Equality cases in the sharp one-dimensional matrix Lieb–Thirring inequality

1 Introduction

Lieb–Thirring inequalities compare the binding energy of a Schrödinger operator with an integral of its potential. Equality asks how the potential must arrange that binding. In one dimension, a scalar one-bound-state optimizer has a hyperbolic-secant profile. For a matrix potential, several such profiles can occupy orthogonal internal channels. The issue is whether equality also permits channels that rotate with position, or several bound states that interact within the same channel. We prove that throughout 1/2<\gamma<3/2 every equality case consists of independent scalar profiles in one fixed basis.

Let m\ge1 be finite, let 1/2<\gamma<3/2, and set p=\gamma+1/2. For a measurable Hermitian positive semidefinite function W:\R\to\C^{m\times m} satisfying

let H_W=-d^2/dx^2-W be the operator associated with

Lemma 2.1 proves that the form is closed and bounded below and that the negative spectrum is discrete, with possible accumulation only at zero. We write

counting multiplicities. The finite-dimensional trace \tr is unnormalized; matrix powers are defined by spectral functional calculus.

The inequality in (1.3) is the sharp matrix inequality of [11, Theorem 1.1]. Our contribution is its equality classification. We retain the analytic constructions in the proof, so that the inequality and its equality conditions are established here under the same measurable potential hypothesis. The scales and centers in (1.4) are independent; repeated scales and identical profiles are allowed.

1.1 History and related results

Lieb and Thirring introduced spectral moment inequalities in their work on the kinetic energy of fermions and stability of matter [8, 9]. Their sharp constants distinguish the regimes in which binding is best concentrated in a single state or distributed among many states. Keller’s optimization of the lowest Schrödinger eigenvalue [6] supplies the one-state problem underlying the scalar profiles in (1.4); their explicit form appears in [9, Section 4(B), Equations (4.19)–(4.20)]. The present question concerns equality in the sum of all negative eigenvalue moments, with a matrix potential that need not commute with itself at different points.

The sharp scalar endpoint \gamma=1/2 was proved by Hundertmark, Lieb and Thomas [5]; Hundertmark, Laptev and Weidl [4, Theorem 3.1] extended it to operator-valued potentials. The semiclassical constant is sharp for \gamma\ge3/2; the lifting of moments of Aizenman and Lieb [1] and the operator-valued inequalities of Laptev and Weidl [7] are central to this development. Benguria and Loss [2] gave a commutation proof of the matrix result at \gamma=3/2. More recently, Read and Schulz [12] proved the sharp scalar inequality at \gamma=1, with constant 4/(3\sqrt3\pi), and its operator-valued extension. The matrix companion [11] proves the sharp one-state constant for the whole open interval considered here.

The endpoint \gamma=3/2 has a different equality structure. In the scalar finite-rank problem, Frank, Gontier and Lewin [3, Theorem 5] identify the optimizers as KdV multisolitons. In particular, several bound states can occur in a single scalar channel at that endpoint. The conclusion of Theorem 1.1, one bound state per nonzero channel, uses the strict inequality r>1 and is asserted only for the open interval.

1.2 Quantitative defects and fixed channels

The proof begins with the finite-interval matrix action method of [11]. For orthonormal real trial functions approximating finitely many negative eigenfunctions, place the functions in the rows of U(x) and choose a lower triangular matrix C. The rows of A=CU remain in the successive trial spans. If K is the diagonal matrix of positive square roots of the trial binding energies, the action involves A', K^2AA^T, and a power of the density AA^T. A symmetric matrix path B(x) connects K to -K and gives a lower bound for that action.

For equality, the lower bound must retain more information. Besides the square \|A'-BA\|_{\HS}^2, we retain a term controlling

Its coefficient depends on the largest binding energy and the exponent, but not on the number of trial functions. This uniformity is the first essential point: an extremal potential is not known in advance to have finitely many negative eigenvalues. The integrated trace identity of the matrix companion supplies the two nonnegative squares from which this estimate follows.

The second point is compactness with a fixed number of columns. As the trial list grows, A has more rows but still only m columns. Its density therefore has rank at most m. The retained defect bounds the tail rows by the small binding energies in that tail. This yields strong pointwise convergence even for an infinite eigenvalue list and gives exact algebraic relations in the limit. Those relations separate the rows with different energies. A differential inequality then makes their row spaces independent of position; their mutual orthogonality leaves at most m nonzero directions. On each resulting finite block, a matrix Riccati equation has a fixed diagonalizing basis and determines the profiles in (1.4).

