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math-ph/0011001 | Let MATH be a solution of REF . Then MATH is the truncation of a convergent series, since there is a constant MATH with MATH for all MATH. Note that MATH hence MATH so that MATH . Since REF are truncations of the series in the LHS of REF , then REF implies REF . |
math-ph/0011001 | CASE: With the substitution MATH the recurrence REF becomes MATH . For MATH and MATH small, positive, define MATH. Let MATH be a small neighborhood of the interval MATH. Consider the NAME space MATH of sequences MATH with MATH analytic on MATH and continuous up to the boundary, with the norm MATH. Direct estimates show that the operator MATH defined by REF takes the ball of radius MATH in MATH into itself (if MATH and MATH are small enough), and is a contraction in this ball. Therefore the equation MATH has a unique solution in MATH, of norm less than MATH. Then MATH for all MATH and all MATH. Since the sequence MATH increases to MATH, REF follows. REF: all MATH are meromorphic on MATH. Since MATH is analytic on MATH, then from REF , MATH is analytic on MATH, having a pole at MATH: MATH (MATH). Iterating REF it follows that MATH are meromorphic on MATH. Since the intervals MATH increase toward MATH it follows that MATH are meromorphic on MATH. REF: there exists MATH and MATH such that MATH. Define MATH; we have (see REF ) MATH . Let MATH be large and denote MATH. Let MATH be large enough so that MATH is in the domain of analyticity of MATH. Iterating REF starting from MATH (and decreasing indices) we get the value MATH. If for some MATH we get MATH, REF is proved. Then assume that MATH. Consider the recurrence MATH where, in fact, MATH. The recurrence REF can be solved explicitly (it is a discrete NAME equation and a substitution MATH transforms it into a linear recurrence with constant coefficients). It has the solution MATH where MATH, MATH, and MATH so that MATH (MATH). We assume, to get a contradiction, that MATH for all MATH. Then MATH which follows immediately by induction using REF , noting that MATH is increasing in MATH. Note that there is a MATH so that MATH . Indeed (from REF ) when MATH decreases from MATH the numerator and denominator in REF increase up to REF, then decrease, until the numerator becomes negative, when MATH equals MATH where MATH is the integer with MATH. Since MATH REF then MATH, and, using REF , clearly MATH (if MATH is sufficiently large). Then REF contradict the assumption that MATH, and REF is proved. CASE: The function MATH is meromorphic on MATH, with MATH. There is a smallest value of MATH in MATH where MATH changes sign: this is either a zero, or a pole. Assume it was a pole. Let MATH be the first pole of MATH. Then MATH is positive and analytic on MATH, and MATH, MATH. Since MATH (see REF ) then MATH, hence MATH changes sign in MATH. But MATH has no zero in MATH (otherwise at that zero MATH would have had a pole, from REF ). Then MATH has a pole, with a change of sign, from MATH to MATH, in MATH. Now the argument can be repeated. It follows that for any MATH, MATH has a pole in MATH, which contradicts the fact that the domain of analyticity of MATH increases to MATH as MATH. Therefore, the first change of sign of MATH is at a zero. Let MATH be the smallest value in MATH such that MATH, MATH. Then from REF we have MATH and MATH changes sign in MATH. Now the argument can be repeated. It follows that MATH has a pole at a point MATH with MATH. CASE: Since MATH takes all the values when MATH there exists MATH such that REF holds. CASE: For MATH, since the solution of REF satisfies MATH we have from REF ,with the notation MATH, that MATH and thus MATH; then from REF MATH. CASE: The substitution (variation of constants) MATH brings the recurrence REF to a first order one: with the notation MATH we have MATH and the rest of the argument consists of straightforward estimates. |
math-ph/0011001 | REF shows that REF has a unique (up to a multiplicative constant) small solution, MATH (MATH), while the general solution behaves like MATH. Since MATH the uniqueness of the MATH solution is proven. |
math-ph/0011001 | CASE: We first show REF for MATH. Separating the positive and negative parts of MATH, MATH write MATH with MATH nonnegative, continuous, nonanalytic only on a discrete set, where the left and right derivatives exist, with MATH. It is enough to show REF for each MATH. Let then MATH be one of the MATH's. Denote MATH. The function MATH on MATH has the same properties as MATH and in addition is decreasing. Then MATH exists a.e. and MATH since MATH . Then MATH therefore MATH which proves the Lemma in this case. CASE: Let now MATH arbitrary. Then REF follows from the result for MATH applied to the function MATH. |
math-ph/0011011 | Both claims are easily verified by noting that MATH if and only if MATH for MATH-vectors MATH and MATH and that MATH is exactly the minimum MATH for which such an equation exists. |
math-ph/0011011 | To say that MATH is equivalent to saying that there exist vectors MATH and MATH such that MATH . (In the case MATH one of these vectors is the zero vector.) Also, merely for the sake of convenience, we introduce the notation MATH and recall that MATH is the classical adjoint matrix (that is, MATH if MATH is invertible). Now, using REF to eliminate MATH, one can rewrite MATH as MATH where MATH. Next, since MATH depends on MATH polynomially, it is enough to prove, that MATH for almost all MATH. Let us assume that MATH. Then we can eliminate reference to MATH by writing MATH where MATH . Let us further re-write MATH as MATH . Finally, denote MATH by MATH. We obtain MATH . Note that MATH and since MATH we also have that MATH . Therefore, MATH . So, using the notation MATH, we see that MATH is a NAME of the form MATH . Substituting MATH into the expression for MATH immediately yields MATH. |
math-ph/0011011 | Given the semi-infinite vector MATH, we use the notation MATH. For an arbitrary constant MATH, we define the semi-infinite vector MATH. Then, it is sufficient to prove that the continuous function MATH defined in REF satisfies the NAME equation in NAME form (compare CITE) MATH uniformly in MATH, MATH and MATH and for all MATH. However, from the definition we see that MATH where we have chosen MATH and used the notation of REF . Similarly, MATH . Consequently, REF is equivalent to demonstrating that the polynomial MATH in REF is zero in the case of this MATH, MATH and MATH. But, according to the second result in REF we have that MATH and so REF demonstrates that the NAME equation is satisfied. Once we know that MATH is a tau-function, the formula for MATH is derived from simply using the ``famous Japanese formula" CITE: MATH . Note that the numerator is simply MATH with MATH. So, again expanding this in terms of the power series for the logarithm we derive the desired expression for MATH. |
math-ph/0011011 | If we consider only the dependence upon MATH and MATH (MATH a multiple of MATH) then MATH . |
math-ph/0011011 | Note that the flows specified have the stated vector fields and that MATH is a rank one matrix if MATH is. |
math-ph/0011013 | Computing the commutator in the definition of MATH and applying the second resolvent formula we have MATH where we have set MATH, MATH and MATH. From the triangle inequality and MATH we obtain MATH . To estimate the operator norms on the right hand side it is sufficient to bound them by the NAME norms MATH. Using bounds REF on the kernels of MATH for MATH, and the properties of the functions MATH, MATH we obtain MATH . For the norms involving the potentials MATH we obtain in a similar way MATH . It is clear that since MATH is bounded, and MATH, MATH do not grow faster than polynomials, the double integral in the right hand side of the last inequality is bounded above by MATH times a constant depending only on MATH and MATH. From this result, REF and MATH we obtain (MATH a constant independent of MATH) MATH where we used the expression for MATH in REF and the fact that MATH. |
math-ph/0011013 | We start by proving REF for MATH. The case MATH is identical. From the decoupling formula we have MATH . Let MATH be a circle of radius MATH in the complex plane, centered at MATH. Because of MATH and MATH, MATH and MATH have no poles in MATH. Moreover the only pole of MATH is precisely MATH. Thus integrating REF along the circle MATH . We proceed to estimate the norms of the three contributions on the right hand side of REF. The norm of the first term is smaller than MATH . Indeed, for MATH we have MATH by construction. For MATH we have MATH. Since MATH we note that in all three REF MATH. Furthermore, since MATH, using REF we get REF. To estimate the second term in REF we note that by the second resolvent formula MATH . Integrating REF along MATH we obtain the identity MATH this implies MATH since the distance (in the MATH direction) between the supports of MATH and MATH is greater than MATH we can proceed in a similar way as in the estimate of REF to obtain MATH where MATH is a constant depending only on MATH. For the third term in REF we use the adjoint of REF MATH to get MATH from which we obtain the same bound as in REF. Combining this result with REF we obtain REF in the proposition. Let us now sketch the proof of REF. From the decoupling formula we have MATH . Given an interval MATH such that MATH, we choose a circle MATH in the complex plane with diameter equal to MATH. Then if we integrate over MATH the last two terms on the right hand side do not contribute while the second and third ones give MATH and MATH. Therefore MATH . From REF , MATH and MATH the first term is bounded above by MATH . In order to estimate the second norm in REF we notice that (in the same way as in REF) MATH thus MATH . Each term of the sum can be bounded in a way similar to REF, and since the number of terms in the sum is equal to MATH we get MATH . The second inequality follows from REF (where we need MATH). For MATH one uses the adjoint of identity REF to obtain the same result. The result REF then follows by combining REF. |
