---
abstract: 'We derive a priori interior Hessian estimates for semiconvex solutions to the sigma-2 equation. An elusive Jacobi inequality, a transformation rule under the Legendre-Lewy transform, and a mean value inequality for the still nonuniformly elliptic equation without area structure are the key to our arguments. Previously, this result was known for almost convex solutions.'
address: |
    University of Washington\
    Department of Mathematics, Box 354350\
    Seattle, WA 98195
author:
- Ravi Shankar
- Yu YUAN
title: 'Hessian estimate for semiconvex solutions to the sigma-2 equation'
---

[^1]

Introduction {#sec:Intro}
============

In this paper, we prove a priori Hessian estimates for semiconvex solutions to the quadratic Hessian equation $$F\left(  D^{2}u\right)  =\sigma_{2}\left(  \lambda\right)  =\sum_{1\leq
i<j\leq n}\lambda_{i}\lambda_{j}=\frac{1}{2}\left[  \left(  \bigtriangleup
u\right)  ^{2}-\left\vert D^{2}u\right\vert ^{2}\right]  =1. \label{Esigma2}$$ Here $\lambda_{i}s$ are the eigenvalues of the Hessian $D^{2}u.$

\[thm:Hess\] Let $u$ be a smooth semiconvex solution to $\sigma_{2}\left(
D^{2}u\right)  =1$ on $B_{R}\left(  0\right)  \subset\mathbb{R}^{n}$ with $D^{2}u\geq-K~I$ for any fixed $K>0.$ Then $$\left\vert D^{2}u\left(  0\right)  \right\vert \leq C\left(  n,K\right)
\exp\left[  C\left(  n,K\right)  \left\Vert Du\right\Vert _{L^{\infty}\left(
B_{R}\left(  0\right)  \right)  }^{2}/R^{2}\right]  .$$

Given the gradient bound in terms of $K$-convex function $u\left(  x\right)  $ (note that Trudinger's gradient estimates for $\sigma_{k}$ equations need no semiconvexity of the solutions \[T\]), we can control $D^{2}u$ in terms of the solution $u$ in $B_{2R}\left(  0\right)  $ as $$\left\vert D^{2}u\left(  0\right)  \right\vert \leq C\left(  n,K\right)
\exp\left[  C\left(  n,K\right)  \left\Vert u\right\Vert _{L^{\infty}\left(
B_{2R}\left(  0\right)  \right)  }^{2}/R^{4}\right]  .$$ One quick application of the above estimates is a rigidity result for entire semiconvex solutions with quadratic growth to (\[Esigma2\]): every such solution must be quadratic.

Recall any solution to the Laplace equation $\sigma_{1}\left(  D^{2}u\right)
=\bigtriangleup u=1$ enjoys a priori Hessian estimates; yet there are singular solutions to the three dimensional Monge-Ampère equation $\sigma
_{3}\left(  D^{2}u\right)  =\det D^{2}u=1$ by Pogorelov \[P\], which automatically generalize to singular solutions to $\sigma_{k}\left(
D^{2}u\right)  =1$ with $k\geq3$ in higher dimensions $n\geq4.$

Sixty years ago, Heinze \[H\] achieved a Hessian bound for solutions to equation $\sigma_{2}\left(  D^{2}u\right)  =1$ in dimension two by two dimensional techniques. More than ten years ago, a Hessian bound for $\sigma_{2}\left(
D^{2}u\right)  =1$ in dimension three was obtained via the minimal surface feature of the gradient graph $\left(
x,Du\left(  x\right)  \right)  $ in the joint work with Warren \[WY\]. Along this integral way, Qiu \[Q\] has proved Hessian estimates for solutions to the three dimensional quadratic Hessian equation with $C^{1,1}$ variable right hand side. Hessian estimates for convex solutions to general quadratic Hessian equations have also been obtained via a new pointwise approach by Guan and Qiu \[GQ\]. Hessian estimates for almost convex solutions to (\[Esigma2\]) have been derived by a compactness argument in \[MSY\]. Hessian estimates for solutions to Monge-Ampère equation $\sigma_{n}\left(  D^{2}u\right)  =\det D^{2}u=1$ and Hessian equations $\sigma_{k}\left(  D^{2}u\right)  =1$ ($k\geq2$) in terms of the reciprocal of the difference between solutions and their boundary values, were derived by Pogorelov \[P\] and Chou-Wang \[CW\], respectively, using Pogorelov's pointwise technique. Lastly, we also mention Hessian estimates for solutions to $\sigma_{k}$ as well as $\sigma_{k}/\sigma_{n}$ equations in terms of certain integrals of the Hessian by Urbas \[U1,U2\], Bao-Chen-Guan-Ji \[BCGJ\].

Note that the almost convexity condition for solutions in \[GQ, (15)\] and \[MSY, Theorem 1.1\] is essential in both the respective arguments toward Hessian estimates for quadratic Hessian equations. The mean value inequality in @WY used the area structure of the equation. For semiconvex solutions, an elusive Jacobi inequality, a transformation rule under Legendre-Lewy transform, and a mean value inequality corresponding to the still nonuniformly elliptic linearized operator without area structure are essential in our proof of Theorem 1.1.

