A Liouville Theorem for the Real k-Hessian Equation with a One-Sided Growth Condition

Introduction

We study the rigidity of entire solutions to the real $k$-Hessian equation

Here $\sigma_j$ denotes the $j$-th elementary symmetric polynomial, evaluated at the eigenvalues of the real Hessian $D^2u$. We use $\sigma_0=1$ and $\sigma_j=0$ when $j$ exceeds the number of variables.

The operators $\sigma_k$ connect the Laplacian and the real Monge–Ampère operator. Their elliptic branches are described by the cones

We write $H\in\Gamma_j$ when the eigenvalues of a symmetric matrix $H$ belong to this cone. For smooth solutions of (1.1), we use $k$-admissible to mean $D^2u\in\Gamma_k$ at every point.

The weak admissibility conventions are specified in Sections 6.1 and 7.1.

A natural hypothesis is positive quadratic lower growth:

This condition makes large sublevel sets bounded. It does not directly bound the oscillation of a rescaled solution on a fixed Euclidean ball, which is needed in the usual interior-estimate argument.

Earlier estimates and rigidity theorems

Liouville theorems for nonlinear partial differential equations have been studied extensively. For Hessian equations, a central question is quadratic rigidity under growth or convexity assumptions. Bao, Chen, Guan, and Ji proved quadratic rigidity for strictly convex entire solutions under (1.3) [1 Theorem 3.2]. For $2 \le k \le n - 1$, Li, Ren, and Wang extended this result to $(k + 1)$-admissible solutions using a Pogorelov estimate [11].

Chu and Dinew treated a general class of Hessian equations through weighted gradient and Hessian estimates. For $k$-admissible solutions of (1.1), their result applies under (1.3) and the additional bound $\sigma_{k+1}(D^2u) \ge-A$ [6 Corollary 1.3].

Within the natural class of $k$-admissible solutions, Wang and Bao proved rigidity under positive two-sided quadratic growth [13 Theorem 1.1]. Their proof combines interior estimates with rescaling and Hölder continuity of the Hessian.

Chen and Xiang proved that every 2-admissible solution $u \in C^4(\mathbb{R}^3)$ of $\sigma_2(D^2u) = 1$ satisfying (1.3) is a quadratic polynomial [4 Theorem 1.4].

Bin Wang proved a lower-growth rigidity theorem for $k$-admissible solutions when $2k > n$ [12 Theorem 1.2]. His argument establishes a Pogorelov estimate independent of the gradient bound.

For the complex Monge–Ampère equation $\det(v_{i\bar{j}}) = 1$, Chu and Liu proved quadratic rigidity for smooth entire plurisubharmonic solutions under positive two-sided quadratic growth [7 Theorem 1.2]. At the same time, using a different method, Hu and Sheng established interior $C^2$ estimates and proved quadratic rigidity for locally bounded entire plurisubharmonic Bedford–Taylor solutions under weak quadratic growth conditions [8 Theorems 1.1 and 1.3]. In particular, the Liouville theorem in that paper permits a positive quadratic lower bound together with an upper bound of order $(1 + |z|^2)\log(e + |z|^2)$. We expect that this upper growth bound can be further weakened.

Motivated by these developments, particularly the recent work of Wang [12], we ask whether the upper growth bound can be removed entirely for the real $k$-Hessian equation, so that (1.3) alone implies quadratic rigidity.

The main results

We first state the Liouville theorem, followed by the interior estimate used in its proof.

The following estimate is essential to the proof of the Liouville theorem. Its constants depend only on the dimension, the order of the equation, and the coefficient of the quadratic lower bound.

and, for some constant $B$,

Then

The Liouville theorem and the interior estimate extend to viscosity solutions, as stated in the following two theorems.

Remark 1.5 (Pogorelov’s example and the growth condition). Pogorelov’s example shows that viscosity solutions can have an unbounded Hessian when the growth hypothesis is absent; see [10 Section 4.4]. We use the construction in $k$ variables and extend it independently of the remaining $n-k$ variables. For $3\le k\le n$, write

Choose a positive smooth solution $h$ on $(-T,T)$, for some $T>0$, of

The function

is defined on the slab $\mathbb{R}^{k-1}\times(-T,T)\times\mathbb{R}^{n-k}$. For $r=|y|>0$, differentiation gives

Here $k\alpha_k-2k+2=0$. The $k-2$ angular eigenvalues are positive, and the radial–$t$ block has positive radial entry and determinant. Thus the active $k\times k$ Hessian block is positive definite.

The remaining $n-k$ eigenvalues vanish, so $\sigma_k(D^2P)=1$ off $\{y=0\}$. Since $\alpha_k>1$, the gradient extends continuously across this set; restriction to lines therefore shows that $P$ is convex.

We verify the viscosity equation on $\{y=0\}$, where $P$ is not $C^2$.

At $y=0$, no $C^2$ upper touching test exists because $1<\alpha_k<2$. A lower touching test has Hessian $A$ with $A|_S\leq0$ on the flat subspace $S=\{y=0\}$. If $A\in\Gamma_k$ and $\sigma_k(A)>0$, Maclaurin’s inequalities give $A\in\Gamma_k$.

Restrict successively to $k-1$ hyperplanes within the preceding subspace, each containing $S$. Equation (62) gives $A|_S\in\Gamma_1^{(n-k+1)}$, so $\operatorname{tr}(A|_S)>0$, contradicting $A|_S\leq0$.

Thus every admissible lower test has $\sigma_k(A)=0$. This verifies the supersolution inequality on the singular set.

