\documentclass[8pt]{beamer}
\usepackage[ansinew]{inputenc}
\usepackage{amssymb, MnSymbol, color}
\usepackage{graphicx}
\usepackage{pgf,tikz}
\usepackage{mathrsfs}
\usetikzlibrary{arrows}
\usepackage[french]{babel}
\usepackage{hyperref}
\usepackage{beamerthemesplit}

\newtheorem{thm}{Theorem}
\newtheorem{dfi}{Definition}
\newtheorem{csq}{Consequence}
\newtheorem{pro}{Proposition}
\newtheorem{exe}{Example}
\newtheorem{exo}{Exercise}

\def\R{\mathbb{R}}
\def\C{\mathbb{C}}
\def\Z{\mathbb{Z}}
\def\n#1{{|\hskip-1pt | #1 |\hskip-1pt |}}
\def\sol{\smallskip\par\textbf{Solution}.~}
\def\a{\mathcal{A}}
\def\f{\mathcal{F}}

\title{Measures constrained by linear PDEs [after de Philippis and Rindler]}
\author{Pierre Pansu, Universit\'e Paris-Saclay}
\date{March 1st, 2021}


\mode<presentation>{\usetheme{Warsaw}}

\begin{document}

\begin{frame}
\titlepage
\end{frame}

\section{Linear PDEs}

\subsection{The problem}

\begin{frame}

A constant coefficient linear PDE on a vector-valued function $u:\Omega\subset\R^d\to\R^N$ takes the form
$$
\a(u)=\sum_\alpha A_\alpha \partial^\alpha u=0,
$$
where $A_\alpha\in Hom(\R^N,\R^n)$ are linear maps. Here $\partial^\alpha u=\partial_1^{\alpha_1}\cdots\partial_k^{\alpha_k}u$ denote partial derivatives.\pause

\begin{exe}
Assume $u\in C^\infty(\Omega,(\R^d)^*)$ is a differential $1$-form on open set $\Omega$. Its exterior differential $du\in C^\infty(\Omega,\Lambda^2(\R^d)^*)$. Then $du=\a(u)$ where $\a$ has order $1$. Putting all $A_\alpha$ together into a single linear map 
\begin{align*}
A&\in Hom(\R^d,Hom((\R^d)^*,\Lambda^2(\R^d)^*))\\
&=Hom((\R^d)^*\otimes(\R^d)^*,\Lambda^2(\R^d)^*),
\end{align*} 
$A$ becomes skew-symmetrization of bilinear forms on $\R^d$.
\end{exe}

\vskip1.3cm

\end{frame}

\begin{frame}

A constant coefficient linear PDE on a vector-valued function $u:\Omega\subset\R^d\to\R^N$ takes the form
$$
\a(u)=\sum_\alpha A_\alpha \partial^\alpha u=0,
$$
where $A_\alpha\in Hom(\R^N,\R^n)$ are linear maps. Here $\partial^\alpha u=\partial_1^{\alpha_1}\cdots\partial_k^{\alpha_k}u$ denote partial derivatives.

\begin{exe}
Assume $u\in C^\infty(\Omega,Hom(\R^d,\R^\ell))$ is a $\R^\ell$-valued differential $1$-form. Its exterior differential $du\in C^\infty(\Omega,Hom(\Lambda^2\R^d,\R^\ell))$. Then $du=\a(u)$ where $\a$ has order $1$. Putting all $A_\alpha$ together into a single linear map 
\begin{align*}
A&\in Hom(\R^d,Hom((\R^d)^*\otimes\R^\ell,\Lambda^2(\R^d)^*\otimes\R^\ell))\\
&=Hom((\R^d)^*\otimes(\R^d)^*\otimes\R^\ell,\Lambda^2(\R^d)^*\otimes\R^\ell),
\end{align*} 
$A$ becomes skew-symmetrization of bilinear maps $\R^d\to\R^\ell$.
\end{exe}\pause

\medskip

\textbf{Question}. Assume a vector-valued measure $\mu$ satisfies $\a(\mu)=0$ in distributional sense. Ellipticity should imply restrictions on the singular part $\mu^s$ of $\mu$. Which restrictions?

\end{frame}

\subsection{Ellipticity}

\begin{frame}

\textbf{The principal symbol}. In Fourier, if $\xi\in(\R^d)^*$, 
$$
\f(\a(u))(\xi)=A(2\pi i\xi)(\f(u)(\xi)),
$$
where $A(\xi)=\sum_{\alpha} A_\alpha\xi^\alpha\in Hom(\R^N,\R^n)$. It is a sum of homogeneous terms, the term of highest degree $A^k(\xi)$ is the \emph{principal symbol} of $\a$ evaluated at $\xi$.\pause

\begin{exe}
For the exterior differential, for $\eta\otimes e\in (\R^d)^*\otimes\R^\ell$, $A(\xi)(\eta\otimes e)=(\xi\wedge\eta)\otimes e$.
\end{exe}\pause

\begin{exe}
For the scalar Laplacian, $A(\xi)=|\xi|^2$.
\end{exe}\pause

Assume $\a$ is \emph{elliptic}, i.e. $A(\xi)$ is injective for all $\xi\not=0$. Then $\a(u)=0\implies \mathrm{supp}(\f(u))=\{0\}$, so $u$ is a polynomial. \pause

\medskip

Conversely, assume that $\a$ is homogeneous. If there exists nonzero $\xi\in(\R^d)^*$ and $\lambda\in\R^n$ such that $\lambda\in\mathrm{ker}(A(\xi))$, then for every distribution $h:\R\to\R$, the plane wave $u=(h\circ\xi)\lambda$ solves $\a(u)=0$. Thus $\a$ has nonsmooth solutions.