The bulk of Section \[sec:Prelims\] is devoted to establishing the Jacobi inequality, Proposition \[prop:Jac\], $\sum F_{ij}b_{ij}\geq\sum F_{ij}b_{i}b_{j}$ with $F_{ij}$ the linearized operator, $b=\frac{1}{4}\ln
\lambda_{\max}\left(  D^{2}u\right)  ,$ and $u\left(  x\right)  $ the semiconvex solution. The difficult nature of the fully nonlinear equation (\[Esigma2\]) is that its linearized operator matrix $\left(  F_{ij}\right)
$ is not uniformly elliptic; see (\[F\_ij\]) and (\[IEDfrange\]). What saves us is that the PDE for the Legendre-Lewy transform of $u(x)$ is uniformly elliptic, found in \[CY\]. By the transformation rule Proposition \[prop:trans\], the subharmonic $b$ in original variables corresponds to a subharmonic $b$ in new variables for the linearized operator of the new, uniformly elliptic equation. In new variables, the local maximum principle implies a mean value inequality for the subharmonic $b,$ which upon pulling back to original variables yields the mean value inequality in Proposition \[prop:mvi\]. The Hessian estimate becomes possible in Section \[sec:Hess\]. The Jacobi inequality combined with the divergence structure of $F_{ij}$ allows us to bound the integral in terms of $\Vert Du\Vert
_{L^{\infty}}.$

The Hessian estimates for general solutions $K=\infty$ to quadratic Hessian equation $\sigma_{2}\left(  D^{2}u\right)  =1$ in higher dimension $n\geq4$ still remain an issue to us.

Preliminaries {#sec:Prelims}
=============

Taking the gradient of both sides of the quadratic Hessian equation (\[Esigma2\]), we have$$\bigtriangleup_{F}Du=0, \label{Egradient}$$ where the linearized operator is given by $$\bigtriangleup_{F}=\sum_{i,j=1}^{n}F_{ij}\partial_{ij}=\sum_{i,j=1}^{n}\partial_{i}\left(  F_{ij}\partial_{j}\right)  , \ \ \label{lin}$$ with$$\left(  F_{ij}\right)  =\bigtriangleup u\ I-D^{2}u=\sqrt{2+\left\vert
D^{2}u\right\vert ^{2}}\ I-D^{2}u>0. \label{F_ij}$$ Here without loss of generality, we assume $\bigtriangleup u>0.$ Otherwise the smooth Hessian $D^{2}u$ would be in the $\bigtriangleup u<0$ branch of the equation . Given the semiconvexity condition, the conclusion in Theorem \[thm:Hess\] would be trivially true.

In passing, we add a quick proof of the quantitative ellipticity for equation (\[Esigma2\]) (again on the positive branch):$$\begin{aligned}
\frac{2}{\left(  n+1\right)  \lambda_{1}}  &  \leq F_{\lambda_{1}}\leq\left(
n-1\right)  \lambda_{1},\label{QEllip}\\
\left(  \sqrt{2}-1\right)  \lambda_{1}  &  \leq F_{\lambda_{i}}\leq\left(
n-1\right)  \lambda_{1}\ \ \text{for }i\geq2,\nonumber\end{aligned}$$ which was first proved by Lin-Trudinger @LT, under the convention $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{n}.$ The upper bound is straightforward. For the lower bound,$$D_{1}\sigma_{2}=\sqrt{\left\vert \lambda\right\vert ^{2}+2}-\lambda_{1}=\frac{\left\vert \lambda^{\prime}\right\vert ^{2}+2}{\sqrt{\left\vert
\lambda\right\vert ^{2}+2}+\lambda_{1}}\geq\frac{2}{\sigma_{1}+\lambda_{1}}\geq\frac{2}{\left(  n+1\right)  \lambda_{1}};$$ and when $i\geq2$,$$D_{i}\sigma_{2}=\sqrt{\left\vert \lambda\right\vert ^{2}+2}-\lambda_{i}>\sqrt{\lambda_{1}^{2}+\lambda_{i}^{2}}-\lambda_{i}\geq\left(  \sqrt
{2}-1\right)  \lambda_{1},$$ since function $\sqrt{\lambda_{1}^{2}+\lambda_{i}^{2}}-\lambda_{i}$ is decreasing in terms of $\lambda_{i}.$

The gradient square $\left\vert \nabla_{F}v\right\vert ^{2}$ for any smooth function $v$ with respect to the inverse metric $\left(  F_{ij}\right)  $ is defined as$$\left\vert \nabla_{F}v\right\vert ^{2}=\sum_{i,j=1}^{n}F_{ij}\partial
_{i}v\partial_{j}v.$$

Jacobi inequality {#sec:Jacobi}
-----------------

Our objective in this subsection is to get a *quantitative* subsolution inequality for the maximum eigenvalue.

\[prop:Jac\] Let $u$ be a smooth solution to (\[Esigma2\]) $\sigma
_{2}\left(  \lambda\right)  =1.$ Suppose that $\lambda_{1}>\lambda_{2}\geq\cdots\geq\lambda_{n}\geq-K$ and $\lambda_{1}\geq\Lambda\left(
n,K\right)  $ for some sufficiently large $\Lambda\left(  n,K\right)  $ at $x=p.$ Set $b=\ln\lambda_{1}.$ Then we have at $p$$$\bigtriangleup_{F}b\geq\varepsilon\left\vert \nabla_{F}b\right\vert ^{2}
\label{Jacp}$$ for $\varepsilon=1/4,$ say.

*Step 1. Differentiation of maximum eigenvalue.*

We derive the following formulas for smooth function $b=\ln\lambda_{1}$ $$\left\vert \nabla_{F}b\right\vert ^{2}=\left(  b^{\prime}\right)  ^{2}\sum_{k=1}^{n}f_{k}u_{11k}^{2} \label{gradientb}$$ and$$\bigtriangleup_{F}b=\left\{
\begin{array}
[c]{c}b^{\prime}\left[  2\sum_{i>j}-u_{ii1}u_{jj1}+\sum_{k>1}\frac{2f_{k}}{\lambda_{1}-\lambda_{k}}u_{kk1}^{2}\right]  +b^{\prime\prime}f_{1}u_{111}^{2}\ \ \ \text{(I)}\\
+\sum_{i>1}\left[  2b^{\prime}+\left(  \frac{2b^{\prime}}{\lambda_{1}-\lambda_{i}}+b^{\prime\prime}\right)  f_{i}\right]  u_{11i}^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \text{(II)}\\
+\sum_{i>j>1}2b^{\prime}\left(  1+\frac{f_{i}}{\lambda_{1}-\lambda_{j}}+\frac{f_{j}}{\lambda_{1}-\lambda_{i}}\right)  u_{ij1}^{2}\ \ \ \text{(III)}\end{array}
\right.  \label{lapb}$$ at $p,$ where $D^{2}u$ is assumed to be diagonalized and $f\left(
\lambda\right)  =\sigma_{2}\left(  \lambda\right)  .$

