paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0008010 | As in the previous theorem, we achieve the upper bound by performing orthogonal flipturns only when no diagonal flipturn is available. However, we also choose orthogonal flipturns carefully if more than one is available. Say that an orthogonal flipturn is good if it can be followed by at least two diagonal flipturns an... |
cs/0008010 | We call an edge of an orthogonal polygon a bracket if both its vertices are convex or both its vertices are concave. An orthogonal MATH-gon has at least four brackets (the highest, leftmost, lowest, and rightmost edges) and at most MATH brackets. We claim that flipturns do not increase the number of brackets, and that ... |
cs/0008010 | In any convexifying sequence, there are exactly MATH diagonal flipturns. Every orthogonal flipturn increases the perimeter of the polygon's bounding box by at least MATH. The initial bounding box has perimeter at least MATH, and the final rectangle has perimeter exactly MATH, so the maximum number of orthogonal fliptur... |
cs/0008010 | This follows directly from REF . |
cs/0008010 | We construct an orthogonal MATH-gon MATH essentially by following the proof of REF . MATH is a rectangle. MATH is an NAME hexagon, which is convexified by one flipturn. MATH is a rectangle with a rectangular orthogonal pocket in one side, which requires three flipturns to convexify. For all MATH, MATH consists of a rec... |
cs/0008010 | Let MATH be a simple polygon and let MATH be the result of one flipturn. As we argued in the proof of REF , it suffices to focus on brackets touching the endpoints of the lid. Let MATH and MATH denote the number of boundary brackets in MATH and MATH, respectively, so that MATH. For nondegenerate flipturns, it suffices ... |
cs/0008010 | We define the potential MATH of a polygon to be its discrete angle plus half the number of brackets, that is, MATH. For the initial polygon MATH, we have MATH and MATH, so the initial potential MATH is at most MATH. For the final convex polygon, we have MATH and MATH, so the final potential MATH is at least MATH. By RE... |
cs/0008010 | It suffices to store only the slopes and lengths of the edges in the proper order, without explicitly storing the vertex coordinates. Any flipturn reverses a contiguous chain of edges, namely, the edges of the flipturned pocket. Our goal, therefore, is to maintain a circular list of items subject to the operation MATH,... |
cs/0008010 | We maintain the polygon vertices in a balanced binary tree, similarly to the proof of REF . The coordinates of the points are represented implicitly by storing a triple MATH at each internal node MATH, encoding an affine transformation to be applied to all edges in the subtree of MATH. Specifically, MATH is a translati... |
cs/0008010 | We can construct the data structures to maintain the polygon and its convex hull in MATH time. In addition to the convex hull itself, we maintain a separate list of the lids of MATH, which requires only trivial additions to our data structures. This allows us to choose a legal flipturn in constant time. By REF , we can... |
cs/0008010 | Fix an index MATH. We prove the theorem by induction on the number of inner regions in the flipturned pocket. If the pocket contains no inner regions, it must be MATH-monotone, and we easily observe that MATH and MATH. The inner regions of MATH have a natural forest structure, defined by connecting each region to the n... |
cs/0008010 | Let MATH denote the vertical width of strip MATH (and strip MATH). REF implies that MATH . Let MATH and MATH denote the MATH-coordinates of the top of MATH and MATH, respectively, and let MATH be the MATH-coordinate of the lid midpoint MATH. We easily observe that MATH . Finally, define MATH and MATH. Combining REF , w... |
cs/0008010 | It suffices to consider the special case of orthogonal polygons. A flipturn sequence for an orthogonal polygon has length greater than MATH if and only if it contains an orthogonal flipturn. Thus, to prove the theorem, we only need to show the NAME of the decision problem Orthogonal NAME: Given an orthogonal polygon, d... |
