---
author:
- '[Nikolai Nadirashvili[^1],.4 cm Serge Vlăduţ[^2] ]{}'
title: '[Singular Solutions of Hessian Elliptic Equations in Five Dimensions]{}'
---

§ Ø

Introduction
============

In this paper we study a class of fully nonlinear second-order elliptic equations of the form $$F(D^2u)=0\leqno(1.1)$$ defined in a domain of ${ \R}^n$. Here $D^2u$ denotes the Hessian of the function $u$. We assume that $F$ is a Lipschitz function defined on the space $ S^2({ \R}^n)$ of ${n\times n}$ symmetric matrices satisfying the uniform ellipticity condition, i.e. there exists a constant $C=C(F)\ge 1$ (called an [*ellipticity constant*]{}) such that $$C^{-1}||N||\le F(M+N)-F(M) \le C||N||\;
\leqno(1.2)$$ for any non-negative definite symmetric matrix $N$; if $F\in C^1(S^2({ \R}^n))$ then this condition is equivalent to $$\frac{1}{ C'}|\xi|^2\le F_{u_{ij}}\xi_i\xi_j\le C' |\xi |^2\;,
\forall\xi\in { \R}^n\;.\leqno(1.2')$$ Here, $u_{ij}$ denotes the partial derivative $\pt^2 u/\pt x_i\pt x_j$. A function $u$ is called a [*classical*]{} solution of (1) if $u\in C^2(\Om)$ and $u$ satisfies (1.1). Actually, any classical solution of (1.1) is a smooth ($C^{\alpha +3}$) solution, provided that $F$ is a smooth $(C^\alpha )$ function of its arguments.

For a matrix $S \in   S^2({ \R}^n)$ we denote by $\lambda(S)=\{
\lambda_i : \lambda_1\leq...\leq\lambda_n\}
 \in { \R}^n$ the (ordered) set of eigenvalues of the matrix $S$. Equation (1.1) is called a Hessian equation (\[T1\],\[T2\] cf. \[CNS\]) if the function $F(S)$ depends only on the eigenvalues $\lambda(S)$ of the matrix $S$, i.e., if $$F(S)=f(\lambda(S)),$$ for some function $f$ on ${ \R}^n$ invariant under permutations of the coordinates.

In other words the equation (1.1) is called Hessian if it is invariant under the action of the group $O(n)$ on $S^2({ \R}^n)$: $$\forall O\in O(n),\; F({^t O}\cdot S\cdot O)=F(S) \;.\leqno(1.3)$$ The Hessian invariance relation (1.3) implies the following:

\(a) $F$ is a smooth (real-analytic) function of its arguments if and only if $f$ is a smooth (real-analytic) function.

\(b) Inequalities (1.2) are equivalent to the inequalities $${\mu\over C_0} \leq { f ( \lambda_i+\mu)-f ( \lambda_i) } \leq C_0 \mu,
 \; \forall  \mu\ge 0,$$ $\forall  i=1,...,n$, for some positive constant $C_0$.

\(c) $F$ is a concave function if and only if $f$ is concave.

Well known examples of the Hessian equations are Laplace, Monge-Ampère, Bellman, Isaacs and Special Lagrangian equations.

Bellman and Isaacs equations appear in the theory of controlled diffusion processes, see \[F\]. Both are fully nonlinear uniformly elliptic equations of the form (1.1). The Bellman equation is concave in $D^2u \in  S^2({ \R}^n)$ variables. However, Isaacs operators are, in general, neither concave nor convex. In a simple homogeneous form the Isaacs equation can be written as follows: $$F(D^2u)=\sup_b \inf_a L_{ab}u =0, \leqno (1.4)$$ where $L_{ab}$ is a family of linear uniformly elliptic operators of type $$L=  \sum a_{ij} {\partial^2 \over  \partial x_i \partial x_j } \leqno (1.5)$$ with an ellipticity constant $C>0$ which depends on two parameters $a,b$.

Consider the Dirichlet problem $$\label{dir}\begin{cases} F(D^2u, Du, u, x)=0 &\text{in}\;
 \Om \cr \quad \quad  u=\vph & \text{on}\; \pt\Om\;,\cr\end{cases}$$ where $\Omega \subset {\R}^n$ is a bounded domain with a smooth boundary $\partial \Omega$ and $\vph$ is a continuous function on $\pt\Om$.

We are interested in the problem of existence and regularity of solutions to the Dirichlet problem (1.6) for Hessian equations and Isaacs equation. The problem (1.6) has always a unique viscosity (weak) solution for fully nonlinear elliptic equations (not necessarily Hessian equations). The viscosity solutions satisfy the equation (1.1) in a weak sense, and the best known interior regularity (\[C\],\[CC\],\[T3\]) for them is $C^{1+\epsilon }$ for some $\epsilon
> 0$. For more details see \[CC\], \[CIL\]. Until recently it remained unclear whether non-smooth viscosity solutions exist. In the recent papers \[NV1\], \[NV2\], \[NV3\], \[NV4\] the authors first proved the existence of non-classical viscosity solutions to a fully nonlinear elliptic equation, and then of singular solutions to Hessian uniformly elliptic equation in all dimensions beginning from 12. Those papers use the functions $$w_{12,\delta}(x)={P_{12}(x)\over |x|^{\delta }},\;w_{24,\delta}(x)= {P_{24} (x)\over  |x|^{\delta }},\:
\delta\in [1,2[,$$ with $P_{12}(x),P_{24}(x)$ being cubic forms as follows: $$P_{12}(x)=Re (q_1q_2q_3),\; x=(q_1,q_2,q_3)\in {\H}^3={ \R}^{12},$$ $  {\H}$ being Hamiltonian quaternions, $$P_{24}(x)={Re((o_1\cdot o_2)\cdot o_3)}={Re(o_1\cdot(o_2\cdot o_3))},\; x=(o_1,o_2,o_3)\in {\O}^3={\R}^{24}$$ $\O$ being the algebra of Caley octonions.

Finally, the paper \[NTV\] gives a construction of non-smooth viscosity solution in 5 dimensions which is order 2 homogeneous, also for Hessian equations, the function $$w_5(x)={P_{5} (x)\over |x|},\;$$ being such solution for the Cartan minimal cubic $$P_{5}(x)=x_1^3+\frac{3 x_1}2\left(z_1^2 + z_2^2-2 z_3^2-2x_2^2\right)+\frac{3\sqrt 3}2\left(x_2z_1^2-x_2z_2^2 + 2z_1z_2z_3\right)$$ in 5 dimensions.

However, the methods of \[NTV\] does not work for the function $w_{5,\delta}(x)=P_{5} (x)/ |x|^{\delta }, \;\delta>1$, and thus does not give singular (i.e. not in $C^{1,1}$) viscosity solutions to fully nonlinear equations in 5 dimensions.

In the present paper we fill the gap and prove

[*The function $$w_{5,\delta}(x)=P_{5} (x)/ |x|^{1+\delta }, \;\delta\in [0,1[$$ is a viscosity solution to a uniformly elliptic Hessian equation $(1.1)$ with a smooth functional $F$ in a unit ball $B\subset {\R}^{5}$ for the isoparametric Cartan cubic form $$P_{5}(x)=x_1^3+\frac{3 x_1}2\left(z_1^2 + z_2^2-2 z_3^2-2x_2^2\right)+\frac{3\sqrt 3}2\left(x_2z_1^2-x_2z_2^2 + 2z_1z_2z_3\right)$$ with $x=(x_1,x_2,z_1,z_2,z_3)$.*]{}

In particular one gets the optimality of the interior $C^{1,\alpha}$-regularity of viscosity solutions to fully nonlinear equations in dimensions 5 and more; note also that all previous constructions give only Lipschitz Hessian functional $F$. Let us recall that in the paper \[NV5\] it is proven that there is no order 2 homogenous solutions to elliptic equations in 4 dimensions which suggests strongly that in 4 (and less) dimensions there is no homogenous non-classical solutions to uniformly elliptic equations.

