paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0010066 | For MATH, we already know this from REF , so fix some MATH. Let MATH be a partition of MATH into positive integers. We can compute REF for each such partition. The maximum value is found to be achieved for MATH and MATH for MATH and for MATH; for MATH and MATH for MATH; and for either MATH and MATH or MATH and MATH for... |
math/0010066 | Define MATH, MATH, etc. analogously to the definitions of MATH, MATH, etc. in the proof of REF . Then MATH for MATH except that MATH, MATH, and, for MATH, MATH (compare CITE); therefore, MATH and MATH . As in REF , MATH . Both numerator and denominator are analytic except for a pole at MATH, so using singularity analys... |
math/0010067 | Given an irredundant primary decomposition of MATH, the ideal MATH is the intersection of the minimal primary components. (See REF; compare REF.) The ideal MATH is the intersection of the primary components contracting to zero. Thus MATH is the intersection of the minimal primary components and those contracting to zer... |
math/0010067 | We may assume that MATH is affine. Thus MATH corresponds to a homomorphism MATH from a discrete valuation ring to an equidimensional ring. Let MATH be the fiber ring, and let MATH be a uniformizing parameter. If MATH is not flat, then MATH is a zero divisor and thus is an element of some associated prime MATH of MATH. ... |
math/0010067 | Assume REF , but that MATH is not flat over MATH. We will first show that MATH does not specialize to MATH, and subsequently that the NAME class fails to specialize (that is, that REF does not hold). Since MATH is equidimensional, so is MATH. And since MATH is internally flat, so is the morphism MATH. Thus by REF , MAT... |
math/0010067 | Let MATH be an elliptic curve. Then the morphism MATH (projection onto the first factor followed by MATH) is a flat projective morphism from an equidimensional scheme. The subscheme MATH, embedded in MATH via MATH, is nowhere dense in MATH, and its normal cone MATH is internally flat over MATH since MATH is. The NAME c... |
math/0010068 | We apply the NAME decomposition to split the expression inside the projection MATH in REF as MATH . As it turns out, all the terms except for REF will be of the form MATH. We first consider the high-frequency contributions REF. We can rewrite REF as MATH . To show that the contribution of this term is MATH, it thus suf... |
math/0010068 | We first observe that REF is easy to show if the derivative is moved to the low frequency term: MATH . Indeed, we simply place MATH in MATH using REF and MATH in MATH using REF. From this and the product rule it thus suffices to show that MATH . Consider the expression MATH. Heuristically, this quantity is approximatel... |
math/0010068 | As noted in the introduction, we have the identity MATH whence MATH for all MATH. We now show inductively that MATH for all MATH. This is clearly true for MATH. Now suppose that MATH and the claim has been proven for all smaller MATH. Then from REF and NAME we have MATH for all MATH. By REF, and REF we thus have MATH a... |
math/0010069 | Let MATH, MATH, MATH, MATH, MATH, MATH, MATH be as above. Define a vertical line segment to be a pair MATH such that MATH. (The notation comes from depicting MATH, MATH as one-dimensional sets, so that MATH becomes a two-dimensional set and MATH is the projection to the first co-ordinate). Let MATH denote the space of ... |
math/0010069 | (Informal) Let MATH be the union of all the tubes MATH as before; this set can be thought of as the union of about MATH-balls. Since each tube occupies about MATH of these balls, we can expect two random MATH, MATH to have a probability of about MATH of intersecting. Define a quadrilateral to be four tubes MATH, MATH, ... |
math/0010070 | Recall that a NAME space MATH is universally negligible if there is no NAME probability measure on MATH that vanishes at singletons. By CITE, there is a universally negligible set MATH of cardinality MATH (see also CITE). Pick a non-null set MATH of size MATH and fix a bijection MATH. If MATH is such that MATH is conti... |
math/0010070 | MATH . Follows from REF MATH (take MATH). MATH . If MATH, then any MATH (for MATH) work. So assume MATH and let MATH. Then MATH. For MATH put MATH . We are going to show that these MATH's are as required. To this end suppose that MATH is such that MATH (for all MATH). Then also MATH (for MATH) and by REF MATH we get MA... |