cs/0008011 | If MATH then dist-prod computes the distance product of MATH and MATH using the naive algorithm that runs in MATH time and we are done. Assume, therefore, that MATH. To see that the algorithm correctly computes the distance product of MATH and MATH in this case, note that for every MATH we have MATH where indices MATH ... |
cs/0008011 | Property MATH clearly holds when MATH is initialized to MATH. In each iteration, the algorithm chooses a set MATH and then lets MATH for every MATH. For every MATH, we have MATH, as follows from the induction hypothesis and the triangle inequality, and thus the new value of MATH is again an upper bound on MATH. REF als... |
cs/0008011 | We prove the lemma by induction of MATH. It is easy to check that the claim holds for MATH. We show next that if the claim holds for MATH, then it also holds for MATH. Let MATH and MATH be two vertices connected by a shortest path that uses at most MATH edges. Let MATH be such a shortest path from MATH to MATH. If the ... |
cs/0008011 | Condition MATH follows, as mentioned, from REF , the fact that in the last iteration MATH, and the fact that if MATH, and if there are no negative weight cycles in the graph, then there is a shortest path from MATH to MATH that uses at most MATH edges. Suppose now that MATH. By REF MATH we get that after the last itera... |
cs/0008011 | For every MATH, let MATH be the number of the iteration of rand-short-path in which MATH was set for the last time. If MATH, let MATH. We need the following claim: If MATH, then MATH. Suppose that MATH was set for the last time at the MATH-th iteration. Let MATH be the elements of the matrix MATH at the beginning of th... |
cs/0008011 | Algorithm wit-to-suc begins by initializing all the elements of the MATH matrix MATH to REF. It then constructs, for each iteration number MATH, the set MATH of pairs MATH for which MATH. It is easy to construct all these sets in MATH by bucket sorting. (In the description of wit-to-suc, MATH denotes the maximal elemen... |
cs/0008011 | The proof is almost identical to the proof of REF . We show again, by induction on MATH, that if MATH, then after the MATH-th iteration of the algorithm we have MATH. The basis of the induction is easily established. Suppose, therefore, that the claim holds for MATH. We show that it also holds for MATH. Let MATH and MA... |
cs/0008011 | By the definition of bridging sets, we get that there exists MATH such that MATH. If MATH, we are done. Assume, therefore, that MATH. Let MATH be next to last vertex on a shortest path from MATH to MATH. Clearly, MATH, MATH and MATH. Thus, there exists MATH such that MATH, and therefore also MATH. There is, therefore, ... |
cs/0008011 | We prove, by induction, that after the MATH-th iteration of short-path we have: CASE: MATH, for every MATH. CASE: If MATH then MATH. Otherwise, MATH and MATH. CASE: If MATH, then MATH. The proofs of REF are analogous to the proofs of REF . We concentrate, therefore, on the proof of REF . It is easy to check that REF ho... |
cs/0008011 | The inequalities MATH follow from the fact that elements are always rounded upwards by scale. We next show that MATH. Let MATH be a witness for MATH, that is, MATH. Assume, without loss of generality, that MATH. Suppose that MATH, where MATH (the cases MATH and MATH are easily dealt with separately). If MATH, then in t... |
cs/0008025 | As described above, we reduce REF-SAT to the given problem by forming a configuration of men with two horizontal lines of men for each variable, and three vertical lines for each clause. We connect these lines by the fan-in and fan-out gadget depicted in REF . If variable MATH occurs as the MATH-th term of clause MATH,... |
cs/0008025 | Piece MATH can king precisely if there is a directed path in MATH from MATH to one of the squares along the opponent's side of the board. A winning move exists precisely if there exists a piece MATH for which MATH includes all opposing pieces and contains an NAME path starting at MATH; that is, precisely if MATH is con... |
