Sharp one-dimensional Lieb–Thirring inequalities for matrix potentials

Proof. With inner products linear in the second argument, let

On the full line its Fourier symbol is X_s+tI, and (3.3) and quadratic growth give X_s+tI\geq c_t(1+s^2)I for some c_t>0. Thus a_t is coercive on H^1(\R;\C^n). Zero extension proves the same assertion on the subspace of H^1 functions on either half-line whose boundary trace is zero.

Choose any H^1 extension \eta of the prescribed boundary value. On the trace-zero space the form a_t is an equivalent Hilbert space inner product. The Riesz representation theorem gives a unique correction u in that space with a_t(\phi,u)=-a_t(\phi,\eta) for every trace-zero \phi. Then \psi=\eta+u solves (3.8) weakly. Conversely, an H^1 solution satisfies this weak identity by density of interior test functions in the trace-zero space, so coercivity proves uniqueness.

The equation implies \psi''\in L^2. In particular the solutions belong to H^2, their first derivatives have boundary traces, and both \psi and \psi' tend to zero at the infinite end of the half-line. Define S and T by (3.9). For the right solutions with data v,w, integration by parts gives a_t(\psi_v,\psi_w)=\langle v,Sw\rangle; on the left it gives a_t(\psi_v,\psi_w)=\langle v,Tw\rangle. Since a_t is Hermitian, both boundary matrices are Hermitian.

Translation and uniqueness imply the same derivative relations at every interior point:

For example, \nabla(-S\psi)=-S\nabla\psi+i[B,S]\psi. Substitution into (3.8), followed by evaluation at zero for arbitrary boundary data, proves the two Riccati identities in (3.10).

Glue the right and left solutions with the same value v at zero. Figure 1 records the boundary signs and the normalization in this gluing identity. The resulting function is continuous, and its derivative has jump -(S+T)v. On the full line it therefore satisfies

Using the Fourier transform with forward kernel e^{-isx} and inverting at zero gives

This inversion may first be taken in tempered distributions; the transformed function is integrable because the inverse symbol is O(s^{-2}), so evaluation at zero is legitimate. Since R(t)>0, the second identity in (3.10) follows.

Finally the right solutions satisfy \psi'=i(B+iS)\psi. If (B+iS)v=zv with v\ne0, uniqueness for this first-order initial-value problem gives \psi(x)=e^{izx}v. It belongs to L^2(0,\infty;\C^n), hence \operatorname{Im}z>0.□

Proof of Theorem 3.1. The convergence assertions were proved above. Let D\geq0 be an n\times n matrix and define

For a scalar eigenvalue d\geq0, its contribution to F_0 is

The integrand is nonnegative. Tonelli’s theorem and (3.2), applied to each eigenvalue of D, give

For R=R(t)>0, the Hilbert–Schmidt square

is valid without a commutation assumption. Indeed the cross term is \tr(tI+D)^{1/2} because the adjacent factors R^{-1/2}R^{1/2} cancel, while cyclicity puts the remaining quadratic term in the form \tr(R(tI+D)). Using the definitions of P,G,F_0, the same square reads

All three terms on the right are integrable against t^\beta\,dt, by (3.7), (3.19), and (3.23). Hence

Now assume N\geq0 and choose D=N^{1/\sigma}. Since p=1+\sigma and q=p/\sigma, we have

Combining (3.25) with (3.22) gives the exact deficit formula

Both integrands are nonnegative, proving (3.6).□

Proof. We first recall a useful estimate for the Hilbert–Schmidt norm. If h is scalar Lipschitz on an interval containing the spectra of Hermitian matrices X and Y, then

Indeed, in eigenbases (e_i) of X and (v_j) of Y, the (i,j) entries of these two differences are respectively (h(\lambda_i)-h(\nu_j))\langle e_i,v_j\rangle and (\lambda_i-\nu_j)\langle e_i,v_j\rangle. The scalar Lipschitz bound, followed by summation of their squared absolute values, proves (4.6).