math/0010070 | Assume toward contradiction that MATH for all MATH. Choose inductively fronts MATH of MATH such that CASE: MATH, CASE: MATH, CASE: MATH for all MATH. Note that then (by REF) for each MATH we have MATH . Since the right hand side of the inequality above approaches MATH (as MATH), we get an immediate contradiction with t... |
math/0010070 | CASE: Let MATH; clearly we may assume that MATH for all MATH. Also we may assume that MATH (remember REF) and MATH. Fix MATH such that MATH for a moment. Let MATH. For each MATH pick a front MATH of MATH such that CASE: if MATH, then MATH, CASE: if MATH, then MATH. Let MATH, MATH, MATH, and MATH . Apply REF MATH for MA... |
math/0010070 | Let MATH (for MATH); note that MATH. Fix MATH for a while. Let MATH be a front of MATH such that MATH . By downward induction, for each MATH, we define MATH and MATH such that CASE: If MATH, then MATH, MATH. CASE: If MATH, then MATH, MATH. CASE: If MATH, MATH, then: if MATH, then MATH, else MATH. Clauses MATH define MA... |
math/0010070 | Let MATH consist of all MATH such that CASE: MATH and there is a normal condition MATH stronger than MATH and such that MATH, MATH, MATH, and for some front MATH of MATH, for every MATH: CASE: MATH and the condition MATH decides the value of MATH, and CASE: no initial segment of MATH has the property stated in MATH abo... |
math/0010070 | We may assume that MATH is normal, MATH, MATH, and MATH for MATH. We build inductively a sequence MATH such that CASE: MATH is a normal condition, MATH, MATH, MATH, CASE: MATH is a front of MATH, MATH, CASE: if MATH, then MATH, and for each MATH such that MATH we have MATH, CASE: if MATH, then MATH, CASE: for each MATH... |
math/0010070 | Should be clear at the moment; compare the proof of CITE. |
math/0010070 | Let MATH, MATH. We try to choose inductively partial functions MATH from MATH to MATH such that CASE: MATH, MATH, CASE: MATH. Note that in REF MATH, the left hand side expression is not more than MATH, so if the inequality holds, then (as MATH) CASE: MATH. Consequently, in the procedure described above, we are stuck at... |
math/0010070 | Clauses REF should be obvious, so let us check REF MATH only. Let MATH, MATH, MATH be as in the assumptions of REF MATH. So in particular MATH . For MATH let MATH. First, we consider the case when both MATH and MATH are not smaller than MATH. Then we may apply REF and get MATH such that MATH, MATH and MATH . Then MATH ... |
math/0010070 | Let MATH, MATH, MATH. First note that MATH . Let MATH. For each MATH we may use REF to pick MATH such that MATH and MATH . Hence, MATH . Consequently, MATH and hence MATH . |
math/0010070 | Assume that MATH is a set of outer REF measure MATH. We have to show that, in MATH, it is still an outer measure one set. Let MATH be a MATH - name for a tree such that MATH and the NAME measure MATH of the set MATH of MATH - branches through MATH is positive, and suppose that some condition MATH forces MATH. Take a co... |
math/0010070 | Let MATH, MATH, MATH. Note that MATH . Let MATH consist of all triples MATH such that MATH and MATH and fix MATH for a moment. Let MATH be such that MATH and MATH. Apply REF (to MATH and MATH for MATH, MATH) to pick MATH such that MATH, MATH and MATH . Next note that MATH, so MATH . Therefore, MATH . Hence, MATH and th... |
math/0010070 | Assume toward contradiction that MATH is a MATH - name for a tree included in MATH, and MATH is a condition such that MATH . (Here, MATH stands for the product measure on MATH.) Passing to a stronger condition and shrinking the tree MATH (if necessary) we may assume that CASE: MATH is special and MATH, and MATH for all... |
math/0010070 | The equivalences MATH are well known (see CITE; also compare with the proof of CITE). MATH: Assume MATH, and suppose that MATH is a non-measurable set. Then we may find a closed set MATH such that CASE: for each MATH, the set MATH is either empty or is a perfect set of positive NAME measure, CASE: for every NAME set MA... |
math/0010070 | Start with universe MATH satisfying CH. Let MATH be countable support iteration such that each iterand MATH is (forced to be) the forcing notion MATH (defined in the second section; of course it is taken in the respective universe MATH). It follows from REF (and CITE) that the limit MATH is proper and MATH - bounding. ... |
math/0010070 | Assume MATH. Let MATH be a NAME isomorphism preserving null sets (see, for example, CITE), and let MATH. Let MATH be given by REF for MATH. Put MATH. We know that MATH is a non-null set (and consequently it is non-measurable), so applying MATH we may pick a NAME function MATH such that the set MATH has positive outer m... |