cs/0008036 | For each minimal model MATH-of MATH: MATH is a model of MATH CASE: for every model MATH of MATH base equivalent to some minimal model MATH-of MATH: MATH is a model of MATH, since MATH by REF CASE: MATH is a logical consequence of MATH. CASE: Only if: MATH is a logical consequence of MATH CASE: every model of MATH is a ... |
cs/0008036 | MATH CASE: MATH for every model MATH of MATH, by REF and transitivity of MATH CASE: for every model MATH of MATH: MATH, since for every model MATH of MATH: MATH CASE: for every model MATH of MATH: MATH is a model of MATH, by REF CASE: MATH is a logical consequence of MATH. |
cs/0008036 | The result is proven by induction on MATH. CASE: Goals with mulitset complexity MATH have to be a satisfiable MATH-constraint MATH. Then MATH and MATH is a MATH-answer of itself. CASE: Suppose the result holds for goals with multiset complexity less than some multiset MATH. CASE: MATH and MATH CASE: there exists a clau... |
cs/0008036 | We have to show that the supremum MATH can be attained for some MATH. CASE: For MATH, we have MATH. CASE: For MATH, we have to show that for any real MATH, MATH, the set MATH is finite. Let MATH be the finite set of real numbers of factors of clauses in MATH, MATH be the greatest element in MATH such that MATH and let ... |
cs/0008036 | We have to show that MATH. We prove by induction on MATH showing for each constraint language MATH, for each quantitative definite clause specification MATH-in MATH, for each MATH-interpretation MATH, for each MATH-chain MATH of MATH-interpretations extending some MATH-interpretation MATH, for each n-ary relation symbo... |
cs/0008036 | CASE: For each minimal model MATH-of MATH: MATH is a model of MATH CASE: for every model MATH of MATH-base equivalent to some minimal model MATH-of MATH: MATH is a model of MATH, since MATH by REF CASE: MATH is a logical consequence of MATH. CASE: MATH is a logical consequence of MATH CASE: every model of MATH-is a mod... |
cs/0008036 | The result is proven by induction on the depth MATH of the quantitative proof tree, where one unit of depth is from max-node to max-node. CASE: We know that quantitative proof trees of depth MATH have to take the form of a single max-node labeled by a satisfiable MATH-constraint MATH with root value REF. Then MATH is a... |
cs/0008036 | The result is proven by induction on MATH. CASE: We know that goals with complexity MATH have to take the form of a satisfiable MATH-constraint MATH. Then there exists a quantitative proof tree for MATH from MATH-consisting of a single max-node labeled with MATH and root value REF. CASE: Suppose the result holds for go... |
cs/0008036 | MATH . |
cs/0008036 | MATH . |
cs/0008036 | MATH . |
cs/0008036 | MATH . |
cs/0008036 | MATH . |
cs/0008036 | MATH . The equality MATH holds iff MATH is a fixed point of MATH, that is, MATH with MATH. Furthermore, MATH is a fixed point of MATH iff MATH, MATH, MATH, MATH, by REF MATH . |
cs/0008036 | Let MATH be a subsequence of MATH converging to MATH. Then for all MATH: MATH and in the limit as MATH, for continuous MATH and MATH: MATH. Thus MATH is a maximum of MATH, using REF , and MATH is a fixed point of MATH. Furthermore, MATH, using REF , and MATH is a critical point of MATH. |
cs/0008036 | MATH . |
hep-lat/0008007 | The result in REF follows from MATH . The result in REF follows from the inequality between the geometric and arithmetic mean: MATH . From the NAME characterization of eigenvalues it follows that MATH for all MATH. Hence MATH holds. Now note that MATH . Thus the result in REF is proved. |
hep-lat/0008007 | The equations in REF can be represented as MATH where MATH is the MATH-th row of MATH. Consider a MATH with MATH. Note that MATH, hence MATH. For the unknown entries in MATH we obtain the system of REF which is equivalent to MATH . The matrix MATH is symmetric positive definite and thus MATH must satisfy MATH . Using M... |
hep-lat/0008007 | Let MATH be the MATH-th basis vector in MATH. Take MATH. The MATH-th rows of MATH and MATH are denoted by MATH and MATH, respectively. Now note MATH . The minimum of the functional REF is obtained if in REF we minimize the functionals MATH for all MATH with MATH. If we write MATH, then for MATH the functional REF can b... |