Section 2 gives the spectral and convexity preliminaries. Section 3 proves the quantitative action estimate, and Section 4 recovers the sharp spectral bound. Section 5 passes to exact relations for an equality case. Section 6 proves that the channels are fixed, solves the Riccati equation, and treats complex potentials and the converse. Appendices A and B prove the trace identity, regularized field construction, and connecting-path theorem used by the action estimate.

1.3 Notation

Throughout the proof,

and

The last identity follows from the beta integral. We use the operator norm \|\cdot\|_{\op} and the Hilbert–Schmidt norm \|A\|_{\HS}^2=\tr(A^*A). The symbol T denotes transpose of a real matrix, and \Sym_n is the space of real symmetric n\times n matrices with inner product \tr(BH). The commutator is [B,H]=BH-HB. The identity matrix, or identity operator, has the size determined by its context.

2 Spectral and convexity preliminaries

Two elementary facts let us work with measurable potentials and identify the equality condition in the potential estimate. The first is the form argument used in the matrix inequality [11, Lemma 7.1]; we give its proof to fix the precise regularity available below.

3 An action estimate that retains the equality defects

The finite-interval argument for the sharp inequality can also measure failure of its equality relations. We retain the first-order square and a quantitative part of the matrix trace deficit. The coefficient of this second defect will be independent of the number of rows. This uniformity allows us to exhaust an eigenvalue list before knowing that it is finite.

The construction adapts the field, connecting paths, and action identity of [11, Sections 3–6]. Their analytic foundations are proved in Appendices A and B; the additional estimate below extracts operator-norm rigidity from the trace deficit.

Fix positive integers n,d, a bounded interval \Omega=(x_-,x_+), and K=\operatorname{diag}(\ell_1,\ldots,\ell_n) with 0<\ell_i\leq\kappa. For A\in H^1_0(\Omega;\R^{n\times d}), set \rho=AA^{\mathsf T} and define

For pointwise matrix values A\in\R^{n\times d} and B\in\Sym_n, put

Thus \Phi_K=0 expresses two algebraic relations, while A'-BA=0 is a first-order differential relation.

3.1 Coercivity at the matrix-field constraint

The algebraic relations \Phi_K(A,B)=0 require [B,K^2]=0 and \rho=(K^2-B^2)^\sigma. For 0<\delta\leq1, we use a locally Lipschitz field M_\delta:\Sym_n\to\Sym_n extending the commuting rule

Thus the constraint M_\delta(B)=\rho is a regularized form of the algebraic equality relation in this case. The extension to arbitrary symmetric B has a C^1 primitive J_\delta:\Sym_n\to\R satisfying

Proposition B.1 constructs these maps and proves their required properties. Along a path B, the derivative identity converts -\tr(M_\delta(B)B') into the total derivative of J_\delta(B), which will supply the boundary term in the action identity. Its endpoint difference is explicit:

The limit follows by dominated convergence and 2\sigma+1=2\gamma.

Define the scalar function

The following lemma bounds this nonlinear term while keeping the algebraic defect. The constraint in its hypothesis will later be enforced by a penalty.

3.2 The action identity and the penalty limit

4 Recovering the sharp spectral bound

The action estimate implies the sharp inequality by the spectral passage of [11, Section 7]. We record it before studying equality, so that the finite spectral moment used below has already been established.

5 From equality to pointwise matrix relations

We now apply Theorem 3.1 to finite collections of negative eigenfunctions. Equality forces each of its nonnegative errors to vanish. The main issue is that the number of eigenfunctions may grow without bound, whereas the potential has only m channels. We will use this fixed number of columns to control the tails of the resulting matrices.

Throughout this section, W is real symmetric, W\geq0 almost everywhere, and 0<\int_{\R}\tr(W^p)<\infty. Suppose that equality holds, in the form

where -k_i^2 are the negative eigenvalues of H_W, repeated according to multiplicity, and

By Lemma 2.1, the index set \mathcal I is a nonempty finite initial segment of \mathbb N or all of \mathbb N, each eigenvalue has finite multiplicity, and k_i\to0 in the infinite case. Fix real orthonormal eigenfunctions \varphi_i\in H^1(\R;\R^m) and regard them as row vectors. Write

For operators between real Hilbert spaces, the superscript T will also denote the adjoint.