The growth condition in Theorem 1.4 excludes this example. Indeed, take $0<R<T$ and $x_*=(0,3R/4,0)\in B_R$. Since $P(0)=P(x_*)=0$, the assumed lower bound would imply

Thus the quantitative lower bound prevents the flat singular axis of Pogorelov’s example. This example is defined on a slab and concerns local regularity.

The results also extend to admissible weak solutions in the sense of Trudinger [15]. We recall his definition in § 7.1.

Organization of the paper

§ 2 constructs the admissible distance power, and § 3 proves the uniform boundary estimate. § 4 proves Theorem 1.2.

§ 5 applies this estimate and the Evans–Krylov theorem to prove Theorem 1.1. § 6 proves the viscosity results, Theorems 1.3 and 1.4. § 7 treats admissible weak solutions and proves Theorems 1.6 and 1.7. Appendix § 8 contains the two algebraic lemmas.

Admissible powers of the distance function

We construct a function that vanishes on the boundary and whose Hessian belongs to the closed admissible cone. At smooth points of the distance function, this follows from Lemma A.2.

For $2 \le k \le n$, fix

These constants satisfy $0 < \theta< 1$ and $0 < \beta< \alpha< 1$.

Let $\Omega\subset\mathbb{R}^{n}$ be a nonempty bounded smooth domain whose outward principal curvatures belong to $\Gamma_{k-1}^{(n-1)}$. We take the outward principal curvatures to be positive on spheres.

Set

A function $f$ is locally semiconvex if, on a convex neighborhood of each point, $f(x) + \frac{C}{2}|x|^{2}$ is convex for some $C \ge0$. Local semiconcavity means that $-f$ is locally semiconvex.

The spectral matrix cone $\Gamma_j$ is convex and is preserved under addition of positive-semidefinite matrices. The same statements hold for $\overline{\Gamma}_j$. These standard properties are summarized in [12 Section 2].

At smooth points of $d$, the normal eigenvalue of $D^{2}v$ is positive. Lemma A.2 shows that it compensates for the possible negative tangential eigenvalues, with a choice of $\alpha$ depending only on $n$ and $k$.

Proof. The distance $d$ is Lipschitz and hence differentiable almost everywhere. If $y$ is a nearest boundary point to such an $x$, then $d(z) \le|z-y|$ with equality at $x$. Differentiation gives

Thus the nearest boundary point is unique. The focal values of the smooth normal map have measure zero by Sard’s theorem, so it remains to compute at points with a unique, nonfocal nearest boundary point.

Let $\nu$ be the outward unit normal. The minimizing segment is normal and points inward, so

Write $\Psi(y,s)=y-s\nu(y)$, and let $e_i$ be a principal direction at $y$. With our sign convention for the curvatures,

Choose a unit-speed boundary curve $\gamma_i$ through $y$ with $\gamma_i'(0)=e_i$. Since $y$ minimizes the distance to $x$, our curvature convention gives

Nonfocality makes these inequalities strict, so the normal map has a smooth local inverse.

Let $x_m\to x$ and choose any nearest boundary points $y_m$. Every convergent subsequence of $y_m$ has a nearest point to $x$ as its limit, because distance is continuous.

Uniqueness of the nearest point and compactness give $y_m\to y$. Also $d(x_m)\to d(x)$, so the pairs $(y_m,d(x_m))$ remain in the local inverse neighborhood for all sufficiently large $m$.

In this neighborhood, the inverse normal coordinate $s$ equals $d$. Thus $d$ is smooth there, with

Differentiating this local gradient identity in each principal direction gives the eigenvalues of $D^2d$:

To compute the Hessian of $v$, apply the chain rule:

The second term acts only in the normal direction, since $Dd$ is a unit normal. Introduce $a_i=d\kappa_i(y)$ and $\mu_i=a_i/(1-a_i)$. The eigenvalues of $D^2v$ are therefore

Here $a\in\Gamma_{k-1}$ and $a_i<1$. Since $1-\alpha=\theta$, Lemma A.2 gives $D^2v\in\Gamma_k$.

The same lemma gives $\sigma_1(\mu)>0$. Using $d\leq D$ and $\alpha-2<0$, we obtain

The distance $d$ is locally semiconcave in $\Omega$; see [3 Proposition 2.2.2]. We can check the needed semiconvexity of $v$ directly on any convex neighborhood $V\Subset\Omega$.

Choose $a>0$ such that $\operatorname{dist}(V,\partial\Omega)\geq a$. For $y\in\partial\Omega$, write $r=|x-y|$ and $e=(x-y)/r$. Since $r\geq a$, differentiation gives

Hence each function $-|x-y|^\alpha+\frac{1}{2}\alpha a^{\alpha-2}|x|^2$ is convex on $V$. Their supremum is convex as well, and

This proves local semiconvexity, including at points where the nearest boundary point is not unique.

The distributional Hessian of the locally semiconvex function $v$ is a matrix-valued Radon measure; see [3 Theorem 2.3.1]. Its Lebesgue decomposition is

The density $H$ agrees almost everywhere with the classical Hessian computed above. Locally, $v+\frac{C}{2}|x|^2$ is convex for some $C\geq0$, so $D^2v+C I\,dx$ is a positive matrix-valued measure.

The added quadratic changes only the absolutely continuous part. Hence the singular part $M_s$ is positive semidefinite.

