paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0010223 | Once we show that REF MATH, then REF the fiber MATH is an AR by REF , so REF the principal bundle has a global section and it is trivial and REF follows from REF . It remains to prove (MATH). CASE: MATH and REF (under REF ): We can apply REF to MATH. We can verify REF as follows: CASE: By the assumption MATH, MATH, MAT... |
math/0010223 | By REF each MATH is an AR. Thus by REF MATH is also an AR and it strongly deformation retracts onto the single point set MATH. Hence the principal bundle MATH is trivial and MATH strongly deformation retracts onto the fiber MATH. In particular, MATH is connected and MATH. |
math/0010223 | By REF , for each MATH there exists a strong deformation retraction MATH REF of MATH onto MATH. A strong deformation retraction MATH REF of MATH onto MATH is defined as follows: MATH . Since MATH, the map MATH is continuous. (In REF , the same conclusion follows from REF by taking the end compactification of MATH.) |
math/0010223 | By REF for each MATH we have the trivial principal bundle MATH . It follows that MATH admits a section MATH, and the map MATH is fiber preserving homotopic to MATH over MATH. Since each fiber of MATH has MATH, this homotopy is a MATH-homotopy. Since MATH is an ANR REF , by REF MATH is also an ANR. |
math/0010223 | Since MATH is locally compact and locally connected, MATH is a topological group and MATH is a closed subgroup of MATH. Since MATH is locally compact and second countable, MATH is also second countable. A complete metric MATH on MATH is defined by MATH for MATH, where MATH is a complete metric on MATH with MATH. Since ... |
math/0010224 | The assertion will be verified in the next four steps: CASE: For each MATH there exists a map MATH such that MATH and MATH. CASE: There exists a homotopy MATH such that REF MATH and MATH and REF each MATH admits a neighborhood MATH such that MATH for some MATH. CASE: There exists a function MATH such that MATH, MATH an... |
math/0010224 | CASE: Since the restriction map MATH is a trivial bundle with an AR fiber and MATH, inductively we can construct a section MATH of MATH such that MATH. The section MATH is defined by MATH. CASE: We replace the interval MATH by MATH. For each MATH let MATH. We have MATH and MATH. Since MATH is a trivial bundle with an A... |
math/0010224 | We verify the following assertion: CASE: Every MATH admits an open neighborhood MATH and a homotopy MATH such that MATH is the inclusion MATH and MATH. This easily implies that MATH itself admits an absorption homotopy into MATH (compare CITE). Consider the bundle MATH (over its image) (compare Proof of CITE, CITE). Fr... |
math/0010224 | CASE: Choose a point MATH and let MATH (MATH). Since MATH is proper, MATH is compact. For MATH and MATH consider the subspace MATH . Since MATH, it suffices to show that each MATH is compact. For each MATH we have MATH, MATH, where MATH. By the NAME theorem, the space of MATH-Lipschitz maps from MATH to MATH, MATH, is ... |
math/0010224 | We define a compatible metric MATH of MATH by MATH where MATH is the length of a path MATH in MATH. First we note that MATH is a locally NAME homeomorphism. This follows from the following observations: CASE: We may assume that MATH is compact. CASE: In general, if MATH and MATH are compact connected Euclidean polyhedr... |
math/0010224 | CASE: The value MATH satisfies the following equations: MATH . Thus MATH and MATH, and so MATH. |
math/0010224 | CASE: The assumption implies that MATH and MATH. CASE: Set MATH and MATH. |
math/0010224 | REF follows from REF and the definition of MATH. REF follows from the next facts: CASE: If MATH (MATH), then MATH is a MATH-Lipschitz homeomorphism CITE. CASE: If MATH, then MATH is a NAME. CASE: MATH. |
math/0010224 | We choose any homeomorphism MATH with MATH and a similar affine embedding MATH, where MATH is a principal simplex of in MATH of MATH. For MATH, MATH and MATH we obtain the stability homeomorphism MATH. The following pairs MATH REF satisfy REF - REF : MATH . In fact, REF follow from REF respectively and REF is obvious. ... |
math/0010224 | Choose a NAME MATH onto a Euclidean polyhedron and let MATH. By REF MATH induces a homeomorphism of tuples, defined by MATH: MATH where MATH. It suffices to show that the latter tuple is MATH-stable. Let MATH, MATH and MATH be the maps given in the proof of REF for MATH. Then MATH and MATH. Since MATH, the data MATH, M... |
