## Notes of Zoltan Balogh’s lecture nr 2

Horizontal convexity in the Heisenberg group

Joint with Andrea Calogero and Alexandru Kristály.

1. Alexandrov’s theorem in Euclidean space

I think of a convex function as a function which stays above its supporting affine function at every point. The set of slopes (covectors) of these affine functions at ${x_0}$ is called the subdifferential ${\partial u(x_0)}$ of ${u}$ at ${x_0}$. The union of all these sets over the domain is denoted by ${\nabla u(\Omega)}$. Its measure ${\mathcal{L}^n(\nabla u(Omega))}$ is sometimes called the Monge-Ampère measure of ${u}$.

Theorem 1 (Alexandrov) There is a dimension dependant constant ${C}$ with the following effect. Let ${\Omega}$ be an open, bounded convex domain in ${{\mathbb R}^n}$. Let ${u}$ be a convex function on the closure of ${\Omega}$, which vanishes on the boundary. Then, for all ${x_0\in\Omega}$,

$\displaystyle \begin{array}{rcl} |u(x_0)|^n\leq C\,d(x_0,\partial \Omega).\mathrm{diameter}(\Omega)^{n-1}.\mathcal{L}^n(\nabla u(Omega)). \end{array}$

This is used in PDE (Caffarelli,…).

2. Convexity in Heisenberg group

Since left translations ar affine is exponential coordinates, Heisenberg group carries an affine structure. Therefore convex domains will simply be Euclidean convex.

Definition 2 (Several competing groups) Say a function on a convex domain ${\Omega}$ of Heisenberg group is ${H}$-convex if its restriction to every horizontal line of ${\Omega}$ is convex.

The subdifferential of ${u}$ at ${x_0}$ is a subset of ${{\mathbb R}^{2n}}$.

Note that there are H-convex functions in ${\mathbb{H}^n}$ which are very irregular (e.g. Weierstrass) in the vertical direction.

2.1. Results

We define a horizontal slicing diameter : this is the maximal diameter of the intersection of ${\Omega}$ with horizontal planes ${H_x}$, ${x\in\mathbb{H}^n}$. We also define a horizontal slicing Monge-Ampère measure

$\displaystyle \begin{array}{rcl} \mathcal{L}_{HS}^{2n}(\nabla_H u(\Omega))=\sup_{x\in\Omega}\mathcal{L}^{2n}(\nabla_H u (H_x \cap\Omega)). \end{array}$

Theorem 3 There is a dimension dependant constant ${C}$ with the following effect. Let ${\Omega}$ be an open, bounded convex domain in ${\mathbb{H}^n}$. Let ${u}$ be a convex function on the closure of ${\Omega}$, which vanishes on the boundary. Then, for all ${x_0\in\Omega}$,

$\displaystyle \begin{array}{rcl} |u(x_0)|^{2n}\leq C\,d(x_0,\partial \Omega).\mathrm{diameter}_{HS}(\Omega)^{2n-1}.\mathcal{L}_{HS}^{2n}(\nabla u(\Omega)). \end{array}$

This improves earlier results by Garofalo et al. where the distance to the boundary appeared with a negative power.

3. Proof

3.1. Back to the Euclidean case

Lemma 4 (Comparison principle) Let ${u}$, ${v}$ be continuous functions on the closure of ${\Omega}$. Assume that ${u\leq v}$. Then

$\displaystyle \begin{array}{rcl} \nabla v(Omega)\subset \nabla u(\Omega). \end{array}$

Indeed, any supporting hyperplane of the graph of ${u}$, when raised, will touch the graph of ${v}$.

