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\title{Sobolev-Poincar\'e inequalities on Carnot groups}
\author{Pierre Pansu}
\date{March 10th, 2023}

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\begin{document}

\begin{frame}
\titlepage
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\section{What are Poincar\'e inequalities?}

\begin{frame}

Sobolev-Poincar\'e inequalities allow to control a function by its first derivatives.

\begin{thm}[Classical Poincar\'e Inequality, global form]
If $1\le q\not=\nu$, there exists a constant $C_q$ with the following effect. For all functions $u:\R^\nu\to\R$ with $Du\in L^q$,
\begin{align*}
\text{if }q<\nu,\text{ let }\frac{1}{p}=\frac{1}{q}-\frac{1}{\nu} ;&\text{ then}\quad \inf_{m\in\R}\|u-m\|_{p}\le C_q\|Du\|_q,\\
\text{if }q>\nu,\text{ let }\alpha=1-\frac{\nu}{q} ;&\text{ then}\quad \sup_{x,y\in\R^n}\frac{|u(x)-u(y)|}{|x-y|^{\alpha}}\le C_q\|Du\|_q.
\end{align*}
\end{thm}\pause

There are local versions too.

\begin{thm}[Classical Poincar\'e Inequality, local form]
Let $B$ denote the unit ball in $\R^\nu$ and $2B$ the twice larger concentric ball. If $1\le q\not=\nu$, there exists a constant $C_q$ with the following effect. For all functions $u:2B\to\R$ with $Du\in L^q(2B)$,
\begin{align*}
\text{if }q<\nu,\text{ let }\frac{1}{p}=\frac{1}{q}-\frac{1}{\nu} ;&\text{ then}\quad \inf_{m\in\R}\|u-m\|_{L^p(B)}\le C_q\|Du\|_{L^q(2B)},\\
\text{if }q>\nu,\text{ let }\alpha=1-\frac{\nu}{q} ;&\text{ then}\quad \sup_{x,y\in B}\frac{|u(x)-u(y)|}{|x-y|^{\alpha}}\le C_q\|Du\|_{L^q(2B)}.
\end{align*}
\end{thm}

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\subsection{Poincar\'e inequalities for mappings}

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Here, we are concerned with mappings $f:B\to G'$, where $B$ is the unit ball in a Carnot group $G$ of homogeneous dimension $\nu$. We care only about the local form, second case, $q>\nu$. What should the condition $Du\in L^q(B)$ be replaced with?\pause

\begin{dfi}
Say that a measurable function $u:B\to\R$ belongs to $W^{1,q}_{\mathrm{loc}}$ if all its directional derivatives $Xu$ along horizontal vectorfields belong to $L^q_{\mathrm{loc}}$ is distributional sense.\pause Then one writes $|D_h u|:=\sup_{|X|\le 1}|Xu|$.\pause

Say that a measurable mapping $f:B\to G'$ belongs to $W^{1,q}$ if there exists a measurable function $g\in L^p_{\mathrm{loc}}$ such that for all $z\in G'$, the composition $u=d(\cdot,z)\circ f$ belongs to $W^{1,q}_{\mathrm{loc}}$, $|D_h u|\le g$ almost everywhere.
\end{dfi}

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\begin{frame}


Here is today's theorem.

\begin{thm}[Hölder bound for $W^{1,q}$ mappings]
If $f\in W^{1,q}(2B, G')$ for $q>\nu$ and $\alpha=1-\frac{\nu}{q}$, then $f$ has a continuous representative and
$$
\sup_{x,y\in B}\frac{d(f(x),f(y))}{|x-y|^{\alpha}}\le C_q\|g\|_{L^q(2B)}.
$$
\end{thm}\pause

\textbf{Remark}. The Theorem follows from the special case $G'=\R$. \pause

\medskip

Indeed, pick a countable dense set of points $z\in G'$ and apply the scalar case to $u_z:=d(\cdot,z)\circ f$. Countable in order to have simultaneously $|D_h u_z|\le g$ for all $z$, almost everywhere. This gives H\"older continuity of $u_z$, hence of $f$.\pause

\bigskip

From now on, $G'=\R$.


