## Tag Archives: fully nonlinear

### Some calculation of sigma2 II

We want find the critical metric of the following functional restricted to the space of conformal metrics of unit volume

$\displaystyle I(g)=\frac{n-4}{2}t\int_M J^2_gd\mu_g+4\int_M\sigma_2(A_g)d\mu_g$

Here ${(M^n,g)}$ is a Riemannian metric with ${n\geq 5}$, ${A}$ is the Schouten tensor and ${J}$ is ${tr_gA}$.

Now let us differentiate the two terms respectively, suppose ${g(s)=(1+s\psi)^{\frac{4}{n-4}}g=u(s)^{\frac{4}{n-2}}g}$. From conformal change property of scalar curvature, we get

$\displaystyle J_{g(s)}=u(s)^{-\frac{n+2}{n-2}}\left(-\frac{2}{n-2}\Delta_{g}u(s)+Ju(s)\right)$

$\displaystyle \dot J=-\frac{2}{n-2}\Delta_g\dot u-\frac{4}{n-2}J\dot u$

Then

$\displaystyle \frac{d}{ds}|_{t=0}\int_{M}J_{g(s)}^2d\mu_{g(s)}=\int_{M}2J\dot{J}+\frac{2n}{n-2}J^2\dot ud\mu$

$\displaystyle =\int_M -\frac{4}{n-2}J\Delta \dot u+\frac{2(n-4)}{n-2}J^2\dot ud\mu$

$\displaystyle =\int_M\left(-\frac{4}{n-4}\Delta J+2J^2\right)\psi d\mu$

where we have used the fact ${\dot u(0)=\frac{n-2}{n-4}\psi}$. Next, it is easy to know

$\displaystyle \frac{d}{ds}|_{t=0}\int_M \sigma_2(A_{g(s)})d\mu_{g(s)}=2\int_{M}\sigma_2(A)\psi d\mu$So

$\displaystyle I'(g)\psi=2t\int_{M}\left(-\Delta J+\frac{n-4}{2}J\right)\psi d\mu_g+8\int_{M}\sigma_2(A)\psi d\mu_g$

Now assume ${g(s)}$ keep the unit volume infinitesimally, that is ${\int_M \psi d\mu=0}$, then ${I'(g)=0}$ under this constraint means

$\displaystyle t\left(-\Delta J+\frac{n-4}{2}J\right)+4\sigma_2(A)=const.$

Remark: M.J. Gursky, F. Hang, Y.-J. Lin, Riemannian Manifolds with Positive Yamabe Invariant and Paneitz Operator, Int. Math. Res. Not.

### An identity related to generalized divergence theorem

I am trying to verify one proposition proved in Reilly’s paper. For the notation of this, please consult the paper.

Propsotion 2.4 Let ${{D}}$ be a domain in ${\mathbb{R}^n}$. ${f}$ is a smooth function on ${{D}}$. Then

$\displaystyle \int_{{D}}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\left(\tilde S_q(z)z_n-\sum_{\alpha\beta}\tilde T_{q-1}(z)^{\alpha\beta}z_{\alpha}z_{n,\beta}\right)dA$

Proof: Take an orthonormal frame field ${\{e_\alpha,e_n\}}$ such that ${\{e_\alpha\}}$ is tangent to ${\partial D}$. Notice

$\displaystyle D^2(f)(X,Y)=X(Yf)-(\nabla_{X}Y)f$

$\displaystyle D^2(f)=\left(\begin{matrix} z_{,\alpha\beta}-z_nA_{\alpha\beta} & z_{n,\alpha}\\ z_{n,\beta} & f_{nn} \end{matrix} \right)$

where ${f_{nn}=D^2(f)(e_n,e_n)}$. It follows from Remark 2.3 that

$\displaystyle \int_{D}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA$

where ${t=(t_1,\cdots,t_n)}$ is the outward unit normal to ${\partial D}$. Changing the coordinates to ${e_\alpha}$ and ${e_n}$, we can get

$\displaystyle \int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA=\int_{\partial D}T_q(f)^{\alpha n}z_{\alpha}+T_{q}(f)^{nn}z_ndA$