math/0010224 | We choose REF a triangulation MATH of MATH with MATH for each simplex MATH of MATH, REF a sequence of principal simplices MATH of MATH such that MATH, MATH and MATH (MATH), and REF similar affine embeddings MATH such that MATH and MATH for some MATH. For MATH, MATH and MATH we obtain the stability homeomorphism MATH. B... |
math/0010224 | We may assume that MATH. For each MATH we choose a MATH-open neighborhood MATH of the origin MATH in MATH such that the exponential map MATH maps MATH diffeomorphically onto a small open neighborhood MATH of MATH in MATH with MATH. Since MATH is isometric at the origin of MATH, if MATH is so small, then MATH is a REF-L... |
math/0010224 | CASE: MATH is a MATH-set of MATH itself since MATH REF are MATH-sets of MATH and MATH. Note that MATH. We identify MATH with MATH by taking any linear ordering of MATH. Take an appropriate sequence MATH of disjoint infinite subsets of MATH so that MATH satisfies REF . It follows that REF MATH and REF MATH. CASE: The as... |
math/0010224 | The proof is a modification of the argument of CITE. CASE: We can write MATH, where MATH is a MATH subset of MATH, and MATH, where MATH is a closed subset of MATH and MATH. By CITE there exist REF a map MATH such that MATH and REF maps MATH such that MATH, MATH and MATH. The required map MATH is defined by MATH. CASE: ... |
math/0010224 | Let MATH. By REF there exists a closed embedding MATH such that MATH, MATH and MATH. Note that MATH. There exists a map MATH such that MATH and MATH CITE. The required closed embedding MATH is defined by MATH, where the homeomorphism MATH is defined by MATH. |
math/0010224 | Since MATH, the product MATH satisfies both MATH- and MATH-stability. Hence by REF MATH is both a MATH- and a MATH-manifold. Thus by REF we have MATH . |
math/0010225 | First of all, if MATH has an atom then MATH is supported on a periodic orbit, hence the result is trivial because in this case MATH for MATH and MATH otherwise. We may assume then that MATH has no atoms, consequently MATH as MATH for all MATH. At first we suppose that MATH and assume that MATH is small enough that MATH... |
math/0010225 | It is enough to check that the variation of MATH is finite. Since MATH is MATH on the interior of MATH, the variation is estimated by MATH . |
math/0010225 | In the proof we use the estimate given by REF to apply REF. We recall the quantities considered there (MATH is any integer). MATH . We denote by MATH the density of the measure MATH with respect to MATH. Let MATH be an interval in MATH and set MATH. We now compute an upper bound for MATH and MATH. Obviously MATH, and f... |
math/0010225 | Let MATH. Obviously MATH. Let MATH be any recurrent point in MATH; by the NAME Recurrence Theorem, this concerns MATH-a.e. MATH. Let MATH be the component of MATH containing MATH and MATH be any integer such that MATH. As MATH, there exists a neighborhood MATH of MATH such that MATH maps MATH monotonically onto MATH. L... |
math/0010225 | Let MATH and for MATH . Fix MATH and let MATH . Let MATH be the first return map to MATH. We then define a partition of MATH by MATH, where MATH. One easily check that MATH is a partition into intervals, and each branch of MATH is monotone and onto. Let MATH be the normalized NAME measure restricted to MATH and let MAT... |
math/0010225 | First let us make the following remark. As MATH is expanding, a straightforward calculation shows that REF implies that MATH and MATH uniformly in MATH. Let MATH be the NAME operator MATH . Let us write MATH for the seminorm MATH. If MATH, we have MATH . Here we summed over the pairs MATH in the same atom MATH with MAT... |
math/0010225 | We can rewrite REF as MATH where MATH is NAME and bounded away from below (see CITE), and MATH is the NAME operator as defined in the previous proof. Recall also that MATH. Since MATH is one-to-one and onto when MATH, the above facts imply that MATH . The lemma follows now from REF by taking MATH and MATH. |
math/0010225 | Let us assume for simplicity that MATH. We denote by MATH the interior of a subset MATH in the induced topology. Since MATH is expanding, there exists MATH such that MATH for any MATH and integer MATH. By the NAME property, the bounded distortion and the conformality of the map MATH we can find some constant MATH such ... |
