paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9904104 | Repeating the construction of REF for the tree , we have MATH as defined above. We also see that the MATH- invariant maps are the collection, MATH . |
math/9904104 | REF applied to MATH gives MATH as desired. |
math/9904104 | It is clear that the right hand side is contained in the intersection, so we need only to prove that an arbitrary element of the intersection is contained in the power series on the right. But the only difference between the power series MATH, and MATH, is that in the former, polynomials in the variable MATH can exist ... |
math/9904104 | Each of these is an element of MATH which agree when MATH (and when MATH). Because each satisfies MATH they are completely determined by their values at MATH (or MATH), and so are equal. |
math/9904104 | Let MATH and MATH. We shall show that MATH and MATH compose to give an element of MATH. To do so, we select a pullback object MATH as in REF . The total ordering of the internal vertices of the tree MATH provides a total ordering of both MATH and MATH as MATH and MATH. Such a pullback object also consists of a choice o... |
math/9904104 | The idea of the proof is that we get a map between pullbacks by showing that there is a map between the spaces over which we are pulling back. There is also a map between the pullback objects, and these maps form a commuting diagram, inducing a map between pullbacks. For more of the abstract details see CITE. Explicitl... |
math/9904104 | In order to show that MATH gives a vertex algebra, we check the axioms for the vertex algebra. The vacuum axioms follow naturally from the discussion of REF, together with previous example, REF, giving MATH . Translation covariance is automatically satisfied because our functions, MATH, are MATH-invariant. And, we saw ... |
math/9904104 | As in the proof for holomorphic vertex algebras REF , the vacuum defines a map MATH for the empty tree. We described in REF how to construct MATH, and REF showed how how to construct the singular function associated to the flat tree with three leaves. Carrying out a similar process leads to the construction of singular... |
math/9904104 | Since MATH can be evaluated at zero, we know that it has no singularities. By the first vacuum axiom we know that MATH, so the translation covariance axiom says that for any MATH, MATH. Letting MATH, and evaluating this at zero we have MATH and the lemma is proved. |
math/9904104 | Recall that the translation covariance axiom says that MATH. So we have MATH . From the locality axiom we know that for some MATH the following holds MATH . Combining this with the previous lemma, it just says that MATH . Setting MATH equal to zero we have MATH, and since this is an equality of NAME series, we may divi... |
math/9904105 | REF is equivalent to MATH where MATH is the refinement of MATH obtained by replacing all components MATH, for MATH, by MATH. Thus the LHS of REF is equal to MATH . Let MATH be a term of this sum, with MATH. This term can only appear in one summand on the Right-hand side of REF , namely MATH with MATH. To show that it d... |
math/9904105 | If all components of a composition MATH, except perhaps the last, are greater than MATH, then the same is true for all compositions MATH and MATH appearing in the Right-hand side of REF . |
math/9904105 | Each difference MATH is in the kernel of MATH as MATH since MATH. In addition, the non-zero differences are linearly independent as they have different leading terms. Letting MATH denote the MATH-th NAME number, there are MATH such differences, and since MATH our result follows. |
math/9904105 | If the first component of a composition MATH is greater than MATH, then the same is true for all compositions MATH and MATH appearing in the Right-hand side of REF . |
math/9904105 | For each MATH, MATH is forbidden in any monomial MATH appearing as a summand of the function MATH. This is equivalent to saying that MATH is a summand of MATH if and only if MATH for all MATH. Therefore at least one of MATH or MATH must be the largest label of a vertex in a connected component in MATH. Now when going f... |