It is easy to see

$\displaystyle T_q(f)^{nn}=\tilde S_{q}(z)$

and

$\displaystyle T_q(f)^{\alpha n}z_\alpha=\sum\delta\binom{i_1,i_2\cdots,n,\alpha}{j_1,j_2\cdots,\beta, n}z_\alpha$

$\displaystyle =-\sum\delta\binom{\alpha_1,\alpha_2\cdots,\alpha}{\beta_1,\beta_2\cdots,\beta}f_{n\beta}z_\alpha=-\tilde T_{q-1}(z)^{\alpha\beta}z_\alpha z_{n,\beta}$

Therefore the proposition is established.

Remark: Robert Reilly, On the hessian of a function and the curvature of its graph

### Some identities related to mean curvature of order m

For any ${n\times n}$ (not necessarily symmetric) matrix ${\mathcal{A}}$, we let ${[\mathcal{A}]_m}$ denote the sum of its ${m\times m}$ principle minors. For any hypersurface which is a graph of ${u\in C^2(\Omega)}$, where ${\Omega\subset \mathbb{R}^n}$. We have its downward unit normal is

$\displaystyle (\nu,\nu_{n+1})=\left(\frac{Du}{\sqrt{1+|Du|^2}},\frac{-1}{\sqrt{1+|Du|^2}}\right).$

The principle curvatures are taken from the eigenvalues of the Jacobian matrix ${[D\nu]}$. One can define its ${m}$ mean curvature using the notation of above

$\displaystyle H_m=\sum_{i_1<\cdots

Now let us consider general matrix ${\mathcal{A}}$,

$\displaystyle A_m=[\mathcal{A}]_m=\frac{1}{m!}\sum \delta\binom{i_1,\cdots,i_m}{j_1,\cdots,j_m}a_{i_1j_1}\cdots a_{i_mj_m}$

where ${\delta}$ is the generalized Kronecker delta.

$\displaystyle \delta\binom{i_1 \dots i_p }{j_1 \dots j_p} = \begin{cases} +1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an even permutation of } j_1 \dots j_p \\ -1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an odd permutation of } j_1 \dots j_p \\ \;\;0 & \quad \text{in all other cases}.\end{cases}$

Then we define the Newton tensor

$\displaystyle T^{ij}_m=\frac{\partial A_m}{\partial a_{ij}}=\frac{1}{(m-1)!}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}a_{i_2j_2}\cdots a_{i_m j_m}.$

For any vector field ${X}$ on ${\mathbb{R}^n}$, ${DX}$ is a matrix, where ${D=(D_{1},\cdots,D_n)}$ and ${|X|\neq 0}$, denote ${\tilde X=X/|X|}$, we have

$\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}X_{j_2}]\cdots [D_{i_m}X_{j_m}].$

Since for any ${1\leq p,k,l\leq n}$

$\displaystyle D_k\tilde X_l=\frac{D_kX_{l}}{|X|}-\frac{\sum_{p=1}^nX_pD_kX_pX_l}{|X|^3}$

$\displaystyle \sum_{i,j,i_2,j_2}\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_jX_pX_{j_2}D_{i_2}X_p=0$

because $\delta$ is skew-symmetric in $j,j_2$. Then

$\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=|X|^{m-1}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}\tilde X_{j_2}]\cdots [D_{i_m}\tilde X_{j_m}].$

$=(m-1)!|X|^{m-1}T_m^{ij}(D\tilde X)X_iX_j=(m-1)!|X|^{m+1}T_m^{ij}(D\tilde X)\tilde X_i\tilde X_j.$

Applying the formula $(T^{ij}_m(\mathcal{A}))=[\mathcal{A}]_{m-1}I-T_{m-1}(\mathcal{A})\cdot \mathcal{A}$(Check [1] Propsition 1.2) and $(D\tilde X)\tilde X=0$, we get

$T^{ij}_m(DX)X_iX_j=|X|^{m+1}[D\tilde X]_{m-1}$

It follows from the result of Reilly, Remark 2.3(a), that

$\displaystyle mA_m[DX]=D_i[T^{ij}_mX_j]$

Suppose ${X}$ is a vector field normal to ${\partial \Omega}$, then

$\displaystyle m\int_\Omega [DX]_m=\int_{\partial \Omega} T^{ij}_m X_j\gamma_i=\int_{\partial \Omega}(X\cdot \gamma)^m[D\gamma]_{m-1}=\int_{\partial \Omega} (X\cdot \gamma)^m H_{m-1}(\partial \Omega)$

where ${\gamma}$ is the outer ward unit normal to ${\partial \Omega}$.