math-ph/0011013 | Let MATH where MATH is the set given in REF . Since for MATH large enough the right hand side of REF is strictly smaller than one the two projectors necessarily have the same dimension. Therefore MATH contains a unique energy level MATH for each MATH of radius MATH. In particular by taking the smallest value MATH we get REF. The number of such levels is MATH since they are in one to one correspondence with the energy levels of MATH. The sets MATH of REF are precisely MATH . The set of all other eigenvalues in MATH, defines MATH, and is necessarily contained in intervals MATH such that MATH. In view of REF this implies REF. Since the two projectors in REF necessarily have the same dimension, the number of eigenstates in MATH is the same than that of MATH. It remains to estimate the probability of the set MATH. The realizations of the complementary set are such that for at least one MATH but from REF this has a probability smaller than MATH where MATH comes from the number of levels in MATH. Thus for MATH large enough MATH . We recall that MATH. |
math-ph/0011013 | Let MATH. The associated current is by REF and will be compared to that of MATH . The difference between these two currents will be estimated by MATH. First we observe that MATH is trace class. Indeed, MATH with MATH bounded and MATH to get the second inequality one has simply added positive terms to MATH. Similarly MATH . The identity MATH implies MATH . From REF we get MATH . Combining this last inequality with REF we get the result REF . |
math-ph/0011013 | The wavefunctions MATH and MATH are defined on MATH, are periodic along MATH and are twice differentiable in MATH. Here we will work with periodized versions of these functions where the MATH direction is infinite (but we keep the same notation). This allows us to shift integrals over MATH from MATH to MATH. We have MATH . An integration by parts yields MATH where MATH is a boundary term given by MATH . We can add a periodized version of MATH and MATH to the kinetic energy operator in both terms on the right hand side of REF and use that MATH and MATH are eigenfunctions of MATH to obtain MATH . From MATH and the NAME inequality we obtain MATH . With the help of REF we get MATH this concludes the proof of REF. |
math-ph/0011013 | Let MATH an interval like in REF . We consider the maximal set of intervals MATH such that MATH and MATH, MATH. Since the number of gaps between the MATH in MATH is less than MATH and MATH, it follows from REF that MATH . Now suppose that MATH is an eigenstate of MATH corresponding to MATH. For a given MATH one can show that MATH is necessarly included in one of the fattened intervals MATH. In order to check this it is sufficient to adapt the estimates REF to REF to the difference of projectors MATH. The main point is to check that with our choice of intervals one is allowed to replace the circle MATH by circles MATH centered at the midpoint of MATH and of diameter MATH. We do not give the details here. One finds MATH . Therefore MATH for some MATH and we have MATH . To estimate the first term on the right hand side of REF we use the spectral decomposition in terms of eigenstates of MATH, MATH . We have MATH . From REF we get MATH . The second term on the right hand side of REF is estimated by the NAME inequality MATH . The third term is treated in a similar way MATH . The last estimate REF then follows from REF. |
math-ph/0011013 | The proof relies on the estimate REF which we state here for convenience. For MATH, MATH (MATH), MATH and MATH . Using this estimate for MATH, MATH and MATH together with MATH we have (MATH a numerical constant) MATH . From REF if MATH, MATH and MATH we deduce the estimate (MATH a numerical constant) MATH . From REF for MATH and REF we get MATH since MATH the last sum can be bounded by a constant, which yields REF for MATH. To estimate the first derivative it is convenient to use the relation CITE MATH which yields MATH . Using REF to bound the two terms on the right hand side of REF we get MATH the result REF for MATH then follows from REF. |
math-ph/0011013 | Let MATH and MATH the middle point of MATH. Using MATH and MATH we can write MATH and thus from MATH, we get MATH . In the last inequality we have assumed that MATH. Using MATH, MATH, and REF we obtain MATH . Now, from MATH it follows that MATH each NAME norm in REF is bounded by MATH. This observation together with REF gives the result of the lemma. |
math-ph/0011013 | Let MATH and MATH for MATH small enough (we require that MATH is contained in MATH). By the NAME inequality we have MATH where MATH is the expectation with respect to the random variables in MATH. To estimate it we use an intermediate inequality of the previous proof MATH . Writing MATH . Since MATH is trace class we can introduce the singular value decomposition MATH where MATH. Then MATH . An application of the spectral averaging theorem of CITE shows that MATH as well as for the term with MATH replacing MATH and MATH replacing MATH. Combining REF we get MATH . |
math-ph/0011016 | We use induction on MATH. The identity holds trivially for MATH, since MATH and MATH. Let MATH and suppose the identity has been verified for MATH. Since MATH is unchanged if we multiply all the permutations on the left by MATH, we have MATH, where MATH (MATH-identity). For MATH, let MATH denote the collection of MATH-cycles of the form MATH. For each MATH, let MATH denote those permutations MATH that fix the elements MATH. We claim that MATH where MATH. To verify REF , we can assume without loss of generality that MATH. Recall that we need only consider permutations MATH that are products of some of the cycles in MATH (and MATH is determined by the pair MATH, since MATH is the product of the other cycles of MATH when MATH). For the MATH-terms of the sum, MATH contains the cycle MATH so that MATH for MATH. (For the MATH-terms, MATH contains MATH.) Hence by REF , we have MATH where MATH is over those MATH with MATH, for MATH. To compute the double sum in the right side of REF , we notice by REF that it equals MATH with MATH replaced by MATH respectively. Hence by our inductive assumption, we have MATH . The computation of the MATH terms is similar, and hence REF holds. We now have by REF , MATH . Noting that MATH and summing over MATH, we obtain the desired formula. |
math-ph/0011016 | Let MATH be the functions given by MATH, where MATH. From REF , we have MATH . We observe from REF that MATH and thus MATH since MATH is homogeneous of order MATH. |
math-ph/0011019 | We consider the operator MATH. The operator MATH may be discussed similarly. Define the function MATH . It is easy to verify that MATH and MATH for any MATH. Suppose first that MATH. From the assumption MATH and the support property of MATH it follows that MATH and thus MATH for any MATH. If MATH then from the assumption that MATH for some MATH and since MATH has compact support it follows that MATH for any MATH. For the case MATH we estimate as follows MATH with MATH being some constant depending on MATH, MATH, MATH, and MATH only. Similarly for the case MATH we have MATH with MATH depending again on MATH, MATH, MATH, and MATH only. To estimate the norm of MATH we use the NAME inequality CITE (see also CITE) MATH and the NAME inequality (see CITE) MATH . Setting MATH and MATH given by REF in these inequalities and then using the estimates REF proves the lemma. |
math-ph/0011019 | First we consider the case MATH. By the resolvent equation MATH . Since the operator norm of MATH is uniformly bounded, we obtain from REF MATH . We turn to the case MATH. For any MATH we have MATH with some coefficients MATH. The operators MATH are given by MATH . Applying REF we obtain MATH with MATH . We turn to the discussion of the operators MATH. Obviously we have MATH . Let MATH, MATH be given. By MATH we denote the set of all multiindices MATH with MATH satisfying the following conditions MATH . Applying REF to REF recursively we obtain that for any MATH the operator MATH can be represented in the form MATH with MATH being some real numbers. We prove now that for any MATH . From REF and using REF we obtain MATH and MATH for MATH and MATH for MATH. Moreover the inequality MATH holds. Thus MATH with MATH and such that MATH for MATH and MATH for MATH. Moreover the estimate MATH holds. In REF the worst case occurs for MATH. Thus the estimate REF is proved. From REF it now follows that MATH with MATH satisfying MATH and the estimate REF holds. Since the inequality MATH is satisfied with any MATH this completes the proof. |
math-ph/0011021 | In MATH change the integration variable to MATH and let MATH, then MATH . By formulae REF MATH for MATH while MATH. By REF MATH . Multiply the two sums and integrate with respect to MATH and by use of the orthogonality relations obtain MATH for MATH; and MATH for MATH. When MATH let MATH then MATH. In the sum replace MATH by MATH (and MATH by MATH and similarly for MATH ) to obtain the claimed result. |
math-ph/0011021 | In MATH the argument of the MATH is MATH . Then MATH and MATH for each MATH as MATH. Since MATH the dominated convergence theorem applies to prove the claim. |
