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Unique continuation property on the boundary

I am writing a theorem proved by Jin Zhiren in his thesis.

Suppose $\Omega$ is a smooth domain in $\mathbb{R}^n$, $x_0\in \partial \Omega$ and $u$ is a harmonic function in $\Omega$. If there exists $A, b>0$ such that

$\displaystyle |u(x)|\leq Ae^{-\frac{b}{|x-x_0|}}\quad x\in \Omega$

for $|x-x_0|$ small, then $u=0$. If $n=2$, the same conclusion holds for the solutions of a general second order linear elliptic equation.

A borderline example for this theorem is $u$ be the real part of $e^{-1/z^\alpha}$, $\alpha\in (0,1)$. $u$ is harmonic in the right half plane and $u\leq Ae^{-1/|x|^\alpha}$ and consequently $D^\beta u(0)=0$.

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.

Self-shrinker and polynomial volume growth

Proposition: If $M$ is an entire graph of at most polynomial volume growth and $H=\langle X,\nu\rangle$, namely $M$ is a self-shrinker. Then $M$ is a plane.

Proof: Suppose

$\displaystyle v=\frac{1}{\langle \nu, w\rangle}$

Then one can derive the following equation

$\displaystyle \Delta v=\langle\nabla v,X\rangle+|A|^2v+2v^{-1}|\nabla v|^2$

Multiplying both sides by $e^{-|X|^2/2}$ and integration on $M$, which makes sense because of the polynomial volume growth, we get

$\int_M (\Delta v-\langle\nabla v,X\rangle)e^{-\frac{|X|^2}{2}}d\mu=\int_M (|A|^2v+2v^{-1}|\nabla v|^2)e^{-\frac{|X|^2}{2}}d\mu$

However, integration by parts shows the LHS is zero. Thus $A\equiv v\equiv 0$, $M$ must be a plane.

Some calculations of sigma_2

On four-manifold ${(M^4,g_0)}$, we define Shouten tensor

$\displaystyle A = Ric-\frac 16 Rg$

and Einstein tensor and gravitational tensor

$\displaystyle E=Ric - \frac 14 Rg\quad S=-Ric+\frac{1}{2}Rg$

Suppose ${\sigma_2}$ is the elemantary symmetric function

$\displaystyle \sigma_2(\lambda)=\sum_{i

Thinking of ${A}$ as a tensor of type ${(1,1)}$. ${\sigma_2(A)}$ is defined as ${\sigma_2}$ applied to eigenvalues of ${A}$. Then

$\displaystyle \sigma_2(A)= \frac{1}{2}[(tr_g A)^2-\langle A, A\rangle_g] \ \ \ \ \ (1)$

Notice ${A=E+\frac{1}{12}Rg}$. Easy calculation reveals that

$\displaystyle \sigma_2(A)=-\frac{1}{2}|E|^2+\frac{1}{24}R^2 \ \ \ \ \ (2)$

Under conformal change of metric ${g=e^{2w}g_0}$, we have

$\displaystyle R= e^{-2w}(R_0-6\Delta_0 w-6|\nabla_0 w|^2) \ \ \ \ \ (3)$

$\displaystyle A=A_0-2\nabla^2_0 w+2dw\otimes dw-|\nabla_0w|^2g_0 \ \ \ \ \ (4)$

$\displaystyle S=S_0+2\nabla_0^2w-2\Delta_0wg_0-2dw\otimes dw-|\nabla_0 w|^2g_0 \ \ \ \ \ (5)$

We want to solve the equation ${\sigma_2(A)=f>0}$, which is equivalent to solve

$\displaystyle \sigma_2(A_0-2\nabla^2_0w+2dw\otimes dw-|\nabla_0w|^2g_0)=f$

This is an fully nonlinear equation of Monge-Ampere type. Under local coordinates, the above equation can be treated as

$\displaystyle F(\partial_i\partial_j w,\partial_kw,w,x)=f$

where ${F(p_{ij},v_k,s,x):\mathbb{R}^{n\times n}\times\mathbb{R}^n\times\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}}$. This equation is elliptic if the matrix ${\left(\frac{\partial F}{\partial p_{ij}}\right)}$ is positive definite. In order to find that matrix, we need the linearized operator

$\displaystyle L[\phi]=\frac{\partial F}{\partial p_{ij}}(\nabla_0^2\phi)_{ij}=\frac{d}{dt}|_{t=0}F(\partial_i\partial_j w+t\partial_i\partial_j\phi,\partial_kw,w,x) \ \ \ \ \ (6)$

Using the elementary identity

$\displaystyle \frac{d}{dt}\rvert_{t=0}\sigma_2(H+tG)=tr_gH\cdot tr_gG-\langle H, G\rangle_g. \ \ \ \ \ (7)$

for any fixed matrix ${H}$ and ${G}$. Now plug in ${H=A}$ is Schouten tensor and ${B=-2\nabla_0^2\phi}$. One can calculate them as

$\displaystyle tr_g H\cdot tr_g G=\langle \frac{1}{3}Rg, G\rangle_g \ \ \ \ \ (8)$

Then we get

$\displaystyle L[\phi]=\langle S,G\rangle_g=-2\langle S,\nabla^2_0\phi\rangle_g$

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.

When the norm map is surjective

$\mathbf{Problem:}$ Suppose $F$ is finite field with $q=p^m$ elements, $E$ an Galois extension of $F$ such that $[E:F]=n$. Prove that $N_{E/F}:E^\ast\to F^\ast$ is surjective.

$\mathbf{Proof:}$ From theorem 4.26 we know that $\text{Gal }E/F$ is cyclic over $F$ generated by $\eta:a\to a^q$.

$\displaystyle N_{E/F}(a)=aa^qa^{q^2}\cdots a^{q^{n-1}}=a^{\frac{q^n-1}{q-1}}$.

$N_{E/F}:E^\ast\to F^\ast$ is a homomorphism. It is well known that  any finite subgroup of the multiplicative group of field is cyclic, which means $E^\ast$ and $F\ast$ is cyclic.

Since $|E\ast|=q^n-1$ and $|F^\ast|=q-1$, and $N_{E/F}(a)=a^m$, where $m={\frac{q^n-1}{q-1}}$, then $|\text{Ker } N|=(q^n-1,m)=m$ and $|\text{Im }N|=\frac{q^n-1}{(q^n-1,m)}=q-1$. Thus $N$ is surjective.

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

$\mathbf{Remark:}$ Jacobson p300.

Welcome, bro.

Hi, everyone.

This is my formula: $f(x) = \int_0^\infty g(s)\, ds$