We divide the proof into the construction of a sequence whose errors vanish, convergence of every fixed row, and control of the remaining rows.

5.1 Finite spectral approximations and vanishing errors

Let n_j=j when \mathcal I=\mathbb N, and let n_j=|\mathcal I| when \mathcal I is finite. Choose 0<\xi_j<1/j so small that

Set K_j=\operatorname{diag}(\ell_{j,1},\ldots,\ell_{j,n_j}). The same shift \xi_j is used for every eigenvalue: this leaves the differences of their squares unchanged.

The approximation leading to (4.1) can be made arbitrarily accurate in H^1. Apply it to the first n_j eigenfunctions, choosing the H^1 errors below 1/j and the form-matrix error below \xi_j. This gives real, smooth, compactly supported functions u_{j,1},\ldots,u_{j,n_j}, orthonormal in L^2(\R;\R^m), such that

Here h_W[\cdot,\cdot] is the polarized quadratic form, and the second inequality is the ordering of real symmetric matrices.

Let U_j be the matrix with rows u_{j,i}, and choose a bounded interval \Omega_j containing their supports in its interior. Apply Theorem 3.1 with K=K_j, d=m, \kappa=k_1, and \eta=1/j. It gives a lower triangular matrix C_j, a symmetric path B_j with \|B_j\|_{\op}\leq\kappa, and A_j=C_jU_j. Extend A_j by zero outside \Omega_j and extend B_j by K_j to the left and by -K_j to the right. All the errors in that theorem vanish outside \Omega_j, so their integrals may be taken over \R.

The lower triangular form of C_j gives a useful additional error. Put Z_j=A_j^TA_j, an m\times m positive semidefinite matrix. By (5.6),

Indeed, the ith row of A_j is \sum_{l\leq i}(C_j)_{il}u_{j,l}. Its quadratic form, plus \ell_{j,i}^2 times its squared L^2 norm, is at most

Summing over i proves (5.7), since A_jA_j^T and Z_j have the same nonzero eigenvalues, with multiplicity. The ordering of the k_i makes every summand of T_j nonnegative.

Recall the nonnegative Young deficit from Lemma 2.2:

Combining Theorem 3.1, (5.1), and (5.7) yields

Here c_*>0 is fixed: its independence of n_j is essential. Thus all three integral errors tend to zero, as does T_j.

We will also need a uniform bound on A_j. The coercive estimate (2.3) gives

Since Z_j has size m, scalar Hölder’s inequality gives

Equations (5.9) and (5.10) therefore bound A_j uniformly in L^{2q}(\R) with the Hilbert–Schmidt norm, independently of its number of rows.

5.2 Convergence of each fixed row

For fixed i\in\mathcal I, the coefficients (C_j)_{il}, l\leq i, are bounded as j\to\infty. To see this, choose a bounded interval J_i so large that the restricted Gram matrix

where I_i denotes the i\times i identity matrix. Such an interval exists because the full Gram matrix is the identity. By (5.6), the corresponding restricted Gram matrix of the u_{j,l} is at least one half of the identity for all sufficiently large j. Hence

The last expression is uniformly bounded. Lower triangularity is used here to keep the number of coefficients in a fixed row finite.

A diagonal subsequence now makes every coefficient (C_j)_{il} converge to a limit c_{il}, for l\leq i. From T_j\to0 and (5.7),

Define

The sum is finite, and (5.6) implies

The ordering of the eigenvalues and (5.11) show that only eigenfunctions with eigenvalue -k_i^2 occur in a_i. This proves (5.2), including the possibility a_i=0. The eigenfunction equations hold distributionally by testing the form equations against smooth compactly supported functions; the potential is locally integrable and the rows are locally bounded.