As in \[NV3\] we get also that $w_{5,\delta}(x), \;\delta\in [0,1[$ is a viscosity solution to a uniformly elliptic Isaacs equation:

[**Corollary 1.2.**]{}

[*The function $$w_{5,\delta}(x)=P_{5} (x)/ |x|^{1+\delta }, \;\delta\in [0,1[$$ is a viscosity solution to a uniformly elliptic Isaacs equation $(1.4)$ in a unit ball $B\subset { \R}^{5}$.*]{}

The rest of the paper is organized as follows: in Section 2 we recall some necessary preliminary results and we prove our main results in Section 3. The proof in Section 3 extensively uses MAPLE but is completely rigorous.

Preliminary results 
====================

Let $w=w_n$ be an odd homogeneous function of order $2-\delta,\: 0\leq \delta <1$, defined on a unit ball $B =B_1\subset { \R}^n$ and smooth in $B \setminus\{0\}$. Then the Hessian of $w$ is homogeneous of order $-\delta$.

Define the map $$\Lambda :B  \longrightarrow \lambda (S) \in { \R}^n\; .$$ $\lambda(S)=\{ \lambda_i :
\lambda_1\leq...\leq\lambda_n\}
 \in { \R}^n$ being the (ordered) set of eigenvalues of the matrix $S=D^2w$. Denote $\Sigma_n$ the permutation group of $\{ 1,...,n\}$. For any $\sigma \in \Sigma_n$, let $T_{\sigma}$ be the linear transformation of ${ \R}^n$ given by $x_i \mapsto x_{\sigma(i)}, \; i=1,...,n.$

Let $a,b\in B$. Denote by $\mu_1(a,b)\leq ...\leq \mu_n(a,b)$ the eigenvalues of $(D^2w(a)-D^2w(b))$.

[**Lemma 2.1.**]{} [*Assume that for a smooth function $g: U  \longrightarrow  { \R}$ where the domain $U$ contains $$M:=\bigcup_{\sigma \in \Sigma_{n} }\   T_{\sigma }\Lambda (B)\subset { \R}^n$$ one has $$g_{|M}=0.$$ Assume also the condition $$\min_{i=1,\ldots,5}\inf_{x\in M}\left\{\frac{\partial g}{\partial \lambda_i}(\lambda )\right\}>0.\leqno (2.1)$$ Assume further that for any $a,b\in B$ either $\mu_1(a,b)= ...= \mu_n(a,b)=0$ or $$1/C\leq -\frac{\mu_1(a,b)}{\mu_n(a,b)}\leq C,\leqno (2.2)$$ where $C$ is a positive constant (may be, depending on $M,g$ but not on $a,b$). If $\delta> 0$ we assume additionally that $w$ changes sign in $B$. Then $w$ is a viscosity solution in $B$ of a uniformly elliptic Hessian equation $(1.1)$ with a smooth $F$. Function $w$ is as well a solution to a uniformly elliptic Isaacs equation.*]{}

[*Proof.*]{} Denote for any $\theta>0$ by $K_{\theta }\subset {  \R}^n$ the cone $\{\lambda \in { \R}^n,\,
\lambda_i /|\lambda | >\theta\},\, $ and let $K^*_{\theta }$ be its dual cone. Let $x,y$ be orthogonal coordinates in ${ \R}^n$ such that $x=\lambda_1+...+\lambda_n$ and $y$ be the orthogonal complement of $x$. Denote by $p$ the orthogonal projection of ${ \R}^n$ on subspace $y$. Denote $$\Gamma =\{ g=0 \} \subset U,$$ $$G=p(\Gamma ),$$ $$m=p(M).$$ From (2.1), (2.2) it follows that the surface $\Gamma $ is a graph of a smooth function $h$ defined on $G$. By $k_{\theta}$ we denote the function on $y$ which graph is the surface $\partial K^*_{\theta }$. We define the function $H(y)$ by $$H(y) = \inf_{z\in G}\{ h(z)+k_{\theta }(y-z)\}.$$ We fix a sufficiently small $\theta >0$. Then from (2.1), (2.2) it follows that $H=h$ on $G$. Denote by $J$ the graph of $H$. It is easy to show, see similar argument in \[NV1\], \[NV3\], that for any $a,b\in J,\, a\neq b$, $$1/C\leq -\min_i(a_i-b_i)/\max_i(a_i-b_i)\leq C.\leqno (2.3)$$ Let $E$ be a smooth function in ${ \R}^{n-1}$ with the support in a unit ball and with the integral being equal to $1$. Denote $E_c(y)=c^{-n+1}E(y/c)$, $c>0$. Set $$H_c=H\ast E_c.$$ Then $H_c$ will be a smooth function such that any two points $a,b$ on its graph will satisfy (2.3). Moreover $H_c \rightarrow H$ in $C({ \R}^n)$ as $c$ goes to $0$, and $H_c \rightarrow h$ in $C^{\infty }$ on compact subdomains of $G$. Thus for a sufficiently small $c>0$ we can easily modify function $H_c$ to a function $\tilde H$ such that $\tilde H$ will coincide with $h$ in a neighborhood of $m$, coincide with $H$ in the complement of $G$ and the points on the graph of $\tilde H$ will still satisfy (2.3) possibly with a larger constant $C$. Define the function $F$ in ${ \R}^n$ by $$F=x-\tilde H(y).$$ Then $w$ is a solution in ${ \R}^n\setminus \{ 0\}$ of a uniformly elliptic Hessian equation (1.1) with such defined nonlinearity $F$. As in \[NV3\], \[NV4\] it follows that $w$ is a viscosity solution of (1.1) in the whole space ${ \R}^n$. In \[NV3\] we have shown that the equation (1.1) for the function $w$ can be rewritten in the form of the Isaacs equation. The lemma is proved.

We will apply this result to the function $w_{5,\delta}(x)=P_{5} (x)/ |x|^{1+\delta } $.

Let then recall some facts from \[NTV\] about the Cartan cubic form $P_5(x)$.

[**Lemma 2.2.**]{}

[*The form $P_{5}(x)$ admits a three-dimensional automorphism group.*]{}

Indeed, one easily verifies that the orthogonal trasformations

$$A_1(\phi):= \left(%
\begin{array}{ccccc}
   {3\cos(\phi)^2-1\over 2}& { \sqrt 3\sin(\phi)^2\over 2}&0&0&{ \sqrt 3\sin(2\phi)\over 2} \\
{ \sqrt 3\sin(\phi)^2\over 2}&{1+\cos(\phi)^2\over 2}&0&\;0&{-\sin(2\phi)\over 2} \\
  0& 0&\cos(\phi)&\sin(\phi)&0 \\
   0&0&-\sin(\phi)&\cos(\phi)&0\\
  {- \sqrt 3\sin(2\phi)\over 2}&{\sin(2\phi)\over 2} &0&0&\cos(2\phi)\\
\end{array}%
\right)$$

$$A_2(\psi):= \left(%
\begin{array}{ccccc}
   1& 0& 0& 0& 0\\
0&\cos(2\psi)& 0& -\sin(2\psi)& 0\\
0&0& \cos(\psi)&0&-\sin(\psi)\\
0&\sin(2\psi)& 0&\cos(2\psi)& 0\\
0&0&\sin(\psi)& 0& \cos(\psi)\\
\end{array}%
\right)$$

$$A_3(\theta):= \left(%
\begin{array}{ccccc}
{3\cos(\theta)^2-1\over 2}& {-\sqrt 3\sin(\theta)^2\over 2}&0& 0&  {-\sqrt 3\sin(2\theta)\over 2} \\ 
{-\sqrt 3\sin(\theta)^2\over 2}& {1+\cos(\theta)^2\over 2}& 0& 0& {-\sin(2\theta)\over 2}\\
0&0&\cos(\theta)&-\sin(\theta)&0\\
0& 0& \sin(\theta)&\cos(\theta)& 0\\
{\sqrt 3\sin(2\theta)\over 2}&  {\sin(2\theta)\over 2}& 0& 0&\cos(2\theta)\\
\end{array}%
\right)$$ do not change the value of $P_5(x)$.