hep-lat/0008007 | The construction of MATH in REF is as in REF with MATH, MATH. Hence REF is applicable with MATH. It follows that MATH is the unique minimizer of MATH . Decompose MATH as MATH with MATH strictly upper triangular. Then MATH and MATH are lower and strictly upper triangular, respectively, and we obtain: MATH . Hence the mi... |
hep-lat/0008007 | From the construction in REF it follows that MATH that is, MATH is such that MATH for all MATH. This is of the form REF with MATH, MATH. From REF we obtain that MATH is the unique minimizer of the functional MATH that is, of the functional REF . From the proof of REF , with MATH, it follows that the minimization proble... |
hep-lat/0008007 | For MATH we use the decomposition MATH, with MATH diagonal and MATH. Furthermore, for MATH, MATH. Now note MATH . The inequality in REF follows from the inequality between the arithmetic and geometric mean: MATH for MATH. For MATH in REF we use the decomposition MATH. For the approximate inverse MATH we then have MATH.... |
hep-lat/0008007 | The right inequality in REF is already given in REF . We introduce the notation MATH for the eigenvalues of MATH. From REF we obtain MATH and from this it follows that MATH holds. Furthermore, MATH yields MATH and thus MATH. We now use the left inequality in REF applied to the matrix MATH. Note that MATH . A simple com... |
hep-lat/0008007 | From the assumptions it follows that MATH is a MATH-matrix. In CITE REF it is proved that then MATH is a MATH-matrix, too. Let MATH. Because MATH has only nonnegative entries it follows that MATH . Hence MATH. Using MATH we obtain the result REF . |
hep-th/0008095 | The representation function is NAME decomposed by inserting MATH twice into the right hand side of MATH. We use hermiticity MATH, MATH-invariance MATH and transversality MATH to obtain MATH . Since the NAME measure is bi-invariant, all terms vanish except those corresponding to the MATH, MATH, which are equivalent to t... |
hep-th/0008095 | Consider an arbitrary lattice point MATH. The gauge transformation REF multiplies all `incoming' links by MATH and all `outgoing' links by MATH. Since MATH in REF is an intertwiner, MATH is unchanged. This holds for all lattice points MATH. |
math-ph/0008002 | Proof can be found in CITE. We sketch only the idea of the proof of REF. Using the known formula MATH where MATH is the transformation kernel corresponding to the potential MATH, MATH see also REF below, and substituting REF into REF , one gets after a change of order of integration a homogeneous NAME integral equation... |
math-ph/0008002 | REF can be proved in several ways. One way CITE is to recover the spectral function MATH from MATH, MATH. This is possible since MATH, MATH, and MATH where MATH are the bound states of the NAME operator MATH in MATH, MATH is the delta-function, and MATH . Note that MATH and the number MATH in REF can be found as the si... |
math-ph/0008002 | This result is due to CITE. We give a new short proof based on property C CITE. We prove that data REF determine MATH uniquely, and then REF follows from REF . To determine MATH we determine MATH and MATH from data REF . First, let us prove that data REF determine uniquely MATH. Suppose there are two different function... |
math-ph/0008002 | Our proof is new and short. We prove that, if MATH is compactly supported or decays faster than any exponential, for example, MATH, MATH, then MATH determines uniquely MATH and MATH, and, by REF , MATH is uniquely determined. We give the proof for compactly supported potentials. The proof for the potentials decaying fa... |
math-ph/0008002 | We claim that MATH determines uniquely MATH if REF holds. Thus, REF follows from REF . To check the claim, note that MATH, so MATH and use REF to get MATH for MATH, so MATH . From REF the claim follows. REF is proved. |
math-ph/0008002 | If MATH for MATH, then MATH for MATH, MATH, so data REF determine MATH and, by REF , MATH is uniquely determined. REF is proved. Of course, this theorem is a particular case of REF . |