\end{frame}

\subsection{The wave cone}

\begin{frame}

Nonelliptic behaviours are concentrated in the \emph{wave cone}.

\begin{dfi}[Wave cone]
The wave cone of an order $k$ operator $\a$ is
$\displaystyle\Lambda_\a:=\bigcup_{\xi\not= 0} \mathrm{ker}A^k(\xi) \subset\R^N$.
\end{dfi}\pause

\begin{exe}
For the exterior differential $d$, 
$\displaystyle\Lambda_\a=\bigcup_{\xi\not= 0} \xi\otimes\R^\ell\subset (\R^d)^*\otimes\R^\ell=Hom(\R^d,\R^\ell)$ consists of all rank $\leq 1$ linear maps $\R^d\to\R^\ell$.
\end{exe}\pause
Indeed, the kernel of $\xi\wedge\cdot:\Lambda^1\to\Lambda^2$ coincides with the image of $\xi\wedge\cdot:\Lambda^0\to\Lambda^1$. This reflects (at the symbol level) the (local) exactness of the de Rham complex.

\end{frame}

\section{The result}

\begin{frame}



\begin{thm}[de Philippis-Rindler 2018]
Let $\mu$ be an $\R^N$-valued measure which solves $\a(\mu)=0$ on open set $\Omega$. Then $\frac{d\mu}{d|\mu|}(x)\in\Lambda_\a$ for $|\mu|^s$-almost every point $x\in\Omega$.
\end{thm}\pause

\begin{exe}
When $\a$ is the exterior differential on $1$-forms, $Hom(\R^d,\R^\ell)$-valued measures solving $d\mu=0$ are differentials of $\R^\ell$-valued $BV$ functions. \pause
According to the Theorem, their singular parts are supported on points where  $\frac{d\mu}{d|\mu|}$ is a rank $1$ linear map $\R^d\to\R^\ell$.
\end{exe}

This is Alberti's Rank One Theorem (1993), solving a conjecture of Ambrosio-De Giorgi (1988).

\end{frame}


\section{Proof of Theorem}

\subsection{Strategy}

\begin{frame}

I stick to degree $1$ operators. Let $\mu$ be an $\R^N$-valued measure solving $\a(\mu)=0$. Let $|\mu|$ denote its total variation. Consider the Radon-Nikodym decomposition 
$$
\mu=\frac{d\mu}{d|\mu|}d|\mu|.
$$
Let $E=\{x\in\R^d\,;\,\frac{d\mu}{d|\mu|}(x)\notin \Lambda_\a\}$. By contradiction, assume that $|\mu|^s(E)>0$.\pause

\medskip

One can pick $x_0\in E$ and a subsequence $r_k$ of radii which satisfy the following. Let $\mu_k$ denote the blowing up of $\mu$ at $x_0$, at scale $r_k$, normalized by $|\mu|(B(r_k))$.
\begin{enumerate}
  \item $|\mu_k|$ converges vaguely to a measure $\nu$.
  \item $\lambda_0=\frac{d\mu}{d|\mu|}(0)\notin\Lambda_\a$.
  \item $\mu_k$ converges vaguely to the measure $\nu\lambda_0$.
  \item $\frac{|\mu|^s(B(r_k))}{|\mu|(B(r_k))}$ tends to $1$.
\end{enumerate}\pause

\medskip

One shows that $\a(\nu\lambda_0)=0$. Since $\lambda_0\notin\Lambda_\a$, this implies that $\mathrm{supp}(\f(\nu))=0$, hence that $\nu$ is absolutely continuous.\pause

\medskip

On the other hand, one shows that $|\mu_k|$ converges to $\nu$ in total variation. This is a contradiction, since $||\mu_k|-\nu|\geq|\mu_k|^s(B(r_k))/|\mu|(B(r_k))$ which tends to $1$.

\end{frame}

\subsection{Absolute continuity of tangent measures}

\begin{frame}

\textbf{First step: absolute continuity of tangent measures}

\bigskip

In distributional sense, $\a(\nu \lambda_0)=\lim\a(\mu_k)=0$. \pause 
Taking Fourier transforms,
$$
\forall \xi\in(\R^d)^*,\quad (\f(\nu)(\xi))A(\xi)\lambda_0=0.
$$\pause
Since $\lambda_0\notin\Lambda_\a$, $A(\xi)\lambda_0\not=0$, hence $\f(\nu)(\xi)=0$. \pause

\medskip

Since $\mathrm{supp}(\f(\nu))=\{0\}$, $\nu$ is a (polynomial) function times Lebesgue measure.