To this end, we start with the partial derivatives of the distinct eigenvalue $\lambda_{1}$ with respect to arbitrary unit vector $e\in\mathbb{R}^{n}$ at $p$$$\begin{aligned}
\partial_{e}\lambda_{1}  &  =\partial_{e}u_{11},\\
\partial_{ee}\lambda_{1}  &  =\partial_{ee}u_{11}+\sum_{k>1}2\frac{\left(
\partial_{e}u_{1k}\right)  ^{2}}{\lambda_{1}-\lambda_{k}},\end{aligned}$$ which can be reached for example by implicitly differentiating the characteristic equation $$\det(D^{2}u-\lambda_{1}I)=0$$ near any point where $\lambda_{1}$ is distinct from the other eigenvalues.

Thus we get (\[gradientb\]) at $p$$$\left\vert \nabla_{F}b\right\vert ^{2}=\sum_{k=1}^{n}F_{kk}\left(  b^{\prime
}\right)  ^{2}u_{11k}^{2}=\left(  b^{\prime}\right)  ^{2}\sum_{k=1}^{n}f_{k}u_{11k}^{2}.$$

From $$\partial_{ee}b=b^{\prime}\partial_{ee}\lambda_{1}+b^{\prime\prime}\left(
\partial_{e}\lambda_{1}\right)  ^{2},$$ we conclude that at $p$ $$\partial_{ee}b=b^{\prime}\left[  \partial_{ee}u_{11}+\sum_{k>1}2\frac{\left(
\partial_{e}u_{1k}\right)  ^{2}}{\lambda_{1}-\lambda_{k}}\right]
+b^{\prime\prime}\left(  \partial_{e}u_{11}\right)  ^{2},$$ and $$\begin{aligned}
\bigtriangleup_{F}b  &  =\sum_{\gamma=1}^{n}F_{\gamma\gamma}\partial
_{\gamma\gamma}b\nonumber\\
&  =\sum_{\gamma=1}^{n}F_{\gamma\gamma}b^{\prime}\left(  \partial
_{\gamma\gamma}u_{11}+\sum_{k>1}2\frac{\left(  u_{1k\gamma}\right)  ^{2}}{\lambda_{1}-\lambda_{k}}\right)  +\sum_{\gamma=1}^{n}F_{\gamma\gamma
}b^{\prime\prime}u_{11\gamma}^{2}. \label{E4thorder}$$

Next we substitute the fourth order derivative terms $\partial_{\gamma\gamma
}u_{11}$ in the above by lower order derivative terms. Differentiating equation (\[Egradient\]) $\sum_{\alpha,\beta=1}^{n}F_{\alpha\beta}u_{j\alpha\beta}=0$ and using (\[F\_ij\]), we obtain $$\begin{aligned}
\bigtriangleup_{F}u_{ij}  &  =\sum_{\alpha,\beta=1}^{n}F_{\alpha\beta
}u_{ji\alpha\beta}=\sum_{\alpha,\beta=1}^{n}-\partial_{i}F_{\alpha\beta
}u_{j\alpha\beta}=\sum_{\alpha,\beta=1}^{n}-\left(  \bigtriangleup
u_{i}\ \delta_{\alpha\beta}-u_{i\alpha\beta}\right)  u_{j\alpha\beta}\\
&  =\sum_{\alpha=1}^{n}-\left(  \bigtriangleup u_{i}-u_{i\alpha\alpha}\right)
u_{j\alpha\alpha}+\sum_{\alpha\neq\beta}u_{i\alpha\beta}u_{j\alpha\beta}=\sum_{\alpha\neq\beta}\left(  u_{i\alpha\beta}u_{j\alpha\beta}-u_{i\beta
\beta}u_{j\alpha\alpha}\right)  .\end{aligned}$$ Plugging the above identity with $i=j=1$ in (\[E4thorder\]), we have at $p$ $$\bigtriangleup_{F}b=b^{\prime}\left[  \sum_{i\neq j}\left(  u_{ij1}^{2}-u_{ii1}u_{jj1}\right)  +\sum_{\gamma=1}^{n}\sum_{k>1}2F_{\gamma\gamma
}\frac{u_{1k\gamma}^{2}}{\lambda_{1}-\lambda_{k}}\right]  +\sum_{\gamma=1}^{n}b^{\prime\prime}F_{\gamma\gamma}u_{11\gamma}^{2}.$$ Regrouping those terms $u_{\heartsuit\heartsuit1}$ (with $u_{111}$), $u_{11\heartsuit},$ and $u_{\heartsuit\clubsuit1}$ in the last expression, noting $F_{\gamma\gamma}=f_{\gamma}$ at $p,$ we obtain (\[lapb\]).