math/0010225 | The proof closely follows the one of REF , we use the same notation MATH and MATH. Let MATH and MATH to be chosen later on. By REF MATH . Similarly, we get by REF and (twice) REF MATH for all MATH. Taking MATH and MATH gives MATH . For all non periodic points MATH we have MATH, which implies MATH as MATH. Since this co... |
math/0010225 | Let MATH be arbitrary, and let MATH be open disks such that MATH and MATH is compactly contained in MATH. Let MATH be the modulus of MATH. Assume also that MATH is convex-like in the sense that MATH, where MATH is as in REF. The strategy is to find a subset MATH of MATH such that MATH for all MATH, and then we can invo... |
math/0010227 | As the algebra is finite dimensional, it follows from the weight space decomposition that there exists at least one weight MATH which is "extreme" in the following sense : MATH . Then for any vector MATH we have MATH, where MATH denotes the adjoint operator in MATH. Thus the vector is central, and the corresponding ver... |
math/0010227 | As MATH for a weight system MATH of MATH, we know that MATH contains at least one isolated point, call it MATH. Thus, in MATH is adjacent to any other point. If MATH for MATH, then the path MATH is a geodesic joining them. Thus MATH. If one of the points equals MATH, then the distance is one. |
math/0010227 | As we have imposed MATH for any weight MATH, we can reorder the weights of MATH in such manner that a relations MATH corresponds to a sum MATH, where MATH. Thus the maximal number of sums of weights equals the number of possibilities MATH with MATH and MATH. It is easily seen that this number is precisely MATH. |
math/0010227 | The assertion is obviously true for MATH. Suppose it is also for MATH. Then we have MATH CASE: if MATH then we have MATH, and as MATH we have MATH . As MATH and MATH , it follows MATH . CASE: if MATH then MATH. In this case MATH and as before MATH. |
math/0010227 | Let MATH with MATH. From the previous result we know that MATH . Now MATH for MATH, and MATH, with equality when MATH is odd. Then MATH which shows that MATH . |
math/0010234 | Simply observe that every pliable foursome from MATH is also a pliable foursome in MATH and hence that the restrictions of the elements of MATH to MATH form a pliable family for MATH. |
math/0010234 | By normality we can extend, for every MATH, any partition in MATH between MATH and MATH to a partition in MATH between MATH and MATH. Now apply the assumption that the full family is essential. |
math/0010234 | REF implies immediately that MATH is hereditarily indecomposable: for every MATH the projection MATH is as required. To see that MATH is infinite-dimensional we observe that MATH is a partition between the even-numbered faces MATH and MATH of MATH - indeed: MATH . By REF this implies that MATH is an essential family in... |
math/0010234 | As observed in the previous proof the traces of the odd-numbered faces of MATH on MATH form an essential family in MATH. One can therefore find a component MATH of MATH such that the traces from that family on MATH also form an essential family. Now let MATH be (the restriction of) the projection onto the first MATH co... |
math/0010234 | We use the partition MATH and the open sets MATH and MATH from REF again. Let MATH be a pliable family for MATH. Choose open neighbourhoods MATH and MATH of MATH and MATH respectively with disjoint closures. Whenever MATH and the open sets MATH and MATH with disjoint closures are found apply NAME 's lemma to get a cont... |
math/0010234 | We use the proof of REF . Let MATH be a pliable family of continuous functions on the NAME cube MATH. As before let MATH, where MATH and MATH is the projection onto the MATH-th coordinate. The odd-numbered faces of MATH induce an essential family on MATH; it is also essential on some component of MATH. This component i... |
math/0010234 | Let MATH and MATH be disjoint open sets around MATH and MATH respectively such that MATH. Let MATH and MATH and MATH. It is clear that MATH and that MATH, so that we are done in case MATH. If MATH then we can find an increasing sequence MATH and a decreasing sequence MATH such that MATH and MATH for all MATH. Because t... |
math/0010234 | Let MATH be a continuum of weight MATH and assume that MATH is embedded into the NAME cube MATH. Let MATH be the hereditarily indecomposable continuum from REF and let MATH be an essential map. For every finite subset MATH of MATH we consider the map MATH and the continuum MATH, where MATH is the projection of MATH ont... |
