Let satisfy the elliptic equation

Assume is jointly concave with respect to and

Then

can not achieve positive maximum in the interior of .

See the paper Korevaar 1983

when time goes to infinity

Let satisfy the elliptic equation

Assume is jointly concave with respect to and

Then

can not achieve positive maximum in the interior of .

See the paper Korevaar 1983

Suppose is the position vector of which flows by mean curvature, that is

It is well known that for mean curvature flow the evolution equation of mean curvature is

Translator is a special type solution of mean curvature flow such that it moves by translation. Namely, if is a translator moves to direction , then satisfies,

The mean curvature of a translator satisfies and

where is called height function.

How do we connect (2) and (3)? Actually it results from the difference of (1) and (3). Recall that (1) follows from (3) composed with some self diffeomorphism. Let us suppose is that diffeomorphism (). Then satisfies (1). Since is also translator, . Using ,

Since , (2) will imply

This just means (4) holds.

Let is the cylinder in . Suppose is a graph on the cylinder can be written as

where is the normal vector of , that is . What is the mean curvature of ?

We can parametrize the cylinder by where , , , . One can write as

Then the induced metric on is

Since

The normal vector of is

The second fundamental form is

Then the mean curvature is

If and , then

I will describe some parts of Troynov’s work on conical surface. For details, check his paper. For simplicity, we consider a closed Riemann surface with a real divisor . A conformal metric on is said to represent the divisor if is smooth Riemannian metric on and near each , there exists a neighborhood of and coordinate function and such that and

where . is called the angle of the conical singularity. We will always assume .

For example equipped with the metric is isometric to an Euclidean cone of total angle .

Suppose now is a closed Riemann surface with conical metric . **Assume that the Gauss curvature extends on as a H\”{o}lder continuous function.**

Suppose is a conformal change of the conical metric. Then necessarily we have . In order to prescribe the Gauss curvature , we need to solve the equation

.

Any reasonable solution of (0) will satisfy

where follows from the Gauss-Bonnet formula for conical surface. We will use variational method to attack it. To that end, define to be Sobolev space of functions with

and functionals

It is easy to verify that the minimizer of the following variation problem will be a weak solution of (0)

As one can see, there are two immediate questions we need to answer,

(a)

(b) minimizer exists

The first question follows from the Trudinger inequality. Namely, define . Then

**Lemma:** For any fixed , there exist constant such that,

for any and and .

**Lemma:** Suppose and . Let . Then for any fixed , there exists constants such that

for any and . Here .

From this key lemma, one can derive that is lower bounded on provided and . To do that, choose and such that , then it follows from the above lemma that

For the second question, we need to prove the embedding is compact(actually it is true for any for any ). Note that this is true if is some smooth metric, however is conical one. Therefore, we want to compare with of some smooth metric.

Suppose is equipped with smooth metric such that . Here is smooth and positive outside the support of and near . If we use the denote the gradient of with respect to , then and

This is only true in two dimension. Now and have inner product

From the above analysis, we need to examine the difference of and . It follows from the singular behavior of that

for any and some . Then for any ,

Then . On the contrary, we need the following inequality

**Lemma: **For conical metric

Then we have . Therefore is compact for . Furthermore, after some effort, one can prove

is also compact for .

Remark:

[1] M. Troyanov, Prescribing curvature on compact surfaces with conical singularities, Trans. Am. Math. Soc. 324 (1991) 793–821. doi:10.1090/S0002-9947-1991-1005085-9.

Suppose is 2-dimensional surface in . If is simply connected, then may not be simply connected, where is ball in . See the following figure.

But if is minimal surface, then must be simply connected. The reason is has convex haul property.

Learn this example from Jacob Bernstein.

Suppose is a radial function on , here .

therefore

If we use the polar coordinates , and and the following fact

then one can calculate the Hessian of under this coordinates

Then

If the metric is , then we will have

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 be a domain in . is a smooth function on . Then

**Proof:** Take an orthonormal frame field such that is tangent to . Notice

where . It follows from Remark 2.3 that

where is the outward unit normal to . Changing the coordinates to and , we can get

It is easy to see

and

Therefore the proposition is established.

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

For any (not necessarily symmetric) matrix , we let denote the sum of its principle minors. For any hypersurface which is a graph of , where . We have its downward unit normal is

The principle curvatures are taken from the eigenvalues of the Jacobian matrix . One can define its mean curvature using the notation of above

Now let us consider general matrix ,

where is the generalized Kronecker delta.

Then we define the Newton tensor

For any vector field on , is a matrix, where and , denote , we have

Since for any

because is skew-symmetric in . Then

Applying the formula (Check [1] Propsition 1.2) and , we get

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

Suppose is a vector field normal to , then

where is the outer ward unit normal to .

**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 is a symmetric matrix in order to use the formula of Reilly. Not sure about this.

Suppose is a column vector and is an invertible matrix. Set , then

This formula has more general forms.

Consider is a domain in the sphere. is a radial graph over .

What is the unit normal to this radial graph in ?

Suppose is a smooth local frame on . Let be the covariant derivative on . Tangent space of consists of which are

In order to get the unit normal, we need some simplification. Let us assume are orthonormal basis of the tangent space of and . Then

Then we obtain an orthonormal basis of the tangent of

We are able to get the normal by projecting to this subspace

After normalization, the (outer)unit normal can be written

**Remark: **Guan, Bo and Spruck, Joel. *Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.*