*Step 2. Convexity of the level set of the equation* $\left\{  \left.
M\right\vert F\left(  M\right)  =0\right\}  .$

We rewrite the cross terms $2\sum_{i>j}-u_{ii1}u_{jj1}=2D^{2}F\left(
D^{2}u_{1},D^{2}u_{1}\right)  $ inside (I) of (\[lapb\]) in a positive way $$2\sum_{i>j}-u_{ii1}u_{jj1}=\sum_{i\neq j}-t_{i}t_{j}=\frac{\left(  \left\vert
\lambda\right\vert ^{2}+2\right)  \left\vert t\right\vert ^{2}-\left\langle
\lambda,t\right\rangle ^{2}}{\sigma_{1}^{2}}, \label{positive balance}$$ where we denoted $t_{i}=u_{ii1}.$ In fact, squaring the equation (\[Egradient\]) $\sum_{i=1}^{n}f_{i}t_{i}=0$ at $p,$ or equivalently $$\sigma_{1}\ \left(  t_{1}+\cdots+t_{n}\right)  =\lambda_{1}t_{1}+\cdots+\lambda_{n}t_{n},$$ we have$$\sigma_{1}^{2}\left(  \left\vert t\right\vert ^{2}+\sum_{i\neq j}t_{i}t_{j}\right)  =\left\langle \lambda,t\right\rangle ^{2}.$$ Hence, the above positive way follows from equation (\[Esigma2\]) $\sigma_{1}^{2}=\left\vert \lambda\right\vert
^{2}+2.$

*Step 3. Consequence of semiconvexity* $\lambda_{i}\geq-K.$

We are ready to prove the Jacobi inequality (\[Jacp\]). Note that all the off-diagonal terms in (III) of (\[lapb\]) are nonnegative; it follows that$$\bigtriangleup_{F}b-\varepsilon\left\vert \nabla_{F}b\right\vert ^{2}\geq(I)-\varepsilon\left(  b^{\prime}\right)  ^{2}f_{1}t_{1}^{2}+(II)-\varepsilon\left(  b^{\prime}\right)  ^{2}\sum_{i>1}f_{i}u_{11i}^{2}.$$

Now$$\begin{gathered}
(II)-\varepsilon\left(  b^{\prime}\right)  ^{2}\sum_{i>1}f_{i}u_{11i}^{2}=\left(  b^{\prime}\right)  ^{2}\sum_{i>1}\left[  2\lambda_{1}+\left(
\frac{2\lambda_{1}}{\lambda_{1}-\lambda_{i}}-1-\varepsilon\right)
f_{i}\right]  u_{11i}^{2}\\
\geq\left(  b^{\prime}\right)  ^{2}\sum_{i>1}\underset{\lambda_{i}\geq-K}{\underbrace{\left(  \frac{2\lambda_{1}}{\lambda_{1}+K}-1-\varepsilon
\right)  }}\ \underset{\text{(\ref{F_ij})}}{\underbrace{\left(  \sqrt
{\left\vert \lambda\right\vert ^{2}+2}-\lambda_{i}\right)  }}u_{11i}^{2}\geq0\end{gathered}$$ for $$\lambda_{1}\geq\frac{1+\varepsilon}{1-\varepsilon}K\ \ \text{with }\varepsilon<1.$$