Let $\varphi_\rho$ be a nonnegative mollifier of integral one. For $v_\rho= v * \varphi_\rho$, distributional differentiation gives

The first term is a nonnegative average of matrices in $\overline{\Gamma}_k$, so it belongs to $\overline{\Gamma}_k$ by convexity. The second term is positive semidefinite and therefore preserves this cone under addition.

Taking the trace in (37) also gives

These arguments apply whenever the support of the translated mollifier is contained in $\Omega$.

A uniform boundary estimate

We perturb the distance power by a quadratic function. The perturbation makes the barrier strictly admissible, while its trace lower bound supplies the strict differential inequality used in comparison.

The constant is independent of the boundary curvature and its smoothness norms. We prove the proposition in the following two steps.

Since $D^2U \in\Gamma_k$, we have $\Delta U > 0$. The maximum principle and the boundary condition give $U \le0$.

Prepare the comparison function

Fix $x_0 \in\Omega$ and $\varepsilon> 0$. Introduce the barrier

where

On $\partial\Omega$, we have $v = 0$ and $|x-x_0| \le D$. Consequently, $w \le0 = U$ there.

For a symmetric matrix $A \in\overline{\Gamma}_k$, expansion of the symmetric polynomial gives

To see the coefficient, choose $j$ eigenvalue factors from $KA$ in each product defining $\sigma_k$; the remaining $k-j$ factors contribute $2\varepsilon$ and can be chosen in $\binom{n-j}{k-j}$ ways.

Since $A \in\overline{\Gamma}_k$, every $\sigma_j(A)$ in this expansion is nonnegative. Keeping the $j=1$ term gives

Apply Lemma 2.1 to $v_\rho$, and set $w_\rho=Kv_\rho+\varepsilon(|x-x_0|^2-D^2)$. Adding $2\varepsilon I$ makes all the symmetric polynomials of orders $1,\ldots,k$ strictly positive, by the same expansion.

The powers of $D$ and $\varepsilon$ in $K$ cancel those in $(2\varepsilon)^{k-1}c_0$. Thus (2.3) and (3.4) give the fixed strict inequality

Exclude the interior maximum

We prove $w\le U$. Suppose $M:=\max_{\overline{\Omega}}(w-U)>0$. Continuity gives $\delta>0$ such that $w-U<M/4$ whenever $d\le2\delta$.

The compact set $K_\delta=\{x\in\Omega:d(x)\ge\delta\}$ contains a point $x_M$ where $w-U=M$. Every point of $\partial K_\delta$ satisfies $d=\delta$, so the collar estimate applies there.

Choose $\rho<\delta/2$ so small that $\lVert w_\rho-w\rVert_{L^\infty(K_\delta)}<M/4$.

At $x_M$ we have $w_\rho-U>3M/4$, whereas on $\partial K_\delta$ we have $w_\rho-U<M/2$. Therefore its maximum on $K_\delta$ occurs at an interior point, where the mollified barrier is smooth.

At that point, $D^2w_\rho\le D^2U$. Monotonicity of $\sigma_k$ on $\Gamma_k$ and (3.7) give

This contradiction excludes a positive maximum and proves $w\le U$.

Evaluating at $x_0$ yields

We balance the two terms by choosing $\varepsilon=L_{n,k}^{1/k}(d(x_0)/D)^{\alpha/k}$. Since $\beta=\alpha/k$, this choice satisfies

Substitution into (3.9) proves (3.2) with $C_0=2L_{n,k}^{1/k}$. The point $x_0$ was arbitrary.

Proof of the interior estimate

We first convert Proposition 3.1 into an interior estimate for the zero-boundary problem. We then apply that estimate to a sublevel component of the solution in Theorem 1.2.

Estimate the solution away from the boundary

and $0 < \delta< 1$,

whenever the set in the supremum is nonempty.

Proof. Translation and the rescaling $U(Dy)/D^2$ reduce the proof to $D=1$. The Hessian is unchanged by this scaling, while the gradient is divided by $D$ and the function by $D^2$.

Proposition 3.1 then gives

Fix $\eta>0$. If $-U(x)\ge\eta/4$, the boundary estimate forces $C_0d(x)^\beta\ge\eta/4$. Solving for $d(x)$ gives

For the constant right-hand side, all derivatives vanish, so condition (3.13) of [5] holds with $h=0$. Apply their interior gradient estimate [CW01, Theorem 3.2]} on balls of radius $r_\eta/2$ centered in this sublevel.

Together with (4.2), it gives

The balls lie inside $\Omega$ by (4.3). Their radius and the oscillation of $U$ depend only on $n,k,\eta$, which explains the uniform constant in (4.4).

If the sublevel is empty, the corresponding estimate is vacuous.

Choose $s\in(\eta/4,\eta/2)$ such that $-s$ is a regular value of $U$, and put

The set $\Omega_s$ lies in $\{-U\ge\eta/4\}$, so (4.3) places it a positive distance from $\partial\Omega$. Regularity of the chosen value makes its boundary smooth.

Adding $s$ leaves the Hessian unchanged. Thus $V$ is admissible, solves the same equation, and vanishes on $\partial\Omega_s$.

The regular level is $(k-1)$-convex, by the tangential Hessian identity (62) below. Equations (4.2) and (4.4) bound the $C^1$ norm of $V$.

Since $\Omega_s\Subset\Omega$, the function $V$ is smooth on a neighborhood of $\overline{\Omega_s}$. The right-hand side is identically one, so the remaining hypotheses of the estimate are satisfied.

