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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... |
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... |
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)... |
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 a... |
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 bound... |
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... |
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 ge... |
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 MA... |
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 MA... |
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 sho... |
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 fo... |
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 RE... |
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 c... |
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-... |
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 assumpti... |
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... |
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 M... |
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 involv... |
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, ther... |
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 ... |
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. Sin... |
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 an... |
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 appropriate... |
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-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 MA... |
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 M... |
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 w... |
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... |
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 const... |
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 b... |
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... |
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. ... |
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 const... |
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 ... |
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 ... |
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 ... |
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 f... |
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 an... |
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 MA... |
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 har... |
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... |
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 ea... |
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 h... |
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 exa... |
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 o... |
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 th... |
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... |
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... |
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 MA... |
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 s... |
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 su... |
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 func... |
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. Th... |
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 sa... |
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,... |
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 rel... |
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. ... |
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, resp... |
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,... |
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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