# Stability and the equality case in the B-theorem

Orli Herscovici, Galyna V. Livshyts, Liran Rotem, Alexander Volberg  School of
Mathematics, Georgia Institute of Technology, Atlanta, GA
<EMAIL_ADDRESS>School of Mathematics, Georgia Institute of
Technology, Atlanta, GA<EMAIL_ADDRESS>Faculty of Mathematics,
Technion – Israel Institute of Technology, Haifa, Israel
<EMAIL_ADDRESS>School of Mathematics, Michigan State University, East
Lansing, MI<EMAIL_ADDRESS>
###### Abstract.

In this paper, we show the stability, and characterize the equality cases in
the strong B-inequality of Cordero-Erasquin, Fradelizi and Maurey [11]. As an
application, we establish uniqueness of Bobkov’s maximal Gaussian measure
position from [5].

##  1\. Introduction

Let $\gamma$ denote the standard Gaussian measure on $\mathbb{R}^{n}$ with
density $(2\pi)^{-n/2}e^{-\frac{|x|^{2}}{2}}$. We say that a set $K$ in
$\mathbb{R}^{n}$ is _symmetric_ if $K=-K$, i.e. for all $x\in K$ we have
$-x\in K$.

The B-theorem of Cordero-Erausquin, Fradelizi and Maurey [11] states that for
every symmetric convex set $K\subset\mathbb{R}^{n}$, and every $a,b>0$,

(1) $\gamma\left(\sqrt{ab}K\right)\geq\sqrt{\gamma(aK)\gamma(bK)}.$

In other words, $\gamma(e^{t}K)$ is log-concave in $t$ when $K$ is a convex
symmetric set.

The inequality (1) was first conjectured by Latała in [19], who attributed the
question to Banaszczyk. It is a strengthening of the inequality

$\gamma\left(\frac{a+b}{2}K\right)\geq\sqrt{\gamma(aK)\gamma(bK)},$

which holds with no symmetry assumption and follows from the Prékopa–Leindler
inequality [25], [21] – see also Borell [6, 7] and Brascamp and Lieb [10].

More generally, for a vector $x\in\mathbb{R}^{n},$ we will use the notation

(2) $e^{x}K=\\{(e^{x_{1}}y_{1},...,e^{x_{n}}y_{n}):\,\,(y_{1},...,y_{n})\in
K\\}.$

The strong version of the B-theorem then states that for any pair
$x,y\in\mathbb{R}^{n},$

(3)
$\gamma\left(e^{\frac{x+y}{2}}K\right)\geq\sqrt{\gamma(e^{x}K)\gamma(e^{y}K)}.$

Nayar and Tkocz [24] showed that the assumption that $K$ is symmetric is
necessary and (1) does not necessarily hold if one replaces it with the weaker
assumption $0\in K$. A natural question, raised in [11], is what other
measures satisfy an inequality such as (1) or (3). In dimension $n=2$ it
follows from the works of Böröczky, Lutwak, Yang and Zhang [9] and of Saroglou
[26] that all even log-concave measures satisfy (1). For uniform measures on
convex bodies this was verified independently by Livne Bar-On [4]. In
arbitrary dimensions, Eskenazis, Nayar and Tkocz [13] proved that (3) holds
for certain Gaussian mixtures, and in [12] it was proved in particular that
all rotation-invariant log-concave measures satisfy (3).

In this paper we will concentrate solely on the Gaussian case, and prove a
stability version of this celebrated inequality. Recall that the in-radius
$r(K)$ of a symmetric convex body $K$ is the largest number $r>0$ such that
$rB^{n}_{2}\subset K$, where $B^{n}_{2}$ denotes the unit Euclidean ball
centered at the origin. Our theorem then reads:

######  Theorem 1.1.

Suppose $0\leq a<b<\infty$ and let $K$ be a symmetric convex body. Suppose
that

$\gamma(\sqrt{ab}K)\leq\sqrt{\gamma(aK)\gamma(bK)}(1+\epsilon)$

for small enough $\epsilon>0$. Then either the in-radius $r(K)$ satisfies

$r(K)\geq\frac{1}{b}\sqrt{\log\left(\frac{c\log(b/a)^{2}}{n^{2}\epsilon}\right)},$

or

$r(K)\leq\frac{C\sqrt{n}}{a}\epsilon^{\frac{1}{n+1}}\left(\log({b}/{a})\right)^{-\frac{2}{n+1}}.$