Restrict B to a bounded set. For sufficiently large |s|, the spectrum of E_s(B)=-2sB+K^2 lies in [-a|s|,a|s|], with a constant a independent of B. On this interval put

By (4.2),

Here the arguments s^2+y are positive for large |s|, so the derivative formula applies. Since h_s(0)=0 and \|E_s(B)\|_\HS=O(|s|), (4.6) gives

where, uniformly on the chosen bounded set,

For the first bound, include the term f_\delta'(s^2)K^2 in T_s; for the second, use E_s(B_1)-E_s(B_2)=-2s(B_1-B_2). The first term in (4.7) integrates to zero on [-r,r], while 2\beta-2=2\sigma-3<-1 makes both remainder bounds integrable at infinity. On bounded s-intervals, (4.6) and the scalar Lipschitz bound on f_\delta give local Lipschitz continuity directly. This proves the convergence and regularity assertions.

To construct the primitive, choose g_\delta with g_\delta'=f_\delta. The trace differentiation formula is

It follows first for polynomials by cyclicity of the trace and then for g_\delta by polynomial approximation in C^1 on compact intervals containing the relevant spectra. For s\ne0, the function

therefore has differential -\tr((f_\delta(Q_s(B))-f_\delta(s^2)I)H). At s=0 the field is constant and has a linear primitive. It follows that the integral of the negative field around every piecewise smooth closed curve is zero at each finite cutoff in (4.4). Local uniform convergence passes this identity to M_\delta. Path integration on the vector space \Sym_n now defines J_\delta, with (4.5); continuity of M_\delta gives J_\delta\in C^1.□

Proof. Scalar concavity, applied in an orthonormal eigenbasis of Q_s(B), gives

Set a=e^{\mathsf T}Be and z=e^{\mathsf T}K^2e-a^2. Then e^{\mathsf T}Q_s(B)e=(s-a)^2+z. Subtract f_\delta(s^2), integrate over [-r,r], and use (4.10) to obtain (4.11).

If Be_i=ye_i, the vector e_i is also an eigenvector of Q_s(B), with eigenvalue (s-y)^2+k_i^2-y^2. Functional calculus and (4.10) prove (4.12). For (4.13), use SK^2S=K^2 to obtain Q_s(SBS)=SQ_s(B)S, and then conjugate inside (4.4).

On the diagonal subspace, (4.12) and (4.5) give

Integrating from (k_1,\ldots,k_n) to (-k_1,\ldots,-k_n) within this subspace proves (4.14).□

Proof. Write M=M_\delta(B). By (4.11), its positivity and the strict increase of \mu_\delta imply

Consequently

and hence Q_s(B)\geq0 for all s\in\R. This is positivity also on \C^n: for z=u+iv with real u,v, symmetry gives z^*Q_s(B)z=u^{\mathsf T}Q_s(B)u+v^{\mathsf T}Q_s(B)v. The same directional estimate, together with (4.9), gives

The last inequality uses (a+b)^\sigma\leq a^\sigma+b^\sigma for a,b\geq0 and 0<\sigma<1.

Put

Then X_s=Y_s+V=Q_s(B)+\delta I\geq\delta I satisfies the hypothesis of Theorem 3.1. Let R(t) and N be its resolvent average and integrated matrix, as defined in (3.4) and (3.5). We claim that

Diagonalize B and apply (4.10) with z=0 to each diagonal entry. This allows f_\delta(s^2)I in (4.4) to be replaced by f_\delta(Y_s). Both Q_s(B) and Y_s are positive semidefinite, so (4.1) expresses the resulting difference as

For fixed B and \delta>0, positivity and quadratic growth in s give a common lower bound c(1+t+s^2)I for the two matrices being inverted. Their difference before inversion is the constant matrix D. The resolvent identity therefore bounds the norm of the weighted difference by

This is integrable on (0,\infty)\times\R: integration in s leaves a constant times t^\beta(1+t)^{-3/2}, integrable because -1<\beta<1/2. Fubini’s theorem now yields

Subtract this identity from the definition of N and apply (3.2) with y=\delta. This proves (4.17).