Plugging (\[positive balance\]) in $(I)$ of (\[lapb\]), we have$$\left(  I\right)  -\varepsilon b^{\prime2}f_{1}t_{1}^{2}=\frac{1}{\lambda
_{1}\sigma_{1}^{2}}\left\{
\begin{array}
[c]{c}\left(  \left\vert \lambda\right\vert ^{2}+2\right)  \left\vert t\right\vert
^{2}-\left\langle \lambda,t\right\rangle ^{2}+\sigma_{1}^{2}\sum_{k>1}\frac{2f_{k}}{\lambda_{1}-\lambda_{k}}\mathbf{t}_{k}^{2}\\
-\left(  1+\varepsilon\right)  \frac{\sigma_{1}^{2}}{\lambda_{1}^{2}}\underset{1+0.5\lambda^{\prime2}}{\underbrace{\lambda_{1}f_{1}}}t_{1}^{2}\end{array}
\right\}  .$$ Observe that $$f_{1}=\sigma_{1}-\lambda_{1}=\sqrt{\left\vert \lambda\right\vert ^{2}+2}-\lambda_{1}=\frac{\left\vert \lambda^{\prime}\right\vert ^{2}+2}{\sqrt{\left\vert \lambda\right\vert ^{2}+2}+\lambda_{1}}<\frac{0.5\left\vert
\lambda^{\prime}\right\vert ^{2}+1}{\lambda_{1}}, \label{ID1sigma2}$$ where $\lambda^{\prime}=\left(  \lambda_{2},\cdots,\lambda_{n}\right)  .$ Also we see that $$\left\vert \lambda_{k}\right\vert \leq C\left(  n,K\right)  \ \ \text{for\ \ }k\geq2 \label{IEallbutonebounded}$$ from (\[ID1sigma2\]) (cf. \[CY, p.663\]). Indeed by the assumption $\lambda_{n}\geq-K$ and $\left\vert \lambda^{\prime}\right\vert \leq
n\lambda_{1},$ we have$$\begin{aligned}
-\left(  n-2\right)  K+\left\vert \lambda_{+}^{\prime}\right\vert  &
\leq\lambda_{n}+\cdots+\lambda_{2}=\frac{\left\vert \lambda^{\prime
}\right\vert ^{2}+2}{\sqrt{\left\vert \lambda\right\vert ^{2}+2}+\lambda_{1}}\\
&  <2+\frac{\left\vert \lambda^{\prime}\right\vert ^{2}}{\left(  1+1/n\right)
\left\vert \lambda^{\prime}\right\vert }\leq2+\left\vert \lambda_{-}^{\prime
}\right\vert +\frac{\left\vert \lambda_{+}^{\prime}\right\vert }{\left(
1+1/n\right)  },\end{aligned}$$ where $\lambda_{+}=\left(  \lambda_{2},\cdots,\lambda_{m}\right)  $ and $\lambda_{-}=\left(  \lambda_{m+1},\cdots,\lambda_{n}\right)  $ for $\lambda_{2}\geq\cdots\geq\lambda_{m}\geq0\geq\lambda_{m+1}\geq\cdots
\geq\lambda_{n}.$ Solving the above inequality for $\left\vert \lambda
_{+}\right\vert ,$ we get $$\left\vert \lambda_{+}\right\vert <\left(  n+1\right)  \left[  2+2\left(
n-2\right)  K\right]  =C\left(  n,K\right)  .$$ Consequently, $\lambda_{1}\left(  x\right)  $ is a distinct eigenvalue, thus smooth near $x=p$ if $$\lambda_{1}\left(  p\right)  >C_{\text{smooth}}\left(  n,K\right)  ;
\label{smoothcut}$$$$c\left(  n,K\right)  \leq\lambda_{1}f_{1}\leq C\left(  n,K\right)
\ \ \text{and for }k\geq2,\ c\left(  n,K\right)  \leq\frac{f_{k}}{\lambda_{1}}\leq C\left(  n,K\right)  \ ; \label{IEDfrange}$$ and also$$\frac{\sigma_{1}^{2}}{\lambda_{1}^{2}}=1+o\left(  1\right)  \ \ \text{and
}\sigma_{1}^{2}\frac{2f_{k}}{\lambda_{1}-\lambda_{k}}=\left[  2+o\left(
1\right)  \right]  \lambda_{1}^{2}$$ for large enough $\lambda_{1}$ and $k\geq2.\ $Denoting $t^{\prime}=\left(
t_{2},\dots,t_{n}\right)  .$ It follows that$$\begin{gathered}
\hspace*{-0.8in}\lambda_{1}\sigma_{1}^{2}\left[  \left(  I\right)
-\varepsilon b^{\prime2}f_{1}t_{1}^{2}\right]  \geq\\
\hspace*{-0.8in}\left.
\begin{array}
[c]{c}\left\{  \underline{\underline{\lambda_{1}^{2}}}+\left[  1-\varepsilon
-o\left(  1\right)  \right]  \left(  1+0.5\left\vert \lambda^{\prime
}\right\vert ^{2}\right)  \right\}  t_{1}^{2}+\left\{  \left[  \mathbf{3}+o\left(  1\right)  \right]  \lambda_{1}^{2}+\underline{\left\vert
\lambda^{\prime}\right\vert ^{2}}+2\right\}  \left\vert t^{\prime}\right\vert
^{2}\\
\\
-\underline{\underline{\lambda_{1}^{2}t_{1}^{2}}}-\underline{\left\vert
\lambda^{\prime}\right\vert ^{2}\left\vert t^{\prime}\right\vert ^{2}}-2\underset{\text{redistribute}}{\underbrace{t_{1}\left\vert \lambda^{\prime
}\right\vert \lambda_{1}\left\vert t^{\prime}\right\vert }}\end{array}
\right. \\
\geq\left.
\begin{array}
[c]{c}\left\{  \left[  1-\varepsilon-o\left(  1\right)  \right]  \left(
1+0.5\left\vert \lambda^{\prime}\right\vert ^{2}\right)  \right\}  t_{1}^{2}+\left[  \left(  \mathbf{3}+o\left(  1\right)  \right)  \lambda_{1}^{2}+2\right]  \left\vert t^{\prime}\right\vert ^{2}\\
-\left[  \left(  1-\varepsilon-o\left(  1\right)  \right)  \left(
1+0.5\left\vert \lambda^{\prime}\right\vert ^{2}\right)  \right]  t_{1}^{2}-\frac{\left\vert \lambda^{\prime}\right\vert ^{2}}{\left(  1-\varepsilon
-o\left(  1\right)  \right)  \left(  1+0.5\left\vert \lambda^{\prime
}\right\vert ^{2}\right)  }\lambda_{1}^{2}\left\vert t^{\prime}\right\vert
^{2}\end{array}
\right. \\
\geq\left\{  \left[  \mathbf{3}+o\left(  1\right)  \right]  -\frac
{2}{1-\varepsilon-o\left(  1\right)  }\right\}  \lambda_{1}^{2}\left\vert
t^{\prime}\right\vert ^{2}\geq0\end{gathered}$$ for $\varepsilon<1/3,$ say $\varepsilon=1/4$ for simple notation and large enough smooth $$\lambda_{1}\geq\Lambda\left(  n,K\right)  >1+C_{\text{smooth}}\left(
n,K\right)  \label{Lambdacut}$$ with $C_{\text{smooth}}\left(  n,K\right)  $ from (\[smoothcut\]).

We have proved the pointwise Jacobi inequality (\[Jacp\]) in Proposition 2.1.

Integral Jacobi inequality
--------------------------

Eventually in the proof of our Theorem 1.1, we use the following integral form of (\[Jacp\]).

From now on, repeated indices represent summation, unless otherwise indicated.

Let $u$ be a smooth $K$-convex (namely, $D^{2}u\geq-KI$) solution to $F\left(
D^{2}u\right)  =\sigma_{2}=1$ on $B_{3}$, and define the Lipschitz quantity $$b=\frac{1}{4}\ln\max(\Lambda,\lambda_{max}),$$ where the sufficiently large $\Lambda=\Lambda(n,K)$ is from (\[Lambdacut\]). Then for all nonnegative $\varphi\in C_{c}^{\infty}(B_{3})$, there holds the inequality $$0\geq\int_{B_{3}}F_{ij}\varphi_{i}b_{j}+\varphi F_{ij}b_{i}b_{j}\,dx.
\label{IJac}$$

It is easy to see that the Lipschitz function $b\left(  x\right)  $ is smooth away from the level set $\left\{  \left.  x\right\vert \lambda_{\max}\left(
x\right)  =\Lambda\right\}  .$ By Sard's theorem, we perturb $\Lambda$ a tiny bit, still denoted by $\Lambda,$ so that the Lipschitz $b\left(  x\right)  $ is smooth away from a zero measure set. Integrating by parts the pointwise Jacobi inequality (\[Jacp\]) multiplied by $\varphi,$ over a family of approximated domains of $B_{3}$ from the complement of the above zero measure set, we reach the integral Jacobi inequality (\[IJac\]).