math-ph/0011029 | The proof is almost literally the same as for REF. The assertions reduce to the following identities for MATH-numbers for MATH, MATH, MATH, MATH and for MATH, MATH . It is a remarkable and non-trivial fact that these identities which are known to hold without the MATH-brackets, are still valid in this form which involves the MATH-numbers. |
math-ph/0011029 | The proof of this proposition is similar to that of REF . It reduces to the following true identities for MATH-numbers where MATH, MATH and MATH, MATH and for MATH, MATH . |
math-ph/0011029 | From an analogous calculation as for the NAME REF we obtain a well-defined MATH if and only if MATH for all integers MATH in the range MATH. We recall that MATH, and the assertion follows. |
math-ph/0011031 | Note that for operators MATH and MATH the inequality MATH holds. Using this with MATH and MATH we get, for any MATH, MATH . Denoting MATH and MATH, we can estimate MATH where we used MATH. Now MATH is relatively MATH-bounded with arbitrary small bound REF , and by REF the same holds with MATH replaced by MATH. |
math-ph/0011031 | We add MATH to MATH in REF , use MATH, note that MATH, take the expectation value with MATH and expand the double commutator. We get, using moreover MATH, MATH . Now assume that MATH is a discrete eigenvalue and let MATH be the corresponding ground state. Since MATH is a closed operator and MATH is a core of MATH, there exist sequences of wave functions MATH with MATH and angular momentum MATH, converging in norm to MATH such that also MATH in norm. Since MATH REF on MATH guarantees that the operators MATH and MATH are bounded relative to MATH on MATH, the subspace of MATH where MATH. It follows that, as MATH, also MATH and MATH converge and are bounded in norm, and the second line in REF , with MATH, converges to MATH. Set MATH. We claim that it is no restriction to assume that the functions MATH have their minimum at MATH. To see this, let MATH, where MATH is chosen such that MATH for all MATH. This is possible since by REF is a discrete eigenvalue, so one can not attain any expectation value in the gap above MATH by shifting MATH to infinity. Now MATH so there exists a subsequence, again denoted by MATH, such that MATH weakly in MATH. Since also the norms converge, there is even strong convergence. In particular, MATH as MATH. Therefore, since MATH is translation invariant in MATH-direction, and since MATH is relatively bounded, MATH . Now denote MATH by MATH, which proves our claim. With MATH chosen as above, we have MATH . Concerning MATH, the NAME continuity of MATH REF and the fact that MATH if MATH allows us to integrate MATH and apply NAME 's Lemma to conclude that MATH where we used that MATH pointwise. (The convergence is even uniform in MATH, since MATH is uniformly bounded by REF .) If MATH somewhere away from the MATH-axis, the fact that MATH away from MATH implies that the right hand side of REF is strictly negative. Using this and REF we can conclude that MATH . Now, since MATH has angular momentum MATH, and MATH has angular momentum MATH, this implies that the spectrum of either MATH or MATH must reach below MATH, that is MATH . To prove the relations REF in their strict form we use the commutation relation REF and observe that MATH. So REF tells us moreover that MATH . In the case MATH, we note that the strict form of REF holds trivially, if MATH. Otherwise, if MATH, REF gives MATH . In the case MATH we use MATH, which is a consequence of MATH. Using MATH, REF gives MATH . So our assertions on the strict inequalities follow. If MATH does not have a discrete eigenvalue at the bottom of its spectrum, or if MATH happens to be zero everywhere off the MATH-axis, add MATH to the potential, with the superharmonic function MATH, and MATH small enough (more precisely, MATH, with MATH as in REF ). The Hamiltonians MATH do have discrete ground states for each MATH REF , and REF , as well as REF hold for the corresponding energies MATH. Strictly speaking, the potentials MATH are not relatively bounded with respect to MATH on MATH, but it is not difficult to see, using boundedness relative to MATH instead, that the conclusions above remain valid. Now taking the limit MATH leads to REF . |
math-ph/0011031 | For MATH in the increasing part of the sequence MATH, REF is the standard statement that a tangent line lies above a concave sequence. Since the linear tangential sequence is increasing, it is trivial to extend this inequality to the MATH in the decreasing part. To prove REF , consider MATH and observe that REF can be written as a monotonicity relation for MATH, MATH . Starting with MATH, which is not positive, this means MATH if MATH, and MATH if MATH. The extension of the latter inequality to the increasing part of MATH is trivial. Summing these inequalities for MATH, either for MATH or MATH, gives REF . Moreover, any strict inequality for one of the MATH gives REF as a strict inequality. |
math-ph/0011031 | REF imply that either MATH or MATH for all MATH. Since MATH is a discrete eigenvalue by assumption, it cannot be infinitely degenerate, so our assertion is proved. |
math-ph/0011031 | The potential MATH is relatively bounded with respect to MATH with relative bound MATH on the entire NAME space. This implies that MATH fulfills the technical condition which is necessary to apply REF , and moreover that the Hamiltonian is bounded below. So the sequence MATH is non-decreasing and concave, and MATH. Since the asymptotic behavior of the potential, as MATH, is MATH, where MATH, MATH, it is easy, if MATH, to construct trial functions for large MATH, such that the expectation value of MATH is lower than the edge of its essential spectrum, which is at MATH. Such trial wave functions are of the form MATH, where MATH is the wave function in MATH with MATH in the lowest NAME band, and MATH has to be small enough. So each MATH is a discrete eigenvalue, and increase and concavity are strict. |
math-ph/0011031 | One can mimic the proof of REF . All the operations we made there commute with the MATH-translation. Since each MATH has discrete spectrum, as we show in REF in the Appendix, REF - REF hold in strict form if MATH. |
math-ph/0011031 | By the NAME inequality on MATH there is a constant MATH such that MATH for all MATH (compare REF ). Therefore we can estimate, for MATH, MATH where we used NAME inequality, with MATH, twice: MATH . Optimizing REF with respect to MATH yields the desired result. |
math-ph/0011031 | We split the potential MATH into MATH . Then we split the Hamiltonian MATH into MATH where MATH, MATH as defined in REF . We do the same splittings for all the MATH. We use that MATH, where MATH denotes the NAME boundary conditions, use REF to find a lower bound to MATH by a constant operator, MATH . Now we add MATH and MATH and use REF : MATH . The operator on the right side obviously has discrete spectrum, and the min-max principle implies that the same is true for MATH. |
math-ph/0011031 | We map each subspace MATH unitarily onto MATH, by writing, for MATH, MATH and mapping MATH. The Hamiltonian MATH is then unitarily equivalent to MATH and the right side of REF is the MATH'th eigenvalue of MATH . This means that MATH, so REF follows with the help of the min-max principle. |
math-ph/0011031 | The following formula can be interpreted as calculating the acceleration of the particle, subtracting the effect of the NAME force, and then taking a linear combination of the MATH- and MATH-components, denoted as MATH: MATH . Now we take the matrix elements between MATH and MATH, MATH and choose the phases appropriately, so that we can write MATH with MATH. We get MATH . The matrix element on the left side is positive, and by REF we know that MATH. So the sign of MATH, which determines the sign of the right side, determines also the sign of MATH on the left side. |
math-ph/0011033 | By the NAME inequality with respect to the NAME measure MATH . For MATH one obviously has MATH. Now suppose that MATH. Then MATH which completes the proof of the lemma. |
math-ph/0011033 | First let us prove REF . We write the operator under the norm in the form MATH with MATH . By the semigroup property MATH and therefore MATH where we have used MATH and the fact that MATH. By the NAME formula MATH if MATH. Thus MATH preserves positivity. The same is obviously valid for the operator MATH. Also MATH and MATH are bounded operators from MATH to MATH CITE. Therefore we can apply REF (see Appendix) to estimate REF thus obtaining MATH . By REF MATH . By the monotonicity REF MATH . Applying the NAME inequality with respect to the NAME measure to the NAME formula we obtain MATH for any MATH. This leads (see the proof of REF in the Appendix) to the inequality MATH and thus MATH . Now we estimate MATH. From the NAME REF with MATH by means of the NAME inequality with respect to the NAME measure we obtain MATH and hence MATH . Now we note that MATH . We turn to the estimate of MATH. To this end we write MATH . This completes the proof of REF . The proof of REF follows along the same lines. Denoting MATH we obtain MATH and therefore MATH . Again by REF MATH . By REF and by the monotonicity REF MATH . By REF and again by the monotonicity REF MATH . By the NAME formula we obtain MATH which immediately gives MATH . Further we consider MATH which completes the proof of REF . |