[**Lemma 2.3.**]{}

*Let $G_P$ be subgroup of $SO(5)$ generated by*

{$A_1(\phi),A_2(\psi), A_3(\theta):\:(\phi,\psi,\theta)\in { \R}^3\}.$ Then the orbit $G_PS^1$ of the circle $$S^1=\{(\cos(\chi),0,\sin(\chi),0,0):\chi\in { \R}\}\subset S^4$$ under the natural action of $G_P$ is the whole $S^4 . $

We need also the following two simple algebraic results ( \[NV3, Lemmas 2.2 and 4.1\]):

[**Lemma 2.4.**]{} [*Let $ A,B$ be two real symmetric matrices with the eigenvalues $
\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_{n} $ and $
\lambda'_1\ge\lambda'_2\ge\ldots\ge\lambda'_{n} $ respectively. Then for the eigenvalues $
\Lambda_1\ge\Lambda_2\ge\ldots\ge\Lambda_{n} $ of the matrix $A-B$ we have*]{} $$\Lambda_1\ge\max_{i=1,\cdots, n}(\lambda_i-\lambda'_i),
 \;\;\Lambda_n\le\min_{i=1,\cdots, n}(\lambda_i-\lambda'_i).$$

[**Lemma 2.5.** ]{}[*Let $\delta\in[0,1),\;
     W,\;\bar W\in { \R}$ with $|W|\le \frac{1}{3\sqrt 3},|\bar W|\le \frac{1}{3\sqrt 3}$ and let $ \mu_1(\delta)\ge \mu_2(\delta)\ge \mu_3(\delta)$ $($resp., $\bar\mu_1(\delta)\ge\bar\mu_2(\delta)\ge\bar\mu_3(\delta)\;)$ be the roots of the polynomial $$T^3+3W(1+\delta) T^2+(3W^2(1+\delta)^2-1)T+W(1-\delta) +W^3(1+\delta)^3$$ $($resp. of the polynomial $$T^3+3\bar W (1+\delta) T^2+(3 \bar W^2(1+\delta)^2-1)T+\bar W(1-\delta)+ \bar W^3(1+\delta)^3\; ).$$ Then for any $K>0$ verifying $ |K-1|+|\bar W-W|\neq 0$ one has $${1-\delta\over 5+\delta}=:\rho\le { \mu_+(K)\over -\mu_-(K)}\le
 {1\over\rho}= {5+\delta\over 1-\delta}$$ where $$\mu_-(K):=  \min\{\mu_1(\delta)-K\bar\mu_1(\delta),\;
\mu_2(\delta)-K\bar\mu_2(\delta),\;
\mu_3(\delta)-K\bar\mu_3(\delta)\} ,$$ $$\mu_+(K):=  \max\{\mu_1(\delta)-K\bar\mu_1(\delta),\;
\mu_2(\delta)-K\bar\mu_2(\delta),\;
\mu_3(\delta)-K\bar\mu_3(\delta)\}\;.$$* ]{}

Proofs
======

Let $w_{5,\delta}=P_5/|x|^{1+\delta}, \: \delta\in [0,1[$. By Lemma 2.1 it is sufficient to prove the existence of a smooth function $g$ verifying the conditions (2.1) and (2.2). We beging with calculating the eigenvalues of $D^2w_{5,\delta}(x)$. More precisely, we need

[**Lemma 3.1.**]{}

*Let $x\in S^4$, and let $x\in G_P(p,0,q,0,0)$ with $p^2+q^2=1$. Then $$Spec(D^2w_{5,\delta}(x))= \{\mu_{1,\delta},\mu_{2,\delta},\mu_{3,\delta},\mu_{4,\delta},\mu_{5,\delta} \}$$ for $$\mu_{1,\delta}= \frac{p(p^2\delta+6-3\delta)}{2},$$ $$\mu_{2,\delta}= \frac{p(p^2\delta-3-3\delta)+3\sqrt{12-3p^2}}{2},$$ $$\mu_{3,\delta}=\frac{p(p^2\delta-3-3\delta)-3\sqrt{12-3p^2}}{2},$$ $$\mu_{4,\delta}= -\frac{p\delta(6-\delta)(3-p^2)+\sqrt{D(p,\delta)}}{4},$$ $$\mu_{5,\delta}= -\frac{p\delta(6-\delta)(3-p^2)-\sqrt{D(p,\delta)}}{4} ,$$ and $$D(p,\delta):=(6-\delta)(4-\delta)(2-\delta)\delta(p^2-3)^2p^2+144(\delta-2)^2>0.$$*

The characteristic polynomial $F(S)$ of $D^2w$ is given by $$F(S)=S^5+a_{1,\delta} S^4+a_{2,\delta}S^3+a_{3,\delta}S^2+a_{4,\delta} S+a_{5,\delta}$$ for $$a_{1,\delta}= \frac{(\delta+1)(\delta-8)b}{2},$$ $$a_{2,\delta}= \frac{(\delta+1)(21\delta+13-4\delta^2)b^2}{4}+9(2\delta-\delta^2-4),$$ $$a_{3,\delta}= \frac{(6\delta^2-31\delta-1)(\delta+1)^2b^3}{8}+
\frac{27(4\delta-2\delta^2+5+\delta^3)}{2},$$ $$a_{4,\delta}= \frac{(2\delta-1)(5-\delta)(\delta+1)^2b^4}{8}+
\frac{9(\delta-1)(\delta^2-2\delta+9)}{2},$$ $$a_{5,\delta}= \frac{b(1-\delta)\left(b^2(\delta+1)^3+ 108(1-\delta)\right)\left(b^2(\delta+1)(\delta-5)+36(\delta-1)\right)}{32},$$ $$where \quad b:=p(p^2-3).$$

Note that the spectrum in this lemma is unordered one.

