\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small Eighth Mississippi State - UAB Conference on Differential Equations and Computational Simulations. {\em Electronic Journal of Differential Equations}, Conf. 19 (2010), pp. 257--266.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \setcounter{page}{257} \title[\hfilneg EJDE-2010/Conf/19/\hfil Lipschitz constants] {Lipschitz constants for positive solutions of second-order elliptic equations} \author[A. Vitolo\hfil EJDE/Conf/19 \hfilneg] {Antonio Vitolo} \address{Antonio Vitolo \newline Department of Mathematics and Computer Science, University of Salerno, p. Grahamstown, I-84084 Fisciano (SA), Italy} \email{vitolo@unisa.it} \thanks{Published September 25, 2010.} \subjclass[2000]{35J60, 45M20, 49L25} \keywords{Fully nonlinear second order elliptic equations; positive solutions; \hfill\break\indent viscosity solutions} \begin{abstract} We are concerned with positive solutions of second order fully nonlinear elliptic equations. Here we present Lipschitz estimates, in the viscosity setting, and bounds for optimal constants. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction and statement of the results} In a series of papers \cite{CDV(CRAS),CDV(Mosco),CDV(Preprint)}, jointly with Italo Capuzzo Dolcetta, the author proved the inequalities $$\label{LiNi} C^{-1}|Du(x)| \le \begin{cases} \sqrt{u(0)M} & \text{if } 2|x| \le \sqrt{\frac{2u(0)}{M}} \le R \$4pt] \frac{u(0)}R + MR & \text{if } 2|x| \le R \le \sqrt{\frac{2u(0)}{M}}\,, \end{cases}$$ for a positive constant C, when u \ge 0 and  |F(x,Du(x),D^2u(x))|\le M  in the ball B_R of radius R>0 centered at the origin. Here F is a second-order uniformly elliptic operator satisfying suitable assumptions, that will be made precise in the sequel. The estimate \eqref{LiNi} extends the Glaeser's inequality \cite{Gl}, $$\label{Glaeser} |u'(0)| \le \begin{cases} \sqrt{2u(0)M} & \text{if } R \ge \sqrt{\frac{2u(0)}M} \\ \frac{u(0)}{R} + MR & \text{if } R < \sqrt{\frac{2u(0)}M} \end{cases}$$ for non-negative C^2-functions u with |u''| \le M in [-R,R], which is a local version of the Landau inequality \cite{L} $$\label{Landau} |u'(0)| \le \sqrt{2\sup_{\mathbb R}|u|\, M}$$ for C^2-functions u such that |u''| \le M on the whole real axis, see \cite{MK} and \cite{MS} for other variants. Applications of this kind of inequalities can be found for instance in \cite{NT} and \cite{CK}. In the case of a linear second-order uniformly elliptic operator  Lu:=a_{ij}(x)D_{ij}u+b_i(x)D_iu  the inequalities \eqref{LiNi} for non-negative functions u \in C^2(\overline B_R) such that  |Lu|\le M  are due to Yan Yan Li and Louis Nirenberg \cite{LN}. They also observe that C=\sqrt2 is optimal for n=1 in the first of \eqref{LiNi} and the best constant in higher dimensions is not known even for the Laplace operator. Concerning this, denote by B_R the ball of radius R>0 centered at the origin in \mathbb R^n and consider the set \mathfrak{F}_{n} of all pairs (u,B_R) such that u \in C^2(\overline B_R) and \[ u \ge 0 \quad \text{\rm in } B_R, \quad \sup_{B_R}|\Delta u| :=M_R \ge \frac{2u(0)}{R^2}\,.