math-ph/0011033 | We prove REF only since the proof of REF follows along the same lines. Again we use the representation of the operator under the norm in the form MATH with MATH being defined by REF . By means of the identity REF we estimate MATH . Choose an arbitrary MATH with MATH and consider MATH . Similarly we have MATH . Since MATH preserves positivity and MATH, MATH are bounded as maps from MATH to MATH CITE we can use REF to estimate REF , which immediately leads to MATH . Since MATH is self-adjoint, by duality (see for example, CITE) we have MATH . Applying REF we obtain MATH . From REF and the monotonicity of the norm REF it follows that MATH . By the semigroup property and by REF MATH . Applying now REF to the right-hand side of this inequality we obtain MATH thus completing the proof of REF . |
math-ph/0011033 | Given MATH by the NAME - NAME theorem we can find polynomials MATH in MATH such that MATH as MATH (see CITE). Indeed, denoting MATH and MATH we can find polynomials MATH such that MATH and MATH. Since MATH depends on the NAME norm of MATH only, the set MATH is bounded below. Let MATH be such that MATH. By REF MATH as MATH. For MATH by the mean value theorem we have MATH as MATH. Let MATH. Obviously MATH . Then MATH . Dividing these inequalities by MATH and taking the limit MATH gives MATH with an appropriate constant MATH independent of MATH. The third term on the right-hand side of REF can be written in the form MATH with MATH being the coefficients of MATH, and thus by REF MATH for any MATH. We have proved that MATH for any MATH. Taking the limit MATH proves the first part of the claim. To prove the second part we write MATH . Here the second and third terms can be considered as above thus giving MATH with an appropriate constant MATH. The fourth term divided by MATH by REF tends to zero as MATH for any MATH. Let MATH. By REF. We write now the first term on the right-hand side of REF in the form MATH where MATH is the spectral shift function for the pair of operators (MATH, MATH). It can be constructed from the spectral shift function for the pair (MATH, MATH) by means of the invariance principle. Thus the absolute value of the first term on the right-hand side of REF can be bounded by MATH . By REF it follows that for any MATH with some constant MATH independent of MATH. Taking the limit MATH completes the proof. |
math-ph/0011033 | We estimate MATH and apply REF . |
math-ph/0011033 | The proof of that MATH implies MATH is trace class was given by CITE. To obtain the estimate REF we simply repeat the arguments of NAME explicitly controlling the constants in the intermediate estimates. We make use of the NAME formula and write MATH which holds initially weakly. However, by means of the estimate REF with MATH and the fact that MATH and MATH are trace class CITE this identity can be seen to hold in the trace norm sense. Therefore we obtain MATH . Now we prove that for any MATH and any MATH with a constant MATH depending on MATH only. We write MATH giving the a priori estimate MATH . From the inequality CITE MATH which is an easy consequence of the NAME formula, we obtain MATH and thus for any MATH we obtain MATH . Taking MATH we obtain MATH . Now consider the operator MATH with an arbitrary MATH. One has MATH . From REF it follows that MATH . Since MATH is translation invariant it suffices to estimate the NAME norm of the integral operator with kernel MATH. From the inequality MATH (see the proof of REF) we obtain MATH which is obviously square integrable with respect to the measure MATH. |
math-ph/0011033 | Without loss of generality we may suppose that MATH. First we consider the case MATH. It suffices to prove that MATH . Obviously MATH where MATH . The function MATH is monotone non-decreasing, MATH, and MATH. Therefore MATH for all MATH and MATH. Since MATH pointwise as MATH by the NAME dominated convergence theorem we obtain REF. Now we turn to the case MATH. According to the decomposition MATH we represent MATH. Obviously, MATH where MATH . By the NAME theorem MATH. Since MATH by REF the claim follows. In the discrete case the claim can be proved in the same way. |
math-ph/0011033 | For simplicity we consider the case MATH. The general case can be considered in the same way. In the estimate REF we set MATH and MATH. By the monotonicity property of the NAME semigroups REF we have MATH for all MATH's. Since MATH the norm MATH is finite for all MATH. Thus it follows that for any MATH there is a constant MATH independent of MATH such that MATH . Obviously we have MATH . Without loss of generality we can choose boxes MATH such that MATH and then we obtain that the right-hand side of this inequality is bounded by MATH where in the last step we have used the invariance of the norm with respect to translations and the fact that MATH. The assumption that the family MATH is uniformly in MATH (see REF) implies that MATH is summable, that is, MATH. Since MATH we can estimate the right-hand side of REF by MATH . Applying now REF we obtain MATH . Now applying the arguments used to prove REF completes the proof. |
math-ph/0011033 | We write MATH, MATH such that MATH and MATH . Consider the first term on the right-hand side of this expression. We represent it in the form MATH . The proof now closely follows along the lines of the proof of REF . Denoting MATH we obtain MATH . For an arbitrary MATH with MATH we have MATH and analogously MATH . Now by REF it follows that MATH . Similarly we obtain MATH . Finally as in the proof of REF we obtain MATH . The other terms on the right-hand side of REF can be estimated in a similar way. |
math-ph/0011033 | As in the proof of REF given MATH we approximate MATH by polynomials MATH in MATH such that MATH as MATH. Then MATH where MATH. By REF it follows that MATH with some MATH independent of MATH and MATH. By REF there is MATH of full measure such that for any MATH the limit MATH exists and is non-random for any finite MATH. Therefore MATH which proves the first part of the claim. The arguments used above in the proof of REF give that if MATH then the relation MATH holds almost surely. |
math-ph/0011033 | The proof is standard (see for example, CITE). Since the one-dimensional case was treated in detail in CITE we consider the case MATH only. From REF it follows that for any MATH supported in MATH almost surely, where MATH . For MATH by the NAME estimate (see for example, CITE) for MATH with some uniform constant MATH. For MATH by REF MATH for any MATH. Note that the quantities on the right-hand side of REF are finite. Indeed for MATH any compactly supported function MATH with some MATH belongs also to MATH. Similarly in the case MATH any square integrable MATH with compact support belongs to MATH with arbitrary MATH. Since MATH are monotone functions these estimates imply that for every MATH the family MATH is of uniformly bounded variation on MATH. By NAME 's Selection Theorem for every MATH there is a sequence MATH, MATH such that MATH for all those MATH which are continuity point of MATH. By NAME 's second theorem it follows from this that MATH for any MATH and any MATH with support in MATH. From REF it follows that MATH for MATH-almost all MATH and all MATH. Hence MATH a.e. with some constant MATH for MATH-almost all MATH. But MATH for sufficiently large negative MATH and thus MATH. Now we note that two monotone functions which are equal almost everywhere can be different only at the points of discontinuity. This remark completes the proof of the corollary. |
math-ph/0011033 | Let MATH denote the complement of MATH in MATH. Also we denote MATH and MATH. Now we write MATH . Repeating the arguments used in the proof of REF we obtain that both MATH and MATH are bounded by MATH where MATH. By REF the expression in the brackets can be bounded by a constant times MATH if MATH and simply by a constant if MATH. The second inequality in the claim of the lemma can be proved similarly. |
math-ph/0011033 | We consider the case MATH only since the proof for the case MATH carries over verbatim. By the monotonicity property of the spectral shift function CITE MATH for NAME almost all MATH, all MATH and all MATH. From this it follows that the functional MATH is positive. As it is noted in CITE NAME 's representation theorem extends to the case of linear positive functionals on MATH and thus defines a positive NAME measure MATH. |
math-ph/0011033 | For almost every MATH, every MATH and for arbitrary MATH by the chain rule for the spectral shift function (see for example . CITE) we have MATH with MATH. Here MATH is the decomposition of MATH into its positive and negative part such that MATH. By the monotonicity property of the spectral shift function we have that the first summand on the right-hand side of REF is a.e. non-negative and the second one is a.e. non-positive. By REF there is a linear functional MATH . By REF there is a negative linear functional which we denote by MATH such that MATH . By REF the limit MATH exists almost surely and defines a non-random linear positive functional which we denote by MATH. Thus MATH that is, is a difference of two positive linear functionals and therefore defines a signed NAME measure MATH. |
math-ph/0011033 | Using the NAME inequality with respect to the NAME measure in the NAME formula we obtain MATH for any MATH. The operator MATH is convolution by the function MATH. Since this function is in MATH, by the NAME inequality we have MATH with MATH. Therefore by REF MATH thus proving the lemma. |
