Two analogs of Pleijel s inequality

Transcription

1 arxiv: v1 [math.ca] 6 Mar 211 Two analogs of Pleijel s inequality Sergey Y. Sadov 1 Formulation of results Let N(λ) be a nondecreasing function defined on R + = (,+ ) such that N(λ) = for small λ and The Stieltjes transform of N(λ) is defined as * S(ζ) = λ dn(λ) <. (1) (λ ζ) dn(λ), ζ R +. (2) Fix a point ζ = λ + iη in the first quadrant of the complex plane C. Denote by a contour that connects the point ζ to ζ = λ iη and does not cross the integration path R + of (2). Åke Pleijel in [1] obtained the inequality N(λ ) 1 S(ζ) dζ 2πi η 1+π 2 S(ζ ) (3) and used it to give a short proof of Malliavin s [2] Tauberian theorem. The inequality(3) found applications in spectral theory of differential and pseudodifferential operators(see e.g.[3, 4]). In the present paper two generalizations of this inequality are derived. Translated by the author from the Russian original: Funkcionalnyj Analiz. Spektral naja teorija. Mezhvuzovskij sbornik nauchnyh trudov. Ul anovsk, 1987, pp (MR 92j:2613) * There exists an alternative convention according to which the Stieltjes transform of f(t) is defined as (ζ +t) f(t)dt. 1

2 For α >, the Riesz mean of order α of N(λ) is λ ( N (α) (λ) = 1 λ) x α dn(x), λ >. (4) Given a power asymptotics of the Stieltjes transform of N(λ) along a certain parabola-like curve in C that avoids R +, the asymptotics of the Riesz means as λ + can be recovered using the following theorem. Theorem 1 Let the function N(λ) be constant in a neigbourhood of λ. Then for any α > N(α) (λ ) 1 ( S(ζ) 1 ζ ) α dζ 2πi λ 1 ( ) α η η S(ζ ). (5) απ For α < 1 the factor (απ) in the right-hand side may be replaced by π 2 +1/4. * Henceforth the branch z α = exp(αlnz) with π < Im lnz π is assumed. If instead of (1) a weaker ** condition with some integer q > 1 λ λ q dn(λ) < (6) holds, then the leading term of the asymptotics of N(λ) can be recovered by means of the next theorem from the behaviour of its generalized Stieltjes transform S q (ζ) = (λ ζ) q dn(λ), ζ R +. (7) Theorem 2 Let the function N(λ) satisfy (6) and be constant in a neigbourhood of λ. There exist constants C,C 1,..., C q 2 (which depend only on q) such that N(λ ) 1 S q (ζ)(ζ λ ) q dζ 2πi q 2 m= C m η q m (8) S q (ζ)(λ ζ) m dζ. * (Added intranslation.)acontinuousdependenceofthe constantonαcanbeachieved by replacing the term 1/4 with (1 α 1+ε )/4, ε ε 1/16 (found numerically). ** Inequality (6) is weaker than (1) since we assume that dn(λ) = near λ =. 2

3 2 Proof of Theorem 1 a. The left-hand side of (5) vanishes if one uses a closed contour of integration consisting of and the segment [ζ, ζ ]. Indeed, since N(λ) is assumed constant in the vicinity of λ, one may change the order of integration: 1 2πi S(ζ) ( 1 ζ λ ) α dζ = dn(λ) 1 2πi ( λ ζ λ ) α dζ λ ζ. The inner integral in the r.h.s. equals 2πi(1 λ/λ ) α when λ < λ, and when λ > λ. Thus the r.h.s. equals N (α) (λ ). In order to prove (5) we have to evaluate R α (ζ ) = 1 ζ ( S(ζ) 1 ζ ) α dζ (9) 2πi λ ζ ( the integration path is the vertical segment). Set ζ = λ +i λ λ τ, s(λ) = sgn(λ λ ), u(λ) = η λ λ. Changing the order of integration in (9), we get R α (ζ ) = ( i)α 2π ( η λ ) α dn(λ) u α (λ) u(λ) u(λ) τ α( s(λ)+iτ ) 1+τ 2 dτ. (1) We will find constants c 1 C and c 2 > so that for any u > and s = ±1 the inequality 1 u τ α (s+iτ) su dτ c u α 1+τ u 2 c u 2 2 (11) 1+u 2 u will hold. Having (11) and the identities u(λ)s(λ) 1+u 2 (λ) dn(λ) = η ReS(ζ ) and u 2 (λ) 1+u 2 (λ) dn(λ) = η ImS(ζ ) 3