Choose the continuous, locally absolutely continuous representatives of these rows. The Sobolev embedding H^1(\R)\subset C_b(\R) shows that (5.12) also gives pointwise convergence at every x. Passing to a further subsequence, we may arrange that the integrals of the nonnegative integrand in (5.9) are summable, and that, for every fixed i, the squared L^2 errors of the derivatives in (5.12) are summable. A diagonal choice achieves both conditions, by making the first j available derivative errors and the integral error less than 2^{-j} along the chosen subsequence. Consequently, on a common set of full measure,

We also exclude the null set where W is not a finite positive semidefinite matrix. It remains to show that convergence of the individual rows is strong enough to pass to the products in these relations.

5.3 Tail control in the row-index space

Pad A_j, B_j, and K_j with zeros in the unused coordinates of \mathscr H. Thus A_j:\R^m\to\mathscr H, and B_j,K_j are self-adjoint finite-rank operators on \mathscr H. Since the omitted k_i tend to zero,

where a supremum over no indices is zero.

Fix a point x in the full-measure set from (5.13) and suppress x for the moment. Write \rho_j=A_jA_j^T. The definition of \Phi and its convergence to zero give

For h\geq1, let P_h be the orthogonal projection onto the first h coordinates of \mathscr H, with all coordinates included if h exceeds |\mathcal I|. Set Q_h=I-P_h and s_h=\sup_{i>h}k_i^2, again with the empty supremum equal to zero. Then s_h\to0. For every unit vector v in the range of Q_h,

Both terms on the left are nonnegative. Jensen’s inequality for the spectral measure of \rho_j, using r>1, implies

It follows that \|Q_h\rho_jQ_h\|_{\op}\leq(s_h+e_j)^{1/r}. Although \mathscr H may be infinite dimensional, Q_h\rho_jQ_h has rank at most m, because A_j has exactly m columns. Its trace is therefore at most m times its operator norm. We obtain the two tail bounds

The first estimate converts decay of the eigenvalue parameters into decay of the total mass of all omitted rows. Its factor is the fixed physical dimension m, rather than the number of retained eigenfunctions; Figure 1 depicts this distinction.

For each fixed h, the first h rows of A_j converge by (5.12). The first bound in (5.17) makes the remaining rows uniformly small by choosing h large and then j sufficiently large. Thus A_j is Cauchy in Hilbert–Schmidt norm and converges to an operator A:\R^m\to\mathscr H whose rows are precisely a_i(x):

This argument applies to the whole subsequence already selected and does not require a further choice depending on x.

For B_j we need only pointwise compactness. Its finite compressions P_hB_jP_h are bounded in finite-dimensional spaces, so a diagonal subsequence makes them converge for every h. Self-adjointness and the second estimate in (5.17) give

The chosen subsequence consequently converges in operator norm to a self-adjoint B with \|B\|_{\op}\leq\kappa. This subsequence, and hence B, may depend on the point x.

We can now pass to all the required products. Hilbert–Schmidt convergence of A_j implies operator norm convergence of A_jA_j^T and of A_j^TA_j. Continuous functional calculus for nonnegative operators then gives convergence of their rth powers. Together with (5.14) and (5.15), this proves

Also B_jA_j\to BA in Hilbert–Schmidt norm. Taking its ith row and using both the derivative convergence and the first-order error in (5.13) gives a_i'(x)=(BA)_{i,\cdot} for every i. Finally, continuity of the finite-dimensional Young deficit and Y(W,A_j^TA_j)\to0 yield Y(W,A^TA)=0. Lemma 2.2 gives W=q(A^TA)^r.

We have proved these statements at every point of one common set of full measure. The construction selects no function x\mapsto B(x): only the existence of a suitable operator at each such point enters the next section. This completes the proof of Proposition 5.1.

6 Fixed channels and the equality profiles

The pointwise relations of Proposition 5.1 still involve an auxiliary operator B that may depend on the point without a prescribed measurable choice. We now extract constant subspaces from those relations. On each nonzero subspace, B becomes a uniquely determined matrix and satisfies an ordinary differential equation. This will identify every real equality potential; realification and a scalar calculation will then complete the proof of Theorem 1.1.

6.1 The channel subspaces are constant

Assume first that W is real and is a nonzero equality potential. Use the rows a_i, numbers k_i, and operator \Lambda=\operatorname{diag}(k_i) from Proposition 5.1. For each distinct value \alpha>0 in the list (k_i), let n_\alpha be its finite multiplicity, and let G_\alpha be the n_\alpha\times m matrix whose rows are the a_i with k_i=\alpha. Each G_\alpha has entries in H^1(\R), and we use their continuous representatives.

