resolvent implies that MATH is a NAME relation of index MATH for all MATH. On the other hand by REF we have MATH for all MATH. Therefore MATH. In particular MATH and by REF we obtain MATH . But since the system is canonical we have MATH with the constant vectors MATH. Thus MATH for MATH . This is equivalent to MATH, MATH that is to REF . |
math/0012080 | It is clear that MATH if the system MATH is quasi - regular. Conversely, if MATH then the relations MATH for MATH, have been established in the proof of REF . |
math/0012080 | It follows from the assumption MATH that there exists a fundamental MATH matrix solution MATH of the homogeneous equation MATH satisfying MATH where MATH is MATH matrix function with entries MATH . This fact is well known and can be easily checked (compare with the proof of REF ). By REF MATH is quasiregular iff MATH . In view of REF the last inequality is equivalent to REF . |
math/0012080 | If MATH then MATH hence MATH . Applying the same argument to MATH one sees that MATH is an isomorphism from MATH onto MATH. |
math/0012080 | CASE: MATH is invariant under complex conjugation and therefore so is MATH. By REF MATH. The other equalities follow from REF. CASE: If MATH then the relations REF are implied by REF. If MATH then by REF MATH . The equality MATH has been established in the proof of REF (see also REF ). |
math/0012080 | Since MATH then by REF MATH . Applying REF we are, in view of REF , left with three possibilities: MATH . We rule out MATH and MATH . REF yields REF with MATH and MATH. So if MATH or MATH then by REF the system REF is quasi - regular, hence MATH . This contradicts REF. Thus MATH . |
math/0012080 | We show that REF is satisfied and apply REF . Since MATH is real so is MATH . On the other hand MATH since MATH is skew-adjoint. Thus MATH. |
math/0012080 | Since the system MATH is definite then MATH. It follows from REF that MATH . By REF we have MATH, too. Hence MATH. CASE: Let MATH. Then by REF MATH and thus MATH . CASE: Conversely, assume that MATH or MATH. Then MATH or MATH. As in the proof of REF one now concludes that the system is quasi - regular and hence MATH. |
math/0012080 | As explained in Example SDM-SREF the system MATH is unitarily equivalent to a first order system MATH, with MATH defined in REF. By Proposition ML-SREF the system MATH is definite. Then the gauge transformation MATH transforms the system MATH into a canonical (and definite) one MATH with MATH and MATH defined by MATH . Since the Hamiltonian MATH is of positive type the first assertion follows from REF . To prove the second assertion we put MATH and MATH . Since MATH one gets MATH . Using this and the equality MATH we get MATH . This proves the last statement. |
math/0012080 | At first we prove that the homogeneous REF (with MATH) has two MATH matrix solutions MATH and MATH satisfying: MATH where as before MATH stands for the MATH matrix function with entries MATH as MATH. Indeed it is clear that each solution MATH of the integral equation MATH is also a solution of REF with MATH . Choose MATH such that MATH . Further, setting MATH and MATH and using REF and the inequality MATH one easily proves by induction that MATH for MATH . Hence the series MATH converges uniformly for MATH and MATH . Moreover, the matrix function MATH defines the unique solution of REF (for MATH) and satisfies the inequality MATH for MATH . Using this estimate one obtains from REF that MATH . Differentiating REF and applying REF and the above estimate MATH one derives the second relation MATH . Thus the existence of the solution MATH satisfying REF is proved. To prove the existence of the solution MATH satisfying REF we recall see REF that for each MATH matrix solution of REF with MATH the matrix function MATH is constant. Turn MATH and taking REF into account one gets that MATH. This means that MATH is a self-adjoint solution in the sence of REF of the homogeneous REF (with MATH). Using REF (with MATH) it is easy to check and it is known (see CITE), that the classical NAME formula remains valid for the matrix case, that is MATH is also a MATH matrix solution of REF (with MATH). The relations REF are implied now by REF. Further, following the proof of Proposition ML-SREF one transforms the system MATH to a first order system MATH with MATH defined in REF. Then the gauge transformation MATH transforms the system MATH into a canonical system MATH with MATH . We note that generally speaking MATH since MATH . By REF MATH iff MATH . By REF this inequality is also equivalent to the property of the system MATH to be quasiregular. In view of REF this inequality is equivalent to REF. |
math/0012080 | CASE: If MATH then, as it is well known, there exist two MATH matrix solutions MATH and MATH of the homogeneous equation MATH satisfying MATH . Following the proof of REF and using the gauge transformation MATH we reduce REF to a canonical system MATH with MATH and MATH defined in REF. In view of REF the inequality MATH takes place iff-MATH . It remains to apply REF . CASE: Now the homogeneous equation MATH has two MATH matrix solutions satisfying MATH . Starting with these solutions one completes the proof in just the same way as in REF . |
math/0012080 | As in the proof of Proposition ML-SREF we transform REF to a canonical system MATH with MATH and MATH defined in REF. One checks that MATH . To complete the proof it remains to apply REF . |
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