## Category Archives: PDE

342394

### Bach flat four dimensional manifold and sigma2 functional

We want to find the necessary condition of being the critical points of ${\int\sigma_2}$ on four dimensional manifold.

1. Preliminary

Suppose ${(M^n,g)}$ is a Riemannian manifold with ${n=4}$. ${P_g}$ is the Schouten tensor

$\displaystyle P_g=\frac{1}{n-2}\left(Ric-\frac{R}{2(n-1)}g\right)$

and denote ${J=\text{\,Tr\,} P_g}$. Define

$\displaystyle \sigma_2(g)=\frac{1}{2}[(\text{\,Tr\,} P_g)^2-|P_g|_g^2]$

$\displaystyle I_2(g)=\int_M \sigma_2(g)d\mu_g$

where ${|P|_g^2=\langle P,P\rangle_g}$. It is well known that ${I_2(g)}$ is conformally invariant.

Suppose ${g(t)=g+th}$ where ${h}$ is a symmetric 2-tensor. We want to calculate the first derivative of ${I_2(g(t))}$ at ${t=0}$. To that end, let us list some basic facts (see the book of Toppings). Firstly denote ${(\delta h)_j=-\nabla^i{h_{ij}}}$ the divergence operator and

$\displaystyle G(h)=h-\frac{1}{2}(\text{\,Tr\,} h) g$

$\displaystyle (\Delta_L h)_{ij}=(\Delta h)_{ij}-h_{ik}Ric_{jl}g^{kl}-h_{jk}Ric_{il}g^{kl}+2R_{ikjl}h^{kl}$

where ${\Delta_L}$ is the Lichnerowicz Laplacian. Then the first variation of Ricci curvature and scalar curvature are

$\displaystyle \dot{R}=\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle \ \ \ \ \ (1)$

$\displaystyle \dot{Ric}=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta G(h))^\sharp}g=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta h)^\sharp}g-\frac{1}{2}Hess(\text{\,Tr\,} h)$

$\displaystyle =-\frac{1}{2}\Delta_Lh-d(\delta h)-\frac{1}{2}Hess(\text{\,Tr\,} h)$

where we were using upper dot to denote the derivative with respect to ${t}$.

2. First variation of the sigma2 functional

Lemma 1 ${(M^4,g)}$ is a critical point of ${I_2(g)}$ if and only if

$\displaystyle \Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g=0 \ \ \ \ \ (2)$

where ${(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}$.

Proof:

$\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\int_M J\dot J-\langle\dot P,P\rangle+\langle h,P\wedge P\rangle+\frac{1}{2}\sigma_2\text{\,Tr\,} h \,d\mu_g$

where ${(P\wedge P)_{ij}=P_{ik}P_{jl}g^{kl}}$. Since we have

$\displaystyle \int_M\langle P,\dot P\rangle\\ =\frac{1}{n-2}\int_M\langle P, \dot Ric-\dot Jg-Jh\rangle =\frac{1}{n-2}[\langle P, \dot Ric\rangle-\dot J J-J\langle h,P\rangle]$

$\displaystyle =\frac{1}{n-2}[-\frac{1}{2}\langle h,\Delta_L P\rangle+\langle h,Hess(J)\rangle-\frac{1}{2}\Delta J \text{\,Tr\,} h-\dot J J-J\langle h,P\rangle]$

Plugging this into the derivative of ${I_2}$ to get

$\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t))\\ =\int_M\frac{1}{2}\langle h,\Delta_L P\rangle-\langle h,Hess(J)\rangle+\frac{1}{2}\Delta J \text{\,Tr\,} h\\$

$\displaystyle \quad +(n-1)\dot J J+J\langle h,P\rangle+(n-2)\langle h,P\wedge P\rangle+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g$

In order to simplify the above equation, we recall the definition of Lichnerowicz Laplacian ${\Delta_L}$

$\displaystyle (\Delta_LP)_{ij}=(\Delta P)_{ij}-2P_{ik}Ric_{jl}g^{kl}+2R_{ikjl}P^{kl}$