4 one can estimate the r.h.s. in (1) by means of the Schwarz inequality: ( ) α η η R α (ζ ) c 1 λ 2π 2 +c 2 2 Re 2 S(ζ )+Im 2 S(ζ ) ( ) α (12) η η = 2π c 3 S(ζ ). λ Our task is thus reduced to establishing the inequality (11) with constants c 1, c 2 such that c 1 2 +c 2 2 2α (and π 2 +4 if α < 1). and b. Using the change of variable τ τ on [ u,], we get u u τ α (s+iτ) dτ = 2e iπα 1+τ 2 2 u (aτ α+1 +sbτ α )(1+τ 2 ) (1+τ 2 ) 2 dτ (13) a = sin πα 2, b = cos πα 2. (14) We will need the integral representations u α+1 1+u = τα+1 u u (α)τ α+2 +(α+1)τ α 2 1+τ 2 = dτ (15) (1+τ 2 ) 2 u α+2 1+u 2 = u ατ α+3 +(α+2)τ α+1 (1+τ 2 ) 2 dτ. (16) Multiplying the inequality (11) by u α and making the substitutions (13), (15), (16), we transform (11) to the equivalent form u aτ α+3 +sk τ α+2 +aτ α+1 +sk + τ α dτ (1+τ 2 ) c 2 u ατ α+3 +(α+2)τ α+1 dτ, (1+τ 2 ) 2 (17) k ± = b c 1 2 e iπα/2 (α±1). In order for (17) to hold for small positive u, the coefficient of τ α must equal, so we set c 1 = 2b α+1 eiπα/2. (18) 4

5 With this value of c 1, the numerator in the l.h.s. of (17) is real. Clearing the absolute value notation, we rewrite (17) as a system of two inequalities, in which we leave the least favorable sign of s (so as to make the coefficient of τ α+2 negative): u (c 2 α 2a)τ α+3 4 b α+1 τα+2 +(c 2 α+2c 2 2a)τ α+1 (1+τ 2 ) 2 dτ. Taking the least favorable sign in front of 2a, we get u P 2 (τ)τ α+1 dτ, (19) (1+τ 2 ) 2 P 2 (τ) = ( c 2 α 2 a ) τ 2 4 b α+1 τ +( c 2 α 2 a +2c 2 ). The inequality (19) with c 1 defined by (18) implies (11). The rest of the proof amounts to finding an appropriate value of c 2. c. ItsufficestoensurethatthequadraticpolynomialP 2 (τ)isnonnegative. Let us choose the value of c 2 that makes its discriminant equal to zero: c 2 = 2 a (α+1)2 + a 2 +α(α+2). (2) α(α+1)(α+2) In view of (14) we have a, b 1, a 2 +b 2 = 1, and the following estimates readily follow: a α c 2 2 a (α+1)+1. (21) α(α+2) The left estimate shows that the coefficient of τ 2 in P 2 (τ) is nonnegative. By our choice of c 2, this leads to (19) and hence to (11). Using the right inequality in (21) together with (14) and (18), we find c 1 2 +c 2 2 4α 2, as required. d. To finish the proof, let us show that if α < 1 then one may use the value c 2 = 2 a α instead of (2). With this new choice of c 2, the constant c 3 in the r.h.s. of (12) becomes (cos(πα/2) ) 2 ( ) 2 sin(πα/2) c 3 = π α+1 α 2, 5

6 as the theorem claims. We have to verify the inequality 4 a α u τ α+1 b dτ 4 (1+τ 2 ) 2 α+1 u τ α+2 (1+τ 2 ) 2 dτ in the interval < α < 1. The left-hand side, as a function of u, is positive for small u and has a unique critical point (maximum) on R +. It remains to check that a α τ α+1 b dτ (1+τ 2 ) 2 α+1 τ α+2 (1+τ 2 ) 2 dτ. Using the substitution τ 2 = t we express the integrals in terms of Euler s Beta function and the last inequality takes the form ( a α+2 α B, 2 α ) b ( α α+1 B, 1 α ). 2 2 Therightandleftsidesareinfactequal: itfollowsfrom(14)andtheidentities ( α+2 B, 2 α ) = πα/2 ( α sin(πα/2), B, 1 α ) = π(α+1)/2 2 2 cos(πα/2). The proof is complete. 3 Proof of Theorem 2 a. The left-hand side of the inequality (8), likewise the l.h.s. of the inequality (5) in Theorem 1, vanishes if the closed contour of integration consisting of and the segment [ζ,ζ ] is used. In the right-hand side of (8), integration over can be replaced by integration over [ζ,ζ ] since S q (ζ)(λ ζ) m dζ = dn(λ) (λ ζ) m (λ ζ) q dζ, and the residue of the integrand at ζ = λ equals as m q 2. Therefore (8) is equivalent to the inequality q 2 V q,q (λ)dn(λ) C m V q,m (λ)dn(λ), (22) m= 6