In particular N\geq0, so Theorem 3.1 applies and gives

Because N=M+\delta^\sigma I, the matrices N and M have a common eigenbasis and \tr(N^q)\geq\tr(M^q). Equation (4.16) bounds \tr M by n\kappa^{2\sigma}, while V=K^2-B^2+\delta I\leq(\kappa^2+\delta)I bounds \tr V by n(\kappa^2+\delta). Substitution proves (4.15).□

Proof. Fix \delta>0. We use the continuous representative of U, available because the interval is bounded and U\in H^1. Choose a smooth nonnegative function \chi on \Sym_n, of compact support, which is one on \{B:\|B\|_{\mathrm{op}}\leq\kappa\} and is invariant under conjugation by diagonal sign matrices. Such a cutoff can be obtained from any cutoff of a neighborhood of this compact set by averaging over the finitely many sign matrices. By Lemma 4.1, the field

is bounded and globally Lipschitz.

The endpoint map. For fixed \varepsilon>0, C\in\mathcal L, and 0\leq\theta\leq1, solve the initial-value problem

and define

The bounded Lipschitz field and the continuous forcing give a unique C^1 solution on the whole interval. Continuous dependence on the parameters makes \Phi continuous. A zero of \Phi(\cdot,1) gives the required endpoints; we will then show that its path remains where \chi=1.

Let \mathcal G be the group of diagonal matrices whose entries are \pm1. It acts on \mathcal L by C\mapsto SC and on \Sym_n by B\mapsto SBS. The field is equivariant under these actions by (4.13), and SKS=K. Uniqueness in (5.3) therefore gives

All zeros of this endpoint map lie in a bounded set. Indeed, integrating (5.3) at a zero and using the orthonormality of the rows of U yields

Consequently, for 0<\varepsilon\leq1,

For each fixed \varepsilon>0, the same calculation gives a finite bound. Thus the joint zero set \{(C,\theta):\Phi(C,\theta)=0\} is compact.

The sign action is free at every such zero: only the identity sign matrix fixes C. To prove this, it suffices to show that no row of C vanishes. If row i vanished, the reflection S_i that changes only its sign would fix C. Equivariance and uniqueness would then force S_iB(x)S_i=B(x), so that B(x)e_i=y(x)e_i for a scalar function y. The coordinate identity (4.12) would give

Here \chi(B(x)) is a fixed continuous coefficient. Since \mu_\delta is locally Lipschitz and \mu_\delta(0)=0, uniqueness for this scalar equation gives y\equiv k_i. This contradicts the terminal condition y(x_+)=-k_i.

At \theta=0 the endpoint map is explicit:

Its zeros in \mathcal L are precisely

For example, the first row of CC^{\mathsf T}=2\varepsilon K determines d_1 and forces the remaining entries in the first column of C to vanish; induction gives the claim. These 2^n zeros form one orbit under \mathcal G. Each is a regular zero, meaning that the derivative with respect to C is onto \Sym_n. In fact, for a lower triangular variation H, this derivative has diagonal entries -2\varepsilon^{-1}d_iH_{ii} and entries -\varepsilon^{-1}d_jH_{ij} when i>j. It is therefore an isomorphism between vector spaces of dimension n(n+1)/2.

Continuation after identifying row signs. We have a compact zero set, lying entirely where the sign action is free, and a single regular zero at the initial parameter after taking the quotient by that action. We now show that these properties force a zero at \theta=1. The details below replace the continuous endpoint map by a smooth one, so no differentiability of the ODE flow with respect to C is required.

Suppose, to the contrary, that \Phi(C,1) never vanishes. Compactness allows us to choose a bounded invariant open set O\subset\mathcal L containing the C-projection of every zero, with \overline O contained in the open set where all rows are nonzero. There is then a positive lower bound for \|\Phi\|_{\HS} on

Reparametrize \theta by a smooth map \rho:[0,1]\to[0,1] that equals zero near 0 and has \rho(1)=1. The resulting map F(C,\theta)=\Phi(C,\rho(\theta)) agrees with the smooth map (5.7) on an initial collar, that is, for all 0\leq\theta\leq a with some a>0. It retains a positive gap on (5.8).

Approximate F uniformly on \overline O\times[0,1] by a smooth equivariant map \Psi that still equals (5.7) on a smaller initial collar. Here is a direct construction. Extend F constantly in \theta outside [0,1], smooth by convolution in the finite-dimensional variables, and use a smooth function of \theta to splice the approximation to F(C,0) inside its initial collar. Averaging the result in the form

makes it equivariant. Uniform approximation can be as close as needed, so the positive gap persists.