At every point in the common full-measure set of that proposition, choose one of the operators B it supplies. Since B commutes with \Lambda^2, it preserves each of its eigenspaces. Moreover,

commutes with both B and \Lambda^2. Writing B_\alpha for the block on the \alpha^2-eigenspace therefore gives

These statements hold almost everywhere and require no compatibility between the choices of B at different points.

The row spaces already split the physical space \R^m into finitely many constant channels. To determine the potential within one of them, we also need a constant range for G_\alpha in the eigenfunction index space \R^{n_\alpha}.

6.2 The matrix Riccati equation separates the profiles

Fix a nonzero block and the matrices of Lemma 6.2. Diagonalize the symmetric matrix B_0(x_*) at one point by a constant orthogonal change of range coordinates. The right-hand side of its Riccati equation is a polynomial in B_0. Uniqueness for this matrix ordinary differential equation therefore preserves the diagonal form. More explicitly, the strict inequality in (6.7) places each initial diagonal entry in (-\alpha,\alpha). There is a unique y_j\in\R such that the scalar solution with that initial value is

The diagonal matrix with these entries solves the same initial-value problem as B_0 on all of \R, so uniqueness gives equality everywhere.

Equation (6.7) implies that DD^T is diagonal in these coordinates. Since D'=\operatorname{diag}(b_j)D, each row of D is a positive scalar multiple of its value at x_*. None of these rows is zero, and their directions are mutually orthogonal because DD^T is diagonal. Taking their normalized directions as a constant orthonormal basis of \mathcal R_\alpha makes D diagonal with positive entries. The potential in this fixed basis is therefore diagonal, with entries

Combining these bases over the finitely many nonzero spaces \mathcal R_\alpha, and adding a basis of their orthogonal complement, proves the required necessity for real W. There are at most m nonzero profiles by (6.4). The zero potential has the same description with no nonzero profile.

6.3 Complex Hermitian potentials

For a complex Hermitian potential W, take the realification \widetilde W from (4.2). The proof of Proposition 4.1 shows that it is equivalent to W\oplus\overline W by a constant unitary matrix and that both the trace-power integral and the negative spectral moment are doubled. Thus equality for W implies equality for the real potential \widetilde W.

The real result provides a constant unitary matrix V such that

where each f_j is either a profile of the form (6.9) or zero. Let P be the projection onto the first summand \C^m and put \widehat P=V^*PV. Since P commutes with W\oplus\overline W,

Consequently \widehat P_{ij}=0 whenever the functions f_i,f_j are not equal almost everywhere. Partition the indices into groups of identical functions, including the possible zero function. The range of \widehat P splits as the orthogonal sum of its intersections with these coordinate groups. On each group the diagonal potential is a scalar multiple of the identity, so any constant orthonormal basis of that intersection diagonalizes its restriction. Transporting these bases by V to the first summand gives a constant unitary basis of \C^m in which W has the asserted profiles. This proves necessity for complex Hermitian potentials.

6.4 Each listed potential attains equality

It remains to verify attainment, including the exact spectral moment, without appealing to a classification of scalar optimizers. Fix a>0 and x_0\in\R, and set

The first-order factorization used in the commutation argument of [2, Section II] also explains why this profile has only one negative eigenvalue. Put

Then Q^*=-\partial_x+w on H^1(\R). Since w and w' are bounded, both product domains are H^2(\R), and direct differentiation yields

Here the last inequality uses r>1. The equation Q\psi=0 has the one-dimensional L^2 solution space

Thus -a^2 is a simple eigenvalue of H, and (6.10) excludes eigenvalues below it. If H\phi=\lambda\phi with -a^2<\lambda<0, put \mu=\lambda+a^2 and g=Q\phi. Then \|g\|_2^2=\mu\|\phi\|_2^2>0. Moreover, Q^*Q\phi=\mu\phi\in\operatorname{Dom}Q, so g\in\operatorname{Dom}(QQ^*) and QQ^*g=\mu g. This contradicts (6.11), since \mu<a^2. By Lemma 2.1, all negative spectrum is discrete. Hence -a^2 is the sole negative eigenvalue.