math-ph/0008002 | First, assume MATH. If there are MATH and MATH which produce the same data, then as above, one gets MATH where MATH, MATH, MATH. Thus MATH . The function MATH is an entire function of MATH of order MATH (see REF with MATH), and is an entire even function of MATH of exponential type MATH. One has MATH . The indicator of... |
math-ph/0008002 | NAME REF - REF to get MATH . Assume that there are MATH and MATH which generate the same data MATH, MATH. Let MATH. Subtract from REF with MATH, similar equations with MATH, and get MATH . Multiply REF by MATH, where MATH, MATH, MATH, integrate over MATH and then by parts on the left-hand side, using REF . The result i... |
math-ph/0008002 | NAME REF to get MATH where MATH . It follows from REF that MATH where MATH is the NAME solution to REF . From REF one gets MATH where MATH. From REF one obtains MATH . From REF one concludes MATH . From REF one gets MATH . Thus MATH is known for all MATH. Since MATH is compactly supported, the data MATH determine MATH ... |
math-ph/0008002 | Let MATH be arbitrary, MATH, MATH if MATH. Suppose MATH . Denote by MATH and MATH the transformation operators corresponding to potentials MATH and MATH which generate spetral functions MATH and MATH, MATH. Then MATH where MATH and MATH are NAME operators. REF implies: MATH where MATH is the adjoint operator and the no... |
math-ph/0008002 | From REF one gets MATH and, using MATH, one gets MATH . Denote MATH where MATH are NAME operators. From REF one gets: MATH or MATH . Since the left-hand side in REF is a NAME operator of the type REF while the right-hand side is a NAME operator of the type REF , they can be equal only if each equals zero: MATH and MATH... |
math-ph/0008002 | If MATH and MATH have the same spectral function MATH then MATH for any MATH, where MATH the function MATH solves REF with MATH, and MATH, satisfies first two REF , and MATH where MATH is the transformation operator: MATH . Note that MATH . From REF it follows that MATH . Since Range MATH, REF implies that MATH is unit... |
math-ph/0008002 | CASE: Step MATH is done by REF . Let us prove MATH. Assume there are MATH and MATH corresponding to the same MATH. Then MATH . Therefore MATH . By REF relation REF implies MATH, so MATH. CASE: Step MATH is done by solving REF for MATH. The unique solvability of this equation for MATH has been proved below REF . Let up ... |
math-ph/0008002 | CASE: The step MATH is done by REF as we have already mentioned. The step MATH is done by finding MATH, MATH and MATH from the asymptotics of the function REF as MATH. As a result, one finds the function MATH . If MATH and MATH are known, then the function MATH is known. Now the function MATH can be found by the formul... |
math-ph/0008002 | Clearly, every MATH solution to REF solves REF . Let us prove the converse. Let MATH solve REF . Define MATH . We wish to prove that MATH solves REF . Take the NAME transform of REF in the sense of distributions. From REF one gets MATH and from REF one obtains: MATH . Add MATH to both sides of REF and use REF to get MA... |
math-ph/0008002 | The solution to REF is MATH where MATH, MATH, MATH, MATH, MATH, MATH is defined in REF , MATH is defined in REF and MATH is defined in REF . The functions MATH are the data REF . Since MATH when MATH, REF implies MATH, so one knows MATH . From REF one derives MATH and MATH . From REF one gets MATH and MATH . Eliminate ... |
math-ph/0008002 | We prove MATH. The proof of the equation MATH is similar. Since MATH equals to the number of zeros of MATH in MATH, we have to prove that MATH does not vanish in MATH. If MATH, then MATH, MATH, and MATH is an eigenvalue of the operator MATH in MATH with the boundary condition MATH. From the variational principle one ca... |
math-ph/0008002 | This is an immediate consequence of the following: Theorem CITE: If MATH is holomorphic in MATH, MATH is of NAME in MATH, that is: MATH and MATH where MATH then MATH. The function MATH maps conformally MATH onto MATH, MATH and if MATH, then MATH is holomorphic in MATH, MATH for MATH and MATH, and MATH . From REF and th... |