\end{frame}

\subsection{Convergence in total variation}

\begin{frame}

\textbf{Second step: from vague convergence to convergence in total variation}

\bigskip



Let $\nu_k=|\mu_k|^s$. 

Recall that $\frac{|\mu|^s(B(r_k))}{|\mu|(B(r_k))}$ tends to $1$. One can furthermore assume that
$\oint_{B(r_k)}| \frac{d\mu}{d|\mu|}-\lambda_0|\,d|\mu|^s$ tends to $0$. It follows that $|\nu_k\lambda_0-\mu_k|(B(1))$ tends to $0$.\pause

\medskip

Pick a smooth cut-off function $\chi$. Then $|\chi \nu_k \lambda_0-\chi \mu_k|(\R^d)$ tends to $0$.\pause

Since $\a(\mu_k)=0$,
$$
\a(\chi\nu_k\lambda_0)=\a(\chi\nu_k\lambda_0-\chi\mu_k)+A(d\chi)\mu_k.
$$
We shall see that ellipticity allows to invert $\a$ with estimates. By density, one can replace measures with smooth functions and total variation with $L^1$ norm. The goal is to show that $\chi\nu_k$ is precompact in $L^1$. The estimates will stem from H\"ormander- Mikhlin's multiplier theorem.\pause

\medskip

Taking Fourier transforms,
$$
\forall \xi,\quad \f(\chi\nu_k)A(\xi)\lambda_0=A(\xi)\f(V_k)(\xi)+\f(R_k)(\xi),
$$
where $V_k=\chi\nu_k\lambda_0-\chi\mu_k$ tends to $0$ in $L^1$ and $R_k=A(d\chi)\mu_k$ is bounded in $L^1$.

\end{frame}

\begin{frame}
Inner multiplying with $\overline{A(\xi)\lambda_0}$, one gets
$$
|A(\xi)\lambda_0|^2 \f(\chi\nu_k)=A(\xi)\lambda_0\cdot A(\xi)\f(V_k)(\xi)+A(\xi)\lambda_0\cdot\f(R_k)(\xi),
$$
and then
$$
(1+|A(\xi)\lambda_0|^2) \f(\chi\nu_k)=A(\xi)\lambda_0\cdot A(\xi)\f(V_k)(\xi)+A(\xi)\lambda_0\cdot\f(R_k)(\xi)+\f(\chi\nu_k).
$$
So $\f(\chi\nu_k)=T_0(V_k)+T_1((Id-\Delta)^{-1/2}R_k)+T_2((Id-\Delta)^{-1}\f(\chi\nu_k))$ is the sum of three terms, involving operators defined by Fourier multipliers
\begin{align*}
m_0(\xi)&=(1+|A(\xi)\lambda_0|^2)^{-1}\overline{A(\xi)\lambda_0}\cdot A(\xi),\\
m_1(\xi)&=(1+|A(\xi)\lambda_0|^2)^{-1}(1+4\pi^2|\xi|^2)^{1/2}\overline{A(\xi)\lambda_0},\\
m_2(\xi)&=(1+|A(\xi)\lambda_0|^2)^{-1}(1+4\pi^2|\xi|^2).
\end{align*}

$(Id-\Delta)^{-1/2}$ is bounded and compact from $L^1$ to some $L^q$, $q>1$. $T_1$ and $T_2$ are bounded from $L^q$ to $L^q$. Therefore $w_k:=T_1((Id-\Delta)^{-1/2}R_k)+T_2((Id-\Delta)^{-1}\f(\chi\nu_k))$ is precompact in $L^1$.

\end{frame}

\begin{frame}

Since $\lambda_0\notin \Lambda_\a$, $|A(\xi)\lambda_0|/|\xi|$ is bounded away from $0$. This implies that $T_0$ is bounded from $L^1$ to the Lorentz space $L^{1,\infty}$ (H\"ormander-Mikhlin). Hence $u_k :=T_0(V_k)$ tends to $0$ in measure.

\medskip

Since $u_k+w_k=\chi\nu_k\geq 0$, $u_k^-$ is dominated by $|w_k|$. By Vitali's convergence theorem (strengthening of dominated convergence theorem where a.e. convergence is replaced with convergence in measure), $u_k^-$ tends to $0$ in $L^1$.

\medskip

Since $V_k\to 0$ in distributional sense, so does $u_k=T_0(V_k)$. For every smooth cut-off function $\phi$, $0\leq\phi\leq 1$,
$$
\int\phi|u_k|=\int\phi u_k+2\int\phi u_k^-\leq \int\phi u_k+2\int u_k^-
$$
tends to $0$, so $u_k$ tends to $0$ in $L^1$.

\medskip

Hence $\chi\nu_k=u_k+w_k$ is precompact in $L^1$. Its vague limit is $\chi\nu$, so it subconverges in $L^1$ to $\chi\nu$. This shows that $|\mu_k|^s$ converges to $\nu$ in total variation.

\medskip

This contradicts the fact that $\nu$ is absolutely continuous, and completes the proof of the theorem.

\end{frame}

\end{document}