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\section{Convolution proof}

\subsection{Kernels of type 1}

\begin{frame}

The trick is to express $u$ in the form $u=(Du)\star k$ for some matrix-valued function $k$. Indeed,

\begin{pro}[Kernels of type 1 map $L^q$ to $C^\alpha$]
Let $k$ be a kernel of type $1$, i.e. a smooth function on $G\setminus\{e\}$ such that\\
\begin{enumerate}
  \item $|k(g)|\le C\,|g|^{1-\nu}$,
  \item $|k(gh)-k(g)|\le C\,|h||g|^{-\nu}$.
\end{enumerate}
If $v\in L^q$, $v\star k\in C^\alpha$ and $\|v\star k\|_{C^\alpha}\le C_q \|v\|_{L^q}$.
\end{pro}\pause

\textbf{Proof}. We need estimate the integral
$$
I=(v\star k)(g_0 h)-(v\star k)(g_0)=\int v(g_0 g^{-1})(k(gh)-k(g))\,dg.
$$
We treat separately the integral $I_1$ over $B(2|h|)$ and the integral $I_2$ over the complement $B^c(2|h|)$.\pause

\begin{align*}
|I_1|&\le\int_{B(2|h|)} |v(g_0 g^{-1})|(|k(gh)|+|k(g)|)\,dg\\
&\le \|v\|_{q}\||k(\cdot h)|+|k|\|_{L^{q'}(B(2|h|))}\\
&\le 2\|v\|_q \|k\|_{L^{q'}(B(C|h|))}.
\end{align*}
\end{frame}

\begin{frame}

By (1), $|k(g)|\le C\,|g|^{1-\nu}$, so $\|k\|_{L^{q'}(B(C|h|))}\le C\,|h|^{\alpha}$.\pause

\bigskip

By (2) and H\"older $(\frac{1}{q}+\frac{1}{q'}=1)$,
\begin{align*}
|I_2|&\le\int_{B^c(2|h|)} |v(g_0 g^{-1})||k(gh)-k(g)|\,dg\\
&\le \|v\|_{q}\left(\int_{B^c(2|h|)}(|h||g|^{-\nu})^{q'}\right)^{1/q'}\\
&\le\|v\|_q |h|(2|h|)^{-\nu+(\nu/q')}\le C\,\|v\|_q |h|^{\alpha}.
\end{align*}

\pause

\medskip

Adding up,
\begin{align*}
|(v\star k)v(g_0 h)-(v\star k)(g_0 )|\le |I_1|+|I_2|\le C\,\|v\|_{q}|h|^{\alpha},
\end{align*}
which proves the H\"older estimate.

\end{frame}


\subsection{The Euclidean case}

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An exact formula like $u=(Du)\star k$ cannot exist for all $u$ on $\R^\nu$ (add a constant to $u$) or on the unit ball $B$ (the boundary interferes). But a slightly deformed version exists, up to an additive constant.

\begin{pro}
Let $u:2B\to \R$. Then, for $x\in B$, 
$$
u(x)-\oint_{2B}u=\int \langle Du(x),k_x(y)\rangle\,dy
$$
where 
$$
k_x(y)=-\frac{y}{|y|}\ell_x(y)\quad\text{and}\quad \ell_x(y)\sim |y|^{1-\nu}.
$$
\end{pro}\pause

Indeed, write
\begin{align*}
u(x)-u(0)&=\int_{0}^{1}\frac{d}{dt}u(tx)\,dt=\int_{0}^{1}\langle Du(tx),x\rangle\,dt.\\
\oint_{2B}u-u(0)&=c\,\int_{2B}\int_{0}^{1}\langle Du(tx),x\rangle\,dt\,dx\\
&=c\,\int_{|y|\le 2t\le 2}t^{-\nu-1}\langle Du(y),y\rangle\,dt\,dy\\
&=\frac{c}{\nu}\,\int_{2B}\langle Du(y),y\rangle((|y|/2)^{-\nu}-1)\,dy.
\end{align*}

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\section{Generalization to Carnot groups}

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In Carnot groups, families of horizontal curves which suitably sweep space exist, this leads to a geometric proof, see \textcolor{red}{Gromov 1993}.