Finally, 2p-1=2\gamma and the substitution s=\tanh y give

where (r+1)^p/r=D_\sigma^{-1} and p-1=\sigma. Consequently

The zero potential contributes zero to both sides. Finite orthogonal direct sums add the spectral moments and trace-power integrals, and a constant unitary conjugation leaves both unchanged. Every potential listed in Theorem 1.1 therefore attains equality, completing its proof.

A The exact integrated trace deficit

The matrix comparison needed in the action argument has an exact remainder. We give the proof from [11, Section 3], retaining both nonnegative terms. Together they control the commutator and the algebraic equality relation used in the action argument. Throughout this appendix, 0<\sigma<1, \beta=\sigma-\tfrac12, and r=1/\sigma, p=1+\sigma, q=1+r. Define c_\sigma>0 by

The integrand is O(t^{\sigma-1}) at zero and O(t^{\sigma-2}) at infinity. Scaling therefore gives

The hypothesis concerns the full symbol X_s; the perturbation V itself need not be positive. The proof first recovers L and Z from solutions on the two half-lines, then evaluates a logarithmic integral, and finally completes a matrix square.

A.1 Half-line boundary maps

Quadratic growth in s, together with (A.3), gives an integrable bound for X_s^{-1}. Hence R(t)>0 and R(t)=O(1) as t\downarrow0. Diagonalization of B gives

For t\geq1, both X_s+tI and Y_s+tI dominate c(s^2+t)I, with c>0 independent of s,t. The resolvent identity bounds their inverse difference by C(s^2+t)^{-2}, and therefore

These bounds prove the asserted convergence of N.

Fix t>0 and put \nabla=\partial_x-iB. We shall use the equation

Matrix Riccati equations also appear in the commutation method for Schrödinger inequalities [2]. Here their role is to compute the integrated resolvent through boundary data.

For each v\in\C^n there is a unique H^1 solution of (A.9) on either half-line with \psi(0)=v. Indeed, the sesquilinear form

with inner products linear in the second variable, is coercive on H^1(\R;\C^n): its Fourier symbol X_s+tI dominates c_t(1+s^2)I. Zero extension gives coercivity on the trace-zero subspace of either half-line. Correcting any H^1 extension of v by the Riesz representation theorem gives the unique weak solution. The equation implies \psi''\in L^2, so this solution lies in H^2; both \psi and \psi' vanish at the infinite end.

Define the right and left boundary matrices S,T by

Integration by parts gives a_t(\psi_v,\psi_w)=\langle v,Sw\rangle on the right and a_t(\psi_v,\psi_w)=\langle v,Tw\rangle on the left. Thus S,T are Hermitian. Translation and uniqueness show that \nabla\psi=-S\psi throughout the right half-line and \nabla\psi=T\psi throughout the left half-line. Substituting these relations into (A.9), and allowing arbitrary boundary data, gives

Glue the two solutions with boundary value v at zero. The resulting continuous function has derivative jump -(S+T)v, and hence satisfies

Fourier inversion, with forward kernel e^{-isx}, gives

The inverse symbol is O(s^{-2}), so its Fourier inverse is continuous and this evaluation at zero is justified. Therefore

satisfy (A.6), by adding and subtracting (A.11). The second equation in (A.6) determines Z uniquely from L, since L>0; in an eigenbasis of L, the map Z\mapsto LZ+ZL multiplies its (j,k) entry by the positive number L_{jj}+L_{kk}. This also proves the continuity of Z(t), because R(t), and hence L(t), is continuous.

The diagonal entries of that same equation give Z_{jj}=0 in an eigenbasis of L, so \tr Z=0. Multiplication on both sides by R=L^{-1} gives i[R,B]=-(ZR+RZ). Cyclicity of the trace now yields

In particular, the correction \tr(RZ^2)=\tr(ZRZ) is nonnegative. We will also need the location of the spectrum of B+iS. The right solutions obey \psi'=i(B+iS)\psi. For an eigenvector with eigenvalue z, the solution is e^{izx}v; its square integrability forces \operatorname{Im}z>0.

A.2 A logarithmic integral

The next step converts the boundary matrices into a scalar identity that can be integrated in t. Define

The determinants are positive, and the logarithmic ratio is O(|s|^{-2}) at infinity. Its derivative in t is the trace of the resolvent difference, with an integrable bound locally uniform for t>0. Thus the integral converges absolutely and F is continuously differentiable.