[*Proof of Lemma 3.1.*]{} Since $w_{5,\delta}$ is invariant under $G_P$, we can suppose that $x=(p,0,q,0,0)$. Then $w_{5,\delta}(x)=\frac{p(3-p^2)}{2}$ and we get by a brute force calculation: $$D^2w_{5,\delta}(x):= \left(%
\begin{array}{cc} M_{1,\delta}&0\\
0&M_{2,\delta}
  \\
\end{array}%
\right)$$ being a block matrix with $$M_{1,\delta}:= \frac{ 1}{2}\left(%
\begin{array}{ccc}
    m_{1,1}&  {m_{1,2}}& {m_{1,3}} \\ 
{m_{1,2}}& {m_{2,2}}&  {m_{2,3}}\\
 {m_{1,3}}& {m_{2,3}}& { m_{3,3}} \\
  
\end{array}%
\right), \;$$

$m_{1,1}:=-(\delta+2)\delta p^5+(\delta+3)\delta p^3+(12-9\delta)p,$

$m_{1,2}:= 3\sqrt 3 p(p^2-1)\delta,$

$ m_{1,3}:= -q\left((\delta+2)\delta p^4+3\delta(1-\delta)p^2+3\delta-6)\right)$

$ m_{2,2}:=\delta p^3-3(\delta+4)p,$

$ m_{2,3}:=3\sqrt 3q(\delta p^2+2-\delta),  $

$m_{3,3}:=(\delta+2)\delta p^5+(5-4\delta)\delta p^3-3(\delta-1)(2-\delta)p,$

$$M_{2,\delta}:=\frac{ 1}{2} \left(%
\begin{array}{cc}
{ \delta p^3+3(2-\delta)p}& 6\sqrt 3 q \\
 6 \sqrt 3 q & { \delta p^3-3(4+\delta)p}\\ 
\end{array}%
\right)$$ which gives for the characteristic polynomial $F(S) =F_1(S)\cdot F_2(S)\cdot F_3(S)$ where $$F_1(S):=S-\frac{p(p^2\delta+6-3\delta)}{2};$$ $$F_2(S)=S^2+\frac{\delta p (p^2-3) (\delta-6) S}{2}+\frac{(2-\delta)\left( (\delta-6)\delta p^6+6(6-\delta)\delta^2 p^4+9(\delta^2-6\delta) p^2+36(\delta-2)\right)}{4};$$ $$F_3(S) := S^2+(3 +3\delta-\delta p^2) p S+ \frac{ (p^2-3) (\delta^2 p^4-3\delta^2 p^2-6\delta p^2+36)}{4};$$ and the spectrum. Developing $F(S)$ we get the last formulas.

This follows immediately from Lemma 3.1 and a simple calculation since $$\Delta(w)=-a_{1,\delta},\;S_2(w)= a_{2,\delta},\: S_4(w)= a_{4,\delta},\:\det(D^2w)=-a_{5,\delta}.$$

Let then determine the ordered spectrum $\{\lambda_1,\lambda_2,\ldots,\lambda_5 \}, \; $ $\linebreak\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_5 $ of $D^2w$.

*Let $\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_5$ be the eigenvalues of $D^2w_{5,\delta}(x).$ Then*

$$\lambda_1= \mu_{2,\delta}, \quad \lambda_5= \mu_{3,\delta},\quad\quad\quad\quad$$ $$\lambda_2 = \begin{cases} \mu_{4,\delta} & \text{for}\; p\in [-1,p_0(\delta)],\cr
\mu_{1,\delta} & \text{for}\; p\in [p_0(\delta),1],  \end{cases}$$

$$\lambda_3 = \begin{cases} \mu_{5,\delta} & \text{for}\; p\in [-1,-p_0(\delta)],\cr
\mu_{1,\delta} & \text{for}\; p\in [-p_0(\delta),p_0(\delta)], \cr 
\mu_{4,\delta} & \text{for}\; p\in [ p_0(\delta),1], \end{cases}$$

$$\lambda_4 = \begin{cases} \mu_{1,\delta} & \text{for}\; p\in [-1,-p_0(\delta)],\cr\mu_{5,\delta} & \text{for}\; p\in [-p_0(\delta),1],\cr \end{cases}$$

where $$p_0(\delta):=\frac{3^{1/4}\sqrt{1-\delta}}{(3+2\delta-\delta^2)^{1/4}}=\frac{3^{1/4}\sqrt{ \varepsilon}}{(4 - \varepsilon^2)^{1/4}}\in ]0,1].$$

[*Proof.*]{} The inequalities $\mu_{2,\delta}(p)\ge\mu_{1,\delta}(p)\ge\mu_{3,\delta}(p)$ are obvious since $\mu_{2,\delta}(p)$ and $\mu_{3,\delta}(p)$ are decreasing in $p$, $\mu_{1,\delta}(p)$ is increasing in $p$, $\mu_{3,\delta}(-1)=\mu_{1,\delta}(-1),$ $\mu_{2,\delta}(1)=\mu_{1,\delta}(1).$

The resultant $$R(\delta,p)= Res(F_2, F_3)= 144(p-1)^2(p+1)^2\left(r_8p^8-r_6p^6+ r_4p^4-r_2p^2 +r_0  \right)$$ where $$r_8=(\varepsilon^2-4)^2,r_6=12 (\varepsilon^2-4)^2,r_4=3 (4-\varepsilon^2)(72-17\varepsilon^2 ),$$ $$r_2=108 (\varepsilon^2-4)^2 ,r_0=144(3-\varepsilon^2)^2$$ is strictly positive for $(\varepsilon,p)\in ]0,1[\times ]-1,1[$. Indeed, let $$r:=\frac{  R}{144(p-1)^2(p+1)^2}=r_8p^8-r_6p^6+ r_4p^4-r_2p^2 +r_0$$ then $$d:=\frac{\partial r}{4\varepsilon\partial \varepsilon}=(\varepsilon^2-4)p^8 +12p^6(4-\varepsilon^2)+3(17\varepsilon^2-70)p^4+
108(4-\varepsilon^2)p^2 +144(\varepsilon^2-3)<0$$ for $(\varepsilon,p)\in ]0,1[\times [0,1[$ since $$\frac{\partial d}{ 4p\partial p}=(4-\varepsilon^2)(-2p^6+18p^4-51p^2+54)-6p^2\ge (4-\varepsilon^2)\cdot 19-6\ge 51,$$ and for $p=1$ one has $d=-166+76\varepsilon^2\le -90.$ For $\delta=0, \varepsilon=1$ we get $$R(\delta,p)\ge R(1,p)=9(1-p^2)(4-p^2))(p^4-7p^2+16)$$ which proves the positivity. Using then the inequalities $$\mu_{2,\delta}(-1)=\mu_{4,\delta}(-1)>\mu_{5,\delta}(-1)>\mu_{3,\delta}(-1),$$ $$\mu_{2,\delta}(1)>\mu_{4,\delta}(1)>\mu_{5,\delta}(-1)=\mu_{3,\delta}(-1)$$ and the postivity of the resultant we get $$\mu_{2,\delta}(p)\ge\mu_{4,\delta}(p)\ge\mu_{5,\delta}(p)\ge\mu_{3,\delta}(p)$$ for any $p\in[-1,1].$

Calculatig then $$R_1(\delta,p)= Res(F_1, F_3)=12(p^2-3) \left( p^4(\varepsilon^2-4)+3\varepsilon^2\right)$$ and taking into account the equalities $$\mu_{4,\delta}(p_0(\delta))=\mu_{1,\delta}(p_0(\delta)),\;\mu_{5,\delta}(-p_0(\delta))=\mu_{1,\delta}(-p_0(\delta))$$ we get the result.

Note the oddness property of the spectrum: $$\lambda_{1,\delta}(-p)=-\lambda_{5,\delta}(p),\;\lambda_{2,\delta}(-p)=-\lambda_{4,\delta}(p),\;\lambda_{3,\delta}(-p)=-\lambda_{3,\delta}(p).$$

Let us now verify the second condition (2.2) of Lemma 2.1, namely the uniform hyperbolicity of $M_{\delta}(a,b,O)$.