Legendre-Lewy transform {#sec:CoV}
-----------------------

In the integral approach of @WY toward Hessian estimates for (\[Esigma2\]) with $n=3$, a mean value inequality, pertaining to the area structure on the Lagrangian minimal surface $(x,Du(x))\in\mathbb{R}^{2n},$ is used to bound $b(x)$ at $x=0$ by its integral. However, for $n>3,$ an area-like structure is unclear to us.

To construct a mean value inequality for subsolution $b$, in principle, we would apply the local maximum principle, but the ellipticity constants for the linearized operator of $\sigma_{2}=1$ are not uniform. To circumvent this, we show that $b$ is a subsolution of a new uniformly elliptic operator after a change of variables, which we describe below.

The $K$-convexity of $u$ ensures that the smallest canonical angle of the Lewy-sheared gradient graph $(x,Du(x)+Kx)$ is uniformly positive, i.e. $\theta_{min}:=\arctan(\lambda_{min}+K)>0$. This means we can make a well defined Legendre reflection about the origin, $$(x,Du(x)+Kx)=(Dw(y),y), \label{LL}$$ where $w(y)$ is the Legendre transform of $u+\frac{K}{2}|x|^{2}$. Note that $y(x)=Du(x)+Kx$ is a diffeomorphism.

We show here that this transformation preserves the linearized operator of any fully nonlinear PDE, *not just* $F\left(
D^{2}u\right)  =\sigma_{2}.$ Geometrically, this is clear for $K=0$ and the special Lagrangian equation $\sum_{i=1}^{n}\arctan(\lambda_{i})=\Theta$, since at the level of gradient graphs, the transformation is just a reflection, or a $\pi/2$-$U\left(  n\right)  $ rotation followed by a conjugation, so it only changes the constant phase $\Theta.$

\[Transformation rule\]\[prop:trans\] Let $u$ solve $F(D^{2}u(x))=0$, and its Legendre-Lewy transform $w(y)$ solve $G(D^{2}w(y)=-F(-KI+D^{2}w(y)^{-1})=0$. Then $L_{F}\approx L_{G}$, in the sense that for all smooth functions $\varphi$, we have $$\frac{\partial F}{\partial M_{ij}}(D^{2}u(x))\frac{\partial^{2}\varphi
}{\partial x^{i}\partial x^{j}}(x)=\frac{\partial G}{\partial N_{ij}}(D^{2}w(y))\frac{\partial^{2}\varphi^{\ast}}{\partial y^{i}\partial y^{j}}(y),
\label{invarianceofL}$$ where $\varphi^{\ast}(y)=\varphi(x(y))$.

Equivalently, the right hand side will not have first order terms $\partial\varphi(x(y))/\partial x^{i}$.

We will transform the left hand side of (\[invarianceofL\]) into its right. First, $$\frac{\partial\varphi}{\partial x^{j}}=\frac{\partial y^{k}}{\partial x^{j}}\frac{\partial\varphi^{\ast}}{\partial y^{k}}=(K\delta_{jk}+u_{jk})\varphi_{k}^{\ast}.$$ Consequently, $$F_{ij}\partial_{i}(\partial_{j}\varphi)=F_{ij}u_{ijk}\varphi_{k}^{\ast}+F_{ij}(K\delta_{jk}+u_{jk})\varphi_{k\ell}^{\ast}(K\delta_{i\ell}+u_{i\ell
}).$$ The first term on the right hand side vanishes via the equation: $$F_{ij}u_{ijk}\varphi_{k}^{\ast}=\varphi_{k}^{\ast}\frac{\partial}{\partial
x^{k}}F(D^{2}u(x))=0.$$ So it remains to verify that $$(K\delta_{i\ell}+u_{i\ell})\frac{\partial F}{\partial M_{ij}}(D^{2}u(x))(K\delta_{jk}+u_{jk})=\frac{\partial G}{\partial N_{ij}}(D^{2}w(y)),
\label{F=G}$$ which is a little clearer in the eigenvalue dependent case $$F(M)=f(\lambda(M)),G(N)=g(\mu(N))=-f(-K+1/\mu(N)),$$ since if the Hessian $D^{2}u(p)$ is diagonal at $x=p$ , then $K\delta_{i\ell
}+u_{i\ell}=(K+\lambda_{i})\delta_{i\ell}$, so that at $p$, the putative equality is $$\label{gi}(K+\lambda_{i})^{2}f_{i}=g_{i}.$$ Since $(\lambda_{i}+K)^{2}f_{i}=(1/\mu_{i})^{2}\partial f/\partial\lambda
_{i}=\partial g/\partial\mu_{i}$, the result follows in this case.

Let us now return to the general situation. Using the chain rule for $F(M-KI)=-G(M^{-1})$, we get $$\begin{aligned}
F_{ij}(M-KI)  &  =\frac{\partial}{\partial M_{ij}}(-G(M^{-1}))\\
&  =-\left.  \frac{\partial G}{\partial N_{k\ell}}\right\vert _{M^{-1}}\frac{\partial(M^{-1})_{k\ell}}{\partial M_{ij}}\\
&  =\left.  \frac{\partial G}{\partial N_{k\ell}}\right\vert _{M^{-1}}(M^{-1})_{ki}(M^{-1})_{\ell j},\end{aligned}$$ so upon multiplying by $K\delta_{i\ell}+u_{i\ell}=M_{i\ell}$, we obtain (\[F=G\]); in turn, the equivariance (\[invarianceofL\]).