$\displaystyle =(\Delta P)_{ij}-2(n-2)P_{ik}P_{jl}g^{kl}-2JP_{ij}+2R_{ikjl}P^{kl}$

Apply (1) to get

$(n-1)\dot J J=\frac12J\dot R=\frac12J[\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle]$
$=\frac{1}{2}[\langle h, Hess(J)\rangle-\text{\,Tr\,} h\Delta J-(n-2)J\langle h, P\rangle-J^2\text{\,Tr\,} h]$

Therefore we can simplify it to be

$\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t)) =\int_M\frac{1}{2}\langle h,\Delta P\rangle-\frac{1}{2}\langle h, Hess(J)\rangle+h^{ij}R_{ikjl}P^{kl}$

$\displaystyle -\frac{n-2}{2}J\langle h,P\rangle-\frac{1}{2} J^2 \text{\,Tr\,} h+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g$

Let us denote ${(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}$. Using the fact ${n=4}$ and the definition of ${\sigma_2}$,

$\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g$

where

$\displaystyle Q=\Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g$

$\Box$

Remark 1 It is easy to verify ${\text{\,Tr\,} Q=0}$, this is equivalent to say ${I_2}$ is invariant under conformal change. More precisely, letting ${h=2ug}$, then

$\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g=\frac{1}{2}\int_M u\text{\,Tr\,} Q d\mu_g=0.$

Remark 2 If ${g}$ is an Einstein metric with ${Ric=2(n-1)\lambda g}$, then ${P=\lambda g}$, ${J=n\lambda}$ and

$\displaystyle \mathring{Rm}(P)=\lambda Ric=2(n-1)\lambda^2 g$

It is easy to verify that ${Q=0}$. In other words, Einstein metrics are critical points of ${I_2}$.

Are there any non Einstein metric which are critical points of ${I_2}$?

Here is one example. Suppose ${M=\mathbb{S}^2\times N}$, where ${\mathbb{S}^2}$ is the sphere with standard round metric and ${(N,g_N)}$ is a two dimensional compact manifolds with sectional curvature ${-1}$. ${M}$ is endowed with the product metric. We can prove ${Ric=g_{S^2}-g_N}$, ${P=\frac{1}{2}g_{S^2}-\frac{1}{2}g_N}$, ${J=0}$, ${\mathring{Rm}(P)=g_{prod}}$ and consequently ${Q=0}$.

Note that the above example is a locally conformally flat manifold. For this type of manifold, we have the following lemma which can say

Lemma 2 Suppose ${g}$ is locally conformally flat and ${Q=0}$, then

Proof: When ${g}$ is locally conformally flat,

$\displaystyle \mathring{Rm}(g)=JP+|P|_g^2g-2P\wedge P$

${Q=0}$ is equivalent to

$\displaystyle \Delta P-Hess(J)+|P|_g^2g-4P\wedge P=0$

Actually this is equivalent to the Bach tensor ${B}$ is zero. $\Box$

3. Another point of view

We have the Euler Characteristic formula for four dimensional manifolds

$\displaystyle 8\pi^2\chi(M)=\int_M (|W|_g^2+\sigma_2) d\mu_g$

therefore the critical points for ${\int_M \sigma_2d\mu_g}$ will be the same as the critical points of ${\int_M |W|_g^2d\mu_g}$. However, the functional

$\displaystyle g\rightarrow \int_M |W|_g^2d\mu_g$

is well studied by Bach. The critical points of this functional satisfy Bach tensor equal to 0.

$\displaystyle B_{ij}=\nabla^k\nabla^l W_{likj}+\frac{1}{2}Ric^{kl}W_{likj}$

Obviously, ${B=0}$ for Einstein metric, but not all Bach flat metrics are Einstein. For example ${B=0}$ for any locally conformally flat manifolds.