$ For $(u,B_R) \in \mathfrak{F}_{n}$ let $C_u(R)$ be the greatest lower bound of the positive real numbers $C$ such that $$\label{interpolation} |Du(0)| \le C\sqrt{u(0)M_R}\,.$$ If $u(0)=0$, then $u$ has a local mininum at $x=0$ so that $Du(0)=0$ and $C_u(R)=0$. The same happens if $M_R=0$. Otherwise $$C_u(R) = \frac{|Du(0)|}{\sqrt{u(0)\,M_R}}\,.$$ Denote by $\overline C_n$ the optimal constant (in the class $\mathfrak{F}_{n}$) $$\overline C_{n}:= \sup_{(u,B_R) \in \mathfrak{F}_n} \,C_u(R)\,.$$ Note that, if $(u,B_R) \in \mathfrak{F}_n$ is such that $M_R>\frac{2u(0)}{R^2}$, by continuity we find $r\in (0,R)$ such that $M_r=\frac{2u(0)}{r^2}$. Since obviously $(u,B_r) \in \mathfrak{F}_n$ and $C_u(r) \ge C_u(R)$, then $\overline C_n$ can be computed as the least upper bound of $C_u(R)$ over all the pairs $(u,B_R) \in \mathfrak{F}_n$ such that $$\sup_{B_R}|\Delta u| := M_R = \frac{2u(0)}{R^2}\,.$$ Note also that we can also include in $\mathfrak{F}_2$ the limit case $u \ge 0$ in $B_\infty=\mathbb R$ and $M_\infty=0$. By the Liouville theorem we shall have $C_u(\infty)=0$. According to the above notations, the quoted result of Li and Nirenberg \cite{LN} can be reformulated saying that $$\overline C_{1}=\sqrt{2}.$$ The following Example shows that for $n \ge 2$ we have instead $$\overline C_{n}\ge 3/2.$$ \begin{example}\label{C=3/2} {\rm For the polynomial $$u(x,y)=x^2-\frac15\,y^2-2x+\frac{10}9,$$ we have $\Delta u = 8/5$ and $u \ge 0$ in $B_R$ with $R=\frac56\sqrt{2}$. Therefore, $u \in \mathfrak{F}_2$ and $$C_u(R)= \frac{|Du(0,0)|}{\sqrt{u(0,0)\,\Delta u}} = \frac 32\,.$$ }\end{example} In Section \ref{constant} we see that this is the optimal constant in the class $\mathfrak{F}_{2,2}$ of the pairs $(u,B_R) \in \mathfrak{F}_{2}$ such that $u$ is a polynomial of degree $\le 2$ in $B_R$, i.e. $$\sup_{(u,B_R) \in \mathfrak{F}_{2,2}}C_u(R) =3/2.$$ From the qualitative viewpoint we will consider fully nonlinear second-order uniformly elliptic operators $$F(x,Du,D^2u)$$ with at most linear growth in the gradient, see the definition in the next Section. Here $Du$ and $D^2u$ denote the gradient vector and the Hessian matrix of $u$. This is the case of linear second-order uniformly elliptic operators $$Lu = \operatorname{Tr}(A(x)D^2u)+b_i(x)D_iu\,,$$ where the $A(x):=[a_{ij}(x)]$ has eigenvalues in $[\lambda,\Lambda]$ and $|b_i(x)| \le b_0$ for positive constants $\lambda \le \Lambda$ (ellipticity constants) and $b_0$ (first order constant). Different examples of fully nonlinear elliptic operators are the upper and lower envelopes of linear uniformly elliptic operators $L_k$ or $L_{hk}$ with (positive) ellipticity constants $\lambda$ and $\Lambda$ ($\ge \lambda$), for instance Bellman operators $$F(x,Du,D^2u) := \inf_{k}L_k u$$ and Isaacs