Instead, we explain an alternative approach, due to \textcolor{red}{Jerison 1986}. \pause

\bigskip

For $v:G\to\R$ and $0<t<1$, let $I_t(v)(g)=t^{-\nu}v\circ\delta_{1/t}$. Let $\phi$ be a smooth positive compactly supported function with integral $1$. Then $I_t \phi$ converges to the Dirac mass at $e$ as $t\to 0$. 

\begin{enumerate}
  \item By expanding $u\star \phi-u=u\star\int_{0}^{1}\frac{d}{dt}I_t \phi\,dt$, one gets an expression 
$$
u\star \phi-u=(D_h u)\star k,
$$
where $k$ is a kernel of type 1. So $u\star \phi-u\in C^\alpha$.\pause

  \item The vertical derivatives of $u\star\phi$ are again expressible as $D_v(u\star \phi)=(D_h u)\star k'$, where $k'$ belongs to $L^1$. \pause
  
  \item Young's inequality then implies that the full Euclidean derivative $D(u\star\phi)$ of $u\star\phi$ satisfies $\|D(u\star\phi)\|_q\le C_{\phi}\|D_h u\|_q$.\pause

  \item Apply the Euclidean Poincar\'e inequality. Since $d_{\mathrm{Eucl}}\le d_{\mathrm{CC}}$, this shows that $u\star \phi\in C^\alpha$, hence $u\in C^\alpha$.
\end{enumerate}

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\subsection{Step 1, Euclidean case}

\begin{frame}

We explain step 1 first in the Euclidean case.
\begin{align*}
\frac{\partial}{\partial x_i}(t^{-\nu}\frac{x_i}{t}\phi(\frac{x}{t}))=t^{-\nu-1}\phi(\frac{x}{t})+t^{-\nu-1}\frac{x_i}{t}\frac{\partial\phi}{\partial x_i}(\frac{x}{t}),
\end{align*}\pause
\begin{align*}
\frac{d}{dt}I_t\phi(x)&=\frac{d}{dt}t^{-\nu}\phi(\frac{x}{t})=-\nu t^{-\nu-1}\phi(\frac{x}{t})-\sum_{i=1}^{\nu}t^{-\nu-1}\frac{x_i}{t}\frac{\partial\phi}{\partial x_i}(\frac{x}{t})\\
&=-\sum_{i=1}^{\nu}\frac{\partial}{\partial x_i}(t^{-\nu}\frac{x_i}{t}\phi(\frac{x}{t}))\\
&=-\sum_{i=1}^{\nu}\frac{\partial}{\partial x_i}I_t(x_i\phi(x)).
\end{align*}\pause
\begin{align*}
u\star\phi-u=u\star\int_{0}^{1}\frac{d}{dt}I_t\phi(x)\,dt=u\star\sum_{i=1}^{\nu}\frac{\partial}{\partial x_i}\psi_i =\sum_{i=1}^{\nu}\frac{\partial u}{\partial x_i}\star\psi_i,
\end{align*}
where $\psi_i(x)=-\int_{0}^{1}I_t(x_i\phi(x))\,dt$ satisfies $|\psi_i(x)|\le C\,|x|^{-\nu+1}$.