######  Remark 1.2.

We note that our estimate for $r$ is essentially sharp: indeed, consider first
a case when $K$ is a symmetric strip $S_{R}$ of width $2R$, that is

$S_{R}=\\{x=(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}\,:\,x_{1}\in[-R,R]\\}.$

Note that

$\gamma(S_{R})=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-\frac{s^{2}}{2}}ds=1-\frac{2}{\sqrt{2\pi}}\int_{R}^{\infty}e^{-\frac{s^{2}}{2}}ds.$

Suppose $0<a\leq b\leq\infty$ and there exists $\epsilon>0$ such that

$\gamma(\sqrt{ab}S_{R})=\sqrt{\gamma(aS_{R})\gamma(bS_{R})}(1+\epsilon).$

Relaxing the upper and lower bound obtained by Szarek and Werner [29] for
Komatsu inequality, we have that

$\frac{1}{R+1}\,e^{-\frac{R^{2}}{2}}\leq\int_{R}^{\infty}e^{-\frac{s^{2}}{2}}ds\leq\frac{1}{R}\,e^{-\frac{R^{2}}{2}}.$

Let $F(t)=\frac{2}{\sqrt{2\pi}}\int_{tR}^{\infty}e^{-\frac{s^{2}}{2}}\,ds$ for
$t>0$. Then

$\epsilon=\frac{\gamma(S_{\sqrt{ab}R})}{\sqrt{\gamma(S_{aR})\gamma(S_{bR})}}-1=\frac{1-F(\sqrt{ab})}{\sqrt{(1-F(a))(1-F(b))}}-1,$

and, since $F(a)>F(\sqrt{ab})>F(b)$, we get

$\epsilon\leq\frac{1-F(\sqrt{ab})}{1-F(a)}-1\leq\frac{F(a)}{1-F(a)},$

which leads to

$\frac{\epsilon}{1+\epsilon}\leq
F(a)=\frac{2}{\sqrt{2\pi}}\int_{aR}^{\infty}e^{-\frac{s^{2}}{2}}ds\leq\frac{2}{\sqrt{2\pi}aR}\,e^{-\frac{a^{2}R^{2}}{2}}.$

Thus,

$\frac{\epsilon}{1+\epsilon}\leq\frac{2}{\sqrt{2\pi}}\frac{1}{aR}e^{-\frac{a^{2}R^{2}}{2}},$

or,

$R\leq C\,\sqrt{\log\left(1+\frac{1}{\epsilon}\right)}.$

However, the upper estimate on $r$ in Theorem 1.1 is likely not sharp: indeed,
by considering $K=rB^{n}_{2}$ to be a very small ball, we only get that
$r(K)\geq C\sqrt{\epsilon}$, where $\epsilon$ is the corresponding deficit in
the B-inequality.

Theorem 1.1 implies:

######  Corollary 1.3.

Suppose $0\leq a<b<\infty$ and $K$ is a symmetric convex set such that
$\gamma(\sqrt{ab}K)=\sqrt{\gamma(aK)\gamma(bK)}$. Then either
$K=\mathbb{R}^{n}$ or $K$ has an empty interior.

###### Proof.

We can apply Theorem 1.1 with an arbitrary $\epsilon>0$. Letting $\epsilon\to
0$ we see that either $r(K)=\infty$ so $K=\mathbb{R}^{n}$, or $r(K)=0$ so $K$
has an empty interior. ∎

We also prove a stability version for the strong B-theorem. Given a convex
body $K$ and a vector $x\in\partial K$, the outer unit normal to $\partial K$
(i.e. the Gauss map) at $x$ will be denoted by
$n_{x}=(n^{1}_{x},...,n^{n}_{x})$. It is well known (see e.g. [27]) that
$n_{x}$ is uniquely defined almost everywhere on $\partial K$. Recall that for
an $(n-1)$-dimensional surface $M,$ $\gamma^{+}(M)$ denotes the Gaussian
perimeter of $M.$ Our theorem reads:

######  Theorem 1.4.

Fix parameters $\delta,\alpha,\beta>0$ Fix $x,y\in\mathbb{R}^{n}$ such that
$|e^{x}|\leq|e^{y}|$ (where we use the notation
$e^{x}=(e^{x_{1}},...,e^{x_{n}})$, as well as (2)). Let $K$ be a symmetric
convex body, and suppose that

$\gamma(e^{\frac{x+y}{2}}K)\leq\sqrt{\gamma(e^{x}K)\gamma(e^{y}K)}(1+\epsilon)$

for small enough $\epsilon>0$. Consider

$\sigma^{\delta}=\\{i\in[n]:\,|x_{i}-y_{i}|\geq\delta\\},$

and let

$\Omega_{\delta,\alpha}(K)=\bigl{\\{}x\in\partial
K:\,\sum_{i\in\sigma^{\delta}}(n^{i}_{x})^{2}\geq\alpha\bigr{\\}}.$

Then either there exists a vector $z\in[x,y]$ such that

$\gamma^{+}(\Omega_{\delta,\alpha}(e^{z}K))\leq\beta\gamma^{+}(\partial(e^{z}K)),$

or

$r(K)\geq|e^{y}|^{-1}\sqrt{\log\frac{\delta^{2}\alpha\beta}{\epsilon n^{2}}},$

or

$r(K)\leq
C|e^{x}|^{-1}\sqrt{n}\epsilon^{\frac{1}{n+1}}\left(\delta^{2}\alpha\beta\right)^{-\frac{1}{n+1}}.$

As a corollary, we shall deduce:

######  Corollary 1.5.

Let $x,y\in\mathbb{R}^{n}$, set $\sigma_{x,y}=\\{j\in[n]:\,x_{j}\neq
y_{j}\\}$, and let $K$ be a symmetric convex set in $\mathbb{R}^{n}.$ Then

$\gamma\left(e^{\frac{x+y}{2}}K\right)=\sqrt{\gamma\left(e^{x}K\right)\gamma\left(e^{y}K\right)},$

if and only if either $K$ has an empty interior, or $K=\mathbb{R}^{n},$ or
(more generally) $K=K_{0}\times H_{x,y},$ with $K_{0}\subset H_{x,y}^{\perp},$
where $H_{x,y}=\\{z\in\mathbb{R}^{n}:z_{j}=0\,\,\forall j\in\sigma_{x,y}\\}.$

We apply Corollary 1.5 to show uniqueness of the Bobkov maximal Gaussian
measure position (MGM from now on) for a convex body $K.$ Recall from [5] that
a symmetric convex body $K$ is said to be in Bobkov’s MGM position if for any
volume preserving linear operator $T$ on $\mathbb{R}^{n},$ we have
$\gamma(K)\geq\gamma(TK).$ Bobkov showed that $K$ is in the MGM position if
and only if the restriction of the Gaussian measure on $K$ is isotropic
(recall that a measure is isotropic if its barycenter is at the origin, and
the covariance matrix is proportional to the identity). Isotropicity often
arises when the measure is placed in some optimizing position, see e.g. [1],
[2], [3], as well as [15], [16]. It is a natural question: _is the Bobkov MGM
position unique for a symmetric convex body?_ We answer this question in the
affirmative:

######  Theorem 1.6.

Let $K$ be a symmetric convex body. The expression $\sup_{T}\gamma(TK)$, where
the supremum runs over all linear volume preserving operators $T$ on
$\mathbb{R}^{n}$, is attained for the unique $T.$

###### Proof.

Without loss of generality, assume that $K$ is in the Bobkov MGM position.
Suppose by contradiction that there exists a non-identity volume preserving
linear map $T$ such that $\gamma(TK)=\gamma(K)$. Then there exists a traceless
matrix $D$ such that $T=e^{D}$ (see e.g. [3] for the details), and by the
rotation-invariance of the Gaussian measure we may assume that $D$ is
diagonal. Let us now consider a function $F:[0,1]\rightarrow\mathbb{R}$ given
by $F(t)=\gamma(e^{tD}K)$. On one hand, by the strong B-property of the
Gaussian measure [11], $F$ is log-concave on $[0,1].$ On the other hand,
$F(0)=F(1)$ are maximal points for $F$, and therefore, the equality should be
attained in the inequality $F(\frac{1}{2})\geq\sqrt{F(0)F(1)}.$ By Corollary
1.5, we get a contradiction with the assumption that $K$ is a convex _body_
(that is, a convex compact set with non-empty interior). ∎