### Bubble functions under different setting

Bubble function can be defined either on $\mathbb{R}^n$, $\mathbb{S}^n$ or $\mathbb{B}^n$. For the following notations, $c_n$ will denote suitable constants which may be different from line to line.

• For every $\epsilon>0$ and  $\xi\in\mathbb{R}^n$, define

$\displaystyle u_{\epsilon,\xi}=c_n\left(\frac{\epsilon}{\epsilon^2+|x-\xi|^2}\right)^{\frac{n-2}{2}}$

It is well know that $-\Delta u= c_nu^{\frac{n+2}{n-2}}$. Moreover $(\mathbb{R}^n,u^{\frac{4}{n-2}}_{\epsilon,\xi}g_E)$ is isometric to the standard sphere minus one point.

• For any $a\in \mathbb{S}^n$ and $\lambda>0$ define

$\displaystyle\delta(a,\lambda)=c_n\left(\frac{\lambda}{\lambda^2+1+(\lambda^2-1)\cos d(a,x)}\right)^{\frac{n-2}{2}}$

where $d(a,x)$ is the geodesic distance of $a$ and $x$ on $\mathbb{S}^n$. Actually $\cos d(a,x)=a\cdot x$

• For each $p\in \mathbb{B}^{n+1}$, define $\delta_p(x):\mathbb{S}^n\to \mathbb{R}$ by

$\displaystyle\delta_p(x)=c_n\left(\frac{1-|p|^2}{|x+p|^2}\right)^{\frac{n-2}{2}}$

Both the second and third one satisfy

$\displaystyle \frac{4(n-1)}{n-2}\Delta_{\mathbb{S}^n}\delta-n(n-1)\delta+c_n\delta^{\frac{n-2}{n+2}}=0$

If we make $p=\frac{\lambda-1}{\lambda+1}a$, the third one will be changed to the second one.

To get the second one from the first one, let us deonte $\Phi_a:\mathbb{S}^n\to \mathbb{R}^n$ be the stereographic projection from point $a$. Then

$\displaystyle \delta(a,\lambda)\circ \Phi^{-1}_a=c_n\left(\frac{\lambda(1+|y|^2)}{\lambda^2|y|^2+1}\right)^{\frac{n-2}{2}}=c_nu_{\lambda,0}u_{1,0}^{-1}$

It should be able to see the third one from hyperbolic translation directly.

### 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 $D^\beta u(0)=0$.

### Interior estimate for Monge Ampere equation

Suppose we have $u$ is a generalized solution of the Monge-Ampere equation

$\det(\nabla^2 u)=1 \text{ in } B_1\subset\mathbb{R}^n$

when $n=2$, Heinz proved

$|\nabla^2 u|_{B_{1/2}}\leq \sup_{B_1}u$

when $n\geq 3$, Pogorelov has a counter example. One can have a solution $u\in C^1(B_1)$, but $u\in C^{1,\beta}(B_1)$ for some $\beta\in (0,1)$. See his book The Minkowski Multidimensional Problem, on page 83.

### 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

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

### Parallel surfaces and Minkowski formula

Suppose ${X:M^n\rightarrow \mathbb{R}^{n+1}}$ is an immersed orientable closed hypersurface. ${N}$ is the inner unit normal for ${X(M^n)}$ and denote by ${\sigma}$ the second fundamental form of the immersion and by ${\kappa_i}$, ${i=1,\cdots,n}$ the principle curvatures at an arbitrary point of ${M}$. The ${r-}$th mean curvature of ${H_r}$ is obtained by applying ${r-}$elementary symmetric function to ${\kappa_i}$. Equivalently, ${H_r}$ can be defined through the identity

$\displaystyle P_n(t)=(1+t\kappa_1)\cdots(1+\kappa_n)=1+\binom{n}{1}H_1 t+\cdots+\binom{n}{n}H_n t^n$

for all real number ${t}$. One can see that ${H_1}$ represents the mean curvature of ${X}$, ${H_n}$ is the gauss-Kronecker curvature. ${H_2}$ can reflect the scalar curvature of ${M}$ on the condition that the ambient manifold is a space form.