7 V q,m (λ) = η q m ζ (λ ζ) m dζ, m =,1,...,q. (23) ζ (λ ζ) q The substitutions ζ = λ +iη τ, µ = (λ λ )η bring (23) to the form V q,m (λ) = ( i) m T q,m (µ), (24) T q,m (µ) = 1 τ m dτ (µ iτ) q. (25) b. Let us study properties of the functions T q,m (µ). 1. The function T q,m (µ) is even if q m is even, and odd if q m is odd. This is verified by changing µ into µ in (25) and τ into τ. 2. If m q 2, then we can write T q,m (µ) = P q,m(µ) (µ 2 +1) q, (26) P q,m (µ) is a polynomial (even or odd depending on the evenness of q m). Indeed, expanding τ m in powers of µ iτ, integrating the resulting linear combination of the functions (µ iτ) n q (n =,...,m) with respect to τ, and taking the common denominator, we obtain (26). 3. From (25) it is easy to find the asymptotics of T q,m as µ + : T q,m (µ) = b q,m µ q + O(µ q 2 ) for m even, T q,m (µ) = b q,m µ q for m odd, (27) b q,m. Comparing to (26), we see that the polynomial P q,m (m q 2) is of exact degree q 2 for m even, and of exact degree q 3 for m odd. c. Let us show that {P q,m }, m =,...,q 2, is a basis in the space of polynomials of degree at most q 2. It suffices to verify that the corresponding functions T q,m are linearly independent. Suppose, to the contrary, that some their linear combination is zero: L(µ) = 1 U(τ) dτ =, (µ iτ) q 7

8 U(τ) is a polynomial of degree at most q 2. Consider L as an analytic function of complex variable µ. It is regular outside the segment [ i, i], therefore, it equals zero identically. Let γ be a closed contour around the segment [ i,i]. For all integer n q we have = γ L(z)z n dz = 1 U(τ)dτ and β n. Hence for any polynomial Ũ(τ) 1 γ z n dz 1 (z iτ) = β q n U(τ)τ n q+1 dτ, U(τ)Ũ(τ)dτ =. Taking Ũ to be the complex-conjugate of U leads to the conclusion U, which proves the linear independence of the functions T q,m. d. Consider first the case of even q. The result of (c) shows that for some C,...,C q 2 or cf. (26) that q 2 m= As µ, we have q 2 m= C m P q,m(µ) = 1+µ q 2, C m T q,m(µ) = 1+µq 2 (1+µ 2 ) q =: H q(µ). (28) T q,q (µ) b q,q µ q = o ( H q (µ) ). The function T q,q (µ) is bounded, while minh q (µ) > on every finite interval. Therefore there exists a positive C such that T q,q (µ) CH q (µ) for all real µ. Using (24) and passing from T q,m to V q,m, then integrating with dn(λ), we get (22). e. Now consider the case of odd q. There exist constants C,...,C such that q 3 m= q 3 C mt q,m (µ) = 1+µq 3 (1+µ 2 ) q =: H q(µ). (29) 8

9 Now T q,q (µ) decays at infinity slower than H q (µ). However, as seen from (27), T q,q(µ) b q,q T q, (µ) b q, = O(µ q 2 ) = o(h q (µ)), µ. Here, like previously in (d), the left-hand side is a bounded function, while minh q (µ) > on every finite interval. Therefore there exists a positive C such that for all real µ. T q,q(µ) b q 3 q,q T q, (µ) b q, C C m T q,m(µ). Passing from T q,m to V q,m, integrating with dn(λ), and bringing the term const V q, (λ)dn(λ) over to the right-hand side, we get (22). The proof is finished. Remark. Note that due to (b1 ) the coefficients of odd functions T q,m in (28), (29) are equal to. Hence in the right-hand side of (8) the actual summation is carried over the values of index m for which q m is even; if q is odd, the value m = is also included. In conclusion I would like to thank M.S. Agranovich for suggesting the problem and for attention to this work. References [1] PleijelÅ.OnatheorembyP.Malliavin.IsraelJ.Math.1, (1963). [2] Malliavin P. Un théorème taubérien avec reste pour la transformation de Stieltjes, C.R. Acad. Sci. Paris, 255, (1962). [3] Agmon S. Asymptotic formulas with remainder estimates for eigenvalues of elliptic operators. Arch. Rational Mech. Anal. 28, (1968). [4] Pham The Laï. Estimation du reste dans la theorie spectrale d une classe d operateurs elliptiques degeneres, Seminaire Goulaouic-Schwartz , expose n X. m= Moscow Institute of Electronic Engineering. 9

10 Remarks added in translation 1. In both Theorems 1 and 2, the function N(λ) does not have to be continuous at λ. If λ is a point of discontinuity of N(λ), Theorem 1 remains valid without change, while in Theorem 2 the value N(λ ) in the left-hand side can be replaced by any value between N(λ ) and N(λ +). 2. The following theorem stands in the same relation to Theorem 2 as Theorem 1 to Pleijel s inequality (3). Theorem 3 Let function N(λ) defined on R + be nondecreasing, equal zero near λ =, and satisfy condition (6). For any α > and any integer q = 2,3,... there exist nonnegative constants C,...,C q 2, depending on q and α, such that for any λ > Nα (λ ) αb(q,α) S q (ζ)(ζ λ ) (1 q ζ ) α dζ 2πi λ q 2 m= ( ) α η C m η q m λ S q (ζ)(ζ λ ) m dζ, B(q,α) = (q)(α) (q +α). Proof: Repeat the proof of Theorem 2 replacing the function T q,q (µ) by the function 1 τ q+α T q,q+α (µ) = (µ iτ) dτ. q 3. An application of Theorem 1 can be found in: Agranovich M.S. Elliptic operators on closed manifolds. Encycl. Math. Sci., 63 (Partial Differential Equations VI), Springer-Verlag, 1994, Theorem