math/9904105 | By direct calculation we obtain that MATH, MATH, MATH, and MATH. To obtain our recurrence, we observe that for each sqs-function, MATH where MATH, we can encode MATH as a binary word of length MATH, by placing a MATH in position MATH if MATH is contained in MATH, and MATH otherwise. By this one-to-one correspondence we... |
math/9904105 | By REF we have that the leading terms of MATH determine the other summands that belong to MATH. Hence by REF it follows that the summands of MATH will be the union of the summands of MATH and MATH. However, those summands that appear in both will be duplicated. By definition these will be the summands of MATH, and the ... |
math/9904106 | We only explain the first part. REF are obvious and REF follows by REF regarding the structure of cocommutative graded connected NAME algebras. First we arrange that the components of MATH project to immersions in MATH with transverse self-intersections and so that the framing has its first componenet vector field poin... |
math/9904106 | For the first part of the theorem, we begin by choosing MATH in MATH to be a generic link. Note that MATH can be recovered from its projection MATH together with a knowledge of the signs at each overcrossing. Given MATH, let MATH be an associated MATH-link in MATH such that MATH. Without loss of generality, we can assu... |
math/9904106 | (of REF ) For a surface MATH of genus MATH with one boundary component, CITE introduced a homomorphism MATH, where, following the conventions of CITE, MATH and MATH is identified with a submodule of the MATH-th tensor power MATH by defining MATH . In subsequent work, NAME showed that modulo MATH-torsion his homomorphis... |
math/9904106 | For a closed surface MATH of genus at least MATH, REF follows from REF and the following REF , perhaps of independent interest. For a closed surface MATH of genus less than MATH, fix a disk and consider an imbedding of its complement to a surface MATH in MATH of genus at least MATH. Choose a lifting of MATH to a diffeo... |
math/9904106 | First of all, recall that MATH if any leaf bounds a disk disjoint from the other leaves of MATH. As explained in CITE, an alternative way of writing REF is as follows: MATH for arbitrary disjoint imbeddings of two based oriented knots in MATH. Given a based knot MATH in MATH, let MATH denote the based knot obtained by ... |
math/9904106 | Fix a graph MATH. Let MATH (respectively, MATH) denote the MATH-links in MATH (with leaves MATH (respectively, MATH)) such that MATH . It follows by definition that MATH and MATH are homologous links in MATH. After choosing a common base point for each pair MATH of components of MATH and MATH, it follows that the conne... |
math/9904106 | This follows easily from CITE. Suppose MATH is some MATH-fold commutator. Then consider the one-relator group MATH and the projection MATH. Consider MATH of length MATH. By induction we can assume that MATH is uniquely defined in MATH and by CITE, it is well-defined in MATH. Moreover, by naturality under MATH, they tak... |
math/9904106 | The ``if" part follows directly from the above proposition. To prove the ``only if" part we proceed by induction on MATH. The inductive step presents us with a map MATH; consider the diagram MATH . The obstruction to lifting this map is the pullback of the characteristic class in MATH of the central extension MATH. But... |
math/9904106 | Apply NAME five-term exact sequence to the short exact sequence of groups MATH to obtain MATH . Combining this with the map MATH gives the commutative diagram MATH . This diagram yields the exact sequence of the corollary, where MATH is defined as the composition MATH . To prove the formula for MATH first note that, fo... |
math/9904106 | We will use NAME 's formulation CITE of the NAME products. Choose cocycles MATH representing MATH, for MATH. Since we are assuming all NAME products of length less than MATH are defined and vanish, we can choose cochains MATH, for MATH, with the exception of the three cases MATH so that MATH and MATH. For two of the th... |
math/9904106 | The first part is immediate from REF , using the NAME duality isomorphism MATH. In local coordinates, it implies (see REF ) that the NAME product MATH is given by MATH where MATH, the summation is over MATH of length MATH, MATH indicates cup product and MATH indicates cap product with the fundamental homology class of ... |
