Notes of Alessio Figalli’s lecture

Quantitative stability estimates for the Brunn-Minkowski inequality

1. The stability problem

1.1. Brunn-Minkowski’s inequality

Let {A\subset {\mathbb R}^n}. Note that {A\subset\frac{A+A}{2}}, so

\displaystyle  \begin{array}{rcl}  |A|\leq|\frac{A+A}{2}|. \end{array}

When does equality hold ? When {|Conv(A)\setminus A|=0}. This is an easy particular case of the following.

Theorem 1 (Brunn-Minkowski) Let {A} and {B} be sets of positive volume in {{\mathbb R}^n}. Then

\displaystyle  \begin{array}{rcl}  |\frac{A+B}{2}|^{1/n}\geq \frac{|A|^{1/n}+|B|^{1/n}}{2}. \end{array}

To get convinced why exponents {1/n} need be there, take balls.

When does equality hold ? When {A}, {B} are convex and homothetic, up to sets of measure 0.

Brunn-Minkowski’s inequality is related, via isoperimetric and Sobolev inequalities, to all functional inequalities.

1.2. Are minimizer stable ?

For instance, for Sobolev inequality

\displaystyle  \begin{array}{rcl}  \|u\|_q \leq C\|\nabla u\|_p, \end{array}


\displaystyle  \begin{array}{rcl}  \leq C\|\nabla u\|_p-\|u\|_q \ll 1 \end{array}

imply that {u} is close to a minimizer ?

We are concerned with a similar question for Brunn-Minkowski’s inequality.

1.3. Quantitative formulation

Let {|A|\geq\lambda} and {|B|\leq\Lambda}. Set

\displaystyle  \begin{array}{rcl}  \delta=|\frac{A+B}{2}|^{1/n}- \frac{|A|^{1/n}+|B|^{1/n}}{2}. \end{array}

Is it true that, as {\delta} tends to 0, {A} and {B} become nearly convex and nearly homothetic ? Of course, we want something quantitative. We separate the convexity issue from the homothety issue.

So we assume first that {A} and {B} are convex. Set

\displaystyle  \begin{array}{rcl}  \gamma=\frac{|A|^{1/n}}{|B|^{1/n}}, \end{array}

so that {|\gamma B|=|A|}. Set

\displaystyle  \begin{array}{rcl}  d(A,B)=\inf_{x\in{\mathbb R}^n}|A\Delta(\gamma B+x)|. \end{array}

Question. Find an estimate of {d(A,B)} in terms of {\delta}.

2. The homothety issue

2.1. A result dating back to 2009

Theorem 2 (Figalli-Maggi-Pratelli 2009) There exists a constant {C=C(n,\lambda,\Lambda)} such that

\displaystyle  \begin{array}{rcl}  d(A,B)\leq C\sqrt{\delta}. \end{array}

Why a square root ? {\delta} behaves as if it were a smooth function on the space of pairs of convex sets of equal volumes, with a nondegenerate minimum at homothetic pairs.

2.2. Proof of Brunn-Minkowski’s inequality

Based on optimal transport. Consider normalized Lebesgue measures on {A} and {B}. Let {T:{\mathbb R}^n\rightarrow{\mathbb R}^n} be the optimal transport map. Optimal transport theory says that {T} exists, it has constant Jacobian on {A} and it is the gradient of a convex function {\phi} on {{\mathbb R}^n}.

{A+B} contains all points {a+Ta}, so

\displaystyle  \begin{array}{rcl}  |A+B|\geq|(Id+T)(A)|=\int_{A}\mathrm{det}(Id+\nabla\phi). \end{array}

Diagonalize {\nabla\phi}. Then

\displaystyle  \begin{array}{rcl}  \mathrm{det}(Id+\nabla\phi)&=&\prod_{i=1}^n(1+\lambda_i)\\ &\geq&(1+(\prod_{i=1}^n \lambda_i)^{1/n})^n\\ &=&(1+\gamma^{-1})^n. \end{array}

It follows that

\displaystyle  \begin{array}{rcl}  |A+B|\geq|A|(1+\gamma^{-1})^n=(|A|^{1/n}+|B|^{1/n})^n. \end{array}

3. The convexity issue

3.1. One dimensional case

Assume first that {A=B}. Does {\delta} control {|Conv(A)\setminus A|} ? No.