math-ph/0011035 | It is known (see CITE, p. REF, where a stronger result is obtained) that a solution to the second order elliptic inequality: MATH where MATH and MATH is a strictly elliptic homogeneous second order differential expression (summation is understood over the repeated indices): MATH cannot have a zero of infinite order at a point MATH provided that: MATH and MATH . By zero of infinite order of a solution MATH of REF a point MATH is meant such that MATH where MATH are positive constants independent of MATH, and MATH is sufficiently small so that the ball MATH. We use this result to prove that the same is true if MATH. Let MATH, be the local equation of MATH in a neighborhood of the point MATH which we choose as the origin, and MATH-axis be directed along the normal to MATH pointed into MATH. Let us introduce the new orthogonal coordinates MATH so that MATH is the equation of MATH in a neighborhood of the origin, and assume MATH. For example, one can use the coordinate system in which the MATH-axis is directed along the outer normal to MATH, and the MATH-coordinates are isothermal coordinates on MATH, which are known to exist for two-dimensional MATH-surfaces in MATH ( and even for MATH-surfaces in MATH, CITE, p. REF). For an arbitrary MATH-surface in MATH one can prove that locally one can introduce (non-uniquely) the coordinates in which the metric tensor is diagonal in a neighborhood of MATH. To do this, one takes two arbitrary linearly independent vector fields tangent to MATH and orthogonalize them with respect to the Euclidean metric in MATH using the NAME procedure, which is always possible. The third axis of the coordinate system, which we are constructing, is directed along the normal to MATH at each point of the patch on MATH. Let MATH and MATH be the resulting orthogonal vector fields tangent to MATH. Then one can find (non-uniquely) a function MATH, defined on the local chart, such that the NAME bracket of the vector fields MATH and MATH vanishes, MATH. Then the flows of the vector fields MATH and MATH commute and provide the desired orthogonal coordinate system in which the metric tensor is diagonal CITE. The condition MATH can be written as the following linear partial differential equation: MATH, where MATH, MATH and MATH are the parameters, and MATH is the local equation of the surface MATH on the chart. The above equation for MATH has many solutions. One can find a solution MATH by the standard method of characteristics. A unique solution is specified by prescribing some NAME data, which geometrically means that a noncharacteristic curve through which the surface MATH passes, should be specified CITE. If MATH the situation is less simple if one wants to use the same idea in the argument: there are many REF NAME brackets to vanish in the case of MATH vector fields tangent to MATH in MATH, and it is not clear for what MATH these conditions can be satisfied. However, for our argument it is sufficient to have the coordinate system in which the metric tensor has zero elements MATH for MATH, and MATH does not vanish. In this case the even continuation REF , that is used below, still allows one to claim that the function REF solves the same equation in the region MATH as it solves in the original region MATH. If the elements MATH for MATH do not vanish, then the equation in the region MATH will have some of the coefficients in front of the second mixed derivatives MATH with the minus sign, while these coefficients in the region MATH enter with the plus sign. So, in this case the principal part MATH of the operator MATH, which is used in REF , will be different in the regions MATH and MATH. Recall that the NAME operator in the new coordinates has the form of the NAME operator: MATH, where MATH, summation is understood over the repeated indices, and MATH. If MATH is a zero of MATH of infinite order in the sense MATH, then REF imply that MATH, so that condition similar to REF holds for the integrals over MATH. Indeed, the derivatives of MATH in the tangential to MATH directions at the point MATH vanish by the assumption. From REF it follows that the normal derivatives of MATH of the first and second order vanish at MATH. Differentiating REF along the normal one concludes that all the normal derivatives of MATH vanish at the point MATH. Thus we may assume that MATH is a zero of MATH of infinite order, that is, REF hold for MATH for the integral over MATH. In the MATH- coordinates one writes the equation for MATH: MATH as follows: MATH . We drop the subscript MATH of MATH and of MATH in what follows. In REF the operator MATH is defined as: MATH over the repeated indices summation is understood, MATH is the metric tensor of the new coordinate system, MATH and the coefficients in front of the second-order derivatives in MATH are extended to the region MATH as even functions of MATH, so that the extended coefficients are NAME in a ball centered at MATH with radius MATH . Let us define MATH in a neighborhood of the origin, MATH, by setting MATH and MATH . The functions MATH defined by REF , are NAME in the ball MATH, if MATH is NAME in MATH, where MATH. Furthermore, MATH where MATH and MATH if one assumes MATH for all sufficiently small MATH. Note that the change of variables MATH is a smooth diffeomorphism in a neighborhood of the origin, which maps the region MATH onto a neighborhood of the origin in MATH in MATH space, and one can always choose a half-ball MATH belonging to this neighborhood, so that REF follows from REF . Therefore MATH in MATH and consequently MATH in MATH. This implies that MATH in MATH by the unique continuation theorem (see CITE and CITE). This is a contradiction since MATH. REF is proved. MATH . This lemma provides a justification of the argument given for REF in the introduction. In this remark we comment on the numerical recovery of the potential from the data MATH. We have explained how to get the data MATH from MATH. Therefore the spectral function MATH and the resolvent kernel MATH are known on MATH. If MATH is known, then the NAME REF map MATH is known. This map MATH at every fixed MATH associates to every MATH a function MATH by the formula MATH . Here MATH . The unique solution to REF is given by REF . We now outline an alternative proof of REF : REF are the same as in REF ; REF consists of the construction of the N-D map by REF and reference to REF below. In CITE and CITE the result similar to REF (see below) is proved for the D-N REF map. Our proof follows the idea of the proof in CITE. As above, MATH stands for the NAME operator MATH. Let us assume that MATH is a regular point for MATH and MATH. If MATH and MATH generate the same N-D map then MATH. Take an arbitrary MATH and consider the problems MATH. Subtract from the equation with MATH the equation with MATH and get MATH where MATH. The condition MATH on MATH follows from the basic assumption, namely the N-D map is the same for MATH and MATH. Let MATH be arbitrary. Multiply REF by MATH integrate by parts using boundary REF , and get the orthogonality relation: MATH . The last inclusion in REF follows since MATH is arbitrary. From REF and property MATH (see CITE) it follows that MATH. REF is proved. In CITE CITE - CITE one finds a discussion of the inversion algorithm. Appendix Lemma. Let MATH be a connected MATH manifold and suppose that a real-valued function MATH does not have zeros of infinite order, that is, for an arbitrary point MATH, MATH for some MATH. Then, up to sign, the function MATH is uniquely defined on MATH by its square MATH. Proof. Let MATH be a real-valued function satisfying the condition MATH. Then the lemma states that either MATH or MATH. Consider the function MATH. If MATH on MATH, then MATH and the lemma is proved. Consider MATH. This is an open set and MATH on this set, since, for a fixed MATH, the condition MATH implies that MATH or MATH. The equality MATH will also hold on the closure MATH. If MATH, then MATH, that is, MATH on MATH, and the lemma is proved. Let MATH be a nonempty open set. Since MATH and MATH is connected, there exists a point MATH. We have MATH on MATH, and the function MATH does not have a zero of infinite order, hence there exists a MATH such that MATH. On the other hand, MATH and MATH on the open set MATH. Thus MATH for all MATH. This and the continuity of MATH imply, as MATH, that MATH. This contradiction proves the lemma. |
math-ph/0011039 | Let MATH defined by MATH . Set MATH. Obviously, MATH and MATH . REF imply that MATH and MATH. Now from the NAME inequality, we have MATH . The Lemma follows from REF . |
math-ph/0011039 | Choosing MATH, by the NAME inequality CITE we have that MATH REF is bounded for some MATH. The Lemma follows from the standard elliptic estimates. |
math-ph/0011039 | Let MATH and MATH for MATH . Set MATH and MATH. Clearly MATH in MATH. By REF below and REF , MATH REF is bounded from above. Thus, MATH for some constants MATH and MATH. Now the Lemma follows from REF below. |
math-ph/0011039 | See CITE. |
math-ph/0011039 | The previous Lemma implies that MATH . Since MATH, from REF , we deduce that MATH . Similarly, we have MATH . |
math-ph/0011039 | From above, we have MATH . The Lemma follows from potential analysis, see for instance CITE. |
math-ph/0011039 | From REF , we have the NAME identities as follows: MATH and MATH . From above, we get the NAME identity for the NAME system REF MATH . Applying REF and letting MATH, we get MATH which is equivalent to REF . This proves the Lemma. |