operators $$F(x,Du,D^2u) := \sup_{h}\inf_{k}L_{hk}u\,,$$ arising in optimal control problems and differential games. Taking the upper and lower envelope of the totality of linear uniformly elliptic operators with (positive) ellipticity constants $\Lambda$ and $\Lambda$ ($\ge \lambda$), we obtain the maximal and minimal Pucci operators ${\mathcal P}^+_{\lambda,\Lambda}(D^2u)$ and ${\mathcal P}^+_{\lambda,\Lambda}(D^2u)$, such that \begin{gather*} {\mathcal P}^+_{\lambda,\Lambda}(Z) :=\Lambda \operatorname{Tr}(Z^+)-\lambda \operatorname{Tr}(Z^-)\,, \\ {\mathcal P}^-_{\lambda,\Lambda}(Z) :=\lambda \operatorname{Tr}(Z^+)-\Lambda \operatorname{Tr}(Z^-) \end{gather*} for all $n \times n$ real symmetric matrices $Z$. Note that ${\mathcal P}^\pm_{1,1}(D^2u)$ is the Laplace operator $\Delta u$. For more examples and results about elliptic differential operators we refer to \cite{GT} and \cite{CC}. In \cite{CDV(Mosco)} and \cite{CDV(Preprint)}, based on the perturbation method of Caffarelli \cite{Caf}, qualitative Glaeser's type results in $B_R$ are obtained by restraining the oscillations of the main term with respect to the $x$-variable. There, letting $$\beta(x,y):=\sup_{X\neq 0}\frac{|F(x,0,X)-F(y,0,X)|}{|X|}\,,$$ it is required, for some $\tau \in (0,\frac12)$, $$\label{Caffarelli} \sup_{y \in B_{R/2}}\Big(-\hskip-0.4cm\int_{B_{\tau R}(y)} |\beta(x,y)|^n\,dx\Big)^{1/n} \le \theta$$ with a suitably small positive constant $\theta$. \begin{remark}\label{Caffarelli-linear} {\rm The integral condition \eqref{Caffarelli} allows a generalization the Cordes-Nirenberg type estimates here below. Let $L$ be a linear second-order uniformly elliptic operator with ellipticity constants $\lambda, \Lambda$ and first order constant $b_0$, and $f \in L^\infty(B_1)$. For any $\alpha \in (0,1)$ there exists $\theta$ such that, if $$|a_{ij}(x)-a_{ij}(y)| \le \theta, \quad x,y \in \overline B_{1/2}, \quad |x-y|< r_0\quad (< \tfrac12)\,,$$ then a bounded solution $u$ of the equation $$Lu =f(x) \quad \text{in } B_1$$ is $C^{1,\alpha}(\overline B_{1/2})$ and $$\|u\|_{C^{1,\alpha}(\overline B_{1/2})} \le C(\|u\|_{L^\infty(B_1)}+\|f\|_{L^\infty(B_1)})\,,$$ see \cite{Caf}. This is the case of continuous coefficients $a_{ij}(x)$ with a sufficiently (uniformly) small modulus of continuity. }\end{remark} The integral condition \eqref{Caffarelli} is used by Caffarelli \cite{Caf} to have $C^{1,\alpha}$ estimates for viscosity solutions, see also \cite{Tr}, \cite{Sw} and \cite{CDV(Preprint)}. Suppose now that $$\beta(x,y)=\mu(|x-y|),$$ where $\mu:[0,2R] \to [0,+\infty)$. Combined with the uniform ellipticity, from which (recall that $F(x,0,0)=0$) \begin{align*} \beta(x,y)&\le \frac{|F(x,0,X)-F(x,0,0)}{|X|} +\frac{|F(y,0,0)-F(y,0,X)|}{|X|} \\ &\le 2\,\frac{{\mathcal P}^+_{\lambda,\Lambda}(X)}{|X|} \le 2\Lambda, \end{align*} this yields $$-\hskip-0.4cm \int_{B_{r_0}(y)} \beta(x,y)^n\,dx \le n\; 2^{n-1} \Lambda^{n-1} \int_0^{r_0} \frac{\mu(\sigma)}\sigma\,d\sigma.