Remark: [1] R.C. Reilly, On the Hessian of a function and the curvatures of its graph., Michigan Math. J. 20 (1974) 373â€“383. doi:10.1307/mmj/1029001155.

[2] N. Trudinger, Apriori bounds for graphs with prescribed curvature.

We probably have to assume $DX$ is a symmetric matrix in order to use the formula of Reilly. Not sure about this.

### Newton tensor

Suppose ${A:V\rightarrow V}$ is a symmetric endomorphism of vector space ${V}$, ${\sigma_k}$ is the ${k-}$th elementary symmetric function of the eigenvalue of ${A}$. Then

$\displaystyle \det(A+tI)=\sum_{k=0}^n \sigma_k t^{n-k}$

One can define the ${k-}$th Newton transformation as the following

$\displaystyle \det(A+tI)(A+tI)^{-1}=\sum_{k=0}^{n-1}T_k(A)t^{n-k-1}$

This means

$\displaystyle \det(A+tI)=\sum_{k=0}^{n-1}T_k(A+tI)t^{n-k-1}$

$\displaystyle =T_0 t^n+\sum_{k=0}^{n-2}(A\cdot T_k(A)+T_{k+1}(A))t^{n-k-1}+T_{n-1}(A)$

By comparing coefficients of ${t}$, we get the relations of ${T_k}$

$\displaystyle T_0=1,\quad A\cdot T_k(A)+T_{k+1}(A)=\sigma_{k+1}I,\, 0\leq k\leq n-2\quad T_{n-1}(A)=\sigma_n$

Induction shows

$\displaystyle T_{k}(A)=\sigma_kI-\sigma_{k-1}A+\cdots+(-1)^kA^k$

For example

$\displaystyle T_1(A)=\sigma_1I-A$

$\displaystyle T_2(A)=\sigma_2-\sigma_1A+A^2$

One of the important property of Newton transformation is that: Suppose ${F(A)=\sigma_k(A)}$, then

$\displaystyle F^{ij}=\frac{\partial F}{\partial A_{ij}}=T_{k-1}^{ij}(A)$

The is because

$\displaystyle \frac{\partial }{\partial A_{ij}}\det(A+tI)=\det(A+tI)((A+tI)^{-1})_{ij}.$

If ${A\in \Gamma_k}$, then ${T_{k-1}(A)}$ is positive definite and therefore ${F}$ is elliptic.

Remark: Hu, Z., Li, H. and Simon, U. . Schouten curvature functions on locally conformally flat Riemannian manifolds. Journal of Geometry, 88(1${-}$2), (2008), 75${-}$100.

### General approach for fully nonlinear elliptic equation

Consider the Dirichlet problem in a bounded domian $\Omega\subset \mathbb{R}^n$ with smooth boundary $\partial \Omega$

$\displaystyle \begin{cases} F(D^2u)=\psi(x) \text{ in } \Omega\\\quad \quad u=\phi\quad\text{ on }\Omega\end{cases}\quad(1)$

The function $F$ are represented by a smooth symmetric function

$\displaystyle F(D^2u)=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$

here $\lambda_1,\lambda_2,\cdots,\lambda_n$ are the eigenvalues of $D^2u$. In order to be elliptic, we require

$\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i>0\quad (2)$

$f$ is defined in an open convex cone $\Gamma\subset \mathbb{R}^n$ with vertex at origin, and

$\displaystyle \bigcap\limits_{i=1}^n \{\lambda_i>0\}\subset \Gamma\subset \left\{\sum\lambda_i>0\right\}$

Since $f$ is symmetric, we also require $\Gamma$ to be symmetric.