math/9904106 | Given an element MATH we construct maps MATH, where MATH corresponds to the canonical projection MATH under the identification of MATH with MATH, and MATH. Since MATH, we have MATH and so we can combine the two maps to define a map MATH, where MATH is the double of MATH. We would like to extend MATH to a map MATH, for ... |
math/9904106 | Let MATH denote the kernel of the natural projection MATH. We first construct a map MATH as follows. If MATH we can write MATH, where MATH. Then, we define MATH, where MATH and MATH is a lift of MATH. Using the isomorphism MATH this defines a map (denoted by the same name) MATH with corresponding description in local c... |
math/9904106 | Let MATH and define MATH, where MATH are two copies of the solid handlebody MATH of genus MATH, which are attached to MATH via the diffeomorphisms MATH so that, referring to a basis MATH of MATH corresponding to a symplectic basis of MATH, the MATH are represented by the boundaries of meridian disks in MATH. Thus MATH,... |
math/9904106 | We apply the constructions and results of CITE. Consider the mapping cone MATH of the natural map MATH of NAME spaces. MATH is constructed from MATH by adjoining MATH-cells MATH along the generators MATH. Then MATH is simply-connected and MATH for MATH. So we have the NAME epimorphism MATH, where MATH denotes the third... |
math/9904106 | It follows from the definitions that we need to establish the commutativity of the following diagram MATH where MATH is defined in REF , MATH is the inclusion map inducing MATH and the NAME homomorphism MATH, and MATH is the NAME map MATH when MATH. Now suppose MATH is represented by a MATH-cycle MATH in MATH . since M... |
math/9904107 | For MATH and MATH the statement is clearly true and we use induction on MATH. Suppose we know the statement for all positive integers smaller than MATH. Then we distinguish two cases: CASE: If REF and MATH are in the same block of MATH, then the construction of MATH starts by putting the entry MATH in the last slot of ... |
math/9904107 | We show that our bijection MATH is an order-reversing map MATH. The conclusion then follows from the self-duality of the lattice of noncrossing partitions. Suppose MATH in MATH. This means MATH is a finer partition than MATH, so every element which is the minimum of its block in MATH is also the minimum of its block in... |
math/9904107 | The properties of the rank sizes of MATH are immediate consequences of REF and the corresponding properties known to hold for MATH. Moreover, every antichain of MATH is, via the bijection MATH, an antichain of MATH, and the strong NAME property of MATH follows from the strong NAME property of MATH. |
math/9904107 | We use induction on MATH. For MATH the statement is true. Now suppose we know it for all positive integers smaller than MATH. Denote by MATH the smallest element of MATH, and let MATH be a REF-avoiding MATH-permutation whose descent set is MATH. CASE: Suppose that MATH. Then we have MATH and, because MATH avoids the pa... |
math/9904107 | It is clear that, in MATH, permutations which have the same descent set will cover the same elements and they will be covered by the same elements. The permutations with a prescribed descent set MATH form an orbit of MATH and they can be permuted among themselves arbitrarily by elements of MATH. On the other hand, REF ... |
math/9904107 | For each MATH, let MATH be the set of REF MATH-permutations with excedence set MATH. Let also MATH be the set of REF MATH-permutations with descent set equal to MATH, the reverse-complement of MATH. Thus, in the notation of the previous subsection, the cardinality of MATH is MATH. We construct a bijection MATH (illustr... |
math/9904107 | It is immediate from its definition and REF that MATH is a ranked poset (namely, MATH) and has rank-sizes given by MATH, equal to the rank-sizes in MATH. Also, MATH is a self-dual poset: clearly, if MATH is the MATH-avoiding signed permutation which corresponds to the pair MATH, then the mapping MATH is an order-revers... |
math/9904107 | Let MATH and MATH be the reverse of MATH. Let MATH be the ``barred complement" of MATH, that is, MATH, and MATH is barred if and only if MATH is not barred. Then it is straightforward to verify that MATH if and only if MATH. Therefore, the reverse complement operation reverses the inclusion of excedence sets for signed... |