The first identity in (A.11) factors the symbol as

If u_j+iv_j are the eigenvalues of B+iS, counted with algebraic multiplicity, then v_j>0 and

The denominator in (A.13) has the same form, with centers the eigenvalues of B and widths \sqrt t. Translations of the individual logarithms change their symmetric-cutoff integrals by o(1): for h(s)=\log(s^2+v^2),

uniformly for bounded a. After centering the factors, the identity

follows by differentiation in v and integration from w to v. Since \sum_jv_j=\tr S=\tr L, we obtain

Set

Both functions are nonnegative. For G, apply \lambda-2\sqrt t+t/\lambda\geq0 to the positive eigenvalues \lambda of L. For F, take the trace of (A.6) and use (A.14) to obtain

Equations (A.12) and (A.14) give the identity

Every term on the right is nonnegative. The bounds (A.8) therefore show that F, G, and \tr(RZ^2) are integrable against t^\beta\,dt, and that t^{\beta+1}F(t) vanishes at both endpoints.

Differentiating (A.13) and using (A.14)–(A.15) gives

Integration by parts, with the endpoint limits just proved, yields \int_0^\infty t^\beta G\,dt =\sigma\int_0^\infty t^\beta F\,dt. Consequently, integration of (A.16) gives

It remains to express the first integral as \tr(N^q)/q plus a nonnegative remainder.

A.3 The noncommutative square

For a positive semidefinite matrix D, let

For a scalar eigenvalue d\geq0, the corresponding summand is \int_0^d(t^{-1/2}-(t+y)^{-1/2})\,dy. Tonelli’s theorem and (A.2) therefore imply

Now expand the Hilbert–Schmidt square

No commutation is used: the cross terms cancel adjacent factors R^{-1/2}R^{1/2}, and cyclicity changes the remaining quadratic term into \tr(R(tI+D)). All three terms in the last line are integrable by (A.8), (A.16), and (A.18). Thus

When N\geq0, choose D=N^r. Since rp=q=1+r, the first two terms on the right equal \tr(N^q)/q. Substitution into (A.17) proves (A.7) and completes the proof of Theorem A.1.

If B and V commute, the symbol hypothesis implies V\geq bI. Simultaneous diagonalization then gives R(t)=(tI+V)^{-1/2}, N=V^\sigma, and Z=0. Both remainders vanish, as required by scalar equality.

B The regularized field and connecting paths

The finite-interval action argument requires a symmetric matrix path from K to -K whose derivative measures the difference between a matrix field and AA^{\mathsf T}. We construct both the field and the path here. These are the constructions of [11, Sections 4–5]; we include their proofs to make the analytic input to the equality argument self-contained. The positivity statement below also identifies the resolvent to which Theorem A.1 applies.

Throughout this appendix, K=\operatorname{diag}(k_1,\ldots,k_n) with k_i>0, \kappa=\max_i k_i, and \Sym_n denotes the real symmetric matrices. We use \beta=\sigma-\tfrac12 and the constant c_\sigma from (A.1).

B.1 A field with a scalar boundary integral

For \delta>0, define on y\geq0

and extend to y<0 by the tangent line f_\delta'(0)y. The resulting function is C^1, strictly increasing, and concave. Indeed, for y\geq0,

Both integrals converge since -\tfrac12<\beta<\tfrac12. The extension allows the functional calculus below even when its matrix argument has negative eigenvalues.

Set

where the symmetric cutoff cancels the leading term -2s f_\delta'(s^2)B, which is odd in s. The scalar counterpart is

The field is defined on all symmetric matrices, but the action argument compares it with a positive matrix AA^{\mathsf T}. At such values its directional bound forces positivity of the entire quadratic symbol Q_s(B). This is the condition needed for the trace deficit formula.

B.2 Choosing the lower triangular coefficient matrix

Let \mathcal L_n be the real lower triangular n\times n matrices. Their dimension equals that of \Sym_n. This dimension match allows us to prescribe a symmetric terminal value by choosing C\in\mathcal L_n. At a simple initial problem there is one choice up to signs of the rows. Continuation preserves the parity of this number, and hence produces the required terminal value.