[*Proof.*]{} The proof depends on the value of $$k:=p_0(\delta):=\frac{3^{1/4}\sqrt{1-\delta}}{(3+2\delta-\delta^2)^{1/4}}=\frac{3^{1/4}\sqrt{ \varepsilon}}{(4 - \varepsilon^2)^{1/4}}.$$ Note that $C=\frac{1000(\delta+1)(3-\delta)}{3(1-\delta)^2}= \frac{1000}{k^4}$. We shall give the proof for $k\in ]0,\frac{1}{2}]$, the proof for $k\in [\frac{1}{2},1]$ is similar, simpler and uses $C=10^4$.

Suppose that the conclusion does not hold, that is for some $a\neq b $ and some $O\in {\hbox {O}}({5} )$ one has $$M_\delta(a,b,O):=M_\delta(a)- {^tO}\cdot M_\delta(b)\cdot O\neq 0,$$ but

$${1\over  C}> - {\Lambda_1\over
\Lambda_{5}}\;\text{\rm or} - {\Lambda_1\over
\Lambda_{5}} >C .$$ We can suppose without loss that $|b|\le 1=|a|\in \S^4_1.$ Let $\overline{b}:=b/|b|\in \S^4_1,$ $W:=W(a),\overline W:=W(\overline b), K:= |b| ^{-1-\delta}.$ Note that since for any harmonic cubic polynomial $Q(x)$ on ${\R}^n$ and any $a\in \S_1^{n-1}\subset{\R}^n$ one has $$Tr\left (D^2\left ({Q(x)\over |x|^{1+\delta}}\right )(a)\right)= (\delta^2-2\delta-3-n) Q(a),$$ we get $Tr\left(M_{\delta}(a,b,O)\right)=(2\delta+8-\delta^2)(K\overline W-W),\; P_5   $ being harmonic. Let us prove the claim for $(K\overline W-W)\ge 0$, the proof for $(K\overline W-W)\le 0$ being the same while permuting $a$ with $b$ and $\Lambda_{1}$ with $\Lambda_{5}$. Since $$Tr\left(M_{\delta}(a,b,O)\right)=(2\delta+8-\delta^2)(K\overline W -W)\ge 0,$$ we get $ 4\Lambda_{1}+\Lambda_{5}\ge 0$ and $-\Lambda_{5}/\Lambda_{1}\le 4$. Therefore, we have only to rule out the inequality $   {1\over  C}> - {\Lambda_1\over\Lambda_{5}}$ i.e. $   -\Lambda_{5}> C {\Lambda_1}.$ Recall that $$W=\frac{3p-p^3}{2},\overline W=\frac{3\overline p-\overline p^3}{2}$$ for some $p, \overline p\in [-1,1].$

We have then 3 possibilities:

1). $p,\overline  p  \in [-k,k]$;

2). $p \in [-k,k], \overline p  \notin  [-k,k]$;

3). $p,\overline p  \notin  [-k,k]$.

In the cases 1) and 3) applying Lemma 2.4 we get $\Lambda_1\ge \mu_+(K)$, $\Lambda_{5}\le \mu_-(K)$ in the notation of Lemma 2.5 which permits to finish the proof as in Proposition 4.1 of \[NV3\]. We thus have to treat the (more difficult) case 2). Lemma 2.4 together with the inequality $   -\Lambda_{5}> C {\Lambda_1}$ gives $$-\min_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}>C\max_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}.$$

Thank to the oddness of the spectrum we suppose without loss that $\overline p >k$. Recall that then by Lemma 3.2 one has $$\lambda_{1,\delta}(\overline p )=\mu_{2,\delta}(\overline p ),\lambda_{1,\delta}( p )=\mu_{2,\delta}(p ),  \;\lambda_{2,\delta}(\overline p )=\mu_{1,\delta}(\overline p ),\lambda_{1,\delta}( p )=\mu_{4,\delta}(p ),$$ $$\lambda_{3,\delta}(\overline p )=\mu_{4,\delta}(\overline p ),\lambda_{3,\delta}( p )=\mu_{1,\delta}(p ), \;\lambda_{4,\delta}(\overline p )=\mu_{5,\delta}(\overline p ),\;\lambda_{4,\delta}( p )=\mu_{5,\delta}(p ),$$ $$\lambda_{5,\delta}(\overline p )=\mu_{3,\delta}(\overline p ),\;\lambda_{5,\delta}( p )=\mu_{3,\delta}(p ).$$

We have then 2 possibilities for $p:$

2a). $p  \in [-k,0]$;

2b). $p  \in ]0,k]$.

Let $p  \in [-k,0]$, then $\mu_{1,\delta}(p )\le 0$ and thus $$C\max_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\ge CK\lambda_{3,\delta}(\overline p ) =CK\mu_{4,\delta}(\overline p )\ge CK\mu_{4,\delta}(p_0(\delta) )=$$ $$=CK\mu_{1,\delta}(k) = CK(k^3(\sqrt {k^4+3}-k^2+3)/\sqrt {k^4+3})\ge 2CKk^{3}$$ since one verifies that the function $\mu_{4,\delta}(p)$ is increasing on $[k,1].$

On the other hand, $$|\min_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}|\le K\max_{i=1,\cdots,5, p}|\{\lambda_{i,\delta}( p )\}|+\max_{i=1,\cdots,5, p}|\{\lambda_{i,\delta}( p )\}|\le 8(K+1).$$

Therefore one gets $8(K+1)\ge  2CKk^{3}$ which clearly is a contradiction for, say, $K\ge 1/4$. For $0<K\le 1/4$ we get $$C\max_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\ge C(K\lambda_{5,\delta}(\overline p )-\lambda_{5,\delta}( p ))\ge C(K(-8)-(-5))\ge 3C$$ which can not be less than $ 8(K+1)\le 10$.

Let finally $p  \in ]0,k]$. We consider then 2 possibilities for $K$:

$(i)$ $K \le  20/31=(1.55)^{-1} $,

$(ii)$ $K>20/31=(1.55)^{-1}$.

In the case $(i)$ one has $$C\max_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\ge C(K\lambda_{5,\delta}(\overline p )-\lambda_{5,\delta}( p ))\ge$$ $$\ge C(K\mu_{3,\delta}(1)-\mu_{3,\delta}( 0 ))\ge C(3\sqrt 3+20(\varepsilon-8)/31)>C/30>8(K+1)$$ since $\lambda_{5,\delta}( p )=\mu_{3,\delta}(p)$ is decreasing, $\mu_{3,\delta}(0)=-3\sqrt 3,\mu_{3,\delta}(1)=\varepsilon-8.$