The following are the assumptions for  (1) to be solvable,

$\psi\in C^\infty(\overline{\Omega})$, $\phi\in C^\infty(\partial \Omega)$

$\displaystyle \psi_0= \min_{\overline{\Omega}}\psi\leq \max_{\overline{\Omega}}\psi=\psi_1\quad (4)$

$f$ is a concave function  $(5)$

$\displaystyle \overline{\lim\limits_{\lambda\to\partial \Gamma}}f(\lambda)\leq \tilde{\psi_0}<\psi_0\quad (6)$

For every compact set $K$ in $\Gamma$ and every constant $C>0$, there is a number $R=R(K,C)$ such that

$\displaystyle f(\lambda_1,\lambda_2,\cdots,\lambda_n+R)\geq C$ for all $\lambda\in K\quad(7)$

$\displaystyle f(R\lambda)\geq C$ for all $\lambda\in K\quad (8)$

Also we need restrict $\partial \Omega$. There exists sufficiently large constant $R$ such that for every point on $\partial \Omega$, if $\kappa_1,\kappa_2,\cdots,\kappa_{n-1}$ are the principle curvatures of $\partial \Omega$

$\displaystyle (\kappa_1,\kappa_2,\cdots,\kappa_{n-1},R)\in\Gamma\quad (9)$

$\mathbf{Thm(CNS):}$ If $(2-9)$ are satisfied, then $(1)$ has a unique solution $u\in C^\infty(\overline{\Omega})$ with $\lambda(D^2u)\in \Gamma$.

$\mathbf{Proof:}$ The existence get from continuity method.

Krylov has shown how from a priori estimates

$|u|_{C^2(\overline{\Omega})}\leq C\quad (10)$

and uniform ellipticity of the linearized opeartor $L=\sum{F_{ij}}\partial_{ij}$ to derive

$\displaystyle |u|_{C^{2,\nu}({\overline{\Omega}})}\leq C$

So we only need to derive $(10)$ for any solution of $(1)$.

Using a maximum principle of fully nonlinear equation and $\psi\leq \psi_1$ and $(9)$, it is possible to construct a subsolution $\underline{u}$of $(1)$. So $u\geq \underline{u}$

If $\lambda(D^2u)\in \Gamma$ then $u$ can be bounded above by a harmonic function $v$ in $\Omega$.

$\underline{u}\leq u\leq v$

Additionally,                                      $|\nabla u|\leq C$ on $\partial \Omega$

Then differentiate $F(D^2u)=\psi$ to get a linear elliptic function. Using maximum principle to get $|u|_{C^1}\leq C$.

For the second derivative estimate, it is also estimate $u_{ij}$ on the boundary first and then differentiate $F(D^2u)=\psi$ another time to get a linear elliptic function of $u_{ij}$. And then apply maximum principle to get interior gradient estimate.

The bound of $u_{\alpha n}$, $\alpha is achieved from  constructing a barrier function.

The estimate of $u_{nn}$ is complicate. Still constructing a barrier function.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$Cafferelli, Nirenberg, Spruck: The Dirichlet problem for nonlinear second order elliptic equations III.

### Monge-Ampere equation and bounday behavior of domain

$\mathbf{Problem:}$ Suppose $\Omega\subset \mathbb{R}^n$ is a bounded domain and $\partial \Omega$ is $C^2$. If there exists a convex function $u\in C^2(\overline{\Omega})$ satisfies

$\displaystyle \begin{cases}\det D^2u=1 \text{ in }\Omega\\ u=0\quad \quad\text{ on }\partial \Omega\end{cases}$

Then $\Omega$ is uniformly convex. In other words, the principle curvature  of every point on $\partial \Omega$, namely $\kappa_1,\kappa_2,\cdots,\kappa_{n-1}$, are positive. Moreover, $\partial \Omega$ is connect.

$\mathbf{Proof:}$ For any boundary point of $\Omega$, we may suppose the point is origin. Since $\partial \Omega\in C^2$, there exists a neighborhood of 0 such that $\partial \Omega$ is represented by $x_n=\rho(x_1,x_2,\cdots,x_{n-1})$ with $\rho\in C^2$. Choosing the principle coordinate system, poositive $x_n$ axis is interior normal at origin and

$\displaystyle \rho(x')=\frac{1}{2}\sum\limits_{i=1}^{n-1}\kappa_ix_i^2+O(|x'|^3)$ here $x'=x_1,x_2,\cdots, x_{n-1}$.