Apply the zero-boundary estimate of Chou and Wang [5 Theorem 1.5] on each component of $\Omega_s$:

The constant in their estimate is independent of the domain. For a point with $-U\ge\eta$, the choice $s<\eta/2$ gives

Dividing (4.6) by this positive fourth power, and using $D^2V=D^2U$, gives

To express this bound in terms of distance, fix $x \in\Omega$ and $0 < r < d(x)$. On $B_r(x)$ introduce

All eigenvalues of $D^2q$ equal $2a_{n,k}>0$, and hence

On $\partial B_r(x)$, we have $U \le0=q$. Both functions are admissible solutions of the same equation, so the comparison principle gives $U \le q$ throughout the ball.

Evaluate at $x$ and let $r \uparrow d(x)$. This yields

Taking $\eta=a_{n,k}\delta^2$ in (53) bounds the Hessian on $\{d\ge\delta\}$.

The bound for $U$ follows from (48). The gradient bound follows from (50), because (55) places $\{d\ge\delta\}$ inside $\{-U\ge\eta\}$.

Scaling back proves (47).

Select the admissible branch

Let $u$ satisfy the hypotheses of Theorem 1.2. Introduce the rescaled function $u_R(y)=R^{-2}u(Ry)$ on $B_1$. Its derivatives satisfy

Thus the equation is unchanged, and the lower bound becomes $u_R(y)\ge b|y|^2-B/R^2$. The hypothesis on $B$ gives $0\le B/R^2\le b/4$.

It therefore suffices to prove (9) when $R=1$. In this normalization, $u(0)=0$, $u(x)\ge b|x|^2-B$, and $0\le B\le b/4$.

The growth bound gives

Choose $r_0\in(1/2,1)$. The quadratic lower bound is positive on $B_1\setminus\overline{B}_{r_0}$, whereas $u(0)=0$. A minimum on $\overline{B}_{r_0}$ is therefore also a minimum in $B_1$.

At such a point $x_*\in B_1$, the second derivative test gives

Since $\sigma_k(D^2u(x_*))=1$, at least $k$ of its nonnegative eigenvalues are positive. Hence $\sigma_j(D^2u(x_*))>0$ for $1\le j\le k$, so $D^2u(x_*)\in\Gamma_k$.

We now use a continuity argument. Fix $x\in B_1$ and follow the Hessian along the segment from $x_*$ to $x$:

Define

Since $H(0)\in\Gamma_k$, we have $0\in G$. If $t_0\in G$, then every number $\sigma_j(H(t_0))$, $1\le j\le k$, is positive. Continuity preserves these finitely many strict inequalities for $t$ sufficiently close to $t_0$, so $G$ is relatively open. To prove closedness, we use Maclaurin’s inequalities: for $\lambda\in\Gamma_k$,

Iterating these inequalities and using $\sigma_k(H(t))=1$, we obtain

The lower bounds in (61) are independent of $t$. Let $t_m \in G$ converge to $t \in[0,1]$. Continuity gives

The equation also gives $\sigma_k(H(t))=1$. Thus all defining inequalities of $\Gamma_k$ remain strict at $H(t)$, which proves $t\in G$ and the relative closedness of $G$.

Since $[0,1]$ is connected and $G$ is nonempty, open, and closed in $[0,1]$, we conclude that $G=[0,1]$. In particular, $D^2u(x)=H(1)\in\Gamma_k$.

As $x\in B_1$ was arbitrary, $u$ is admissible throughout the ball. We can now apply admissible comparison and the interior estimates.

Describe the regular sublevel

We first explain the curvature of an admissible level set. For a symmetric matrix $H$ and $0\le j\le n-1$, its $j$-th Newton tensor is the symmetric matrix $T_j(H)$ defined by

If $H\in\Gamma_k$ and $E=\nu^\perp$ for a unit vector $\nu$, then

To see the equality, choose orthonormal coordinates with last vector $\nu$. Differentiating $\sigma_{j+1}(H)$ with respect to $H_{nn}$ leaves the sum of all $j$-dimensional principal minors of $H|_E$. Since $H\in\Gamma_k\subset\Gamma_{j+1}$, ellipticity makes $T_j(H)$ positive definite. Evaluating its quadratic form on $\nu$ proves positivity in (62).

By Sard’s theorem, choose a regular value $t\in(b/4,b/2)$. Since $u(0)=0<t$, the set $\{u<t\}$ has a component $\Omega_t$ containing the origin.

For $x\in\Omega_t$, the lower bound gives $b|x|^2-B\le u(x)<t$. Using $t<b/2$ and $B\le b/4$, we obtain

Its boundary is smooth and compactly contained in $B_1$. The outward normal and second fundamental form are

The second identity follows by differentiating $\nu=Du/|Du|$ in tangent directions. The term from differentiating $|Du|^{-1}$ vanishes when paired with a tangent vector.

Equation (62) places the tangential Hessian in $\Gamma_{k-1}$. Division by the positive number $|Du|$ preserves this cone, so the principal curvatures belong to $\Gamma_{k-1}$. The function $U=u-t$ is admissible, satisfies $\sigma_k(D^2U)=1$, and vanishes on $\partial\Omega_t$. It therefore satisfies the hypotheses of Proposition 3.1 and Lemma 4.1.

Apply the domain estimate

Set $d_t(x)=\operatorname{dist}(x,\partial\Omega_t)$. At the origin, the boundary estimate and (63) imply

Define constants depending only on $n,k,b$ by

Then $d_t(0)\ge\gamma$, so $B_\gamma\subset\Omega_t$. Indeed, a segment from the origin to a point outside $\Omega_t$ must first meet $\partial\Omega_t$, at distance at least $d_t(0)$.