Alexandrov compares the graph of ${u}$ with the cone on ${\partial \Omega}$ with vertex at ${(x_0,u(x_0))}$. Its subdifferential is concentrated at the vertex. Let ${x}$ be the nearest point in the boundary. In the subdifferential ${\partial v(x_0)}$, there is a covector ${p_1}$ of size ${\sim|u(x_0)|/d(x,x_0)}$. All othe covectors ${p}$ in ${\partial v(x_0)}$ satisfy

$\displaystyle \begin{array}{rcl} |p|\geq \frac{|u(x_0)|}{\mathrm{diameter}(\Omega)}. \end{array}$

3.2. Failure of comparison principle in Heisenberg group

There exists functions ${u,v}$ on a cyclinder ${\Omega}$, which are equal n the boundary and ${u\leq v}$, but ${\nabla_H v(\Omega)\not\subset \nabla_H u(\Omega)}$.

Indeed, set ${v(x,y,t)=t}$. Check that ${\partial _H v(x,y,t)=(2y,-2x)}$, so that ${\nabla_H v(\Omega)=B(0,2)}$ contains the origin. Modify ${v}$ in an annulus,

$\displaystyle \begin{array}{rcl} u(x,y,t)=t-(1-t^2)g(x,y) \end{array}$

where ${g}$ has support in an annulus. Assume that ${0\in \nabla_H u(\Omega)}$. Then ${0\in \partial_H u(q)}$, and ${u}$ achieves its minimum on ${H_q}$ at point ${q}$. One can achieve that this never happens.

3.3. Comparison for convex functions

What saves us is that comparison holds for convex functions.

Theorem 5 Let ${\Omega}$ be a convex domain in ${\mathbb{H}^n}$, let ${u,v}$ be convex functions on ${\Omega}$ that are equal on the boundary. Assume that for some ${x_0\in\Omega}$, there exists ${p\in\partial_H v(x_0)}$ such that, for al ${x\in H_{x_0}\cap\Omega}$ different from ${x}$,

$\displaystyle \begin{array}{rcl} v(x) > v(x_0)+p.(\pi(x)-\pi(x_0)). \end{array}$

Then ${p\in \partial_H u(x_0)}$.

The proof uses degree theory for set valued maps. For simplicity, let us assume that ${u}$ is smooth, and ${x_0=0}$. Let ${U=H_{x_0}\cap \Omega}$ projected to ${{\mathbb R}^{2n}}$. We view ${\partial_H u}$ as a mapping of ${U}$ to ${{\mathbb R}^{2n}}$. To show that ${p}$ belongs to its image, it suffices to show that the degree of ${\partial_H u}$ on ${U}$ at ${p}$ is non zero. We check that this is the case when ${\partial_H u}$ is replaced with ${\partial_H v}$. Then a linear homotopy allows to conclude. Indeed, assume by contradiction that the homotopy hits ${p}$ along ${\partial U}$, i.e. there exists a point ${x\in\partial\Omega\cap H_{x_0}}$ and ${t\in[0,1]}$ such that

$\displaystyle \begin{array}{rcl} t\partial_H u(x)+(1-t)\partial_H v(x)=\partial_H v(x_0). \end{array}$

Along the horizontal line from ${x_0}$ to ${x}$,

$\displaystyle \begin{array}{rcl} u(x_0) \geq u(x)+\partial_H u(x).(\pi(x_0)-\pi(x)),\quad v(x_0) \geq v(x)+\partial_H v(x).(\pi(x_0)-\pi(x)). \end{array}$

Take the convex combination of these two inequations, get and inequality that contradicts the assumption ${v(x) > v(x_0)+p.(\pi(x)-\pi(x_0))}$.

Computation of the index for ${v}$.

3.4. End of the proof

One gets

$\displaystyle \begin{array}{rcl} |u(x_0)|^{2n}leq C\,d(x_0,H_{x_0}\cap\partial \Omega).\mathrm{diameter}_{HS}(\Omega)^{2n-1}.\mathcal{L}_{HS}^{2n}(\nabla u(Omega)). \end{array}$

There remains to replace ${d(x_0,H_{x_0}\cap\partial \Omega)}$ with ${d(x_0,\partial \Omega)}$. This relies of an Harnack inequality, which allows to replace ${x_0}$ with a nearby point where the horizontal plane is tilted and hits the boundary at a distance comparable to the distance of ${x_0}$ to the boundary.

Next sessions : April 4th and May 14th.