math/0010070 | It is a standard application of fusion arguments somewhat similar to the proof of CITE; compare also the proof of preservation of ``proper+MATH - bounding" in CITE or CITE. Of course, we use here REF . |
math/0010070 | For a set MATH, MATH, and MATH we let MATH . We may assume that MATH (otherwise the lemma is easier and actually included in this case). Fix an enumeration MATH such that MATH, and let MATH . We are going to define inductively a decreasing sequence MATH of closed (non-empty) subsets of MATH such that MATH and CASE: for... |
math/0010070 | For MATH let MATH be defined by MATH . This defines a function MATH on MATH. Plainly, MATH is a nice measured tree creating pair (we are going to use it to simplify notation only). Let MATH be a tree such that MATH and MATH. For MATH let MATH be such that MATH . Let MATH. It should be clear that (as MATH has positive N... |
math/0010070 | For MATH let MATH be a MATH - name for the generic real added at stage MATH (so it is a member of MATH if MATH, and a member of MATH if MATH). Suppose that MATH is a MATH - name for a function from MATH to MATH, and MATH. For each MATH pick a template with a name MATH, an enumeration MATH of MATH, and a condition MATH ... |
math/0010073 | Let MATH be a simple polytope. Choose a linear function MATH which is generic, that is, it takes different values at all vertices of MATH. For this MATH there is a vector MATH in MATH such that MATH. Note that MATH is parallel to no edge of MATH. Now we view MATH as a height function on MATH. Using MATH, we make REF-sk... |
math/0010073 | In CITE the statement was proved directly, using MATH-vectors. Instead, we use MATH-vectors, which somewhat simplifies the proof. It is sufficient to prove that the affine hull of the MATH-vectors MATH of simple MATH-polytopes is a MATH-dimensional plane. Set MATH, MATH. Since MATH, REF gives MATH . It follows that MAT... |
math/0010073 | For any subset MATH denote by MATH the square-free monomial MATH. Let MATH be a square-free monomial ideal. Set MATH . Then one easily checks that MATH is a simplicial complex and MATH. |
math/0010073 | Let MATH be a regular sequence of degree-two elements of MATH. Then MATH is a graded algebra generated by degree-two elements, and MATH. Now the result follows from REF . |
math/0010073 | Since MATH is a MATH-vector, there exists a graded algebra MATH generated by degree-two elements such that MATH REF . In particular, MATH. Since MATH is generated by MATH, the number MATH can not exceed the total number of monomials of degree MATH in MATH variables. The latter is exactly MATH. |
math/0010073 | Using the NAME resolution MATH for the definition of MATH we calculate MATH . |
math/0010073 | Set MATH, MATH, MATH, MATH. Then MATH is a free MATH-module and MATH. Therefore, REF , gives a spectral sequence MATH . Since MATH is a regular sequence, MATH is a free MATH-module. Therefore, MATH . It follows that MATH for MATH. Thus, the spectral sequence collapses at the MATH term, and we have MATH which concludes ... |
math/0010073 | We apply REF to a minimal resolution of MATH. It follows from REF that the numerators of the summands in the right hand side of REF are exactly MATH, MATH. Hence, MATH . Using REF we get MATH . |
math/0010073 | This follows from REF . |
math/0010073 | This follows from the fact that the MATH-faces of MATH are in one-to-one correspondence with pairs MATH of faces of MATH such that MATH. |
math/0010073 | Subdividing each cube of MATH as described in Construction REF we obtain a simplicial complex, say MATH. Then applying Construction REF to MATH we get a cubical complex that subdivides MATH and embeds into some MATH (see REF ) as the subcomplex MATH. |
math/0010073 | Realise the simple polytope as a lattice polytope MATH. Let MATH be the corresponding toric variety. REF is already proved REF . It follows from REF that the multiplication by MATH is a monomorphism MATH for MATH. This together with REF gives MATH, MATH, thus proving REF . To prove REF , define the graded commutative M... |
math/0010073 | The MATH-th NAME number equals the number of MATH-dimensional cells in the cellular decomposition constructed above. This number equals the number of vertices of index MATH, which is MATH by the argument from the proof of REF . |