We want to study the consequence of moving the hypersurface parallel. Namely, define ${X_t}$ to be

$\displaystyle X_t= X-tN.$

When ${t}$ is small enough, ${X_t}$ is well defined immersed hypersurface. Suppose ${e_1,\cdots, e_n}$ are principle directions at a point ${p}$ of ${M}$, then

$\displaystyle \quad(X_t)_*(e_i)=(1+\kappa_it)e_i$

here we identify ${X_*(e_i)=e_i}$ as abbreviation. This implies that ${N_t= N\circ X_t^{-1}}$ is also an unit normal field of ${X_t}$. The area element ${dA_t}$ will be

$\displaystyle dA_t=(1+t\kappa_1)\cdots(1+t\kappa_n)dA=P_n(t)dA.$

The second fundamental form of ${X_t}$ with respect to ${N}$ will be

$\displaystyle \sigma_t(v,w)=\langle N_t,\nabla^{\mathbb{R}^{n+1}}_vw\rangle=-\langle \nabla^{\mathbb{R}^{n+1}}_vN_t,w\rangle$

for all ${v,w}$ tangent vector fields on ${X_t(M)}$. Plugging in ${v=(X_t)_*(e_i)}$ and ${w=(X_t)_*(e_j)}$, we get

$\displaystyle (\nabla^{\mathbb{R}^{n+1}}_vw)(X_t(p))=(\nabla^{\mathbb{R}^{n+1}}_{e_i}e_j)(X(p))$

$\displaystyle \nabla_{v}^{\mathbb{R}^{n+1}}N_t=-\frac{\kappa_i}{1+t\kappa_i}v$

So ${e_1,\cdots, e_n}$ are also principle directions for ${X_t}$ and principle curvatures are

$\displaystyle \frac{\kappa_i}{1+t\kappa_i}$

Another way to see this is by choosing a geodesic local coordinates such that ${\partial_iX}$ are the principle directions of ${X}$ at ${p}$. Then

$\displaystyle \partial_j\partial_iX=\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij}$

$\displaystyle \partial_iN=-\kappa_i\partial_iX$

$\displaystyle \partial_i X_t=\partial_i X-t\partial_i N=\partial_i X+t\kappa_i\partial_iX$

$\displaystyle \partial_j\partial_iX_t=(1+t\kappa_i)\partial_j\partial_iX=(1+t\kappa_i)(\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij})$

Since ${g^{ij}_t=(1+\kappa_it)^{-2}\delta_{ij}}$ at ${p}$. Therefore we get the principle curvature are ${\frac{\kappa_i}{1+t\kappa_i}}$.

Therefore the mean curvature ${H(t)}$ for ${X_t}$ is

$\displaystyle H(t)=\frac{1}{n}\frac{P_n'(t)}{P_n(t)}$

Since we have identity

$\displaystyle \Delta|X_t|^2=2n(1+H\langle X_t,N\rangle)$

which implies

$\displaystyle \int_M\left(1+H(t)\langle X_t,N\rangle\right)dA_t=0$

Plugging in all the information,

$\displaystyle \int_M\left(nP_n(t)+P_n'(t)\langle X,N\rangle-tP_n'(t)\right)dA=0$

Reorder the terms in the above identity by the order of ${t}$, we get

$\displaystyle \int_M (H_{r-1}+H_r\langle X,N\rangle )dA=0$

One can use this to prove Heintze-Karcher inequality. There are Minkowski formula in Hyperbolic space and $\mathbb{S}^n$ also.