We suppose then $K>20/31=(1.55)^{-1}$. Then if $p\le 3k/4$ one has $$C\max_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\ge CK(\lambda_{3,\delta}(\overline p )-\lambda_{3,\delta}( p )/K) \ge CK(\mu_{4,\delta}(\overline p)-\mu_{1,\delta}( p )/K)\ge$$ $$\ge CK(\mu_{4,\delta}(k)-\mu_{1,\delta}( 3k/4 )/K)=CK(\mu_{1,\delta}(k)-\mu_{1,\delta}( 3k/4 )/K)>$$ $$> CK\mu_{1,\delta}\left( \frac{3k}{4}\right)\left(\frac{\mu_{4,\delta}(k))}{\mu_{1,\delta}\left( \frac{3k}{4}\right)}-\frac{31}{20}\right)\ge$$ $$\ge CK\frac{\mu_{1,\delta}\left( \frac{3k}{4}\right)}{ 100}  \ge CK \frac{2k^3}{100 }>20K > 8(K+1),$$ contradiction for $k\in [0,1/2]$ since $\frac{\mu_{4,\delta}(k))}{\mu_{1,\delta}\left( \frac{3k}{4}\right)}>1.56,\mu_{1,\delta}\left( \frac{3k}{4}\right)>2k^3$ there. Thus we can suppose that $p\in ]\frac{3k}{4}, k] .$ One notes then that $\mu_{4,\delta}(p)\le \mu_{4,\delta}(k)$ for$p\in [\frac{k}{4},1].$ This permits to rule out the case $\overline p\ge \frac{3k}{2}$. Indeed, one has in this case $$C\max_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\ge CK(\lambda_{2,\delta}(\overline p )-\lambda_{2,\delta}( p )/K) \ge CK(\mu_{1,\delta}(\overline p)-\mu_{4,\delta}( p )/K)\ge$$ $$\ge CK(\mu_{1,\delta}(3k/2)-\mu_{1,\delta}( k)/K)=CK(\mu_{1,\delta}(3k/2)-\mu_{1,\delta}( k)/K),$$ and one gets a contradiction as above since $$\frac{\mu_{1,\delta}\left( \frac{3k}{2}\right)}{\mu_{1,\delta}(k)}  >2$$ for $k\in [0,1/2]$.

The last case to rule out is thus $ K\ge 20/31, p\in [\frac{3k}{4},k], \overline p \in [k,\frac{3k}{2}]$. Let then $$\alpha:=k-p\in [0,\frac{k}{4}]
\subset [0,\frac{1}{8}],\;\beta:=\overline p -k\in [0,\frac{k}{2}]\subset [0,\frac{1}{4}],\; a:=\max \{\alpha,\beta\}.$$

It is easy to verify that on $\left[\frac{3k}{4},\frac{3k}{2}\right]$ one has the following inequalities:

$$\frac{\partial \mu_{1,\delta}(p)}{\partial p}\ge   3k^2, \;  \frac{11k^3 }{4}\ge \mu_{1,\delta}(k)\ge \frac{5k^3 }{2}\:;$$ $$\frac{\partial \mu_{2,\delta}(p)}{\partial p}\ge -5, \;4k-5 \ge\mu_{2,\delta}(k)\ge 4k-\frac{11}{2} \:;$$ $$\frac{\partial \mu_{3,\delta}(p)}{\partial p}\ge -\frac{ 9}{2}, \; -5-3k\ge\mu_{3,\delta}(k)\ge -\frac{11+7k  }{2}\:;$$ $$\frac{\partial \mu_{4,\delta}(p)}{\partial p}\ge -\frac{k }{29}, \; \frac{11k^3 }{4}\ge \mu_{4,\delta}(k)\ge \frac{5k^3 }{2}\:;$$ $$\frac{\partial \mu_{5,\delta}(p)}{\partial p}\ge 10k^2-12, \; -10k\ge\mu_{5,\delta}(k)\ge -12k \:.$$ Let then $ K\in \left[\frac{20}{31},1\right].$ Therefore,

$$C\max_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\ge$$ $$\ge C\max \{K\mu_{1,\delta}(k+\alpha )-\mu_{1,\delta}( k-\beta), K\mu_{3,\delta}(k+\alpha ) -\mu_{3,\delta}( k-\beta)\}\ge\max\left\{ M_1,M_2\right\}=$$ $$=C\max\left\{3(K-1)k^3a+3(K+1)k^2, (1-K) (5+3k)a-\frac{ (11+7k ) (K+1)}{10}\right\}$$ for linear forms $M_1,M_2$ in $K$. Note that the minimal value of $\max\{ M_1,M_2\}$ as a function of $K$ equals (recall that our $C=1000/k^4$): $$\frac{1500a(40-9k)}{k^2(12k^2a+18a+11k^3+20+12k)}>\frac{1250a}{k^2}>0$$ attained for $K=K_0:= (11k^3+20+12k)/(12k^2a+18a+11k^3+20+12k)<1$.

On the other hand, $$-\Lambda_5\le -\min_{i=1,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\le \max\{ l_1,l_2,l_3,l_4,l_5\}$$ for the following linear forms (in $K$) $$l_1:=-k^2\left(\frac{11a}{4}+3k\right)K-\frac{11a}{4}k^2+3k^3,$$ $$l_2:=\left(5a+4k-\frac{11}{2}\right)K+5a+\frac{11}{2}-4k,$$ $$l_3:= (5a+5+3k)K+5a-5-3k,$$ $$l_4:=\left(\frac{ak}{29}-\frac{11k^3}{2}\right)K+\frac{ak}{29}+\frac{11k^3}{2},$$ $$l_5:=((12-10k^2)a+12k)K+(12-10k^2)a-12k.$$ To refute our inequality it is sufficient to prove that $M_i(K_{j,k})> 0$ for any triple $(i,j,k)$ with $i,j\in \{1,2\},\; i\neq j, \; k\in \{1,2,3,4,5\}$ where $ l_k(K_{j,k})=M_j(K_{j,k}).$

Explicit calculations give (for the values $m_{ijk}:=\frac{M_i(K_{j,k})}{500ak^2}$) $$\frac{m_{121}}{3} = \frac{(9k^4+6k^6)a+10000+5k^7+20k^4+3k^5-2250k}{k^2((3k^4+3000)a+3k^5+2750k)}>0,$$ $$\frac{m_{211}}{3} = \frac{(9k^4+6k^6)a+10000+5k^7+20k^4+3k^5-2250k}{(4500-3k^6)a+5000+3000k-3k^7}>0,$$ $$m_{122} = \frac{60000-(60k^4+90k^2)a-192k^3+66k^4-13500k-101k^2-158k^5}{k^2((6000-10k^2)a-11k^2+8k^3+5500k)}>0,$$ $$m_{212} = \frac{60000-(60k^4+90k^2)a-192k^3+66k^4-13500k-101k^2-158k^5}{(10k^4+9000)a+11k^4-8k^5+10000+6000k}>0,$$ $$m_{123} = \frac{30000-(45k^2+30k^4)a-55k^2-6750k-33k^3+30k^4-37k^5} {k^2((3000-5k^2)a+2750k-5k^2-3k^3)} >0,$$ $$m_{213} = \frac{30000-(45k^2+30k^4)a-55k^2-6750k-33k^3+30k^4-37k^5} {(5k^4+4500)a+5k^4+3k^5+5000+3000k} >0,$$ $$m_{124} = \frac{3480000-(36k^3+24k^5)a-t(k)} {k^2((348000-4k^3)a+319000k+319k^5)} >0,$$ $$m_{214} = \frac{3480000-(36k^3+24k^5)a-t(k)} {(522000+4k^5)a+580000+348000k-319k^7} >0,$$

$$m_{215} = \frac{(9k^4+30k^6-54k^2)a+15000-u(k)} {5k^4a+1500a-6k^3+1375k-6k^2a} >0,$$ $$m_{125} = \frac{(9k^4+30k^6-54k^2)a+15000-u(k)} {(2250+6k^4-5k^6)a+2500+1500k+6k^5 } >0,$$ where $t(k):=783000k+ 80k^3+48k^4+44k^6+2871k^5+1914k^7<4\cdot 10^5$, $u(k):=120k^2-30k^5+3375k+18k^3-
100k^4-55k^7<2000$.