math/0010073 | At the beginning we choose signs of the edge vectors at MATH arbitrary, and express MATH as in REF . Then MATH is the edge vector corresponding to the edge MATH opposite to MATH, MATH. It follows that MATH and MATH for MATH. Hence, MATH (see REF ). Since MATH is a primitive vector, it follows from REF that MATH. Changi... |
math/0010073 | The orientation of MATH is determined by the canonical orientation of the almost complex manifold MATH and the orientation of the torus REF . Since the almost complex structure on MATH is MATH-invariant, it induces almost complex structures on the MATH-fixed submanifolds MATH. It follows that for any vertex REF the vec... |
math/0010073 | Since the one-dimensional subtori MATH, MATH, generate MATH, we may define MATH as any automorphism of MATH that maps MATH to MATH, MATH. |
math/0010073 | The characteristic map for MATH has the form MATH where MATH, MATH, is the diagonal subgroup in MATH. Let MATH be a quasitoric manifold over MATH with characteristic map MATH. We may assume that MATH, MATH, by REF . Then it easily follows from REF that MATH where MATH, MATH. Now define the automorphism MATH by MATH . I... |
math/0010073 | In order to construct a required embedding we provide a smooth function MATH such that the equation MATH defines a hypersurface diffeomorphic to MATH. For MATH we take MATH; then we proceed by induction on MATH. Suppose we have a function MATH such that MATH defines a smooth embedding MATH and MATH for any MATH satisfy... |
math/0010073 | Construction REF provides the atlas MATH for MATH as a manifold with corners. The set MATH is based on the vertex MATH and is diffeomorphic to MATH. Then MATH. We claim that MATH can be realised as an open set in MATH, thus providing a chart for MATH. To see this we embed MATH into MATH as a closed hypersurface MATH RE... |
math/0010073 | It follows from REF that MATH meets every isotropy subgroup only at the unit. This implies that the action of MATH on MATH is free. By definitions of MATH and MATH, the projection MATH descends to the projection MATH which displays MATH as a principal MATH-bundle over MATH. |
math/0010073 | Recall that the cubical decomposition of MATH consists of the cubes MATH based on the vertices MATH. Note that MATH is contained in the open set MATH (see Construction REF). The inclusion MATH is covered by an equivariant inclusion MATH, where MATH is a closed subset homeomorphic to MATH. Since MATH and MATH is stable ... |
math/0010073 | In this proof we identify the polyhedrons MATH and MATH with their images MATH and MATH under the map MATH, see REF . For each vertex MATH denote by MATH the union of MATH-cubes of MATH that contain MATH. Alternatively, MATH is MATH. These MATH are analogues of facets of a simple polytope. Moreover, if MATH for some MA... |
math/0010073 | Indeed, MATH is a union of ``moment-angle" blocks REF , and each MATH is the closure of the cell MATH. |
math/0010073 | Since MATH is the closure of the cell MATH, the statement follows from REF . |
math/0010073 | To prove that MATH is a cellular subcomplex of MATH we just mention that it is the closure of the MATH-cell MATH. So, it remains to prove that MATH is contractible within MATH. To do this we show that the embedding MATH is homotopic to the map to the point MATH. On the first step we note that MATH contains the cell MAT... |
math/0010073 | Indeed, REF-skeleton of our cellular decomposition of MATH is contained in the torus MATH. |
math/0010073 | Consider the decomposition REF . Since each MATH is MATH-stable, the NAME construction MATH is patched from the NAME constructions MATH, MATH. Suppose MATH; then MATH (see REF ). By definition of NAME construction, MATH. The space MATH is the total space of a MATH-bundle over MATH. It follows that there is a deformatio... |
math/0010073 | Note that MATH and REF-skeleton of MATH coincides with that of MATH. If MATH is MATH-neighbourly, then it follows from REF that the MATH-skeleton of MATH coincides with that of MATH. Now, both statements follow easily from the exact homotopy sequence of the map MATH with homotopy fibre MATH REF . |
math/0010073 | Since MATH acts freely on MATH, we have MATH . |
math/0010073 | The differentials of the spectral sequence are all trivial by dimensional reasons, since both MATH and MATH have cells only in even dimensions (see REF ). |
math/0010073 | The monomorphism MATH takes MATH to MATH, MATH. Since MATH is a free MATH-module (note that this follows from REF , so we do not need to use REF ), MATH is a regular sequence. It follows that the kernel of MATH is exactly MATH. |