Remark: S. Montiel and Anotnio Ros, compact hypersurfaces: the alexandrov theorem for higher order mean curvatures. Differential Geometry, 52, 279-296

### f-extremal disk

In the last nonlinear analysis seminar, Professor Espinar talked about the overdetermined elliptic problem(OEP) which looks like the following

$\Delta u+f(u)=0\quad\text{ in }\Omega$

$u>0\quad \text{ in }\Omega$

$u=0 \quad \text{on }\partial \Omega$

$\frac{\partial u}{\partial\eta}=cst\quad\text{on }\partial \Omega$

There is a BCN conjecture related to this

BCN: If $f$ is Lipschitz, $\Omega\subset \mathbb{R}^n$ is a smooth(in fact, Lipschitz) connected domain with $\mathbb{R}^n\backslash\Omega$ connected where OEP admits a bounded solution, then $\Omega$ must be either a ball, a half space, a generalized cylinder or the complement of one of them.

BCN is false in $n\geq 3$. Epsinar wih Mazet proved BCN when $n=2$. This implies the Shiffer conjecture in dimension 2. In higher dimension of Shiffer conjecture, if we know the domain is contained in one hemisphere of $\mathbb{S}^n$, then one can use the equator or the great circle to perform the moving plane.

### Compensated compactness

Suppose $T$ is a vector field and $\nabla\cdot T = 0$. $E= \nabla \psi$ and $\psi$ is a scalar function. We have following theorem(Coifman-Lions-Meyers-Semmes)

Theorem: If $T\in L^2(\mathbb{R}^n)$ and $T\in L^2(\mathbb{R}^n)$, then $E\cdot T\in \mathcal{H}^1(\mathbb{R}^n)$, which is the hardy space.
Given $f(x)\in L^1(\mathbb{R}^n)$, it has harmonic extension $\mathbb{R}^{n+1}_+=\{(x,t)|x\in\mathbb{R}^n, t>0\}$

$\tilde{f}(x,t)=c_n\int_{\mathbb{R}^n}\frac{ f(x-y)t}{(t^2+|x|^2)^{\frac{n+1}{2}}}dy$

Definition: the non-tangential maximal function

$N(f)=\sup_{(\xi,t)\in \Gamma(x)}|\tilde f(\xi, t)|$

It is easy to prove that $N(f)\leq c_n f^*(x)$ the Hardy-Littlewood maximal function. From this we can Hardy norm as

$||f||_{\mathcal{H}^1}=||f||_{L^1}+||N(f)||_{L^1}$

Hardy space consists of all $f$ having finite hardy norm. There is well know fact that the dual space of $\mathcal{H}^1$ is BMO, which is defined as the following.

Define $f\in L^1_{loc}(\mathbb{R}^n)$, if for any cube $Q$,

$\sup_Q\frac{1}{|Q|}\int_Q|f-f_Q|<\infty,\quad \text{where }f_Q=\frac{1}{|Q|}\int_Qf$

then $f\in BMO$. $L^\infty \subset BMO$ and $\log|x|\in BMO$ but not in $L^\infty$.

Let us see how do we use the main theorem. Suppose on $\mathbb{R}^2$, $u$ is the solution of the following elliptic equation

$\displaystyle\frac{\partial}{\partial x_i}\left(a_{ij}(x)\frac{\partial u}{\partial x_j}\right)=\frac{\partial f}{\partial x_1}\frac{\partial g}{\partial x_2}-\frac{\partial f}{\partial x_2}\frac{\partial f}{\partial x_2}$

where $||\nabla f||_{L^2}<\infty$, $||\nabla g||_{L^2}<\infty$ and $(a_{ij})$ is uniform elliptic. YanYan Li and Sagun Chanillo proved that the green function of this elliptic operator belongs to BMO. The right hand side of this equation can be rewritten as $T\cdot E$, where

$T=\left(\frac{\partial f}{\partial x_2}, -\frac{\partial f}{\partial x_1}\right),\quad E=\left(\frac{\partial g}{\partial x_1},\frac{\partial g}{\partial x_a}\right)$

therefore the right hand side belong to $\mathcal{H}^1$. Since

$u(x)=\int G_x(y)T\cdot E(y)dy$

therefore from the theorem we stated at the beginning, we get

$||u||_\infty\leq C||\nabla f||_{L^2}||\nabla g||_{L^2}$