Example 1 Let {A} be the union of two remote intervals of length {1/2} in {{\mathbb R}}.

Then {\frac{A+A}{2}} contains a third interval of length {1/2} in the middle, so {|\frac{A+A}{2}|=\frac{3}{2}}, {\delta(A)=\frac{1}{2}}.

It turns out that the 1-dimensional case has been known for a while. It follows from Freiman’s theorem in additive combinatorics.

Theorem 3 (Freiman 1959) Let {A\subset {\mathbb R}} have {|A|=1} and {\delta(A)\leq\frac{1}{2}}. Then

\displaystyle  \begin{array}{rcl}  |Conv(A)\setminus A|\leq 2\delta(A). \end{array}

In fact, Freiman’s theorem is much stronger. It deals with subsets in {{\mathbb Z}}, and shows that {|A+A|} close to {|A|} implies that {A} is close to an arithmetic progression.

3.2. Higher dimensions

Theorem 4 (M. Christ, 2012) {|Conv(A)\setminus A|} tends to 0 as {\delta(A)} tends to 0.

Theorem 5 (Figalli-Jerison, 2013) If {\delta(A)\leq\delta_n},

\displaystyle  \begin{array}{rcl}  |Conv(A)\setminus A|\leq C_n\delta(A)^{\alpha_n}. \end{array}

This uses induction on dimension, based on the 1-dimensional result.

3.3. Sketch of proof

\subsubsection{Step 1}

Slice {A} with parallel lines {D_y}, {y\in{\mathbb R}^{n-1}}. Show that {A_y} is close to a segment for most {y}. Furthermore, the projection of {A} to {{\mathbb R}^{n-1}} is close to be convex.

\subsubsection{Step 2}

{y\rightarrow|A_y|} is close to an affine function.

\subsubsection{Step 3}

Let {f:K\rightarrow{\mathbb R}} be a Lipschitz function on a set {K}. Assume that

\displaystyle  \begin{array}{rcl}  f(\frac{x+y}{2})\leq\frac{f(x)+f(y)}{2}+\gamma, \end{array}

where {\gamma=|Conv(A)\setminus A|}. Then {f} is {L^1} close to a concave function {F},

\displaystyle  \begin{array}{rcl}  \int_{K}|f-F|\leq C\,\gamma^\alpha. \end{array}

3.4. General case

I.e. {B\not=A}.

In 1 dimension, Freiman’s theorem does the job. Michael Christ’s qualitative result also extends.

Theorem 6 (Figalli-Jerison, 2014) If {\delta(A,B)\leq\delta_n}, there exists a convex set {K} such that

\displaystyle  \begin{array}{rcl}  |K\Delta A|+|K\Delta B|\leq C_n\delta(A,B)^{\beta_n}. \end{array}

Let {|A|=|B|=1}. Since

\displaystyle  \begin{array}{rcl}  \frac{A_y+B_y}{2}\subset (\frac{A+B}{2})_y, \end{array}

Fubini seems to show that

\displaystyle  \begin{array}{rcl}  |\frac{A+B}{2}|-\frac{|A|+|B|}{2}\geq 0, \end{array}

which was the first step of our proof when {A=B}. But this argument collapses when {B\not=A}, since it happens that {A_y} is empty and {B_y} is not, in which case Brunn-Minkowski’s inequality fails!

To avoid this, we perform two preliminary symmetrizations, Schwarz and Steiner, in orthogonal directions. This preserves {\delta} and also volumes of superlevel sets of {y\mapsto |A_y|}.

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