Let, finally $K\ge 1$, then $$C\Lambda_1\ge C(K\mu_{1,\delta}(k+\alpha )-\mu_{1,\delta}( k-\beta))\ge \frac{5C }{2}(K-1)k^3a+3(K+1)Ck^2=$$ $$=L_1:= \frac{500(5kK+6a-5k)}{k^2}=\frac{2500(K-1)}{k}+\frac{3000a}{k^2},$$ and $$-\Lambda_5\le -\min_{i=2,\cdots,5}\{K\lambda_{i,\delta}(\overline p )-\lambda_{i,\delta}( p )\}\le\max\left\{L_2, L_3,L_4,L_5\right\}$$ for the following linears forms in $K$: $$L_2:=5(K+1)a+(1-K)(5-4k) = (5a-5+4k)(K-1)+10a,$$ $$L_3:=\frac{9}{2}(K+1)a+\frac{11+7k}{2}(K-1) =\frac{9a+11+7k}{2} (K-1)+ 9a,$$ $$L_4:=(K+1)a\frac{k}{29}-\frac{5k^3}{2}(K-1)=\left(\frac{ak}{29}-\frac{5k^3}{2}\right)(K-1)+\frac{2ak}{29},$$ $$L_5:=12k(K+1)a+\frac{9}{2}(K-1)=3\left(4ak+\frac{3}{2}\right)(K-1)+24ak.$$ One immediately sees that both the slope and the value at $K=1$ of $L_1$ are (much) bigger than those of $L_i, i=2,3,4,5$ which finishes the proof.

[*Proof of Theorem 1.1.*]{} To prove the result it is sufficient to verify the condition (2.1) in Lemma 2.1, namely, that the five partial derivatives $\frac{\partial g}{\partial \lambda_i}, i=1,\ldots,5$ are strictly positive (and bounded which is automatic thank to compacity) on the symmetrized image $$M:=\bigcup_{\sigma \in \Sigma_{n} }\   T_{\sigma }\Lambda (B)\subset { \R}^n$$ of the unit ball under the map $\Lambda$, $$g(\lambda_1,\ldots,\lambda_5)=det(D^2w)- e_5(\Delta(w))^5  -e_3(\Delta(w))^3 S_2(w)-e_1\Delta(w) S_4(w)$$ being our equation. By homogeneity it is sufficient to show this on $M':=\Lambda (\S_1^4)$ which is an algebraic curve, the union of 120 curves $T_{\sigma }\Lambda (\S_1^4)$ and that it is sufficient, by symmetry, to verify the condition on the curve $\Lambda (\S_1^4)$ only. A brute force calculation shows then that $$g_1(p,\varepsilon ):=\frac{\partial g}{\partial \lambda_1}=\sum_{i=0}^{12} m_{i}p^{i}  =$$ $$=m_{12}p^8(p^4-12p^2+54)+m_9 b^3+m_6p^4(p^2-\frac{3}{4})+m_2+m_0,$$ with $$m_{12} =3(\varepsilon^4+3\varepsilon^3-20\varepsilon^2
+12\varepsilon-56)(\varepsilon-2)^2(\varepsilon+2)^2,  m_{10} =-12m_{12}, m_{8} =54m_{12},$$ $$m_{11}=m_{1}=0, m_9=3D(p,\varepsilon)(\varepsilon+7)(\varepsilon+2)(\varepsilon^2+2)(\varepsilon-2)^2,m_7=-9m_9,m_5=27m_9,$$ $$m_6= 108(2-\varepsilon)(3\varepsilon^7+17\varepsilon^6-54\varepsilon^5 -152\varepsilon^4+72\varepsilon^3-42\varepsilon^2+ 384\varepsilon+1344), m_3=27m_9,$$ $$m_4=-\frac{3m_6}{4},\;\;m_2=
1944 \varepsilon^2 (2-\varepsilon) (\varepsilon^2  - 7) (\varepsilon^2  +3),\;\; m_0 = -7776 \varepsilon^2  (\varepsilon^2  + 3)$$ for $D(p,\varepsilon):=\sqrt{(16-\varepsilon^2)(4-\varepsilon^2) b^2+144\varepsilon^2}, \; b=(p^2-3)p;$

$$g_2(p,\varepsilon ):=\frac{\partial g}{\partial \lambda_2}=\sum_{i=0}^{12} n_{i}p^{i}$$ with $$n_{12} =(\varepsilon+4)(\varepsilon+1)(4-\varepsilon^2)^2 ,  n_{10} =-(\varepsilon+2)(\varepsilon^4+19\varepsilon^3+86\varepsilon^2+182\varepsilon+96)(2-\varepsilon)^2,$$ $$n_{11}=n_{1}=0, \;\; n_{8} =9(\varepsilon+2)(\varepsilon^2+10\varepsilon+6)(\varepsilon^2+3\varepsilon+8)(2-\varepsilon)^2,$$ $$n_9=\varepsilon(\varepsilon+7)(\varepsilon+2)(\varepsilon^2+2)(2-\varepsilon)^2 \sqrt{3(4-p^2)},\; n_7=-9n_9,\; n_5=27n_9,$$ $$n_{6} =3(2-\varepsilon)(13\varepsilon^6+115\varepsilon^5+218\varepsilon^4+
170\varepsilon^3-876\varepsilon^2-2856\varepsilon-1152),$$ $$n_4=9(\varepsilon-2)(11\varepsilon^6+62\varepsilon^5+33\varepsilon^4
-24\varepsilon^3-348\varepsilon^2-1176\varepsilon-288),$$ $$n_3=\varepsilon^3(2-\varepsilon)(\varepsilon+7)(3\varepsilon^2+2)\sqrt{3(4-p^2)},$$ $$n_2=108\varepsilon^2(2-\varepsilon)
(\varepsilon^2+3)(\varepsilon^2+4\varepsilon-3), n_0=1296\varepsilon^2(\varepsilon^2+3);$$

$$g_3(p,\varepsilon ):= \frac{\partial g}{\partial \lambda_3}=\sum_{i=0}^{6} h_{2i}p^{2i}$$ with

$$h_{12} = (\varepsilon + 4) (\varepsilon + 1) (4-\varepsilon^2) ^2, h_{10}  = 2 (\varepsilon + 2)(\varepsilon^4  + \varepsilon^3  - 40 \varepsilon^2 - 70\varepsilon - 48)(2-\varepsilon )^2$$ $$h_{8} = -18 (\varepsilon + 2)(\varepsilon^4  + 4\varepsilon^3  - 19\varepsilon^2  - 28\varepsilon - 24)(2-\varepsilon )^2,$$ $$h_6 = 6 (\varepsilon - 2) (7\varepsilon^6  + 37\varepsilon^5  - 136 \varepsilon^4 - 274 \varepsilon^3  + 330\varepsilon^2  + 672\varepsilon + 576),$$ $$h_4 = 9 (\varepsilon - 2) (2 \varepsilon^6  - \varepsilon^5 + 27\varepsilon^4- 66\varepsilon^3  - 348\varepsilon^2  - 1176\varepsilon - 288),$$ $$h_2=108\varepsilon(2-\varepsilon)(\varepsilon^3+4\varepsilon^2- 15\varepsilon-84) (\varepsilon^2+3),$$ $$h_0 = -1296\varepsilon(\varepsilon^2  + 3) (\varepsilon^2  + 4\varepsilon - 14);$$

$$g_4(p,\varepsilon ):=\frac{\partial g}{\partial \lambda_4}=g_1(-p,\varepsilon ),$$ $$g_5(p),\varepsilon :=\frac{\partial g}{\partial \lambda_5}=g_2(-p,\varepsilon ),$$ and thus we need to consider only the functions $ g_1(p,\varepsilon ), g_2(p,\varepsilon ), g_3(p,\varepsilon )$ on the set $[-1,1]\times(0,1].$ We have to prove that for any fixed $\varepsilon\in (0,1]$ they are strictly positive.