math/0010073 | Let us consider the NAME - NAME spectral sequence of the commutative square MATH where the left vertical arrow is the pullback along MATH. REF shows that MATH is homotopy equivalent to MATH. By REF , the map MATH induces the quotient epimorphism MATH, where MATH denotes the cellular cochain algebra. Since MATH is contr... |
math/0010073 | The spectral sequence under consideration converges to MATH and has MATH . The differential in the MATH term acts as in REF . Hence, MATH (the last identity follows from REF ). |
math/0010073 | A routine check shows that the cochain homotopy operator MATH for the NAME resolution (see the proof of REF) establishes a cochain homotopy equivalence between the maps MATH and MATH of the algebra MATH to itself. That is, MATH . We just illustrate the above identity on few simple examples. MATH . |
math/0010073 | It follows directly from the definitions of MATH and MATH that the map is an isomorphism of graded algebras. Thus, it remains to prove that it commutes with differentials. Let MATH, MATH and MATH denote the differentials in MATH, MATH and MATH respectively. Since MATH, MATH, we need to show that MATH, MATH. We have MAT... |
math/0010073 | We make calculations with the cochain complex MATH. The module MATH has the basis consisting of monomials MATH with MATH and MATH. Since MATH, MATH, the bigraded component MATH is generated by monomials MATH with MATH and MATH. In particular, MATH if MATH or MATH, whence REF follows. To prove REF we observe that MATH i... |
math/0010073 | It follows from REF that MATH . Then MATH . Denote MATH. Then it follows from REF that MATH . Substituting MATH for MATH above, we finally obtain from REF MATH which is equivalent to REF . |
math/0010073 | We have MATH . Now the statement follows from REF . |
math/0010073 | Since MATH and MATH, we have MATH . Combining REF we get MATH . Hence, MATH where we used REF . |
math/0010073 | This follows from REF . |
math/0010073 | It follows from REF that MATH. By definition, the module MATH is spanned by monomials MATH such that MATH, MATH, MATH. Any such monomial is a cocycle. Suppose that MATH are two MATH-simplices of MATH that share a common MATH-face. We claim that the corresponding cocycles MATH, MATH, where MATH, MATH, represent the same... |
math/0010073 | Every subgroup of MATH of dimension MATH intersects non-trivially with any MATH-dimensional isotropy subgroup, and therefore can not act freely on MATH. |
math/0010073 | Since any isotropy subgroup for MATH is coordinate (see REF ), it intersects with MATH only at the unit. |
math/0010073 | The map MATH defines the epimorphism of tori MATH. It is easy to see that if MATH is a regular colouring, then MATH acts freely on MATH. |
math/0010073 | It follows from REF that the orbits of the MATH-action on MATH corresponding to the vertices of MATH have maximal (rank MATH) isotropy subgroups. The isotropy subgroup corresponding to the vertex MATH is the coordinate subtorus MATH. The subgroup REF acts freely on MATH if and only if it intersects each isotropy subgro... |
math/0010073 | REF shows that if MATH admits a characteristic map MATH, then the MATH-dimensional subgroup MATH acts freely on MATH, whence MATH. Now suppose MATH, that is, there exists a subgroup REF of rank MATH that acts freely on MATH. The corresponding MATH-matrix MATH defines a monomorphism MATH whose image is a direct summand.... |
math/0010073 | REF shows that MATH . The quotient MATH is identified with MATH. |
math/0010073 | Since MATH and MATH, we have MATH . By REF , MATH . Combining the above two identities with REF we get MATH, which concludes the proof. |
math/0010073 | The inclusion of the subgroup MATH defines the map of classifying spaces MATH. Let us consider the commutative square MATH where the left vertical arrow is the pullback along MATH. It can be easily seen that MATH is homotopy equivalent to the quotient MATH. The NAME - NAME spectral sequence of the above square converge... |
math/0010073 | The map MATH takes the cohomology ring MATH to the subring MATH of MATH. This subring is isomorphic to the quotient MATH. Now the assertion follows from the fact that a polytope MATH is MATH-neighbourly if and only if the ideal MATH does not contain monomials of degree MATH. |
math/0010073 | To prove the first assertion we construct homotopy equivalence MATH as it is shown on REF . This map is covered by an equivariant homotopy equivalence MATH. The second assertion follows easily from the definition of MATH. |
math/0010073 | We need only to check that the differential MATH adds MATH to bidegree. This follows from REF and the following formulae: MATH . |