math/9904109 | The first relation in REF is trivial on MATH, so we only need to show it for MATH since MATH. Note that MATH, therefore REF yields MATH hence MATH . For the second relation we use the fact that MATH for MATH: MATH for all MATH. |
math/9904109 | Let MATH and MATH be isometries. Then MATH and MATH. Now MATH. Inserting this in REF yields the statement. |
math/9904109 | Since MATH must be a subsector of MATH for MATH a conjugate of MATH, there is an isometry MATH. Note that then MATH. Hence by naturality and REF we compute MATH and hence also MATH. We also obtain MATH and MATH by REF . Note that MATH by restriction. REF -REF follow now by naturality, REF . Next, we note that MATH, and... |
math/9904109 | With the normalization convention as in REF , this is just the expansion of the identity in REF , and this certainly holds as well using similar expansions with other intertwiner bases. |
math/9904109 | That each MATH is a minimal central projection and that all minimal central projections arise in this way is obvious from the description of the matrix units. The vertical product MATH is given graphically by the left-hand side of REF . We can use the expansion of REF for the two parallel wires MATH and MATH in the mid... |
math/9904109 | By REF , we compute MATH . Since the horizontal unit MATH is given by MATH we find that MATH. As MATH sends off-diagonal matrix units to zero and the diagonal ones to strictly positive numbers, this proves that MATH is a faithful state. Obviously also MATH sends off-diagonal matrix units (with respect to MATH) to zero ... |
math/9904109 | We only show the statement for the MATH-sign; the other case is analogous. First we fix MATH and MATH. For each MATH we choose orthonormal bases of isometries MATH, MATH, so that MATH. Using NAME reciprocity, we obtain an orthonormal basis of isometries MATH. Here we chose an isometry MATH such that there is an isometr... |
math/9904109 | From REF we obtain MATH . Hence MATH . Application of the horizontal state MATH of REF and multiplication by MATH yields REF since MATH and MATH decompose into sectors MATH with MATH, and by REF . Now the right-hand side of REF is given graphically by the diagram on the left in REF , and we can slide around the trivale... |
math/9904109 | Using the diagram for the matrix elements MATH in REF , the sum MATH can be represented by the diagram on the left-hand side of REF . Using REF and also the trick to turn around the small arcs given in REF , we obtain the right-hand side of REF . We can now slide around the lower trivalent vertex of the wire MATH to ob... |
math/9904109 | The sum MATH is given graphically by the left-hand side of REF . By using REF for the two parallel vertical wires MATH on the bottom and the IBFE moves we obtain the right-hand side of REF . For the summation over the thin wire MATH we can use REF again to obtain the left-hand side of REF . Now we can slide around the ... |
math/9904109 | As in particular MATH, we can write MATH with MATH according to the direct sum structure of MATH, MATH. Assume MATH. Then clearly MATH for all MATH. Now the MATH part of MATH is given by MATH, hence MATH. A similar argument applies to MATH, and hence the element MATH is independent of the linear expansions of the MATH'... |
math/9904109 | Using REF we can replace the left-hand side of REF by the left-hand side of REF . Next we can slide one of the trivalent vertices of the wire MATH around the wire MATH. Using the identity of REF , we obtain a factor MATH, and we can now proceed with the summation over MATH, again using REF . Using also REF for the para... |
math/9904109 | Using REF and also the trick of REF , we can draw the diagram on the left-hand side in REF for MATH. Now let us look at the part of this picture above the dotted line. In a suitable NAME annulus, this part can be read for fixed MATH and MATH as MATH, and the sum runs over a full orthonormal basis of isometries MATH in ... |
math/9904109 | It follows from REF that MATH unless MATH and MATH. We now show that MATH. (We denote MATH.) The sum is given graphically by the left-hand side in REF . A twofold application of REF yields the right-hand side in REF . Applying REF twice again, we obtain the left-hand side of REF . We can now slide the upper trivalent v... |