$$ for $r_0 \in (0,R)$. Hence, according to condition \eqref{Caffarelli}, in order to have $C^{1,\alpha}$-estimates we should require the latter integral to be small. However, if we only ask for Lipschitz estimates, then by Ishii-Lions \cite{IL} we may just assume that integral to be finite. \begin{theorem}\label{main} Let $F$ be uniformly elliptic with ellipticity constants $\lambda >0$, $\Lambda \ge \lambda$ and first order constant $b_0 >0$. Suppose that $F$ is continuous, $F(x,0,0)=0$ and $$\label{Ishii-Lions} |F(x,\xi,X)-F(y,\xi,X)| \le \mu(|x-y|)|X|+\omega(|x-y|)|\xi|$$ for $x,y \in B_R$, $\xi \in \mathbb R^n$, $X \in {\mathcal S}^n$, where $\mu, \omega$ are non-negative real functions such that $$\int_{0}^{2R}\frac{\mu(\sigma)}{\sigma}\,d\sigma<+\infty$$ and $\lim_{\sigma\to 0^+}\omega(\sigma)=0$. Let $u \in C(\overline B_R)$ be a viscosity solution of the equation $$F(x,Du,D^2u)=f(x)$$ in $B_R$ with $f \in C(B_R)$ such that $\|f\|_{L^\infty(B_R)}=M < +\infty$. Then $u$ is locally Lipschitz continuous in $B_R$. If in addition we assume that $u \ge 0$, then there exists a positive constant $C=C(n,\lambda,\Lambda, b_0R, \mu, \omega)$ such that $$\label{LiNi-Lip} C^{-1}\limsup_{y \to x}\frac{|u(x)-u(y)|}{|x-y|} \le\begin{cases} \sqrt{u(0)M} & \text{if } 2|x| \le \sqrt{\frac{2u(0)}{M}} \le R\\ \frac{u(0)}R + MR & \text{if } 2|x| \le R \le \sqrt{\frac{2u(0)}{M}}\,. \end{cases}$$ \end{theorem} \begin{remark} {\rm By Rademacher's Theorem it follows from Theorem \ref{main} that the inequalities \eqref{LiNi} hold true almost everywhere for $x \in \overline B_{R/2}$ with a positive constant $C$ depending on $n,\lambda,\Lambda,b_0R,\mu,\omega$, and everywhere if $u$ is differentiable in $\overline B_{R/2}$. }\end{remark} This Theorem is based on a result of Ishii-Lions \cite[Theorem VII.2]{IL} which yields the Lipschitz continuity of viscosity solutions under assumptions which are guaranteed by \eqref{Ishii-Lions}. The proof will be given in Section \ref{proof} after a few preliminaries about viscosity solutions. In Section \ref{constant} we will be concerned with a lower bound for the optimal constant, as mentioned in advance. \section{Lipschitz inequalities for viscosity solutions} \label{proof} We start with the basic notations and definitions. Throughout the paper $B_r(y)$ will be a ball of radius $r >0$ centered at $y \in \mathbb R^n$ and $B_r:=B_r(0)$. By ${\mathcal S}^n$ we denote the set of $n \times n$ real symmetric matrices endowed with the partial ordering induced by semidefinite positiveness. Let $\Omega$ be a domain (open connected set) of $\mathbb R^n$, then $F:\Omega \times \mathbb R^n \times {\mathcal S}^n \to \mathbb R$ is said uniformly elliptic in $\Omega$ with ellipticity constants $\lambda>0$ and $\Lambda \ge \lambda$ and first order constant $b_0>0$ if $${\mathcal P}^-_{\lambda,\Lambda}(Z)-b_0|\zeta| \le F(x,\xi+\zeta,X+Z)-F(x,\xi,X) \le {\mathcal P}^+_{\lambda,\Lambda}(Z)+b_0|\zeta|$$ for all $x \in \Omega$, $\xi,\zeta \in \mathbb R^n$ and $X,Z \in {\mathcal S}^n$. A viscosity subsolution $u$ of the equation $F(x,Du,D^2u)=f(x)$ is a function $u \in C(\Omega)$ such that for all $(y, \varphi) \in \Omega \times C^2(B_r(y))$ a local maximum for $u-\varphi$ at $y$ implies $$F(y,D\varphi(y),D^2\varphi(y))\ge f(y).