\end{frame}

\subsection{Steps 1 and 2, general case}

\begin{frame}

In a general Carnot group, each adapted coordinate $x_i$ has a weight $w_i$, and 
\begin{align*}
\frac{d}{dt}I_t\phi(x)&=\frac{d}{dt}t^{-\nu}\phi(\ldots,\frac{x_i}{t^{\textcolor{red}{w_i}}},\ldots)\\
&=-\nu t^{-\nu-1}\phi(\delta_{1/t}x)-\sum_{i=1}^{\nu}t^{-\nu-1}\textcolor{red}{w_i} \frac{x_i}{t^{\textcolor{red}{w_i}}}\frac{\partial\phi}{\partial x_i}(\delta_{1/t}x)\\
&=-\sum_{i=1}^{\nu}\frac{\partial}{\partial x_i}(t^{-\nu}\textcolor{red}{w_i} \frac{x_i}{t^{\textcolor{red}{w_i}}}\phi(\delta_{1/t}x))\\
&=-\sum_{i=1}^{\nu}\frac{\partial}{\partial x_i}I_t(\textcolor{red}{w_i}x_i\phi(x)).
\end{align*}\pause

Whence an expression like
$$
\sum_{i=1}^{\mathrm{dim}(G)}\frac{\partial u}{\partial x_i}\star\psi_i.
$$
\end{frame}

\begin{frame}

The other change derives from noncommutativity. If $Y$ is a left-invariant vectorfield, let $Y^R$ denote its image by $g\mapsto g^{-1}$. It is a right-invariant vectorfield. Then
$$
Y(u\star v) = u\star Yv, \quad (Yu)\star v=-u\star Y^R v.
$$\pause

We need replace each $\frac{\partial}{\partial x_i}$ with a sum of products $Y_j^R D_{ij}$ where $Y_j^R$ are horizontal right-invariant vectorfields and $D_{ij}$ are differential operators. This is a matter of linear algebra. One gets
$$
u\star\phi-u=\sum_{i,j}u\star Y_j^R D_{ij}\psi_i=\sum_{j}Y_j u\star (-\sum_i D_{ij}\psi_i),
$$
and checks that the right multipliers are kernels of type 1.\pause

\bigskip

To perform step 2, i.e. express vertical derivatives, one needs a similar algebraic trick, with $\frac{\partial}{\partial x_i}$ replaced with a left-invariant vectorfield $Y_i$: $Y_i=\sum_{j}Y_j^R D'_{ij}$ implies that
\begin{align*}
Y_i(u\star\phi)=u\star Y_i \phi=\sum_{j}u\star (\sum_{j}Y_j^R D'_{ij}\phi)=\sum_{j}Y_j u\star D'_{ij}\phi.
\end{align*}

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\subsection{Algebraic tricks}

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I explain the algebraic trick on the example of Engel's group, whose Lie algebra has basis $X,Y,Z,T$ with $Z=[X,Y]$ and $T=[X,Z]$. $X,Y$ are homogeneous of degree $-1$, $Z$ of degree $-2$ and $T$ of degree $-3$.\pause

\medskip

Since $T$ is central, 
$$
T=-T^R=-[X^R,Z^R]=-X^R Z^R +[X^R,Y^R] X^R = X^R(-Z^R+Y^R X^R)-Y^R(X^R X^R).
$$\pause


Then $Z=-Z^R+aT$ where the vectorfield $aT$ is homogeneous of degree $-2$, so is the function $Ta$. Since $Ta$ is smooth, $Ta=0$, so $aT=T\circ a$. Hence
$$
Z=-X^R Y^R+Y^R X^R+T^R\circ a.
$$\pause

Finally, $X=-X^R+bZ+cT$ where the vectorfields $bZ$ and $cT$ are homogeneous of degree $-1$, so are the functions $Zb$ and $Tc$, so both vanish. Thus
$$
X=-X^R+Z \circ b+T\circ c
$$
again has the required form thanks to above expressions for $Z$ and $T$.

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\end{document}

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\begin{pro}[Young's inequality]
Assume $\frac{1}{p}=\frac{1}{q}+ \frac{1}{r}+1$ and $p>1$, $r>1$. Then convolution maps $L^q \otimes L^r$ to $L^p$.
\end{pro}