Since $u(x',\rho(x'))=0$ near the origin, it following, on differentiation,

$\displaystyle u_{ij}=-u_n\rho_{ij}=-u_n\kappa_i\delta_{ij}$    for $i,j.

So at origin, we get

$\displaystyle D^2u=\left( \begin{array}{cccc} \kappa_1 & 0 &\cdots & u_{1n} \\ 0 & \kappa_2 & \ldots & u_{2n} \\ \vdots &\vdots & \ddots &\vdots\\ u_{1n}&u_{2n}& \cdots & u_{nn} \end{array} \right)$

which means

$\displaystyle 1=\det D^2u=|u_n|^{n-2}\prod\limits_{i=1}^{n-1}\kappa_i\left\{|u_n|u_{nn}-\sum\limits_{i=1}^{n-1}\frac{(u_{in})^2}{\kappa_i}\right\}.$

So $\kappa_i\neq 0$ on $\partial \Omega$ and for all $\, i=1,2,\cdots,n$. However, $\kappa_i$ is continuous function on $\partial \Omega$, which means $\kappa_i$ does not change sign on $\partial \Omega$.

Since $u$ is convex and $u=0$ on $\partial \Omega$, then $u\leq 0$ in $\Omega$. Also we know that $D^2u$ is positive definite matrix. Thus $\Delta u>0$ in $\Omega$, by the hopf lemma,

$u_n>0$ at $\partial \Omega$

So near origin $\kappa_iu_n$ is almost $u_{ii}$, which is positive by the property of $D^2u$. So $\kappa_i$ is positive.

For the argument $\partial \Omega$ is connect, see the paper in the following remark.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Caffarellli, Nirenberg and Spruck. The Dirichlet problem for nonlinear second order elliptic equations,III:Functions of the eigenvalues of the Hessian.

Also refer to the formula [GT, p471].

### Fully nonlinear ellipticity and conformal invariancy

Let $\mathcal{S}^{n\times n}$ be the space of symmetric $n\times n$ matrices and $\mathcal{O}(n)$ is the orthogonal matrices. Suppose $U\subset \mathcal{S}^{n\times n}$ satisfies

$\displaystyle O^{-1}UO=U$, $\forall\, O\in \mathcal{O}(n)$

Let $F\in C^1(U)$ satisfy

$\displaystyle F(O^{-1}MO)=F(M)$, $\forall\, M\in U$, and $O\in \mathcal{O}(n)\quad$      (1)

The fully nonlinear equation $F(D^2u)=\psi$ is called elliptic in $U$ if

$(F_{ij})(D^2u)>0$ for any $D^2u\in U\quad$      (2).

From (1), we know there exists $f\in C^1$ such that $F(M)=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$, where $\lambda_1,\lambda_2,\cdots,\lambda_n$ are the eigenvalues of $M$. And $f$ must be symmetric.

$\mathbf{Problem:}$ $F$ is elliptic is equivalent to $\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i$

$\mathbf{Proof:}$ Firstly, let us assume $F$ is elliptic, i.e. $F_{ij}(M)>0$, $\forall M\in U$.

Then in particular choose $M=\Lambda=diag\{\lambda_1,\lambda_2,\cdots,\lambda_n\}\subset U$,

$f(\lambda_1,\lambda_2,\cdots,\lambda_n)=F(\Lambda)$

Then                                             $\displaystyle \frac{\partial f}{\partial \lambda_i}=F_{ii}(\Lambda)>0$

Secondly, assume $\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i$.

$\forall\, M\in U$, there exists $O\in \mathcal{O}(n)$ such that $M=O\Lambda O^{-1}$, here we denote $O=(O_{ij}), O^{-1}=(O^{ij})$.