math/0010073 | The definition of MATH (see REF ) shows that MATH (see REF) is a cell of MATH if and only if the following two conditions are satisfied: CASE: The set MATH is a simplex of MATH. CASE: MATH. Let MATH denote the number of cells MATH with MATH, MATH, MATH, MATH, MATH, MATH. It follows that MATH where MATH is the MATH-vect... |
math/0010073 | By REF , MATH and MATH. Moreover, it can be seen in the same way as in REF that relative NAME duality isomorphisms REF regard the bigraded structures in the (co)homology of MATH and MATH. It follows that MATH . Using REF we calculate MATH . Substituting the formula for MATH from REF and the above expression into REF we... |
math/0010073 | Since MATH, MATH, we have MATH (the coefficient MATH can be dropped since for odd MATH the left hand side is zero). |
math/0010073 | Suppose MATH is a coordinate subspace arrangement in MATH. Define MATH . Obviously, MATH is a simplicial complex. By the construction, MATH depends only on MATH (that is, on MATH) and MATH, whence the proposition follows. |
math/0010073 | Take MATH. Let MATH be the maximal subset of MATH such that MATH (that is, MATH). Then it follows from the definition of MATH (see REF ) that MATH is a simplex of MATH. Hence, MATH and MATH. Thus, the first statement is proved. Since MATH is the quotient of MATH, the second assertion follows from the first one. |
math/0010073 | First, we construct a deformation retraction MATH. This is done inductively. We start from the boundary complex of a MATH-simplex and remove simplices of positive dimensions until we obtain MATH. On each step we construct a deformation retraction, and the composite map would be a required retraction MATH. If MATH is th... |
math/0010073 | If MATH then vectors MATH (see REF ) are linearly dependent, which is impossible. |
math/0010073 | A point from MATH has the non-trivial isotropy subgroup with respect to the action of MATH on MATH if and only if at least one of its coordinates vanishes. It follows from REF that if a point MATH has some zero coordinates, then the corresponding facets of MATH have at least one common vertex MATH. Let MATH. The isotro... |
math/0010073 | This follows from REF . |
math/0010073 | Let us consider the NAME - NAME spectral sequence of the NAME fibration MATH with fibre MATH, where MATH is the NAME - Reisner space REF and MATH is the path space over MATH. By REF , MATH and the spectral sequence converges to MATH. Since MATH is contractible, there is a cochain equivalence MATH. We have MATH. Therefo... |
math/0010073 | Consider the bundle MATH with fibre MATH. It is not hard to prove that the corresponding loop bundle MATH with fibre MATH is trivial (note that MATH). To finish the proof it remains to mention that MATH. |
math/0010075 | We prove the theorem using the atomic decomposition of MATH. See CITE and Latter CITE. Since finite sums of atoms are dense in MATH we will work with such sums and we will obtain estimates independent of the number of terms in each sum. The general case will follow by a simple density argument. Write each MATH, MATH as... |
math/0010075 | Let us set MATH for all MATH. Given MATH, we define MATH to be the unique MATH such that MATH. We also set MATH . We now observe that if MATH and MATH, then MATH is constant on MATH. Therefore for MATH the sets below are well-defined MATH . For MATH and any MATH, we let MATH . Note that MATH is a constant since both fu... |
math/0010075 | Given the cubes MATH we can find a finite collection of dyadic cubes MATH with MATH and MATH where MATH. We apply REF to the functions MATH. (We collapse terms when the same dyadic cube is used twice). We obtain MATH where MATH. Inserting this estimate in REF gives MATH and the required conclusion follows from the last... |
math/0010075 | We discuss the multilinear interpolation needed to prove this theorem for all indices MATH. REF is valid when all the MATH's satisfy MATH as proved in CITE. In REF we considered the case when all MATH. We now fix indices MATH so that some of them are bigger than MATH and some of them are less than or equal to MATH. We ... |
math/0010075 | The proof is similar to that for MATH. First we consider the case where all the MATH's are less than or equal to one. It follows from CITE that MATH is bounded on the same range as MATH with bound at most a multiple of MATH. Thus the estimates in REF follow as before. Next observe that the kernels MATH satisfy REF unif... |
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