$$ Similarly for a viscosity supersolution $u \in C(\Omega)$, a local minimum for $u-\varphi$ at $y$ will imply $$F(y,D\varphi(y),D^2\varphi(y))\le f(y).$$ The viscosity solutions are both viscosity subsolutions and supersolutions.\\ We recall that in the above definition we may equivalently require that the local maximum or minimum is equal zero, so that the graph of the test function $\varphi$ touches above or below, respectively, that one of the solution $u$. For a widespread treatment of viscosity solutions we refer to \cite{CIL} and \cite{CC}. \begin{proof}[Proof of Theorem \ref{main}] It is sufficient to consider the case $R=1$, we can use a rescaling argument for arbitrary $R >0$ . \\ From \cite[Theorem VII.2]{IL}, which provides the Lipschitz continuity of $u$, we deduce for $\|u\|_{L^{\infty}(B_{3/4})}, \|f\|_{L^\infty(B_{3/4})} \le 1$ the inequality $$\label{Lip-anorm} |u(x)-u(y)| \le C|x-y|, \quad x,y \in \overline B_{1/2},$$ with a positive constant $C$ depending on $n$, $\lambda$, $\Lambda$, $b_0$, $\mu$ and $\omega$. For general $u$ and $f$, we set $$K:= \|u\|_{L^{\infty}(B_{3/4})}+\|f\|_{L^\infty(B_{3/4})}$$ and $v = K^{-1}u$. Since $F(x,Du,D^2u)=f(x)$ then $$G(x,Dv,D^2v)=g(x),$$ where $$G(x,\xi,X)= K^{-1}F(x,K\xi,KX), \ \ g(x)=K^{-1}f(x).$$ Since $G$ satisfies, as $F$ does, the structure conditions of \cite[Theorem VII.2]{IL} with $\|v\|_{L^{\infty}(B_{3/4})}, \|g\|_{L^\infty(B_{3/4})} \le 1$, then from \eqref{Lip-anorm} we have $$\label{Lip-norm} \sup_{x,y\in B_{1/2},\, x\neq y} \frac{|u(x)-u(y)|}{|x-y|} \le C(\|u\|_{L^{\infty}(B_{3/4})} +\|f\|_{L^\infty(B_{3/4})})\,.$$ For $u \ge 0$ in $B_1$, using the Harnack inequality (see for instance \cite{CC,KT,ARV}) we get $$\label{Lip-norm-u0} \sup_{x,y\in B_{1/2},\, x\neq y} \frac{|u(x)-u(y)|}{|x-y|} \le C(u(0)+\|f\|_{L^\infty(B_1)})$$ with a possibly different constant $C >0$. Next, for each point $x_0 \in \overline B_{1/2}$ we localize the equation in $B_{r}(x_0)$ with $r \in (0,\frac12)$ and rescale setting $$u(x)=v(r^{-1}(x-x_0)), \quad x \in B_{r}(x_0).$$ Then $v \in C(\overline B_1)$ satisfies the equation $$\label{local-expanded} G(y,Dv,D^2v)=g(y)$$ where $$G(y,\eta,Y):= r^2F(x_0+ry,r^{-1}\eta, r^{-2}Y), \quad g(y) = r^{2}f(x_0+ry).$$ Note that $G$ is uniformly elliptic in $B_1$ with elliptic constants $\lambda,\Lambda$ and first order constant coefficient $b_0r$. Moreover $G$ satisfies \eqref{Ishii-Lions} with $R=1$ and slightly modified $\mu$ and $\omega$, namely $$|G(x,\xi,X)-G(y,\xi,X)| \le \mu(r|x-y|)|X| + r\omega(r|x-y|)|\xi|.