We next make zero a regular value in the interior by a small equivariant perturbation. The freeness of the action on \overline O is exactly what permits this step. Around any C_0\in\overline O, choose a ball whose translates by distinct sign matrices are disjoint, and a smooth bump function b supported in that ball with b(C_0)=1. For H\in\Sym_n, the map

is smooth and equivariant, and V_H(C_0)=H. Taking H in a basis of \Sym_n gives maps whose values span the target near C_0. A finite covering of \overline O therefore supplies smooth equivariant maps V_1,\ldots,V_N whose values span \Sym_n at every point of \overline O. The action on the target need not be free: the disjoint supports in the domain allow an arbitrary target value at a representative of each orbit.

Choose a smooth function h:[0,1]\to[0,\infty) that vanishes near 0 and is positive thereafter, with its zero set contained in the collar where \Psi(C,\theta)=\Phi(C,0). For a parameter \alpha=(\alpha_1,\ldots,\alpha_N)\in\R^N, put

Restrict \alpha to a sufficiently small open ball about zero so that the gap on (5.8) remains. At an interior zero of the map of all three variables (C,\theta,\alpha), its differential is onto: if h(\theta)>0, the parameter directions span the target; if h(\theta)=0, the derivative in C is the isomorphism already computed for (5.7).

The interior zero set of this parameter family is thus a smooth manifold. Apply Sard’s theorem [13] to its projection onto the parameter ball. For a regular value \alpha of this projection, zero is a regular value of (C,\theta)\mapsto\Psi_\alpha(C,\theta). To see the implication, write the differential of the parameter family as (L_1,L_2) in the (C,\theta) and \alpha directions. Surjectivity of the differential of the projection means that, for every w, there is a v with L_1v+L_2w=0. Thus \operatorname{ran}L_2\subset\operatorname{ran}L_1; since (L_1,L_2) is onto, L_1 is onto as well. Sard’s theorem supplies such parameters arbitrarily close to zero.

For this choice of \alpha, the zero set of \Psi_\alpha in O\times[0,1] is a compact smooth one-dimensional manifold with boundary. Its dimension follows from \dim\mathcal L=\dim\Sym_n. There are no zeros on (5.8), and near \theta=0 the map is the initial map, independent of \theta, with invertible derivative in C. The boundary therefore consists precisely of the 2^n initial zeros.

The sign group acts freely on this manifold. Its quotient is again a compact one-dimensional manifold with boundary: around each point, choose a neighborhood disjoint from its nontrivial translates, so that the quotient has the same local interval or half-interval coordinate. All initial zeros belong to one orbit, so the quotient has exactly one boundary point. This is impossible. Every compact one-dimensional manifold is a finite disjoint union of circles and closed intervals, and hence has an even number of boundary points. The contradiction proves that the original map \Phi(\cdot,1) has a zero. This is the usual regular-value proof of invariance of parity, applied after quotienting by row signs; see also [10, Sections 2–4].

Confinement of the path. Let C and B be given by such a zero. For any fixed real unit vector e, set y(x)=e^{\mathsf T}B(x)e. By (4.11), whenever |y(x)|>\kappa,

Since \chi\geq0 and (CU)(CU)^{\mathsf T}\geq0, the differential equation gives y'(x)\leq0 on this set. Both endpoint values of y lie in [-\kappa,\kappa]. A connected interval on which y>\kappa cannot begin at the level \kappa while y is nonincreasing. Likewise, an interval on which y<-\kappa cannot end at the level -\kappa while y is nonincreasing. The first endpoint excludes the former excursion, and the second endpoint excludes the latter. Thus |e^{\mathsf T}B(x)e|\leq\kappa for every unit vector e and every x.