math-ph/0008002 | Subtract from REF with MATH this equation with MATH and get: MATH where MATH . Multiply REF by MATH, integrate over MATH, and then by parts on the left, and get MATH . By the assumption MATH if MATH, so MATH and MATH vanish at infinity. At MATH the left-hand side of REF vanishes since MATH . Thus REF implies REF . |
math-ph/0008002 | One can prove CITE that the kernel MATH of the transformation operator must solve the NAME problem MATH and conversely: the solution to this NAME problem is the kernel of the transformation operator REF . The difficulty in a study of the problem comes from the fact that the coefficients in front of the second derivativ... |
math-ph/0008002 | The idea of the proof is to consider MATH in REF as a parameter and to reduce REF to a NAME equation with constant integration limits and kernel depending on the parameter MATH. Let MATH, MATH, MATH . Then REF can be written as: MATH REF is equivalent to REF , it is a NAME equation with kernel MATH which is an entire f... |
math-ph/0008002 | First, we prove convergence of the process REF in MATH. The proof makes it clear that this process will converge in MATH and that in final number of steps one recovers MATH uniquely on CITE. Let MATH, MATH, MATH. Let us start with The map MATH maps MATH into itself and is a contraction on MATH if MATH, MATH. Let MATH, ... |
math-ph/0008002 | It is sufficient to prove that, for any MATH, the function MATH . Since MATH, and since MATH provided that MATH (see REF ), it is sufficient to check that MATH . One has MATH, thus MATH where MATH . From REF one obtains REF since MATH. REF is proved. |
math-ph/0008002 | The proof goes as above with one difference : if MATH then MATH is present in REF and in REF with MATH one has MATH . Thus, using REF , one gets MATH where MATH is a constant. Similarly one checks that MATH if MATH. REF is proved. |
math-ph/0008002 | Write MATH . Clearly MATH . By the NAME theorem CITE, one has MATH . Actually, the NAME theorem yields MATH. However, since MATH, one can prove that MATH. Indeed, MATH and MATH are related by the equation: MATH which implies MATH or MATH where MATH is the convolution operation. Since MATH and MATH the convolution MATH.... |
math-ph/0008002 | One has MATH . From REF one gets: MATH . If MATH, then MATH . Here by MATH we mean the right-hand side of REF since MATH is, in general, not analytic in a disc centered at MATH, it is analytic in MATH and, in general, cannot be continued analytically into MATH. Let us assume MATH. In this case MATH is continuously diff... |
math-ph/0008003 | The idea of the proof in the MATH direction is as follows. One constructs a functor MATH by taking tensor products: on objects one has MATH for MATH, and on arrows one puts, in obvious notation, MATH. To go in the opposite direction, one repeats the above procedure, in defining a functor MATH by means of MATH, etc. Usi... |
math-ph/0008003 | The MATH claim is part of ``NAME I", compare no. REFEF for REF, and REFEF for REF. The converse follows from nos. REF |
math-ph/0008003 | This is essentially REF (Schweizer works with the category of MATH-algebras with equivalence classes of NAME bimodules as arrows, rather than with the bicategory whose arrows are the NAME bimodules themselves, but his proof may trivially be adapted to our situation). We are indebted to NAME for drawing our attention to... |
math-ph/0008003 | This follows from a combination of REF above with REF. A direct proof would be desirable. |
math-ph/0008003 | The second part of the proposition, which at the same time proves the MATH claim in the first part, is proved by the argument following REF. For the MATH claim in the first part, we are given a regular bibundle MATH, with regular inverse MATH. This leads to two isomorphisms, as displayed in REF for rings. From MATH as ... |
math-ph/0008003 | This follows from REF . |
math-ph/0008003 | The equivalence of REF is REF. By REF is equivalent to MATH in REF , which by REF means that there is an invertible bibundle MATH in REF . By REF this is equivalent to the invertibility of the associated symplectic bimodule MATH in CITE, which by REF is REF . |