math/0010234 | We now assume that our continuum MATH is embedded in MATH, which we consider to be a subset of MATH. We take the continuum MATH from REF and let MATH be an essential map. For every finite subset MATH of MATH we let MATH denote the projection of MATH onto MATH. An application of REF yields for every finite set MATH a su... |
math/0010234 | This is a straightforward generalization of the proof of REF. Let MATH denote the standard NAME set in MATH. For MATH put MATH-if MATH then MATH and set MATH. Just as in CITE one verifies that MATH is a closed and connected subset of MATH; to see that MATH is one-dimensional one only has to realize that every basic ope... |
math/0010234 | By the previous lemma we can find a one-dimensional continuum MATH of the same weight as MATH and a monotone surjection MATH. Next find a hereditarily indecomposable continuum MATH and a weakly confluent surjection MATH. As in the proof of REF we take the monotone-light factorization of MATH, that is, a space MATH, a m... |
math/0010234 | Consider the continuum MATH constructed in REF and its essential family MATH. Partition MATH into MATH many sets MATH of size MATH and let MATH be a base for MATH. For each MATH let MATH be the union of all components of MATH on which MATH is not essential. Observe that MATH is open in MATH and that the family MATH is ... |
math/0010234 | We only sketch the argument. Let MATH be an embedding and define MATH by ``MATH is the unique point in MATH". It is straightforward to check that MATH is onto and that MATH for all MATH. |
math/0010234 | Let MATH be a lattice-base for the closed sets of MATH (of minimal size) and identify it with its copy MATH in MATH. By the NAME theorem CITE there is a lattice MATH, of the same cardinality as MATH, such that MATH and MATH is an elementary substructure of MATH. The space MATH is as required. |
math/0010234 | That MATH is a chain follows from hereditary indecomposability of MATH. It is clear that MATH is complete: if MATH then MATH is the supremum of MATH in MATH. To see that MATH has no jumps take MATH and MATH in MATH with MATH and fix an open set MATH such that MATH and MATH. Now the component MATH of MATH that contains ... |
math/0010234 | Apply hereditary indecomposability. |
math/0010234 | The function MATH is continuous and, for every MATH, the set MATH is an arc that connects MATH and MATH; it follows that MATH. |
math/0010234 | Striving for a contradiction, assume that for MATH such as in the formulation of the lemma, every continuum in MATH has diameter less than MATH. Since MATH is finite, there clearly is a finite disjoint collection MATH of closed subsets of MATH of mesh less than MATH such that MATH is contained in the interior MATH of M... |
math/0010234 | Put MATH and assume that MATH is a finite closed cover of MATH with mesh less than MATH such that each element of MATH meets at most MATH other elements of MATH. We shall associate to each element MATH a compact subset MATH of MATH such that CASE: MATH meets every element of MATH, CASE: MATH, CASE: MATH whenever MATH a... |
math/0010234 | Let MATH be an essential family in the normal space MATH. Apply NAME 's lemma to get continuous functions MATH such that MATH and MATH for all MATH and take the diagonal map MATH. If MATH is a partition between the faces MATH and MATH of MATH for each MATH, then MATH is a partition between MATH and MATH and so MATH; bu... |
math/0010236 | Denote MATH. Then we have the triple MATH of cell complexes. The lemma immediately follows from the long exact homological sequence for this triple: MATH . Since MATH, MATH is an injection. If MATH is connected then MATH is REF-dimensional, and, from comparing dimensions, we see that MATH is a surjection. Hence MATH an... |
math/0010236 | Notice first that the map MATH has at least one basis of cardinality MATH. To construct it, take a spanning tree MATH in the graph MATH (MATH can be empty), then MATH is rather obviously a basis of MATH and has cardinality MATH CITE. If now MATH are the (co)edges in MATH then the linear functionals MATH on MATH are lin... |
math/0010236 | Indeed, let MATH be a small cycle around a covertex MATH (for vertices the proof is analogous). Let MATH be the oriented edges exiting from or entering MATH; notice that we count loops with both ends at MATH twice, but with opposite signs. The corresponding edges MATH form the boundary of the country MATH around MATH; ... |