It follows that \|B(x)\|_{\mathrm{op}}\leq\kappa throughout the interval, so \chi(B(x))=1 and (5.3) becomes (5.1). Finally, (5.6) gives the uniform bound on C_\varepsilon in (5.2) for fixed \delta. This completes the construction.□

Proof of Theorem 2.1. Fix \delta>0 and, for each 0<\varepsilon\leq1, choose C_\varepsilon, A_\varepsilon=C_\varepsilon U, and B_\varepsilon as in Proposition 5.1. Temporarily suppress the subscript \varepsilon. Since A\in H^1_0 and B\in C^1, the function

is absolutely continuous. The primitive identity (4.5) and the path equation give, almost everywhere,

The last line is the Hilbert–Schmidt Cauchy–Schwarz inequality followed by 2ab\leq a^2+b^2. Since A vanishes at both endpoints, integration and the definition of the action yield

We next bound the integral uniformly as \varepsilon\downarrow0. The bounds in (5.2), together with the boundedness of U, place all pairs

in one compact set \mathcal K_\delta\subset \R^{n\times d}\times\Sym_n. On this set define the continuous functions

If r_\delta(A,B)=0, then M_\delta(B)=AA^{\mathsf T}\geq0, and Proposition 4.3 gives F(A,B)\leq E_\delta. This is the only point at which positivity of the field is needed; the constructed paths themselves need not have this property.

Compactness now implies

Indeed, a failure would give \eta>0, a sequence \varepsilon_j\downarrow0, and points (A_j,B_j)\in\mathcal K_\delta at which the expression exceeds E_\delta+\eta. Boundedness of F then forces r_\delta(A_j,B_j)=O(\varepsilon_j). A convergent subsequence has limit (A_*,B_*) with r_\delta(A_*,B_*)=0 and, by continuity, F(A_*,B_*)\geq E_\delta+\eta. This contradicts the preceding comparison.

Apply (6.2) in (), bound \mathcal E(A_\varepsilon) by the supremum over lower triangular C, and let \varepsilon\downarrow0. We obtain

All compactness estimates so far were for fixed \delta. This is sufficient: the supremum on the right is independent of \delta, and the remaining limit concerns only the explicit endpoint values and the error term.

By (4.14),

For z\geq0, the formula \mu_\delta(z)=(z+\delta)^\sigma-\delta^\sigma gives

Dominated convergence therefore sends the endpoint difference to

The error E_\delta in (4.15) tends to zero. Letting \delta\downarrow0 in (6.3) proves the finite-interval action inequality.□

Proof. Put w(x)=\|W(x)\|_{\op}. Positivity gives w^p\leq\tr(W^p), so w\in L^p(\R). The one-dimensional Sobolev estimate

holds also for vector-valued functions, by applying the scalar estimate to their norm. Hölder’s inequality and interpolation therefore give

The last step is Young’s inequality. Consequently a sufficiently shifted form norm is equivalent to the H^1 norm. The potential form is continuous on H^1, so the form is closed and bounded below.

For compactness, truncate the multiplier both in space and in size:

Restriction to [-R,R] followed by the compact embedding H^1([-R,R];\C^m)\hookrightarrow L^2([-R,R];\C^m) and bounded multiplication shows that T_{R,L} is compact. If T\psi=W^{1/2}\psi, Hölder’s inequality and the same Sobolev embedding imply

The coefficient tends to zero as R,L\to\infty, proving that T is compact.

Finally, fix a>0. If the spectral subspace of H_W in (-\infty,-a] were infinite-dimensional, it would contain an L^2-orthonormal sequence (\psi_j). Semiboundedness and the spectral theorem place this sequence in the form domain with uniformly bounded shifted form norm, hence uniformly bounded H^1 norm. A weakly convergent subsequence in H^1 has limit zero, because an orthonormal sequence converges weakly to zero in L^2. Compactness of T would then give \|T\psi_j\|_2\to0, whereas

a contradiction. Every such spectral subspace is therefore finite-dimensional, which proves the assertion about the negative spectrum.□

Proof of Theorem 1.1. Recall that \sigma=\gamma-\tfrac12, p=1+\sigma and q=1+1/\sigma. We first prove the inequality for real symmetric potentials. If there are no negative eigenvalues, it is immediate. Otherwise fix any finite nonempty list of negative eigenvalues, respecting their multiplicities. Since complex conjugation preserves each eigenspace, we may choose corresponding orthonormal eigenfunctions \phi_1,\ldots,\phi_n that are real-valued.