math-ph/0011039 | Here, for simplicity we only give a proof of the Theorem when MATH. In view of REF , we may assume that there exist two nonnegative bounded measures MATH and MATH such that MATH for every smooth function MATH with support in MATH and MATH. A point MATH is called a MATH-regular point with respect to MATH if there is a function MATH , MATH, with MATH in a neighborhood of MATH such that MATH . We define MATH . By definition and REF , it is clear that MATH and MATH are finite. And MATH is independent of MATH for small MATH, see below. We divide the proof into REF steps. CASE: For MATH, MATH provided MATH . First from REF , we know that for any point MATH, there is some MATH such that MATH . Here MATH. The argument in CITE implies directly that MATH . Hence, MATH and MATH are both finite. Let MATH. Assume by contradiction that MATH. Thus, MATH for any small constant MATH. Note that MATH. Define MATH by MATH . The maximum principle implies that MATH. Since MATH is finite, we may assume that MATH is uniformly bounded in MATH. In view of MATH, a result of CITE implies that MATH, which in turn implies that MATH, a contradiction. Hence, MATH. MATH follows from the argument in CITE. Similarly, we have MATH. CASE: MATH implies REF . MATH and MATH imply REF . MATH means that MATH and MATH are bounded in MATH. Thus, MATH and MATH are bounded in MATH for any MATH, which implies that MATH. Applying the NAME inequality as in CITE, we have REF or REF . The second statement follows similarly. CASE: MATH and MATH imply REF . We need the following lemma. REF is true for any MATH. We use a blow-up argument to prove this Lemma. Assume by contradiction that REF were false for some MATH, that is, there exists a sequence of solutions MATH of REF with MATH REF for some MATH such that MATH . Without loss of generality, we may assume that MATH and MATH. We may also assume that there exists a sequence of points MATH such that MATH . Let MATH. Define MATH for MATH with MATH. Clearly, MATH as MATH and MATH satisfies MATH where MATH. Applying REF -REF , we have two possibilities: CASE: MATH converges to MATH in MATH which satisfies the NAME system REF in MATH. (In this case MATH is bounded in MATH.) CASE: MATH converges to MATH in MATH and MATH tends to MATH uniformly in any compact subset in MATH. Moreover, MATH satisfies MATH with MATH . In view of REF and a classification result of REF obtained by CITE, in these two cases MATH which is a contradiction. This proves the Lemma. Now we continue to prove REF . As in REF , we know that either CASE: MATH is bounded on any subset of MATH, or CASE: MATH on any subset of MATH. In view of REF implies that MATH for any MATH and any small MATH. Now we can follow the argument in CITE to exclude REF . |
math-ph/0011039 | Set MATH . Since MATH is positive definite, it is easy to see that MATH from the ordinary NAME inequality CITE. In fact, one can show easily that MATH and MATH REF from the ordinary NAME inequality and the NAME inequality. Clearly, MATH preserves a partial order of MATH, namely, if MATH, then MATH provided that MATH and MATH. The Theorem is equivalent to MATH . Assume by contradiction that REF is false. We may assume that there is a point MATH such that CASE: For any MATH, MATH, CASE: For any MATH, MATH. We first need several lemmas. For any MATH with MATH and MATH, there exists a constant MATH such that MATH . Moreover, MATH admits a minimizer MATH. Choose a small number MATH such that MATH and MATH. By the definition of MATH, we know that MATH, that is, there is a constant MATH such that MATH . It follows that MATH . This inequality means that MATH satisfies the coercivity condition. Now it is standard to show that MATH admits a minimizer, for MATH is weakly lower semi-continuous. There exists a sequence MATH such that MATH and MATH where MATH. Assume by contradiction that for any sequence MATH with MATH, MATH . Then we can show that MATH satisfies the coercivity condition, MATH for some constant MATH, which implies that there is a small MATH such that MATH, a contradiction. CITE For any two sequences MATH and MATH satisfying MATH there exists a smooth function MATH satisfying MATH and MATH for some subsequence MATH. We give the proof for completeness, though it is rather elementary. If there is a subsequence MATH of MATH with property that MATH, we can choose MATH . So we may assume that MATH and MATH. NAME, we assume more that MATH is non-increasing. Choose another sequence MATH with MATH . It is easy to find a non-increasing smooth function MATH with the property that MATH . Clearly, this function MATH satisfies all conditions of the Lemma. Applying REF to sequences MATH where MATH is obtained in REF , we can find a function MATH satisfying REF . For any small MATH, define a perturbed functional by MATH where MATH. Let MATH. We have CASE: for MATH, the infimum MATH, and it is achieved by MATH satisfying MATH with MATH where MATH . CASE: for MATH, MATH has no lower bound, that is, MATH . CASE: As in REF , we can show that MATH REF satisfies the coercivity condition from the construction of MATH. It is also easy to check that MATH is weakly lower semicontinuous. Therefore, MATH has a minimizer which satisfies REF . CASE: From the construction of MATH, we have MATH as MATH, where the sequence MATH was obtained in REF . We now continue to prove REF by considering the sequence MATH obtained in the previous Lemma. We claim that MATH blows up, that is, MATH . Otherwise, we can show that MATH converges to some MATH in MATH that is a minimizer of MATH. On the other hand, by REF , MATH has no minimizer, a contradiction. Let MATH. We have MATH as MATH . Define a new sequence MATH by MATH near a small, but fixed neighborhood MATH of MATH, where MATH. Here, we have abused a little bit the notation. (For simplicity, we consider MATH as a domain in MATH.) Clearly, MATH satisfies on MATH that MATH . Recall that MATH as MATH and MATH. By using the same argument as in the proof of REF , we have that one of MATH and MATH is larger than or equal to MATH, which is a contradiction. This completes the proof of REF . |
math-ph/0011039 | REF implies that MATH not only has a lower bound but also satisfies the coercivity condition. Since MATH is weakly lower semi-continuous, there is a minimizer that satisfies REF which is the NAME equation of MATH. |
math-ph/0011039 | NAME, assume that MATH and MATH contains a flat disk MATH for a small constant MATH. Let MATH . We estimate MATH which implies that MATH . The Proposition follows from the previous formula by letting MATH . |
math-ph/0011039 | By REF , we have MATH. Let MATH and choose MATH so small that MATH. One can use the blow-up argument as in the previous section to show that MATH which implies that MATH by REF , because of REF . Assume by contradiction that MATH. Then there exists MATH but MATH. Similarly, we can show that there is a small constant MATH satisfying MATH and MATH . It follows, together with REF that MATH . By REF , we have MATH, a contradiction. |
math-ph/0011039 | For small MATH, let MATH be a function defined by MATH . Since MATH is uniformly bounded from above, MATH is bounded in MATH by elliptic estimates. Let MATH. Clearly, MATH satisfies REF with MATH, which is bounded in MATH. By definition, MATH. Thus MATH is also bounded in MATH. In view of REF , the function MATH is harmonic, thus MATH for some constant MATH. By REF , MATH is uniformly bounded. Hence, MATH is bounded. This proves the Lemma. |
math-ph/0011039 | First we show that for any MATH, MATH . Let MATH be determined by MATH. By definition, we know MATH . The NAME embedding theorem implies that for any MATH with MATH, MATH for some constant MATH. Hence, MATH . It follows that MATH. Similarly, we have MATH. This proves REF . By REF , we can show that MATH in the sense of measures as MATH. Like REF , we have MATH . Therefore, we have MATH . It follows that MATH as MATH. Hence, we have MATH in MATH. Similarly, MATH in MATH. Now it is easy to show REF . This proves the Lemma. |
math-ph/0011039 | Define a function MATH satisfying MATH . Since MATH is bounded from above in MATH, MATH is bounded in MATH. Now it is easy to see that MATH is also bounded in MATH. Hence the Lemma follows. |
math-ph/0011039 | Note that MATH satisfies MATH . Clearly, MATH as MATH. For any large, but fixed constant MATH, we consider MATH on MATH. Define MATH by MATH . The elliptic estimate implies that MATH in MATH. Set MATH. It is clear that MATH . Since MATH, MATH and MATH is bounded, MATH is bounded from above and MATH is bounded. It is easy to show that MATH, hence MATH, converges in MATH. (See, for example, CITE.) Now by a diagonal argument, we have that MATH converges to MATH in MATH, where MATH satisfies MATH with MATH. A classification of CITE implies that MATH . |
math-ph/0011042 | The map MATH maps MATH to the right half-plane MATH and MATH to MATH. On MATH we have MATH and MATH. The result follows using the transformation REF . |
math-ph/0011042 | From CITE we have MATH where MATH . Note that MATH is real when MATH or MATH and pure imaginary when MATH or MATH. Furthermore on a fundamental domain for MATH, MATH is zero or infinite only at the corners of MATH (see CITE). So the argument of MATH is constant on each edge of the rectangle MATH. At a corner MATH, we have MATH and so MATH changes by MATH at each corner of MATH when going counterclockwise around MATH. |
math-ph/0011042 | From the differential equation MATH (where MATH) we obtain (upon differentiating the logarithms of both sides) MATH from which we get MATH with similar expressions for MATH. Their product is MATH and by CITE this equals MATH . |
math-ph/0011042 | If MATH is a MATH rectangle, this follows from REF upon setting MATH. |