Distance to a fixed closed set is 1-Lipschitz. Consequently, for $x\in B_\rho$ we have

Apply Lemma 4.1 with $\delta=\rho/\sqrt{3}$. Since $D_t\le\sqrt{3}$ and $u=U+t$ with $t<b/2$, we obtain

Apply (68) to $u_R$ and use the derivative identities from § 4.2. The three terms become $R^{-2}|u|$, $R^{-1}|Du|$, and $|D^2u|$ on $B_{\rho R}$, which proves Theorem 1.2.

Proof of the Liouville theorem

We apply Theorem 1.2 on expanding balls. The resulting Hessian bound makes the equation uniformly elliptic. The Evans–Krylov estimate then forces the Hessian to be constant.

Obtain a global Hessian bound

Let $u$ satisfy the hypotheses of Theorem 1.1. Normalize its value at the origin by setting

The last inequality follows from (5) at $x=0$. For every sufficiently large $R$, we have

Theorem 1.2 therefore gives

For each fixed $x\in\mathbb{R}^n$, choose $R$ large enough that $\widetilde{B}\le bR^2/4$ and $|x|<\rho R$. The first estimate then bounds $|D^2u(x)|$ by the same constant $M$.

The branch argument in § 4.2 applies on each of these balls. Since the balls exhaust $\mathbb{R}^n$, it also gives $D^2u\in\Gamma_k$ everywhere.

Rescale the uniformly elliptic equation

Set $F(H)=\sigma_k(H)^{1/k}$ and introduce the set containing all Hessian values:

The Maclaurin inequalities (4.14) extend by continuity to $\overline{\Gamma}_k$. Since $\sigma_k(H)=1$ on $\mathcal{K}$, they give $\sigma_j(H)>0$ for every $H\in\mathcal{K}$ and $1\leq j<k$.

Thus $\mathcal{K}$ is a compact subset of $\Gamma_k$. The derivative matrix is

It is positive definite on $\Gamma_k$ and continuous on $\mathcal{K}$. Compactness therefore gives constants $0<\lambda\leq\Lambda$, depending only on $n$, $k$, $M$, such that $\lambda I\leq(F^{ij}(H))\leq\Lambda I$.

For completeness, extend the operator to all symmetric matrices by

The matrices $A$ and $(F^{ij}(A))$ are bounded on $\mathcal{K}$, so this infimum is finite for every $H$. An infimum of affine functions is concave.

For every positive-semidefinite matrix $P$, each affine function in (5.5) increases by a number between $\lambda\operatorname{tr}P$ and $\Lambda\operatorname{tr}P$.

Taking infima preserves these bounds, so

Thus $\widetilde{F}$ is uniformly elliptic on all symmetric matrices. For $H,A\in\mathcal{K}$, concavity of $F$ gives

Every expression in the infimum is therefore at least one. Choosing $A=H$ attains one, which proves $\widetilde{F}(H)=1$ on $\mathcal{K}$.

Subtract the affine Taylor polynomial at the origin and rescale:

The normalization gives $w_R(0)=0$ and $Dw_R(0)=0$. Also, $D^2w_R(y)=D^2u(Ry)$, so (5.3) bounds its Hessian independently of $R$.

Integrating along the segment from $0$ to $y$ yields

Hence $\|w_R\|_{C^{1,1}(B_1)}\leq C(n)M$. Its Hessian takes values in $\mathcal{K}$, and therefore $\widetilde{F}(D^2w_R)=1$.

To use the convex formulation of [2 Theorem 1], consider $-w_R$ and the operator $H\mapsto1-\widetilde{F}(-H)$. This operator is convex and uniformly elliptic.

Its value on $D^2(-w_R)$ is zero. The cited Evans–Krylov estimate therefore gives constants $\zeta\in(0,1)$ and $C>0$, independent of $R$, such that

For fixed $x,y\in\mathbb{R}^n$, both points lie in $B_{R/2}$ for all sufficiently large $R$. The second estimate then gives

Letting $R \to\infty$ makes the right-hand side zero. Thus $D^2u(x) = D^2u(y)$ for every pair of points. Write this constant Hessian as $A$. Integration gives $u(x) = \frac{1}{2}x^T A x + \ell\cdot x + c$, and the equation gives $\sigma_k(A) = 1$.

Finally, take $x = re$ in (1.5), where $|e| = 1$. Substitution of the quadratic expression and division by $r^2$ give

Letting $r \to\infty$ removes the affine and constant terms, yielding

Hence $A \ge2bI$, proving Theorem 1.1.

Viscosity solutions

We specify the viscosity formulation and prove automatic admissibility. Approximation by smooth solutions then gives the interior estimate in Theorem 1.4 and the Liouville theorem in Theorem 1.3.

The viscosity formulation

Let $\Omega\subset\mathbb{R}^n$ be open. A proper upper semicontinuous function $u : \Omega\to[-\infty,\infty)$ is $k$-convex if every $C^2$ function touching $u$ from above has Hessian in $\overline{\Gamma}_k$ at the contact point.

Here proper means that $u$ is not identically $-\infty$ on any component. We denote this class by $\Phi_k(\Omega)$ and use its upper semicontinuous representatives throughout. For weak functions, the admissibility condition means membership in $\Phi_k(\Omega)$.