$$ So by \eqref{Lip-norm-u0} we can infer that $$\label{Lip-local} \sup_{x,y\in B_{r/2}(x_0) ,\, x\neq y} \frac{|u(x)-u(y)|}{|x-y|} \le C\Big(\frac{u(x_0)}{r}+r\|f\|_{L^\infty(B_1)}\Big).$$ From this, again by the Harnack inequality, it follows that $$\label{Lip-normb} \limsup_{{x \to x_0}}\frac{|u(x)-u(x_0)|}{|x-x_0|} \le C\Big(\frac{u(0)}{r}+\frac{Mr}2\Big), \quad r\le 1\,,$$ for a possibly larger constant $C$. Therefore, minimizing the right-hand side by the choice $r=\sqrt{\frac{2u(0)}M}$ if $u(0)\le \frac M2$, $r=1$ otherwise, we obtain the result in the case $R=1$, as it suffices. \end{proof} \section{Computation for optimal constants} \label{constant} Let $u$ be a $C^2$-function in $B_R$ and $M_u(x)$ be the maximum modulus of the eigenvalues of $D^2u(x)$. The natural extension of the Glaeser's one-dimensional inequality \eqref{Glaeser} to higher dimensions would be the following. Suppose that $u(x) \ge 0$, $\sup_{x\in B_R}M_u(x)= M \ge \frac{2u(0)}{R^2}$. Using the Taylor's formula for $|x|1$. This is not the case of convex or concave functions, because $|\Delta u | \le M$ implies $M_u(x)\le M$ and hence again $C_u(R) = \sqrt2$, but it may happen as soon as the eigenvalues of $D^2u$ have different sign, as Example \ref{C=3/2} shows. The remaining part of this Section is essentially devoted to prove that $C_u:=C_u(R) \le \frac32$ for any $(u,B_R) \in \mathfrak{F}_{2,2}$. Indeed, let $$u(x,y) = a_{11}x^2+2a_{12}xy+a_{22}y^2+2a_{13}x+2a_{23}y+a_{33}$$ be a quadratic polynomial in $\mathfrak{F}_{2}$, namely $$u \ge 0 \quad \text{in } B_R, \quad |\Delta u| = 2 |a_{11}+ a_{22}| = M, \quad a_{33} = \tfrac12 M R^2.$$ We may suppose $a_{11}+ a_{22} \ge 0$, otherwise we consider the polynomial $$v(x,y) = u(x,y) - 2(a_{11}+a_{22})x^2.$$ Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of the quadratic form $$q(x,y) = a_{11}x^2+2a_{12}xy+a_{22}y^2.$$ Since $\lambda_1\lambda_2 \ge 0$ implies either the convexity or the concavity of $u$, from the above discussion we still have in this case $C_u \le \sqrt2$. Thus we are left with $\lambda_1\lambda_2 < 0$, and we may assume, eventually interchanging the axes with each other: $$- \lambda_1 < \lambda_2 < 0 < \lambda_1\,.$$ Introducing the parameter $\alpha =-\lambda_{2}/\lambda_{1}$, the above reads $0 < \alpha < 1$. \begin{remark}\label{alpha-forbid} {\rm As a matter of fact, since $u \in \mathfrak{F}_{2}$, we cannot have $$\frac12 < \alpha < 1.$$ To see this, it is convenient to put $u$ in the form $$u(x,y) = \lambda_1(x-x_0)^2+\lambda_2(y+y_0)^2+c$$ with $x_0 \ge 0$, $y_0 \ge 0$. By positivity in polar coordinates $x=r \cos \theta$, $y=r \sin \theta$ we have \label{positivity} \begin{aligned} &\lambda_{1}r^2\cos^2\theta+\lambda_{2}r^2\sin^2\theta -2\lambda_{1}x_0r\cos\theta\\ &+\lambda_{2}y_0r\sin\theta+\lambda_{1}x_0^2+\lambda_{2}y_0^2+c\ge 0 \end{aligned} for all $r \in (0,R)$ and $\theta \in [0,2\pi]$. Since $u \in \mathfrak{F}_2$, $$\label{u0} \lambda_{1}x_0^2+\lambda_{2}y_0^2+c =: u(0,0)= \frac12MR^2:=(\lambda_1+\lambda_2)R^2,$$ and substituting in \eqref{positivity} we obtain \label{compatibility} \begin{aligned} &\lambda_{1}R^2\cos^2\theta+\lambda_{2}R^2\sin^2\theta -2\lambda_{1}x_0R\cos\theta\\ &+2\lambda_{2}y_0R\sin\theta+(\lambda_{1}+\lambda_{2}) R^2\ge 0. \end{aligned} Dividing both the sides of \eqref{compatibility} by $\lambda_{1}$, we obtain $$\label{compatibility-a} (1-\alpha+\cos^2\theta-\alpha \sin^2\theta)R \ge 2\,(x_0\cos\theta +\alpha y_0\sin\theta)\,.$$ Since $y_0 \ge 0$, computing the above for $\theta = \pi/2$ we deduce that $\alpha \le \frac12$, as claimed. }\end{remark} For $\alpha =1/2$, again taking $\theta = \pi/2$ in \eqref{compatibility-a}, we have $y_0=0$ and \eqref{compatibility-a} implies $$\frac32\cos^2\theta - \frac{2x_0}R\cos\theta \ge 0, \quad \theta \in [0,2\pi],$$ which implies $x_0=0$. Therefore, the standard equation is $$u(x,y)=\lambda_1x^2+\lambda_2y^2+c,$$ so $Du(0,0)=0$ and consequently $C_u=0$. Hence, by Remark \ref{alpha-forbid}, we are left with $\alpha \in (0,\frac12)$. In this case we derive from \eqref{compatibility-a} the lower bound $$\label{R} R \ge 2 \max_{\theta \in [0, \pi/2]} \frac{x_0\cos\theta +\alpha y_0\sin\theta}{1-\alpha+\cos^2\theta-\alpha \sin^2\theta}.$$ to estimate $$\label{Cu} C_u = \frac{|Du(0,0)|}{\sqrt{u(0,0)\Delta u}} = \frac{\sqrt2}R\,\frac{\sqrt{\lambda_{1}^2x_0^2+\lambda_{2}^2y_0^2}}{\lambda_{1}+\lambda_{2}}= \frac{\sqrt2}R\,\frac{\sqrt{x_0^2+\alpha^2y_0^2}}{1-\alpha}\,.$$ If $x_0=0$ (with $y_0\not=0$, otherwise $C_u=0$), using the above inequalities with $\theta=\pi/2$ and recalling that $\alpha < 1/2$, we have $$C_u \le \frac{\sqrt2}2\,\frac{1-2\alpha}{1-\alpha} \le \frac 1 {\sqrt2}\,.$$ Next, we set $\xi=y_0/x_0$ and $t=\tan \theta$. If $x_0>0$ then, combining inequalities \eqref{R} and \eqref{Cu} once more, we obtain \begin{align*} C_u &\le \sqrt2 \inf_{\theta \in (0,\pi/2)}\, \frac{1-\alpha+\cos^2\theta-\alpha \sin^2\theta} {2(x_0\cos\theta +\alpha y_0\sin\theta)}\, \frac{\sqrt{x_0^2+\alpha^2y_0^2}}{1-\alpha}\\ &\le \sqrt2 \sup_{0<\alpha<1/2,\,\xi >0} \inf_{t >0} \frac{\sqrt{1+\alpha^2\xi^2}}{1-\alpha}\, \frac{2-\alpha +(1-2\alpha) {t^2}}{2(1+\alpha \xi t)\sqrt{1+t^2}}\\ &\le \sqrt2\quad \inf_{t >0}\,\sup_{0<\alpha<1/2,\, \xi >0} \frac{\sqrt{1+\alpha^2\xi^2}}{1-\alpha}\, \frac{2-\alpha +(1-2\alpha) {t^2}}{2(1+\alpha \xi t)\sqrt{1+t^2}}\\ &\le \sqrt2\quad \inf_{t >0}\,\frac{\sigma(t)}{2\sqrt{1+t^2}}\,, \end{align*} where $$\sigma(t):=\sup_{0<\alpha<1/2,\, \xi >0} \frac{2-\alpha+(1-2\alpha) {t^2}}{1-\alpha}\, \frac{\sqrt{1+\xi^2}}{1+\xi t}.$$ Observing that $$\sigma(t)\le \begin{cases} 3/t & \text{if } t \le 1\\ 2+t^2 & \text{if } t > 1 \end{cases}$$ and choosing $t=1$ we finally get $$C_u \le \sqrt2\,\inf_{t}\frac{\sigma(t)}{2\sqrt{1+t^2}} \le \frac32\,.$$ The above discussion shows that $C=3/2$ is an upper bound for the optimal constant in $\mathfrak{F}_{2,2}$. 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