math/9904109 | The vector MATH is given graphically by the left-hand side of REF . Now note that the upper part of the diagram represents an intertwiner in MATH. Therefore it vanishes unless MATH and then it must be a scalar multiple of MATH. Hence we can insert a term MATH which corresponds graphically to the disconnection of the wi... |
math/9904109 | For MATH and isometries MATH and MATH we define a vector MATH by the diagram in REF . Using again intertwiner bases, we also put MATH with some multi-index MATH. It follows from the right-hand side in REF that MATH. Conversely, we obtain by REF that MATH unless MATH, hence MATH. With MATH, closing the wires on the bott... |
math/9904110 | The assertion MATH is proved in REF in the form MATH and MATH for MATH. Let us prove MATH. We write MATH for MATH, and MATH for MATH for convenience. By NAME vanishing theorem (see CITE) MATH for MATH. The sequence REF remains exact after shifting by MATH . Note, that MATH is a free MATH-module of rank MATH and MATH is... |
math/9904110 | The statement follows immediately from REF by taking the restriction of the sheaf MATH to MATH. |
math/9904110 | The formula MATH is the assertion of REF. We prove this statement independently. Making use of the exact sequence REF , we get MATH . Now the requested formula follows from REF and the vanishing of MATH for MATH. Let us prove assertion MATH. The short exact sequence REF remains exact after tensoring by MATH. As in the ... |
math/9904110 | We need a special form of the NAME duality (see REF) MATH for any locally free sheaf of the MATH-modules MATH with the dual MATH on a complete simplicial toric variety MATH. Take MATH, MATH. From the NAME duality we have the isomorphism MATH . Hence, using the vanishing of MATH for MATH, we get the equality MATH which ... |
math/9904110 | We have the chain of equalities MATH where we have used the identity MATH which is proved in the Appendix. |
math/9904110 | Using the equality of REF , we get MATH as required. |
math/9904110 | Indeed, from the long exact sequence of REF , we have MATH . Since MATH is ample, MATH applying the generalized NAME formula given in REF , we prove the statement. |
math/9904110 | It follows from REF that MATH and MATH . Each MATH can be uniquely written as MATH, where MATH, MATH. The support polytope MATH is defined by the inequalities MATH in MATH. Let MATH. Then the inequalities above are equivalent to MATH . The same arguments as in REF show that MATH, where MATH, and MATH follows immediatel... |
math/9904110 | From the MATH-th exterior power of the sequence REF , we have MATH . Since MATH is ample, for all MATH, we obtain the exact sequence of global sections MATH REF implies the isomorphisms MATH and consequently, the isomorphisms MATH . Hence we obtain the relations MATH . By induction we get the requested formula. |
math/9904110 | Since MATH is a simple, there are precisely MATH-faces of MATH containing a given MATH-face MATH of MATH. Hence, the requested formula is equivalent to the combinatorial identity MATH or MATH . We give a proof of this identity based on the method of integral representations. Rewrite the sum MATH as MATH where the cycle... |
math/9904111 | Observe that MATH and MATH are well defined as functions on the MATH-interval MATH since MATH. The formula for MATH follows by observing that MATH is of the form REF with MATH and MATH given by MATH . By inserting the explicit functions MATH and MATH, we see that the corresponding MATH and MATH coincide with the ones g... |
math/9904111 | Since MATH and MATH (respectively MATH and MATH) are real valued on the MATH-interval MATH, we may restrict the proof to real valued functions MATH and MATH. Then we derive by a direct computation using REF that MATH . The lemma is now an easy consequence of this formula since the finite sums become telescoping. |
math/9904111 | Using the formula MATH it follows that the behaviour of the weights of MATH at infinity is given by MATH where MATH is the positive constant MATH . This implies that MATH for all MATH. The proof follows now from the definition of the Wronskian REF and the fact that MATH . |
