Negative spectra of elliptic operators

We establish different estimates for the sums of negative eigenvalues of elliptic operators. Our proofs are based on a property of the eigenvalue sums that might be viewed as a certain convexity with respect to the perturbation.


However, condition (1.1) is violated and the operator
is only bounded, then we study the bottom of the spectrum of the operator H which might be either an eigenvalue or a point of accumulation of the set σ (H ). Let now V 1 and V 2 be two positive selfadjoint operators, such that (1.2) Lemma 1.1 (Convexity) Let V, V 1 and V 2 satisfy (1.2). Let λ j (V ) and λ j (V n ) be enumerated in the increasing order: Then for any p ≥ 1, and ε ∈ (0, 1), the eigenvalue sum |λ j | p has the property that If (1.1) does not hold, then Proof Consider the case p = 1. In this case, where the minimum is taken over all possible orthogonal projections P n of rank n that have the property Ran P n ⊂ D(H ). Since we obtain that This implies (1.3) for p = 1. A standard convexity argument allows one to pass from the case p = 1 to the case p > 1. We have established already that where (·) + denotes the operation of taking the positive part of a number. It remains to observe that The proof is completed.
Applications. Consider some applications of Lemma 1.1. Let A be the selfadjoint operator in L 2 (R d ) defined by and let V be the operator of multiplication by a positive function V (x) ≥ 0.
In particular, the negative eigenvalues λ j of the operator Proof Let us prove this statement for p = 1. The case p > 1 is similar. Assume for Using the convexity of the eigenvalue sum we obtain: we derive the inequality Repeating this operation over and over, we will obtain a similar inequality with φ(x) = δ(x). In this case S(αφ) can be easily computed, because the operator (− ) l − αδ(x) has exactly one eigenvalue below zero.
Remark 1 If d = l = 1, this statement follows from the celebrated result of Hundertmark et al. [3] saying that , then the operator (− ) l − V has exactly one eigenvalue λ 1 such that Consider the matrix-valued case, when A is the selfadjoint operator in and V is an n × n-matrix valued potential. We shall assume that (V (x)η, η) ≥ 0 for each x ∈ R d and each vector η ∈ C n . The importance of such problems with a matrix-valued potentials was discovered by Laptev and Weidl [4]. Using the technique of [4], one can reduce the study of the Schrödinger operator in the multi-dimensional case to the case d = 1. As a result, one obtains the best constants in the Lieb-Thirring inequalities.
Proof We apply the same method as before, but take into account the fact that V (x) is a matrix. Again, we prove this statement only for p = 1, because the case p > 1 is similar. Assume for the moment that V = j h j φ(x − x j )P j , where the positive function φ has the property (1.6) and P j are orthogonal projections in C n of rank 1.
Using the convexity of the eigenvalue sum we obtain: we derive the inequality and P is an orthogonal projection of rank 1. Repeating this operation over and over, we will obtain a similar inequality with φ(x) = δ(x)P. In this case S(αφ P) can be easily computed, because the operator (− ) l I − αδ(x)P has exactly one eigenvalue below zero.

with Dirichlet boundary conditions on the boundary of the interval [0, L]. Suppose that
is the function satisfying the equation The proof is very simple. One has to minimize the lowest eigenvalue of the operator Proof Again, one has to minimize the lowest eigenvalue of the operator by varying x 0 . However, the optimal choice of x 0 will be now different because of the Neumann boundary conditions. By the Birman-Schwinger principle the lowest eigenvalue λ of this operator operator is the solution of the equation where G λ (x, y) is the integral kernel of the operator (−d 2 /dx 2 − λ) −1 . The minimum of λ is attained at the edges of the interval [0, L]. It remains to note that (1.10) with λ = −γ and x 0 = 0 is equivalent to (1.9) with t = α.

Birman's theorem
As we mentioned before, the bottom of the spectrum of the operator H does not have to be an eigenvalue. It might be a point of the essential spectrum of H . Still, λ 1 = inf σ (H ) has the same property as S(V ): it is convex in the following sense: Combining this property with the possibility to impose Neumann conditions, we obtain the following statement.

Theorem 2.1 Let H be the one-dimensional Schrödinger operator
Then the bottom of the spectrum of H satisfies the estimate Proof By imposing the Neumann boundary conditions at integer points n ∈ Z, we reduce this theorem to the estimate of the lowest eigenvalue of the operator on the interval [n, n + 1] with Neumann boundary conditions at the edges. But this has been already done in Theorem 1.4.
Theorem 2.1 could be viewed as a version of the following result proven by Birman. Such a statement was mentioned in [2], but we believe that the full credit can be already given to the earlier paper [1].

Theorem 2.2 Let H be the one-dimensional Schrödinger operator
The proof of Theorem 2.1 seems to be simple and natural. Besides that, this theorem is sharp, which means that the function γ [α] can not be made smaller.
Proof Indeed, set This potential is an even periodic function, therefore the ground state ψ, defined as the unique periodic solution of the equation is also even. Consequently, it satisfies the Neumann boundary condition at x = 0. Moreover, ψ(x − 1) is even as well. It implies that ψ (n) = 0 at all points n ∈ Z. Therefore we conclude that λ 1 coincides with the lowest eigenvalue of the operator −d 2 /dx 2 − V (x) with Neumann boundary conditions at the edges of the interval [0, 1]. This eigenvalue was studied in the proof of Theorem 1.4, according to which λ 1 → γ [α] as → 0.