math/0010236 | REF depict the process of contraction. When contracting the edge MATH we wish to keep all changes restricted to the closure of the union of two countries of the dual map MATH containing the endpoints MATH and MATH of the contracted edge MATH. REF is pre-contraction and REF is post-contraction. The horizontal edge MATH ... |
math/0010237 | Two vertices of a polytope are adjacent exactly if there exists a linear functional which takes equal, maximal values on the vertices, and smaller values on all other vertices of the polytope. Such a functional in the first case is MATH by inspection. In the second case, we have the condition that not both of the two s... |
math/0010237 | Choose a fundamental basis. Orient it with positive sign, and where MATH is the height function relative to this basis, assign each other basis MATH the sign MATH. Note that this is MATH, as MATH is always even. REF is satisfied, since horizontal edges connect only bases of the same height; REF is satisfied as vertical... |
math/0010237 | This can be checked, case by case, by considering the restrictions that the axioms place on each of the possible face types listed in the previous section. We exhibit the first the case of the rectangle nsRect. Notice that short edges connect two bases whose heights differ by REF; thus, the four vertices of the face ta... |
math/0010237 | First cut MATH by the hyperplane MATH which is constant for MATH and passes through MATH, letting MATH be the resulting polytope containing MATH and MATH. Now let MATH be the vertex figure of MATH at MATH. Now MATH has a facet MATH contained in MATH, and vertices MATH and MATH corresponding to the original edges MATH a... |
math/0010237 | Set MATH to be the assertion that the theorem holds for some particular MATH as above, and now attempt to prove MATH for every vertex in the polytope (which is exactly the theorem) by an induction we shall call MATH. We must prove: CASE: MATH holds. CASE: Given that MATH holds for all vertices MATH with MATH we may ded... |
math/0010237 | Notice first that upon obtaining the form MATH as above, bases of maximal height correspond to non-zero minors of MATH of maximal size. This means exactly that rank of MATH is this height. We utilise two results from the standard theory of quadratic forms (see, for example, CITE for a particularly clear treatment. Note... |
math/0010237 | We first consider contradiction-freeness. The result is trivial when the Lagrangian matroid polytope is one-dimensional or smaller. If not, it is enough to show the procedure contradiction - free on REF - dimensional faces, which is a simple check. For part two, as observed when REF - REF were stated, it is enough to c... |
math/0010239 | By the inductive construction in REF above we showed that any MATH is of the form MATH for MATH as in the proposition (and that these MATH's all lie in MATH). We also showed that MATH for all MATH, so in particular MATH, since MATH. Therefore MATH is precisely the set of these MATH. |
math/0010241 | We prove REF together by showing that for MATH as in REF and MATH, if MATH then MATH and MATH for some MATH. With these hypotheses, choose any vertex MATH such that MATH is maximum. Then, since MATH we see that MATH . Since there are MATH terms on the right side (considering MATH as the multiplicity of the term MATH) a... |
math/0010241 | If MATH has a single vertex then the result is trivial, so assume MATH. Now, since MATH is strongly connected, the matrix MATH is invertible. Since MATH is one-dimensional, MATH is also one-dimensional; hence there is a unique (nonzero) vector MATH such that MATH and MATH. Then MATH, so MATH is the stationary distribut... |
math/0010241 | Let MATH be a list of the strong components of MATH such that if there is a directed edge from MATH to MATH, then MATH. We prove the claim by induction on MATH. For the basis of induction (MATH), and for the induction step (MATH), we may assume that MATH for all MATH such that MATH is a non-terminal strong component of... |
math/0010241 | Let MATH. By REF , MATH if MATH is not a terminal strong component of MATH. Hence, if MATH is a terminal strong component of MATH, then MATH so that MATH . Conversely, if MATH for each terminal strong component of MATH then the MATH defined by MATH is in the kernel of MATH. |
math/0010241 | Let MATH be a reduction sequence in MATH, and let MATH be the Laplacian matrix of MATH. The first claim is that we may apply elementary row and column operations to MATH, involving only rows MATH and columns MATH, so that these rows and columns of the resulting matrix induce a MATH-MATH identity matrix. We prove this b... |