math-ph/0011042 | Let MATH be a domino tiling of MATH. Let MATH and MATH be the neighboring squares of MATH which are in MATH. NAME 's trick assigns to a vertex MATH an outgoing edge MATH if and only if MATH is a domino of MATH. NAME 's trick, applied to MATH, gives a set of directed edges of the graph MATH such that each vertex has exactly one outgoing edge, except for MATH and MATH which have no outgoing edges. Furthermore the union of these edges is a forest, that is, each component is a tree. There are exactly two components to this graph since each component is rooted at exactly one of MATH or MATH. We claim that MATH and MATH are in different components of this forest. To see this, first note that the directed branch from MATH cannot pass through MATH, for otherwise the path from MATH to MATH followed by the segment from MATH to MATH to MATH would be a lattice path in MATH enclosing an odd number of squares of MATH and so could not arise from a domino tiling of MATH. Similarly the paths from MATH and MATH cannot end both at MATH or both at MATH because their union would contain a closed lattice path from MATH to MATH. So the paths from MATH and MATH end one at MATH and one at MATH. On the path which ends at MATH, shift the dominos along it by one square towards MATH, and add an extra domino from MATH to its freed neighbor. This makes a domino tiling of MATH in which the tree branch from MATH passes through MATH (shifting dominos by MATH along a tree branch has the effect of changing the direction of the edges on that branch). This process is reversible: from a tiling of MATH whose branch from MATH to MATH passes through MATH, shift the dominos by one up to MATH to get a tiling of MATH. |
math-ph/0011042 | As noted in the first two paragraphs of the proof of REF , we must compute the discrete NAME 's function MATH on MATH, that is, the function of MATH harmonic on MATH with NAME boundary values MATH on the boundary MATH, which satisfies MATH for MATH, and asymptotically MATH when MATH. Let MATH be the harmonic function on MATH whose boundary values are MATH for MATH and MATH for MATH. Let MATH . By definition, MATH is harmonic on MATH with boundary value MATH at MATH and boundary value MATH for MATH. Therefore MATH is a constant times the desired NAME 's function MATH. This constant is MATH over the Laplacian of MATH at MATH, that is, MATH (note that MATH is harmonic at MATH and MATH is zero at MATH and MATH). So we have MATH . The asymptotic values of MATH for large MATH and MATH are MATH. By REF applied to the function MATH, we have MATH and MATH . Also when MATH we have MATH . Therefore when MATH, and in the limit as MATH, we have MATH . Scaling everything by MATH (replacing MATH with MATH) completes the proof. |
math-ph/0011045 | Consider the product MATH and define the map MATH by the formula: MATH. Clearly, MATH is a cobordism between the tamed MATH-manifold MATH and the empty set. Let MATH be the lift of MATH to MATH. Define the NAME module structure MATH by the formula MATH . Then MATH is a cobordism between MATH and the NAME module over the empty set. |
math-ph/0011045 | Consider the product MATH. Let MATH be a smooth increasing function, such that MATH for MATH and MATH for MATH. Define the map MATH by the formula MATH. Then MATH is a cobordism between MATH and MATH. Let MATH be the lift of MATH to MATH, endowed with the NAME module structure defined in the proof of REF . Then MATH is a cobordism between MATH and MATH. |
math-ph/0011045 | Consider a smooth function MATH such that CASE: MATH, for all MATH, and MATH; CASE: there exists a smooth function MATH, such that MATH for all MATH. Then the set MATH is compact for all MATH. Let MATH denote the vector field induced by MATH on MATH. Recall that the function MATH is defined in REF. Let MATH be a smooth strictly increasing function, such that MATH . Let MATH be a smooth function, such that MATH . Set MATH . The integral converges, since MATH. Moreover, MATH . Let MATH be a smooth function such that CASE: MATH for MATH; CASE: MATH for MATH. Let MATH and let MATH be a smooth function, such that MATH for MATH and MATH for MATH. Then MATH. Recall that MATH is an admissible function for MATH. This function induces a function on MATH, which, by a slight abuse of notation, we will also denote by MATH. Set MATH . Clearly, MATH for any MATH. We have to show that MATH tends to infinity as MATH. Consider, first, the case MATH. Then MATH and MATH. Hence, from the definition of the functions MATH we get MATH . From REF, we obtain MATH as MATH. Note that REF holds even if MATH, though in this case MATH. Consider now the case MATH. Then MATH . Hence, MATH . When MATH, the expression MATH tends to infinity by definition of MATH, while MATH tends to infinity by REF is proven. |
math-ph/0011045 | Note, first, that MATH on MATH. Thus, since MATH anti-commutes with MATH, we have MATH. Hence, the identity of the lemma holds, when restricted to MATH. We now consider the restriction of MATH to the cylinder MATH. Recall that MATH denotes the projection and that the function MATH was defined in Subsection REF. Then MATH . Using REF, we obtain MATH . The operators MATH and MATH commute with MATH. Also the operators MATH and MATH anti-commute. Hence, we obtain MATH . Since MATH, it follows, that the statement of REF holds with MATH. Since MATH on MATH, the restriction of MATH to this cylinder equals MATH. Finally, since MATH is uniformly bounded on MATH, so is the bundle map MATH. |
math-ph/0011045 | We shall use the notation introduces in Subsection REF. In particular, MATH denotes the isotipic component of MATH, corresponding to an irreducible representation MATH of MATH. As in Subsection REF, it is enough to prove that the spectrum of the restriction of MATH to MATH is discrete. The arguments of Subsection REF show that there exists a smooth function MATH such that MATH on MATH and the following REF conditions hold CASE: MATH as MATH and MATH; CASE: MATH uniformly in MATH, as MATH and MATH. Let MATH denote the norm of the bundle map MATH and let MATH. Set MATH . Then MATH as MATH. Also, by REF , we have MATH . By CITE, the spectrum of MATH is discrete. Hence, REF implies that so is the spectrum of the restriction of MATH to MATH. |
math-ph/0011045 | Each summand in REF is the index of the operator MATH. Thus, since MATH is bounded operator depending continuously on MATH, the index MATH is independent of MATH. Therefore, it is enough to prove the proposition for one particular value of MATH. Recall that the norm MATH was defined in the proof of REF . Choose MATH such that MATH. It follows now from REF, that MATH, so that MATH. Hence, MATH. |
math-ph/0011045 | The same calculations as in the proof of REF , show that MATH . Thus, we obtain the following formulas for the restrictions of MATH to the spaces MATH: MATH . Here the first summand coincides with the lift of MATH to MATH, while the second summand may be considered as an operator acting on the space of MATH-valued functions on MATH. Also, both summands in the right hand side of REF are non-negative. Hence, the kernel of MATH equals the tensor product of the kernels of these two operators. The space MATH is one dimensional and is spanned by the function MATH. Similarly, MATH is one dimensional and is spanned by the one-form MATH, where we denote by MATH the generator of MATH. It follows that MATH . |
math-ph/0011045 | Using the equality MATH we can write MATH . Similarly, MATH . Summing these identities and dividing by REF, we come to REF. |
math-ph/0011045 | Note that MATH for any MATH in the support of MATH. Hence, if MATH, we have MATH. Recall that the norm MATH was defined in the proof of REF . Set MATH and let MATH. Using REF , we obtain MATH . |
math-ph/0011045 | Since MATH is a NAME operator, it follows from CITE, that MATH . The lemma follows now from the obvious identities MATH . |
math-ph/0011046 | By the third assumption, MATH for all MATH. By the first assumption, MATH is continuous in MATH. Since MATH by the second assumption, the above two facts imply that MATH cannot enter the forbidden interval MATH when MATH, and hence MATH for all MATH. |
math-ph/0011046 | We prove the desired bound for MATH, because the bound for large MATH follows from that for small MATH. In the following, let MATH denote the smallest constant that can be used in the error bound of REF, that is, MATH. NAME walk and percolation. We will prove that MATH obeys the upper bound of REF uniformly in MATH. This is sufficient, by the monotone convergence theorem. Let MATH . For self-avoiding walk and percolation, we will employ REF with MATH, MATH and MATH chosen arbitrarily in MATH. We verify the assumptions of REF one by one, with the bound REF then following immediately from REF . In the course of the proof, the desired upper bound on MATH will be shown to be a consequence of a weaker bound than REF, in REF. Since the proof actually establishes REF then follows. CASE: Continuity of each MATH on MATH is immediate from the fact that MATH is a power series with radius of convergence MATH, and from the continuity in MATH of MATH proved in CITE. We need to argue that the supremum of these continuous functions is also continuous. For this, it suffices to show that the supremum is continuous on MATH for every small MATH. It is a standard result that MATH and MATH decay exponentially in MATH, with a decay rate that is uniform in MATH (though not in MATH) CITE. Thus MATH can be made less than any MATH, uniformly in MATH, by taking MATH larger than some MATH. However, choosing MATH such that MATH, we see that MATH. Hence the supremum is attained for MATH, which is a finite set, and hence the supremum is continuous and the first assumption of REF has been established. CASE: For the second assumption of the lemma, we note that MATH and apply the uniform bound of REF to conclude that MATH. Since we have restricted MATH to be