A continuous function $u$ is a viscosity solution of $\sigma_k(D^2u) = 1$ if every $C^2$ upper touching test $\phi$ satisfies

and every $C^2$ lower touching test $\phi$ with $D^2\phi(x) \in\overline{\Gamma}_k$ satisfies

Touching from above means that $u - \phi$ has a local maximum zero at the contact point; touching from below means a local minimum zero. The lower tests are taken on the elliptic branch.

The following lemma shows that the upper-test inequality already forces admissibility, giving the usual formulation in [16 Section 2].

Proof. Fix an upper touching test $\phi$ at $x_0$, and put $A = D^2\phi(x_0)$. Choose $T > 0$ such that $A + TI > 0$, and define

The matrix $A + TI$ is positive definite, so $T \in G$. For $t_0 \in G$, continuity keeps all the finitely many inequalities $\sigma_j(A + tI) > 0$ valid when $t$ is sufficiently close to $t_0$. Thus $G$ is relatively open in $[0,T]$.

For every $t \in[0,T]$, introduce

The added term is nonnegative and vanishes at $x_0$, so $\phi_t$ still touches $u$ from above. Applying (77) to this test gives $\sigma_k(A+tI)\ge1$ for the whole interval.

For $t\in G$, Maclaurin’s inequality (60) yields

To prove closedness, take $t_m\in G$ with $t_m\to t\in[0,T]$. Continuity of each symmetric polynomial preserves the positive lower bounds in (80).

Together with $\sigma_k(A+tI)\ge1$, these bounds imply $A+tI\in\Gamma_k$. Hence $t\in G$, and $G$ is relatively closed.

Connectedness gives $G=[0,T]$, so $A\in\Gamma_k$. Since the upper touching test was arbitrary, $u\in\Phi_k(\Omega)$. $\square$

This argument does not use the growth condition. For the classical theorems, admissibility follows instead from the growth condition and continuity of the Hessian, as in Section 4.2.

A margin in the smooth estimate

We need a small margin in the growth condition to accommodate the error in approximation. The proof of Theorem 1.2 gives the following variant with constants depending on the same parameters.

Lemma 6.2. Let $u\in C^\infty(B_R)$ satisfy $\sigma_k(D^2u)=1$ and $u(0)=0$. Suppose $b>0$ and

There are constants $\eta=\eta(n,k,b)\in(0,1)$ and $C=C(n,k,b)$ such that

Proof. Rescale to $R=1$ and choose $r_0\in(1/\sqrt{2},1)$. For $r_0\le|x|<1$, the lower bound gives

Hence a minimum on $\overline{B}_{r_0}$ is an interior minimum in $B_1$. The continuity argument in Section 4.2 then gives $D^2u\in\Gamma_k$ throughout $B_1$.

Choose a regular value $t\in(b/8,b/4)$, and let $\Omega_t$ be the component of $\{u<t\}$ containing the origin. Then

As in Section 4, its boundary is $(k-1)$-convex. For $U=u-t$ and $d_t(x)=\operatorname{dist}(x,\partial\Omega_t)$, Proposition 3.1 gives

Choose

Then (84) gives $d_t(0)\ge\gamma_0$. The segment argument following (66) gives $B_{\gamma_0}\subset\Omega_t$.

Thus $B_\eta\subset\Omega_t$. Since distance is 1-Lipschitz and $\eta=\gamma_0/2$, every $x\in B_\eta$ satisfies $d_t(x)\ge\eta$. Writing $D_t = \operatorname{diam}\Omega_t \le\sqrt{3}$, we therefore have $d_t(x) \ge(\eta/\sqrt{3})D_t$ on $B_\eta$. Lemma 4.1 applies with $\delta= \eta/\sqrt{3}$.

Finally, $u = U + t$ and $t < b/4$, so the resulting bounds for $U$, $DU$, and $D^2U$ give (6.6) when $R = 1$. The change of variables $x = Ry$ restores the stated powers of $R$.

Approximation by solutions of the equation

Proof of Theorem 1.4. By Lemma 6.1, the continuous viscosity solution $u$ is admissible. Rescale to $R = 1$ and set $r = \sqrt{3}/2$. The smaller ball has closure inside $B_1$, so $u$ is continuous on $\overline{B_r}$ and its boundary values can be approximated uniformly.

Choose smooth functions $\varphi_m$ on $\partial B_r$ with

The sphere $\partial B_r$ is connected, and its principal curvatures are all $1/r$. Consequently,

The smooth Dirichlet theorem of Caffarelli, Nirenberg, and Spruck [14 Theorem 3] therefore supplies admissible solutions, smooth up to the boundary, satisfying

Maclaurin’s inequality (60) bounds every lower symmetric polynomial of $D^2u_m$ away from zero. Continuity preserves these bounds on $\partial B_r$.

For fixed $m$, it follows that $D^2u_m(\overline{B_r}) \Subset\Gamma_k$. Thus there exists $\delta_m > 0$ such that perturbations by $\pm2\delta I$ preserve admissibility whenever $0 < \delta< \delta_m$.

We explain the comparison that transfers the boundary error into the ball. Define

Take $0 < \delta< \delta_m$. For any symmetric matrix $A$,

Apply this identity on the admissible segments from $D^2u_m$ to $D^2u_m \pm2\delta I$. Since $\sigma_k(D^2u_m) = 1$, it gives $\sigma_k(D^2q_m^+) < 1 < \sigma_k(D^2q_m^-)$.