The technique of the proof is identical for all three derivatives, and we begin with $g_3$ wich is slightly simpler since it is a polynomial in two variables. One can rearrange it in the form $$g_3(p,\varepsilon )= g_{37}\varepsilon^7+g_{36}\varepsilon^6+g_{35}\varepsilon^5+
g_{34}\varepsilon^4+g_{33}\varepsilon^3+g_{32}\varepsilon^2+
g_{31}\varepsilon+g_{30}$$ with $$g_{37}=2 q^5+42 q^3-18 q^4-108 q+18 q^2,\; g_{36}=138 q^3-2 q^5-
216 q+q^6-36 q^4-45 q^2,$$ $$g_{35}=-92 q^5+558 q^4+2160 q+5 q^6-1296-1260 q^3+261 q^2,$$ $$g_{34}=-4 q^6+5184 q-12 q^3-
5184-1080 q^2+28 q^5-36 q^4,$$ $$g_{33}=-1944 q^2-2520 q^4-40 q^6
+14256-10692 q+5268 q^3+520 q^5,$$ $$g_{32}=-144 q^4+72 q^3+112 q^5+17496 q-4320 q^2-16 q^6-15552,$$ $$g_{31}=-736 q^5+80 q^6+2304 q^4-54432 q+54432-4608 q^3
+18576 q^2,\:g_{30}=64q^2(q-3)^4\ge 0$$ for $q=p^2\in [0,1]$. Therefore, $$g_3(p,\varepsilon )\ge \varepsilon(\bar g_{37}\varepsilon^6+\bar g_{36}\varepsilon^5+\bar g_{35}\varepsilon^4+
\bar g_{34}\varepsilon^3+ \bar g_{33}\varepsilon^2+\bar g_{32}\varepsilon+
\bar g_{31})$$ where $\bar g_{3i}:=\min_{q\in [0,1]}g_{3i}(q)$, and an elementary calculation gives $$\frac{g_3(p,\varepsilon )}{\varepsilon}\ge   -64\varepsilon^5-160\varepsilon^4- 5184\varepsilon^3+ 4848\varepsilon^2-15552\varepsilon+15616 >1620$$ for $\varepsilon\in (0,\frac{9}{10}]$. For $\varepsilon\in [\frac{9}{10},1]$ we have ${g_3(p,\varepsilon )}\ge\sum_{i=0}^{6}\bar h_{2i}q^{i}$ where $\bar h_{2i}:=\min_{\varepsilon\in [\frac{9}{10},1]} h_{2i}$ and thus $${g_3(p,\varepsilon )}\ge -736q^5+80q^6+2304q^4-54432q+54432-4608q^3+18576q^2>4840.$$ Thus, finally ${g_3(p,\varepsilon )}\ge \min \{ 1620\varepsilon,4840\}.$

The function ${g_1(p,\varepsilon )}=s_1-t_1$ with $$s_1:=(q^3-6q^2+9q) \varepsilon^7+(5q^3-30q^2+45q-72)\varepsilon^6+
(108q^2-18q^3-162q)\varepsilon^5+$$ $$(-432q+288-48q^3+288q^2)
\varepsilon^4+(24q^3-144q^2+216q)\varepsilon^3+$$ $$1512\varepsilon^2+
(1152q-768q^2+128q^3)\varepsilon+448q^2(q-3)^4\ge$$ $$-72\varepsilon^6-72\varepsilon^5+96\varepsilon^4+1512\varepsilon^2\ge 1440\varepsilon^2>0,$$ and $$t_1:=( \varepsilon^2-4)(\varepsilon+7) (\varepsilon^2+2)bD(p,\varepsilon);$$ simplifying $s_1^2-t_1^2=(s_1-t_1)(s_1+t_1)=g_1(p,\varepsilon )(s_1+t_1)$ one finds $$(540 q^2-1296 q-216 q^4-4 q^6+288 q^3+48 q^5) \varepsilon^{13}+$$ $$(-1944 q^4+3024 q^3+432 q^5-7776 q+5184+2268 q^2-36 q^6) 
\varepsilon^{12}+$$ $$(-2052 q^2+3240 q^4-5328 q^3-720 q^5+60 q^6+10368 q)
\varepsilon^{11}+$$ $$(936 q^6+50544 q^4+55944 q^2-11232 q^5-41472-97776 q^3
+29808 q)\varepsilon^{10}$$ $$+(26136 q^2+120 q^6-15696 q^3+6480 q^4-24624 q
-1440 q^5) \varepsilon^9+$$ $$(57456 q^5+156816 q-492372 q^2-4788 q^6-134784
+534528 q^3-258552 q^4)\varepsilon^8+$$ $$(180576 q^2-352 q^6-313632 q+
3168 q^3+4224 q^5-19008 q^4)\varepsilon^7+$$ $$(54432 q^4-238464 q^3+
859248 q^2-1166400 q+1008 q^6+870912-12096 q^5)\varepsilon^6+$$ $$(1118592 q^3-518400 q^4-9600 q^6-1268352 q^2+736128 q +115200 q^5)
 \varepsilon^5+$$ $$(1524096 q^4+207360 q+2286144+28224 q^6-338688 q^5+
2147904 q^2-3025152 q^3) \varepsilon^4+$$ $$(10752 q^6+580608 q^4-
903168 q^3-677376 q^2+2322432 q-129024 q^5)\varepsilon^3+$$ $$(3640320 q^3-25344 q^6+304128 q^5-1368576 q^4+ 8128512 q-
7471872 q^2)\varepsilon^2+$$ $$(3096576 q^4+4644864 q^2+57344 q^6-688128 q^5-6193152 q^3) \varepsilon\ge$$ $$64\varepsilon^4(-10\varepsilon^9+ 18\varepsilon^8- 648\varepsilon^6- 141\varepsilon^5-2214\varepsilon^4- 2266\varepsilon^3+5760\varepsilon^2 +35721)\ge 2.2\cdot 10^6\varepsilon^4.$$ Since $s_1+t_1\le 10^6$ one gets ${g_1(p,\varepsilon )}>2\varepsilon^4.$

Similarly, ${g_2(p,\varepsilon )}=s_2-t_2$ with a polynomial $s_2\ge 3000\varepsilon^2$ and $$t_2 =\varepsilon (\varepsilon+7) (\varepsilon-2)t(\varepsilon,q) p^3\sqrt{3(4-q)}$$ where $$t(\varepsilon,q):=  (\varepsilon^4-2 \varepsilon^2-8) q^3+9(- \varepsilon^4+2\varepsilon^2+8) q^2
+27 (\varepsilon^4-2\varepsilon^2-8) p^2-9(3\varepsilon^2+2) \varepsilon^2 .$$ Simplifying $s_2^2-t_2^2 =g_2(p,\varepsilon )(s_2+t_2)$ one gets a polynomial $\ge$ $$(-2560\varepsilon^9-18176\varepsilon^8-325632\varepsilon^6 -1254656\varepsilon^4  +2202112\varepsilon^2+15116544) \varepsilon^4\ge 1.5\cdot 10^7 \varepsilon^4$$ and $g_2(p,\varepsilon )\ge 15\varepsilon^4$ which finishes the proof.

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[^1]: CMI, 39, rue F. Joliot-Curie, 13453 Marseille FRANCE, nicolas@cmi.univ-mrs.fr

[^2]: IML, Luminy, case 907, 13288 Marseille Cedex FRANCE and IITP RAS, B.Karetnyi,9, Moscow, RUSSIA, vladut@iml.univ-mrs.fr