math/0010241 | To show that MATH is injective, we must show that if MATH is such that MATH, then MATH. Accordingly, assume that MATH and MATH satisfy MATH. Let MATH. Then MATH . The entries of MATH are rational numbers, but we need to find MATH such that MATH. To do this, notice that MATH so that MATH. Since MATH is strongly connecte... |
math/0010241 | Let MATH be the vector of vertex activities of MATH. Use elementary column operations to add column MATH to column MATH for all MATH. Then use elementary row operations to add MATH times row MATH to row MATH for all MATH. The resulting matrix MATH is equivalent to MATH and has the required form, since MATH. |
math/0010241 | Let MATH be the submatrix of MATH obtained by deleting the row and column indexed by MATH; so we have MATH for some row vector MATH and column vector MATH. Since MATH, we have MATH . Similarly, let MATH be the submatrix of MATH obtained by deleting the row and column indexed by MATH; so we have MATH for some row vector... |
math/0010241 | For MATH, let MATH be the row vector with entries MATH for each MATH, and let MATH be the column vector with entries MATH for each MATH. For MATH, let MATH be the row vector with entries MATH for each MATH, and let MATH be the column vector with entries MATH for each MATH. Then the matrices MATH and MATH have the block... |
math/0010241 | In particular, the naturality condition must hold when MATH and MATH is in the automorphism group MATH of MATH. In this situation, the assignment MATH gives a representation of MATH acting on MATH and the assignment MATH gives a representation of MATH acting on MATH. (All the representation theory we need is in REF.) C... |
math/0010241 | If MATH is nonnegative and MATH is legal for MATH, then MATH is nonnegative. From this it follows that if MATH is nonnegative then every configuration in MATH is nonnegative. More generally, for any configuration MATH, define the configuration MATH by MATH and MATH. If MATH is a legal sequence for MATH which does not c... |
math/0010241 | This follows immediately from the facts that the off-diagonal elements of MATH are nonpositive, and that a vertex MATH is legal for a configuration MATH if and only if MATH. |
math/0010241 | We proceed by induction on MATH, the length of MATH. For the basis of induction, MATH, assume that MATH is legal for MATH. If MATH then the empty sequence is legal for MATH, as required. Otherwise, write MATH for some sequence of vertices MATH. Since MATH, MATH does not occur in the sequence MATH. The previous lemma an... |
math/0010241 | Let MATH be the graph with vertex-set MATH and directed edges MATH when MATH is legal for MATH. Then MATH is a nonempty graph, and, by the proof of REF , since MATH for all MATH contains no directed cycles. Therefore, MATH has at least one sink vertex, which is a stable configuration on MATH. Now suppose that MATH and ... |
math/0010241 | For REF , let MATH be a critical configuration, and let MATH be a nonempty legal sequence of vertices for MATH, such that MATH. Then MATH is a nonzero vector of nonnegative integers such that MATH, so that MATH. By REF , it follows that MATH is a positive integer multiple of the vector MATH of vertex activities. Every ... |
math/0010241 | That coevalence is an equivalence relation is easy to see. Let MATH be a critical configuration on MATH, let MATH be a nonempty sequence of vertices which is legal for MATH, such that MATH, and as short as possible subject to these conditions, and let MATH. As in the proof of REF , we see that MATH for some positive in... |
math/0010241 | Let MATH be the vector of vertex activities of MATH. Let MATH be any critical configuration of MATH, and let MATH be a nonempty sequence of vertices which is legal for MATH and such that MATH. Let MATH be the multiplicity vector of MATH; as in REF we have MATH for some positive integer MATH. Since MATH is stable, MATH.... |
math/0010241 | For MATH the element MATH of MATH is in MATH if and only if MATH for some positive integer MATH. Since every column MATH of MATH satisfies MATH, if MATH is in MATH then MATH. Conversely, if MATH then MATH for some rational MATH-indexed vector MATH, since the rank of MATH is MATH. Hence, MATH for some positive integer M... |