math/9904111 | REF follows immediately from REF . For the proof of REF , we observe first that the weights of MATH around zero behave like MATH . Define now the function MATH by MATH if MATH, MATH if MATH and MATH if MATH. Then MATH follows from REF, and MATH since MATH if MATH. Furthermore, MATH and MATH, hence MATH. By construction... |
math/9904111 | We have to show that MATH . By REF and the fact that MATH for MATH, it suffices to show that MATH for all MATH. Since MATH, we have for all MATH, MATH as desired. |
math/9904111 | Since MATH is symmetric, it suffices to prove the inclusion MATH. Let MATH. Then by REF, MATH. For any MATH we have, using REF , that MATH . Now using REF , one derives from the existence of the limits MATH for all MATH that MATH and MATH. It follows that MATH, as desired. |
math/9904111 | We start with computing MATH. The coefficient MATH in REF of MATH is non-zero for MATH since MATH. It follows that solutions of MATH on MATH are in one to one correspondence with solutions MATH of a recurrence relation of the form MATH with MATH by definition. The correspondence is obtained by associating the sequence ... |
math/9904111 | Suppose that MATH, then we claim that the map MATH is injective. Suppose that the map is not injective. Then there exist two linearly independent functions MATH such that MATH are linearly dependent. The Wronskian MATH is not identically zero as function on MATH by the linear independence of MATH and MATH. Since MATH i... |
math/9904111 | The proof for MATH is a direct consequence of REF and the explicit REF for MATH. The proof for MATH follows then from REF. |
math/9904111 | First of all, observe that if MATH and MATH, then MATH, hence MATH is well defined. Let MATH and let MATH, MATH be such that MATH. Using the asymptotics REF for the weights of the inner product MATH . , it follows from REF that MATH because MATH. It remains to show that MATH. For the proof we use some well known result... |
math/9904111 | The explicit expression for the Wronskian follows by computing the limit MATH using the first expression of REF . Since the Wronskian is non-zero, it follows that MATH and MATH are linear independent, hence they form a basis of MATH by REF . |
math/9904111 | We first prove the connection coefficient formula for MATH. Observe that MATH is well defined for MATH since MATH. We fix MATH with MATH such that MATH. Furthermore, we assume that MATH and we fix MATH such that MATH and MATH. By the assumptions on MATH and MATH, we may apply the three term recurrence relation CITE wit... |
math/9904111 | Using the minimal solution CITE of the three term recurrence relation CITE, it follows that MATH is a solution of MATH on MATH, where MATH is the second order MATH-difference operator defined by REF. In particular, we have MATH since MATH. It follows that MATH. We have MATH by REF and MATH. It follows that MATH for a u... |
math/9904111 | We first assume that MATH. Using REF we then have MATH where MATH and MATH are given by REF, respectively. By REF , the right hand side of REF is equal to MATH where MATH is the positive constant defined by REF. It follows by direct computation that MATH . Now we can apply the MATH-product identity MATH see CITE, with ... |
math/9904111 | We fix MATH satisfying REF. Then the Wronskian MATH is non-zero since MATH, see REF . Hence MATH is a linear basis of MATH by REF . For the proof of REF , we observe first that MATH satisfies the three properties as stated in REF , see REF. If MATH is another function satisfying the same three properties, then MATH sin... |
math/9904111 | This is well known, see for instance CITE and CITE. The connection between MATH and MATH with the MATH in the last equality of REF follows from CITE. |
math/9904111 | We first prove REF for MATH. By REF , there exist unique MATH such that REF holds for all MATH. These coefficients can be expressed in terms of NAME by MATH . By REF we have MATH where MATH is the positive constant defined by REF now follows by substituting REF, the explicit REF for the Wronskian MATH, and the explicit... |
math/9904111 | Immediate from REF . |
math/9904111 | The proof for MATH is trivial since MATH, MATH, MATH and MATH are regular at MATH. For MATH (MATH), observe that MATH and MATH have simple poles at MATH and that MATH and MATH are regular at MATH. It follows from REF and the first equality of REF that the singularity of MATH at MATH is removable for MATH if the (at mos... |