larger than MATH, this implies MATH. CASE: Fix MATH. We assume that MATH, which implies MATH . We will apply REF with MATH and MATH. Since we have taken MATH, we have MATH and MATH for sufficiently large MATH depending on MATH. REF then implies that MATH where MATH was defined in REF. It addition, for percolation, as argued at the end of REF, the remainder term MATH vanishes in the limit MATH under REF , yielding the form REF of the expansion. Summing REF over MATH gives MATH which is finite for MATH. The numerator is positive by REF, and hence the denominator is also positive. Therefore, since MATH by our assumption that MATH, REF implies that MATH . Since MATH, this implies that MATH for all MATH, when MATH is large. Thus, to prove that MATH, it suffices to show that MATH. The bound REF also implies that MATH and MATH are well-defined by REF - REF, and that MATH uniformly in MATH. Using the convolution bound of REF , and the first bound of REF, it then follows that MATH . By REF with MATH and MATH, for MATH large we have MATH . Using the first bound for MATH and the second bound for MATH, we conclude from this that MATH . By REF , it then follows that MATH where we have used REF to bound the first term in the first inequality. Using the fact that MATH as MATH, and the definition of MATH, it then follows from the identity REF that for MATH sufficiently large we have MATH . This yields MATH, and completes the proof for self-avoiding walk and percolation. Lattice trees and lattice animals. We will first prove that MATH obeys the upper bound of REF uniformly in MATH. By REF and the fact that MATH is bounded, there is a MATH such that MATH for all MATH. The number of MATH-bond lattice trees or lattice animals containing the origin is less than the number MATH of MATH-bond lattice trees on the NAME lattice of coordination number MATH (the uniform tree of degree MATH), which contain the origin. A standard subadditivity argument, together with the fact that, as MATH, MATH (see, for example, CITE), implies that MATH. Therefore, for lattice trees or lattice animals, MATH . Let MATH. We use MATH in REF . Note that MATH is well-defined, since REF gives MATH for MATH. In addition, REF implies that MATH, so MATH. We again fix MATH, and we use the function MATH of REF , taking now MATH . The desired bound on MATH, for MATH, together with the desired bound on MATH, will follow once we verify the three conditions of REF . We verify these conditions now. CASE: Continuity of MATH follows from the exponential decay of MATH for MATH, as in the previous discussion, together with the continuity of MATH for MATH. CASE: By the remarks surrounding the definition of MATH, we have MATH. Moreover, this implies MATH. It remains to show that MATH . Since MATH, we have MATH. Each lattice tree or lattice animal containing MATH and MATH can be decomposed in a non-unique way into a walk from MATH to MATH with a lattice tree or lattice animal attached at each site along the walk. Therefore MATH. Using MATH, it follows from REF that MATH, which implies REF. CASE: Fix MATH. The assumption that MATH implies the bound MATH, and we take MATH in REF . We then proceed as in the discussion for self-avoiding walk and percolation. We obtain REF as before, so that MATH as required. The proof of REF also proceeds as before. The above discussion proves that MATH is bounded by the right side of REF and that MATH, uniformly in MATH, and that MATH. The proof is then completed by observing that MATH, by monotone convergence. |
math-ph/0011046 | By the MATH-symmetry of MATH, MATH . The expression in brackets of the third term is bounded in absolute value both by MATH and MATH. The third term of REF is therefore bounded by MATH . Using the assumed upper bound on MATH then gives REF and completes the proof. |
math-ph/0011046 | The proof is by induction on MATH. To deal with the fact that MATH is not defined literally by a convolution of MATH with MATH, we proceed as follows. Let MATH be the quantity defined by replacing MATH by MATH in the definition of MATH. Because all the constituent factors in the definitions of MATH and MATH obey the same bounds, it suffices to prove that MATH obeys REF with MATH replaced by MATH. We prove this by induction, with the inductive hypothesis that MATH obeys REF with MATH replaced by MATH and MATH replaced by MATH on the right side. For MATH, let MATH . This quantity is depicted in REF . By definition, and using REF - REF, MATH . By the induction hypothesis, REF, MATH . It therefore suffices to show that MATH . To prove REF, we write MATH, with MATH defined to be the contribution to MATH arising from each of the following four cases. In the discussion of these four cases, MATH denotes a generic constant whose value may change from line to line. CASE: MATH and MATH. This implies MATH and MATH, so that MATH . CASE: MATH and MATH. This implies MATH and MATH. Then MATH . The second term of REF is bounded above by MATH, as required. We bound the first term using REF to estimate the two convolutions, obtaining a bound MATH. (Here we used the assumption MATH to ensure that MATH, as required to apply REF .) It follows from REF that MATH, which gives the desired result. CASE: MATH and MATH. This implies MATH and MATH, and hence MATH . Each term is bounded by MATH, as required. CASE: MATH and MATH. This implies MATH and MATH, and hence MATH . Adding the contributions in the four cases yields REF and completes the proof. |
math-ph/0011046 | By virtue of its definition in REF, we can write MATH with MATH . To derive bounds on MATH and thus on MATH, we first derive bounds on MATH and MATH. Assuming MATH, we have MATH . Also, by the formula for MATH of REF and our assumption that MATH, it follows that MATH . The bounds REF imply MATH and hence MATH . By REF, this proves REF for MATH. For MATH, we note that MATH when MATH. For such MATH, REF implies MATH, and therefore MATH . |
math-ph/0011046 | The proof proceeds as in the proof of REF . Since MATH, as in REF - REF we have MATH . A calculation using REF then implies that for MATH, MATH . The above bounds imply REF for MATH. The case MATH is bounded simply as MATH which satisfies REF for MATH. |
math-ph/0011050 | According to REF one gets for all MATH with MATH . |
math-ph/0011050 | (compare CITE REF .) Recall that, by REF , MATH is strictly convex, hence MATH is strictly convex. Let MATH be a minimizing sequence. There exists a constant MATH such that MATH . By the NAME theorem we can extract a subsequence, still denoted as MATH, with MATH . Since by use of NAME 's inequality MATH and MATH is relatively compact in MATH, if MATH is a bounded smooth domain, MATH has a subsequence converging in MATH. Using NAME 's diagonal trick on a sequence of increasing MATH's we arrive at MATH . By NAME 's Lemma we get MATH . Since MATH norms are weakly lower semicontinuous, MATH . Moreover, one can show, in analogy to REF , that MATH so we altogether arrive at MATH . The uniqueness follows from the strict convexity of MATH. |
math-ph/0011050 | (compare CITE REF .) Note that MATH implies MATH, which gives REF a meaning in the sense of distributions. Consider the set MATH . If MATH then MATH and MATH . We find MATH for all MATH. Let MATH. Using the fact that MATH, we easily arrive at MATH . |
math-ph/0011050 | (compare CITE REF .) Since REF yields MATH, with MATH, one gets MATH. Again using REF the proposition follows by means of standard elliptic regularity theory. |
math-ph/0011050 | (compare REF .) Assume, by contradiction, MATH. Then we can choose a MATH, such that MATH for some MATH. Therefore MATH which gives us MATH with MATH. Thus there exists a MATH, such that for MATH . From REF we get MATH for MATH. Now we choose a comparison density MATH which satisfies MATH . Hence by REF MATH for MATH. We fix MATH such that MATH . If MATH for MATH, we immediately get MATH from the maximum principle. The fact that MATH and MATH contradicts our assumption. Unfortunately, we only know that MATH as MATH in a weak sense, namely MATH, so the authors in CITE used a variant of NAME 's method to verify the statement of the Lemma, which also works in our case. |
math-ph/0011050 | The derivation of REF works analogously to REF apart from the difference that MATH is the NAME parameter for the restriction MATH. We can infer MATH which implies by means of convexity of the functional MATH or equivalently for every MATH . On the other hand we derive from REF MATH which yields with MATH and MATH, respectively, MATH . Hence REF together imply REF . |
math-ph/0011050 | Note, that we have not yet shown that MATH as MATH in a strong sense, for we only know MATH. From REF we derive MATH, which implies MATH. Since we know MATH, recall that MATH and MATH, we conclude from MATH that MATH at infinity, for example, from the well known fact REF that the convolution MATH of two functions MATH, with MATH, goes to MATH at infinity. REF (compare REF .) Let MATH. From REF we get MATH and MATH . Since MATH as MATH, there is a MATH, such that MATH for MATH. This implies MATH and REF with REF This follows directly from REF and the fact that MATH as MATH. |
math-ph/0011050 | (compare CITE REF .) Let MATH be the normalized eigenvector of MATH belonging to the lowest eigenvalue MATH and let MATH. Then MATH . On the other hand we have MATH which altogether implies MATH . Taking into account that MATH this is equivalent to REF . |
math-ph/0011050 | This is a consequence of REF, which tells us that for every MATH there exists a constant MATH so that MATH for every MATH. |
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