math/0010242 | Note that MATH is invertible on MATH and MATH. Let MATH with MATH, then MATH. In the same way we get MATH, hence the matrices MATH are invertible and uniformly bounded by MATH and MATH . REF yields that MATH strongly. |
math/0010243 | Since MATH is positive definite we have MATH and by REF operators CITE MATH . We denote by MATH the set of all formal series MATH for which MATH . It follows from REF that MATH is a NAME algebra with respect to the multiplication (discrete convolution) MATH where MATH and MATH. By REF on page REF an element of MATH has... |
math/0010243 | We have MATH . To prove REF we note that by REF there exists a MATH such that MATH with a constant MATH depending only on MATH and on the condition number of MATH. Since MATH we get MATH for all MATH. That means we can choose MATH independently of MATH. Write MATH and note that MATH for MATH. Since the non-zero entries... |
math/0010243 | CASE: It is well-known CITE that the eigenvalues of MATH are given by MATH . For REF that means they are regularly spaced samples MATH of the function MATH, which immediately yields the left hand side of REF . In order to prove the right hand side of REF it is sufficient to estimate MATH where MATH and MATH. Define the... |
math/0010243 | CASE: Set MATH and MATH. By REF operators CITE MATH is given by MATH . By REF we can easily find a MATH such that MATH is invertible for all MATH, which implies that MATH. In this case by the properties of circulant matrices CITE the entries MATH of MATH can be computed as MATH . Now consider MATH . We estimate the exp... |
math/0010243 | Similar to REF we write MATH . Using the NAME complement CITE we can write MATH as MATH . We will first show that the entries of the matrix MATH are exponentially decaying off the corners of the matrix. Note that REF implies MATH where MATH. We analyze the decay behavior of MATH in two steps by considering first MATH a... |
math/0010243 | First note that by REF we can always find a MATH such that MATH exists for all MATH. There holds MATH where MATH is a MATH finite section of MATH. The result follows now by applying REF . |
math/0010243 | The proof is similar to that of REF . To avoid unnecessary repetitions we only indicate the modifications, that are required. By REF MATH is invertible. Note that MATH is a matrix with three bands, one band is centered at the main diagonal, and the two other bands of width MATH are located at the lower left and upper r... |
math/0010243 | By definition of MATH we have MATH where MATH. We know from REF in the proof of REF that the entries of MATH can be bounded by MATH . For MATH and MATH we define the orthogonal projection MATH by MATH and identify the image of MATH with MATH. We set MATH. In words, MATH is obtained from MATH by taking only the central ... |
math/0010244 | First we show that REF implies that the entries of MATH satisfy MATH for some MATH and some constant MATH. It is clear that MATH and MATH . Now MATH for some MATH. Set MATH, then MATH for some constant MATH depending on MATH. Similarly MATH . By combining REF and taking the square root we get MATH . This together with ... |
math/0010244 | Set MATH and MATH. Since MATH is positive definite we have MATH and by REF operators CITE MATH . The property MATH is equivalent to MATH. By REF on page REF an element of MATH has an inverse in MATH if it is not contained in a maximal ideal. Any maximal ideal of MATH consists of elements of the form (compare REF) MATH ... |
math/0010244 | The assumption MATH implies MATH and MATH for MATH. Denote MATH . MATH is periodic in MATH with period MATH. Since MATH we also get MATH, compare REF . In words, MATH is a block NAME operator. The frame property implies that MATH is hermitian positive definite and MATH . The submultiplicativity of the function MATH imp... |
math/0010244 | CASE: The proof of this result is similar to the proof of REF . We only indicate the necessary modifications. Let MATH be the matrix-valued defining function of a hermitian positive definite NAME operator MATH and denote MATH. It follows from the basic properties of block NAME operators that MATH for all MATH and MATH ... |
math/0010245 | Without loss of generality assume MATH, otherwise we can always set MATH. Note that MATH where MATH and MATH. Define MATH and MATH. Observe that MATH and MATH . For fixed MATH there holds MATH as MATH. Furthermore, for each MATH we have that MATH for MATH. This together with REF implies that MATH strongly for MATH. |
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