The action theorem requires functions supported on a bounded interval. Choose real smooth compactly supported approximations to the \phi_i in H^1. Their Gram matrices converge to I_n, and their form matrices converge to the diagonal matrix of the selected eigenvalues, by (7.1). Multiplying each approximating family by the inverse square root of its Gram matrix makes its rows orthonormal without changing their common compact support. The form matrices still converge to the same diagonal matrix. For sufficiently good approximations they are negative definite; diagonalize them by a real orthogonal change of rows.

For one such family, let U\in H^1_0(\Omega;\R^{n\times m}) be the matrix whose rows are the resulting functions u_i, where the bounded interval \Omega contains their supports in its interior. Extend these rows by zero when evaluating the whole-line form. They satisfy

where h_W[\cdot,\cdot] is the polarized form and the numbers -k_i^2 converge, along the approximations, to the selected eigenvalues in increasing order. Set K=\diag(k_1,\ldots,k_n).

Let C be any real lower triangular matrix and write A=CU. Its ith row is a_i=\sum_{j\leq i}C_{ij}u_j, so

Summing over i bounds the quadratic terms of the action. Since AA^{\mathsf T} and A^{\mathsf T}A have the same nonzero eigenvalues,

The right side has a pointwise bound that does not require its two matrices to commute. Indeed, scalar maximization gives, for y,z\geq0,

To apply it, diagonalize an arbitrary positive semidefinite matrix Z in an orthonormal basis (v_j), with eigenvalues z_j. Set y_j=\langle v_j,Wv_j\rangle. Convexity of t\mapsto t^p, applied to the spectral measure of W in v_j, gives y_j^p\leq\langle v_j,W^pv_j\rangle. Hence

Using Z=A^{\mathsf T}A in (7.3), we obtain

All terms are finite for each C: the rows of A are bounded and supported on \overline\Omega, and W is locally integrable.

Theorem 2.1 supplies the reverse bound in terms of the k_i. Writing \mathrm B for Euler’s beta function, the substitution s=k_it gives

We conclude that

The coefficient is exactly the one in the theorem:

Letting the approximations converge proves the bound for the chosen finite list of eigenvalues. The sum of all negative eigenvalue moments is the supremum of these finite sums. Thus the same bound holds for \Tr(H_W)_-^\gamma, and in particular this trace is finite.

For a complex Hermitian potential, write W=E+iF with real matrices E,F. Hermiticity means E^{\mathsf T}=E and F^{\mathsf T}=-F, so

is real symmetric. The constant complex unitary

satisfies \mathcal U\widetilde W\mathcal U^* =W\oplus\overline W. It follows that \widetilde W\geq0 and \tr(\widetilde W^p)=2\tr(W^p). Since \mathcal U also commutes with differentiation, the same change of coordinates gives

Complex conjugation intertwines the two summands and preserves eigenvalue multiplicities. The spectral moment for \widetilde W is consequently twice that for W. Applying the real inequality in dimension 2m and dividing both sides by two proves the asserted inequality for all Hermitian W.

It remains to verify sharpness. Put r=1/\sigma>1 and consider the scalar potential and function

The identity \psi'/\psi=-\tanh(rx) gives H_V\psi=-\psi, and \psi\in L^2(\R). We determine the full negative spectrum by the standard factorization method; see [2, Sections II–III] for its use in Lieb–Thirring inequalities. Define Q=\partial_x+\tanh(rx) on H^1(\R). Direct multiplication yields

Thus H_V\geq-1, and its eigenspace at -1 is \ker Q, which is spanned by \psi. If \lambda\in(-1,0) were an eigenvalue with eigenfunction u, then Qu\ne0. The bounded potential gives u\in H^2, so Qu\in L^2. The intertwining identity (QQ^*-1)Qu=\lambda Qu holds in distributions. Boundedness of the partner potential then implies Qu\in H^2, so this identity contradicts nonnegativity. Therefore -1 is the only negative eigenvalue and is simple. Finally,

The scalar inequality is an equality for V. Embedding V in one diagonal channel and setting the other channels to zero gives equality in every matrix dimension, proving that C_\gamma is optimal.□