math-ph/0008010 | Note that MATH, MATH, where MATH are defined in REF above. Without loss of generality take MATH, let MATH. One has MATH . Choose MATH such that REF holds with an arbitrary fixed MATH. Then MATH . Since the set MATH is total in MATH, MATH, MATH is a bounded domain, the conclusion of REF follows. |
math-ph/0008010 | The conclusion of REF follows from REF . |
math-ph/0008010 | The function MATH is analytic with respect to MATH and MATH on the variety REF . Therefore its values on MATH extend uniquely by analyticity to MATH. In particular MATH is uniquely determined in MATH. By REF one gets: MATH . By REF and by REF , the orthogonality relation REF implies MATH. |
math-ph/0008010 | The proof is the same as the proof of REF - REF and is based on the following estimate CITE, CITE: MATH . The proof of REF is not simple CITE. It is given in REF. |
math-ph/0008010 | Multiply both sides of REF by MATH, where MATH, MATH, MATH, and integrate with respect to MATH and MATH over MATH, to get: MATH . Choose MATH and MATH such that MATH where MATH, and note that MATH as MATH, MATH, MATH, MATH, MATH is an arbitrary large but fixed number. From REF one gets MATH where MATH. One can choose M... |
math-ph/0008010 | MATH . The rest of the proof consists of the following steps: CASE: We prove that MATH where the norm is defined in REF . This estimate and REF imply (see the proof of REF ) that MATH . CASE: We prove that MATH where MATH, and the pair MATH solves REF approximately in the sense specified above. (See REF ). This estimat... |
math-ph/0008010 | Using REF , one gets: MATH . As stated below REF , one has MATH . From REF one gets: MATH . It follows from REF that MATH . From REF one gets MATH where we have used the monotone decrease of MATH as a function of MATH. Using estimate REF in order to estimate MATH, MATH one gets: MATH . Minimization of the right-hand si... |
math-ph/0008010 | As in the proof of REF , the data determine uniquely MATH on MATH so that MATH. If one has already proved that MATH, that is, MATH, then the boundary condition on MATH is uniquely determined because the scattering solution MATH is uniquely determined by the scattering amplitude in MATH (and is analytically determined b... |
math-ph/0008010 | Let us sketch the steps of the proof. CASE: MATH . This follows from the uniquness REF and from the compactness of the set MATH in the space MATH, MATH. For simplicity of the presentation we assume that MATH, that is, there is just one patch in the covering of MATH, for example, MATH is star-shaped. CASE: There exists ... |
math-ph/0008012 | If MATH: MATH is compact, then MATH is a composition of a bounded linear operator MATH and a compact operator MATH, so MATH is compact. |
math-ph/0008012 | Choose a sequence MATH such that MATH. Denote by MATH the projection MATH. If MATH is compact then MATH implies MATH and for any MATH there exists a subsequence MATH and a number MATH such that MATH for any MATH. Without loss of generality we can suppose that the sequence MATH is a subsequence of MATH and MATH for MATH... |
math-ph/0008012 | One takes the NAME domain MATH such that MATH. By the known embedding theorem for NAME domains the embedding MATH is compact. Since MATH, one obtains the conclusion of the lemma. |
math-ph/0008012 | It is obvious that MATH is a standard elementary domain. Fix MATH. The open set MATH is a finite union of domains MATH and domains MATH, MATH, MATH, MATH. Join any two points MATH, MATH by a smooth curve MATH and any pair MATH, MATH by a smooth curve MATH . The set MATH is a closed NAME curve that is the boundary of a ... |
math-ph/0008012 | Because MATH is continuous in MATH for any MATH the open set MATH is a finite union of domains of the same type as in REF . Therefore the domain MATH is a standard elementary domain of the class MATH. |
math-ph/0008012 | Since smooth functions are dense in MATH it is sufficient to prove the desired estimate only for smooth functions MATH. Integrating the inequality MATH with respect to MATH over the segment MATH and using the NAME inequality we obtain MATH . For any normed space MATH and any MATH the following inequality holds MATH . C... |
math-ph/0008012 | Using REF , one gets: MATH . |
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