**Remark 2.1.** The disappearance of gradient terms $D\varphi^{\ast}\left(  y\right)  $ in the right hand side of (\[invarianceofL\]) depends on $u$ solving $F(D^{2}u)=0.$ For comparison, without any equation for function $u\left(  x\right)  ,$ the Laplace-Beltrami operator$$\Delta_{g(x)}\varphi(x):=\frac{1}{\sqrt{\det g(x)}}\frac{\partial}{\partial
x^{i}}\left(  \sqrt{\det g(x)}\,g^{ij}(x)\frac{\partial}{\partial x^{j}}\varphi(x)\right)$$ corresponding to the induced metric $$g(x)=\left.  dx^{2}+dy^{2}\right\vert _{L}=\left(  I+(D^{2}u(x))^{T}D^{2}u(x)\right)  dx^{2}$$ on the gradient graph $L=\left(
x,Du(x)\right)  \in\mathbb{R}^{n}\times\mathbb{R}^{n}$ is invariant under any rotation in $\mathbb{R}^{2n},$ in particular the Legendre transform,$$\Delta_{g(x)}\varphi(x)=\Delta_{g(y)}\varphi^{\ast}(y).$$ This is because, by design, the invariant Laplace-Beltrami operator$$\Delta_{g}=g^{ij}\partial_{ij}+g^{ij}\Gamma_{ij}^{k}\partial_{k},$$ carries over those first order derivative terms.

We now prove the mean value inequality using a transformation argument. We suppose $Du(0)=0$ and $K$ is $K+1$ in transform for simplicity.

\[Mean value inequality\]\[prop:mvi\] Let $u$ be a smooth $K$-convex solution to on $B_{3}(0)$. If $b\in C(B_{3})$ is a viscosity subsolution of the linearized operator , then the following inequality holds: $$\begin{aligned} b(0)\le C(n,K)\int_{B_1}b(x)\Delta u(x)\,dx. \end{aligned}$$

Let us first verify that transformed viscosity subsolution $b^{\ast
}(y):=b(x(y))$ is a viscosity subsolution of the transformed linearized operator . We denote $y(B_{3}):=(K(\cdot)+Du)(B_{3})$ the dual domain under the coordinate inversion. Suppose that $\psi(y)\in
C^{2}(y(B_{3}))$ touches $b^{\ast}(y)$ from above near $y_{0}\in y(B_{3}).$ Then $\psi_{\ast}(x):=\psi(y(x))$ touches $b(x)$ from above near $x_{0}=x(y_{0})$, so $$F_{ij}\frac{\partial^{2}\psi_{\ast}}{\partial x^{i}\partial x^{j}}(x_{0})\geq0.$$ We recall $x\mapsto Kx+Du(x)$ is a diffeomorphism: letting $\varphi
(x)=\psi(y(x))\in C^{2}$, it follows from transformation rule that $\varphi^{\ast}(y)=\psi(y)$ satisfies the desired inequality at $y_{0}$.

It was first shown in @CY that the equation solved by the vertical coordinate Lagrangian potential $w(y)$, $$G(D^{2}w)=-F(D^{2}u)=-\sigma_{2}(-KI+(D^{2}w)^{-1})=-1,$$ is conformally, uniformly elliptic for $K$-convex solutions $u$, in the sense that for $H_{ij}:=\sigma_{n}(\lambda(D^{2}w))G_{ij}$, the operator $H_{ij}\partial_{ij}$ is uniformly elliptic: $$c(n,K)I\le(H_{ij})=\sigma_{n}(1/\lambda)(G_{ij})\le C(n,K)I.$$ This can also be seen from using the change of variables .

Since $u\in C^{\infty}(B_{3})$ is $K$-convex, the gradient map $y(x)=Du(x)+Kx$ is uniformly monotone, and we have a lower bound $|y(x_{I})-y(x_{II})|\geq|x_{I}-x_{II}|$ for each $x_{I},x_{II}\in B_{3}$, if, abusing notation for simplicity, $K$ is $K+1$ in our Legendre-Lewy transform . It follows that the dual domain contains the unit ball: $$y(B_{1})=(K(\cdot)+Du)(B_{1}(0))\supset B_{1}(0).$$ Since $b^{\ast}(y)$ is a subsolution of a uniformly elliptic operator, it follows from the local maximum principle @CC [p.36] that $b^{\ast}$ satisfies a mean value inequality: $$b^{\ast}(0)\leq C(n,K)\int_{B_{1}^{y}}b^{\ast}(y)dy.$$ Returning to $x$ variables and using $x(B_{1}^{y})\subset B_{1}$, we obtain $$b(0)\leq C(n,K)\int_{B_{1}}b(x)\det(D^{2}u(x)+KI)dx.$$ Using $\lambda_{i}\leq C(n,K)$ with $i\geq2$ (for the small eigenvalues) from (\[IEallbutonebounded\]), as well as $c\left(  n\right)  \leq\lambda
_{max}=\lambda_{1}<\Delta u,$ we get $$b(0)\leq C(n,K)\int_{B_{1}}b(x)\lambda_{max}(x)\,dx\leq C(n,K)\int_{B_{1}}b(x)\Delta u(x)\,dx,$$ as required.

**Remark 2.2.** Without going through the Legendre-Lewy transform, we do not see a direct proof for Proposition \[prop:mvi\]. In the original $x$-coordinates, in general, without the K-convexity assumption on the solution $u\left(  x\right)  ,$ we have a weaker-quadratic-weight mean value inequality than the one with the linear weight $\bigtriangleup u$ in Proposition \[prop:mvi\]. In fact, given any *smooth* positive subsolution $a\left(  x\right)  $, such as $\bigtriangleup u$, of linearized operator , an easy modification of the local maximum principle @GT [Theorem 9.20] yields the weighted mean value inequality $$a(0)\leq C(n)\int_{B_{1}}\left(  \frac{\Vert DF\Vert^{n}}{\det DF}\right)
a(x)dx,$$ where $\Vert DF\Vert$ is the maximum eigenvalue of $(F_{ij})$. By the eigenvalue bounds (\[QEllip\]) of $(F_{ij})$, we have $$\frac{\Vert DF\Vert^{n}}{\det DF}\leq C(n)\lambda_{1}^{2}<C(n)(\Delta u)^{2},$$ leading to the (ineffective!) mean value inequality $$a(0)\leq C(n)\int_{B_{1}}a(x)(\Delta u)^{2}dx.$$ Still, there follows an $L^{\infty}$ Hessian bound for the solutions $u\left(
x\right)  $ to (\[Esigma2\]) in terms of the $L^{3}$ norm of the Hessian $D^{2}u$, improving a result in @U2.