The boundary approximation gives $q_m^- \le u \le q_m^+$ on $\partial B_r$. If $u-q_m^+$ had a positive interior maximum, a constant translate of $q_m^+$ would violate the upper-test inequality. Similarly, a positive interior maximum of $q_m^- - u$ would provide an admissible lower touching test with $\sigma_k > 1$, contradicting (78). Hence $q_m^- \le u \le q_m^+$ in the ball.

Letting $\delta\downarrow0$ proves

Normalize at the center by putting $v_m = u_m-u_m(0)$. The equation is unchanged by subtraction of this constant. Since $u(0)=0$, we have $|u_m(0)| \le\varepsilon_m$, and hence $\|v_m-u\|_{L^\infty(B_r)} \le2\varepsilon_m$.

Combining this estimate with the lower bound for $u$, we obtain

for all sufficiently large $m$: the assumption $B \le b/4$ leaves a margin $3b/8-b/4=b/8$, which absorbs $2\varepsilon_m$. Lemma 6.2, applied on $B_r$, now gives

The Hessians lie in a common compact subset of $\Gamma_k$, by (60) and $\sigma_k(D^2v_m)=1$. The construction in § 5 gives a common concave, uniformly elliptic extension of $\sigma_k^{1/k}$ on this Hessian range.

The Evans–Krylov estimate gives a uniform $C^{2,\zeta}$ bound on $B_{\eta r/2}$ for some $\zeta>0$. Compactness gives a subsequence converging in $C^2$ on every compact subset of this ball.

Its limit is $u$, because $v_m\to u$ uniformly. Continuity of $\sigma_k$ therefore gives $\sigma_k(D^2u)=1$ classically on $B_{\eta r/2}$.

Pass to the limit at each point of this smaller ball in (89), and then take the supremum. This gives

For smoothness, differentiate the equation of the smooth approximants. With $F=\sigma_k^{1/k}$, each coordinate derivative satisfies

The coefficients have uniform positive ellipticity bounds and uniform $C^{0,\zeta}$ norms on smaller balls. Interior Schauder estimates therefore give uniform $C^{3,\zeta}$ bounds for $v_m$ there.

Iterating on compactly contained balls gives bounds of every order, so $u\in C^\infty(B_{\eta r/2})$. Rescaling gives (1.15) with $\rho=\eta r/2$.

The Liouville theorem

Proof of Theorem 1.3. Let $u$ be the given entire viscosity solution. Normalize by $v=u-u(0)$ and put $\widetilde B=B+u(0)\ge0$. For all sufficiently large $R$, we have

Theorem 1.4 gives $u\in C^\infty(B_{\rho R})$, where $\rho>0$ depends only on $n,k,b$. These balls exhaust $\mathbb{R}^n$, so $u\in C^\infty(\mathbb{R}^n)$.

The quadratic lower bound is unchanged. Theorem 1.1 therefore gives (1.12).

Admissible weak solutions

We use Trudinger’s definition of admissible weak solutions by approximation. His viscosity characterization gives the two weak-solution theorems. We then explain the integration-by-parts identities satisfied by these solutions.

Trudinger's definition

Following [15 p. 1252, (1.4)], a function $u \in C(\Omega)$ is an admissible weak solution of $\sigma_k(D^2u)=1$ if there exist $u_m \in C^2(\Omega)$ satisfying

The approximants are $k$-admissible in the weak sense, including points where their Hessians lie on the boundary of $\Gamma_k$. Their locally uniform limit belongs to $\Phi_k(\Omega)$.

The Liouville theorem and the interior estimate

For a continuous right-hand side, Trudinger characterizes admissible weak solutions by the viscosity inequalities [15 p. 1255]. We apply this characterization with right-hand side equal to one.

Proof of Theorem 1.6. The function $u$ is continuous by definition. Trudinger's characterization makes it a viscosity solution of $\sigma_k(D^2u)=1$ throughout $\mathbb{R}^n$. The lower bound is unchanged, so all hypotheses of Theorem 1.3 hold. That theorem gives (1.22).

Proof of Theorem 1.7. Trudinger's characterization makes $u$ a continuous viscosity solution on $B_R$. The value $u(0)=0$ and the lower bound with $0 \le B \le bR^2/4$ are unchanged. Apply Theorem 1.4 to obtain $u \in C^\infty(B_{\rho R})$ and (1.25). Its constants depend only on $n,k,b$.

Integration by parts and Hessian measures

For smooth $u$, write $T^{ij}_{k-1}(D^2u)$ for the entries of the Newton tensor $D\sigma_k(D^2u)$, as in § 4. Its divergence-free property and the homogeneity of $\sigma_k$ give

The second identity follows by differentiating $\sigma_k(tH)=t^k\sigma_k(H)$ at $t=1$. The first identity removes the derivative of the Newton tensor in the product rule, giving

Multiply by $\phi\in C^\infty_c(\Omega)$ and integrate first in $x_i$. Then integrate in $x_j$; symmetry of the Newton tensor gives the corresponding divergence identity in that index. Thus

Both boundary terms vanish because $\phi$ has compact support. The divergence identities and their mixed versions are discussed in [TW02, Section 2].

The identity extends to $C^2$ functions by local smooth approximation. Apply it to each $u_m$ in (91). The convergence of the Hessian densities in $L^1_{\mathrm{loc}}$ yields

Only the contracted integral is passed to the limit. No pointwise second derivatives of $u$ or separate limits of the products $u_mT_{k-1}^{ij}(D^2u_m)$ are needed.