math/9904111 | REF follow from REF . CASE: Let MATH with MATH, then MATH, hence MATH is well defined. Then MATH follows from REF and from REF . CASE: We first prove the connection coefficient formula for MATH. Then the connection coefficient formula is valid for all MATH, see REF . Since MATH, it follows by the uniqueness property of... |
math/9904111 | We first prove that MATH . For the proof we need to consider the two cases MATH and MATH seperately. CASE: MATH. Observe that the product REF for the MATH-derivative MATH and REF imply the following product rule for MATH: MATH . Combined with the easily verified formulas MATH and REF of MATH, we obtain MATH for MATH. B... |
math/9904111 | Recall that MATH is a generalized eigenfunction of MATH with generalized eigenvalue MATH if MATH, MATH and MATH in MATH. This can be checked for all MATH by an elementary computation using REF . Let MATH. Then MATH is part of the spectrum MATH of MATH since there exists a generalized eigenfunction of MATH with generali... |
math/9904111 | For MATH we write MATH for the unique complex number with modulus less than MATH such that MATH. Fix MATH. By a straightforward computation using REF , we have for MATH and MATH that MATH where MATH is the NAME. Let MATH and let MATH such that MATH and MATH for MATH. Then we have MATH . Using the connection coefficient... |
math/9904111 | We assume first that MATH. In view of REF applied to the special case MATH and MATH, it suffices to observe that MATH, which is a consequence of REF and CITE. Finally, observe that the inequality MATH for MATH and MATH, where MATH is the norm of MATH, holds for all MATH by continuity. Hence MATH can be uniquely extende... |
math/9904111 | For MATH we have that MATH on MATH. The proof follows now from REF since MATH is real valued for MATH. |
math/9904111 | Observe that there exists a constant MATH such that MATH for all MATH and for all MATH with MATH. The first equality follows then from NAME 's dominated convergence theorem. The second equality follows from REF , using that MATH since MATH is continuously differentiable at the origin, see REF . |
math/9904111 | We only sketch the proof, since it is similar to the little MATH-Jacobi case, see CITE and CITE. Let MATH be the algebra of complex valued, continuous, MATH-invariant functions on MATH. We fix MATH such that MATH is not in the support of MATH and MATH. It suffices to give a proof of REF for such functions MATH and MATH... |
math/9904111 | It follows from REF that MATH for all MATH. Hence MATH is the identity on MATH and MATH. Consequently, we have MATH. CASE: Assume that MATH. Let MATH. Observe that MATH by the previous paragraph and by REF , where MATH is the norm of MATH. By REF , this implies that MATH. Let MATH. We have seen that MATH, hence there e... |
math/9904111 | Let MATH. Then MATH by REF . Furthermore, if MATH, then MATH where the first equality follows from REF . Since MATH and MATH are regular at MATH, it follows that REF is valid for all MATH. Hence MATH and MATH on MATH for all MATH, as desired. |
math/9904111 | Throughout the proof, we use the notations introduced in the proof of REF . CASE: Let MATH and fix MATH satisfying the properties as stated in REF . It suffices to prove that MATH. Observe that REF is still valid in the present setting, but that the analogue of REF is now given by MATH where MATH and where MATH is the ... |
math/9904111 | It follows from REF that MATH. Observe now that MATH by REF and MATH by REF . It follows that MATH and MATH, since MATH is the disjoint union of MATH and MATH, see CITE. |
math/9904111 | Orthogonality is clear since the functions MATH REF are eigenfunctions of the self-adjoint operator MATH with mutually different eigenvalues MATH (MATH), see REF . It remains to derive the explicit expression for the quadratic norm MATH. We first assume that MATH. Observe that MATH for MATH by REF . Since MATH and MATH... |
math/9904111 | We write MATH for the inner product of MATH. Suppose that MATH and let MATH. Applying REF , we obtain MATH. In order to extend this result to parameters MATH in MATH, we show that MATH and MATH depend continuously on MATH. This is clear for MATH, while for MATH it is clear except for the term MATH where we use the conv... |
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