Proof of Theorem \[thm:Hess\] {#sec:Hess}
=============================

By scaling $v\left(  x\right)  =u\left(  Rx\right)  /R^{2},$ we assume $R=3$, and we assume $Du(0)=0$ and $K$ is $K+1$ in for simplicity. By Proposition \[prop:Jac\], $b(x)$ is a smooth subsolution of linearized operator when $\lambda_{max}(x)\geq\Lambda(n,K)$ is sufficiently large. Redefining it as $$b\left(  x\right)  =\max\left\{  \frac{1}{4}\ln\lambda_{\max}\left(  x\right)
,\frac{1}{4}\ln\Lambda\left(  n,K\right)  \right\}  ,$$ we see that $b\left(  x\right)  ,$ as the maximum of two smooth subsolutions, is a viscosity subsolution of linearized operator . By Proposition \[prop:mvi\], we conclude it satisfies the mean value inequality $$b(0)\leq C(n,K)\int_{B_{1}}b(x)\Delta u(x)dx.$$ Our next step is to apply the integral Jacobi inequality. Introducing a cutoff function $\varphi=1$ on $B_{1}$ and $\varphi=0$ outside $B_{2}$, we integrate by parts: $$\begin{aligned}
\int_{B_{1}}b(x)\Delta u(x)\,dx  &  \leq\int_{B_{2}}\varphi^{2}b(x)\Delta
u(x)\,dx\\
&  \leq-\int_{B_{2}}\varphi^{2}Db\cdot Du\ dx-2\int_{B_{2}}(\varphi
b)D\varphi\cdot Du\,dx.\end{aligned}$$ The second term is easy if we invoke $b\leq C(n,K)\ln\lambda_{max}\leq
C(n,K)\lambda_{max}\leq C(n,K)\Delta u$: $$-2\int_{B_{2}}(\varphi b)D\varphi\cdot Du\,dx\leq C(n,K)\Vert Du\Vert
_{L^{\infty}(B_{2})}\int_{B_{2}}\Delta u\,dx\leq C(n,K)\Vert Du\Vert
_{L^{\infty}(B_{2})}^{2}.$$ For the first term, we start with $$-\int_{B_{2}}\varphi^{2}Db\cdot Du\,dx\leq C(n)\Vert Du\Vert_{L^{\infty}(B_{2})}\int_{B_{2}}|Db|\,dx.$$ Next, the idea is to bound $\left\vert Db\right\vert $ by $F_{ij}b_{i}b_{j}.$ Assume at a point $p\in B_{2}$ that $D^{2}u(p)$ is diagonal, with $u_{ii}=\lambda_{i}$ and $\lambda_{1}\geq\cdots\geq\lambda_{n}$. Write $|Db(p)|\leq C(n)\sum_{i=1}^{n}|b_{i}(p)|$. For $i=1$: $$|b_{1}|\leq f_{1}b_{1}^{2}+1/f_{1}\leq f_{1}b_{1}^{2}+C(n,K)\lambda_{max},$$ since $f_{1}\geq c(n,K)/\lambda_{1}$ from (\[IEDfrange\]). For *each fixed* $i\geq2$: $$|b_{i}|\leq f_{i}b_{i}^{2}+1/f_{i}\leq f_{i}b_{i}^{2}+C(n,K),$$ since $f_{i}\geq c(n,K)\lambda_{1}\geq c(n,K)>0$ from (\[IEDfrange\]). We conclude that in $B_{2}$, $$|Db|\leq C(n)F_{ij}b_{i}b_{j}+C(n,K)\Delta u,$$ where we used $\Delta u\geq\sqrt{2n/\left(  n-1\right)  }.$ Therefore, we see there is one term left to estimate: $$\int_{B_{2}}|Db|\,dx\leq C(n)\int_{B_{2}}F_{ij}b_{i}b_{j}\,\mathrm{d}x+C(n,K)\Vert Du\Vert_{L^{\infty}(B_{2})}.$$ Let $\Phi$ be another cutoff, defined by $\Phi(x)=1$ on $B_{2}$, and $\Phi=0$ outside $B_{3}$. Applying the integral Jacobi inequality (\[IJac\]), we can write $$\begin{aligned}
\int_{B_{3}}\Phi^{2}F_{ij}b_{i}b_{j}\,dx  &  \leq-2\int_{B_{3}}F_{ij}\Phi
_{i}(\Phi b_{j})\,dx\\
&  \leq\frac{1}{2}\int_{B_{3}}\Phi^{2}F_{ij}b_{i}b_{j}\,dx+2\int_{B_{3}}F_{ij}\Phi_{i}\Phi_{j}\,dx.\end{aligned}$$ Thus, it remains to estimate the final term. Assume again that $D^{2}u(p)$ is diagonal at $x=p$. Then at $p$, it is easy to estimate the integrand: $$F_{ij}\Phi_{i}\Phi_{j}=f_{i}\Phi_{i}^{2}\leq C(n,K)\sum_{i=1}^{n}f_{i}=C(n,K)(n-1)\Delta u.$$ We conclude the final integral has the desired bound: $$\int_{B_{3}}F_{ij}\Phi_{i}\Phi_{j}\,dx\leq C(n,K)\Vert Du\Vert_{L^{\infty
}(B_{3})}.$$ Putting all the pieces together, we conclude $$b(0)\leq C(n,K)\Vert Du\Vert_{L^{\infty}(B_{3})}^{2}.$$ which completes the proof of Theorem \[thm:Hess\].

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[^1]: RS and YY are partially supported by NSF Graduate Research Fellowship Program under grant No. DGE-1762114 and NSF grant DMS-1800495 respectively.

{'source': 'arXiv_out/1911'}