For $k=2$, the Newton tensor has the explicit form $T_1^{ij}(D^2v)=\Delta v\,\delta_{ij}-v_{ij}$. For smooth $v$, integrate the two terms separately:

The last terms cancel. Dividing the resulting identity by two gives

If $v\in W^{1,2}_{\mathrm{loc}}$, ordinary mollifications converge strongly in this space. Their quadratic gradient products consequently converge in $L^1_{\mathrm{loc}}$, by the Cauchy–Schwarz inequality.

Thus the right-hand side is the pairing $\langle S_2[v],\phi\rangle$ used in Remark 1.8.

We also use the Hessian measure associated with $u\in\Phi_k(\Omega)$ by Trudinger and Wang [16 Theorem 1.1]. Its classical density and weak continuity are

Here all functions in the convergence statement belong to $\Phi_k(\Omega)$. Weak convergence is tested against continuous compactly supported functions.

For the approximants in (91), the densities also converge to one in $L^1_{\mathrm{loc}}$. Hence every admissible weak solution in Trudinger’s sense satisfies

Why almost-everywhere equality is insufficient

The necessary condition $\mu_k[u]=dx$ detects singular Hessian mass that is invisible in an almost-everywhere equation. For example, set

This function is convex, has positive quadratic lower growth, and satisfies $\sigma_k(D^2w)=1$ almost everywhere. It is not a quadratic polynomial.

Indeed, on either half-space $\{x_1>0\}$ or $\{x_1<0\}$, the term $|x_1|$ is affine. Thus $D^2w=cI$ there and $\sigma_k(D^2w)=\binom{n}{k}c^k=1$.

Its Hessian measure contains a nonzero singular part:

To compute it, let $\chi_\varepsilon$ be a one-dimensional mollification of $|\cdot|$ and set $w_\varepsilon=c|x|^2/2+\chi_\varepsilon(x_1)$. Then

Since $(|t|)''=2\delta_0$ in distributions, $\chi_\varepsilon''(t)\,dt\rightharpoonup2\delta_0$. Integration in the remaining coordinates gives the hypersurface measure in (98).

Weak continuity identifies this limit with $\mu_k[w]$. Since the hypersurface term is nonzero, $\mu_k[w]\ne dx$. Thus $w$ is not an admissible weak solution in Trudinger’s sense.

Two algebraic lemmas

We prove the two algebraic inequalities used to construct the distance barrier. The first controls a quadratic symmetric polynomial under a reciprocal constraint.

The second transfers this inequality to the fractional transformation of the principal curvatures. Its conclusion gives the admissibility of the distance power in § 2.

Proof. When $N=1$, the denominator $1+\eta_1$ is positive, so the assumption gives $\eta_1\ge0$. Since $\sigma_2(\eta)=0$ in one variable, the conclusion follows.

Assume $N\ge2$. We compare the sum of squares with the square of the sum. Set $S=\sum_i\eta_i$ and $E=\sum_i\eta_i^2/(1+\eta_i)$. The identities

convert the hypothesis into

In particular, $S\ge0$. Fix an index $i$ and apply Cauchy–Schwarz to the $N-1$ positive numbers $1+\eta_j$, with $j\ne i$:

The second factor is at most $N$ by (A.3). Dividing by this positive factor gives

Subtracting from $\sum_j(1+\eta_j)=N+S$, we find $1+\eta_i\le S+2-N^{-1}$for every$i. Multiply this bound by the nonnegative number $\eta_i^2/(1+\eta_i)$ and sum over $i$. Using $E \le S$ and $S+2-N^{-1}>0$, we obtain

Finally, expand $S^2$ into its diagonal and off-diagonal terms:

Dividing by two and using $S=\sigma_1(\eta)$ proves (A.2).

Since $Q' > 0$ on $[0,1]$, the polynomial $Q$ is nonconstant. In particular, $\deg P \ge k-1$, so differentiation reduces its degree by exactly $k-2$. Hence

Let $r_1,\ldots,r_{\deg Q}$ be the roots of $Q$, counted with multiplicity. Since $Q > 0$ on $[0,1]$, each root lies in $(-\infty,0) \cup(1,\infty)$.

For $\eta_j=-1/r_j$, a negative root gives $\eta_j>0$, while a root greater than one gives $-1<\eta_j<0$. Factoring $Q$ and normalizing its value at zero yields

If $\deg Q<N$, we add zero entries to $\eta$.

Logarithmic differentiation at $s=1$ yields

Thus $\eta$ satisfies both hypotheses of Lemma A.1. To express its conclusion in terms of $\mu$, differentiate (A.13) at zero:

Since $P(0)=1$, formula (A.11) evaluates these three derivatives. We obtain

The denominator $\sigma_{k-2}(\mu)$ is positive. Substitute these identities into (A.2) and multiply by $2\sigma_{k-2}(\mu)/(k(k-1))$ to obtain

This is (A.7). When $k=n$, the second identity in (A.15) has both sides zero, since $\deg Q\le1$ and $\sigma_n(\mu)=0$.

Since $\mu\in\Gamma_{k-1}$ and $\theta>0$, for $1\le j<k$ we have

For every $\varepsilon>0$, (A.7) also gives

Thus $(\mu,\theta+\varepsilon)\in\Gamma_k$. Letting $\varepsilon\downarrow0$ proves the closure statement. The formula for $\theta$ gives $0<\theta<1$. $\square$

Acknowledgements

The research is assisted by ChatGPT 6.

Jingchen Hu: [email protected], [email protected].

Heming Jiao: [email protected].