Then $F(M)=F(O\Lambda O^{-1})=F((O_{ik}\lambda_{k}O^{kj}))=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$

Differentiating (1), we get       $\displaystyle \sum_{i,j}O^{ik}F_{ij}(O^{-1}MO)O_{lj}=F_{kl}(M)$

That is                       $\displaystyle \sum_{i,j}O^{ik}F_{ij}(\Lambda)O_{lj}=F_{kl}(M)\quad (3)$

Since $F(\Lambda)=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$, then $\displaystyle F_{ii}(\Lambda)=\frac{\partial f}{\partial \lambda_1}$, $\forall\, i>0$

For $F_{ij}(\Lambda)$, when $i\neq j$, by definition and use the fact that $\boldsymbol{F_{ij}=F_{ji}}$(possible to remove?)

$\displaystyle 2F_{ij}(\Lambda)=\lim\limits_{a \to 0}\frac{F(\Lambda+aE_{ij})-F(\Lambda)}{a}$

here $E_{ij}$ is symmertric and has entries which are 0 except at $(i,j)$-th and $(j,i)$-th positions are 1.

$\displaystyle \Lambda+tE_{ij}$ has eigenvalue $\lambda_k$ with $k\neq i,j$ and $\displaystyle \frac{\lambda_i+\lambda_j+\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2}$, $\displaystyle \frac{\lambda_i+\lambda_j-\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2}$.

So $\displaystyle F_{ij}(\Lambda)=\lim\limits_{a \to 0}\frac{F(\Lambda+aE_{ij})-F(\Lambda)}{2a}=\lim\limits_{a \to 0}\frac{f(\lambda_1,\lambda_2,\cdots,\lambda_n)-f(\lambda_1,\lambda_2,\cdots,\lambda_n)}{2a}$

$\displaystyle =\lim\limits_{a \to 0}\frac{f(\lambda_1,\cdots, \frac{\lambda_i+\lambda_j+\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2},\cdots,\frac{\lambda_i+\lambda_j-\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2},\cdots,\lambda_n)-f(\lambda_1,\lambda_2,\cdots,\lambda_n)}{2a}\quad (4)$

• If $\lambda_i=\lambda_j=\mu$, then (4) becomes

$\displaystyle F_{ij}(\Lambda)=\lim\limits_{a \to 0}\frac{f(\lambda_1,\cdots, \mu+a,\cdots,\mu-a,\cdots,\lambda_n)-f(\lambda_1,\lambda_2,\cdots,\lambda_n)}{2a}$

$\displaystyle =\frac{1}{2}\left(\frac{\partial f}{\partial \lambda_i}(\lambda_1,\lambda_2,\cdots,\lambda_n)-\frac{\partial f}{\partial \lambda_j}(\lambda_1,\lambda_2,\cdots,\lambda_n)\right)=0,$

because $f$ is symmetric and $\lambda_i=\lambda_j$.

• If $\lambda _i\neq \lambda_j$, (4) becomes

$\displaystyle F_{ij}(\Lambda)=\frac{\partial f}{\partial \lambda_i}(\lambda_1,\lambda_2,\cdots,\lambda_n)\lim\limits_{a\to 0}\frac{\lambda_j-\lambda_i+\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2a}+\frac{\partial f}{\partial \lambda_j}(\lambda_1,\lambda_2,\cdots,\lambda_n)\lim\limits_{a\to 0}\frac{\lambda_i-\lambda_j-\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2a}$

$\quad =0$

Combing the above results, we get $\displaystyle (F_{ij}(\Lambda))=diag\left\{\frac{\partial f}{\partial \lambda_1},\frac{\partial f}{\partial \lambda_2}\cdots,\frac{\partial f}{\partial \lambda_n}\right\}$.

Applying $(F_{ij}(\Lambda))$ to (3), it becomes $\displaystyle \sum_{i}O^{ik}\frac{\partial f}{\partial \lambda_i}(\lambda_1,\lambda_2,\cdots,\lambda_n)O_{li}=F_{kl}(M)$

This means $\displaystyle O\left(diag\left\{\frac{\partial f}{\partial \lambda_1},\frac{\partial f}{\partial \lambda_2}\cdots,\frac{\partial f}{\partial \lambda_n}\right\}\right)O^{-1}=(F_{kl}(M))^T$.

$(\displaystyle F_{kl}(M))$ is a positive definite matrix because  $\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$