We remark that the above proof is inspired by the works of Artstein-Avidan,
Katzin [2] and Artstein-Avidan, Putterman [3], where in particular the authors
consider _maximal intersection position_ : a symmetric convex body $K$ is said
to be in the maximal intersection position if for any volume preserving linear
operator $T$ on $\mathbb{R}^{n},$ we have $\mu(K)\geq\mu(TK),$ where $\mu$ is
the uniform measure on the centered euclidean ball of the same volume as $K.$
In both of the aforementioned papers, this position is viewed as part of
different families of positions. It was conjectured in [2], and reiterated in
[3], that the maximal intersection position is indeed unique, and they
explained (along the lines of the argument outlined above) that this
conjecture would follow from the fact that the strong B-property for the
uniform measure on the ball does not have non-trivial equality cases. We
expect some of our ideas to be useful for studying this question, and leave it
for future research.

In Section 2 we outline some preliminaries. In Section 3 we discuss some
estimates concerning special functions related to the Gaussian measure. In
Section 4 we outline the stability for the Gaussian Poincaré inequality
restricted to a convex set, which is the result similar to what was obtained
in [23], however we prove stability in a stronger distance. In Section 5 we
outline the stability in the “even version” of the Gaussian Poincaré
inequality restricted to a convex set, for quadratic forms – this corresponds
to stability in the “local versions” of the Theorems 1.1 and 1.4. Lastly, in
Section 6 we prove Theorems 1.1 and 1.4 and Corollary 1.5.

Acknowledgements. The second named author is supported by NSF DMS-1753260. The
third named author is supported by ISF grant 1468/19 and BSF grant 2016050.
The second and third authors are supported by NSF-BSF DMS-2247834. The fourth
named author is supported by NSF DMS 2154402. The authors are grateful to
ICERM for hospitality during the program “Harmonic Analysis and Convexity”.

##  2\. Preliminaries

Given a convex set $K$ in $\mathbb{R}^{n}$ and the standard Gaussian measure
$\gamma$ on $\mathbb{R}^{n},$ we shall use the notation

$\frac{1}{\gamma(K)}\int_{K}d\gamma=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}d\gamma.$

We will denote by $L^{2}(K,\gamma)$ the class of functions
$u:\mathbb{R}^{n}\to\mathbb{R}$ such that $\int_{K}|u|^{2}d\gamma<\infty$,
with the normalized $L2$ norm
$\|u\|^{2}_{L^{2}(K,\gamma)}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|u|^{2}d\gamma$.

Given a convex set $K$ in $\mathbb{R}^{n}$, for a point $x\in\partial K$ we
denote by $n_{x}$ the outward unit normal at $x$; the vector field $n_{x}$ is
uniquely defined almost everywhere on $\partial K$. We say that $K$ is of
class $C^{2}$ if its boundary is locally twice differentiable. In this case,
$n_{x}$ is well-defined for all $x\in\partial K$. For a $C^{2}$ convex set,
consider the second fundamental form of $K$ to be the matrix
$\rm{II}=\frac{dn_{x}}{dx}$ (with a “plus” because the normal is outer) acting
on the tangent space at $x$. The Gauss curvature at $x$ is $\det(\rm{II})$ and
the mean curvature is $\operatorname{tr}(\rm{II}).$

We denote by $L$ the Ornstein-Uhlenbeck operator
$L:C^{2}(\mathbb{R}^{n})\rightarrow C(\mathbb{R}^{n})$, given by

$Lu=\Delta u-\langle\nabla u,x\rangle.$

The operator $L$ satisfies the following integration by parts identity
whenever it makes sense (as follows immediately from the classical divergence
theorem):

$\int_{K}vLu\,d\gamma=-\int_{K}\langle\nabla v,\nabla u\rangle
d\gamma+\int_{\partial K}v\langle\nabla u,n_{x}\rangle d\gamma_{\partial K}.$

Here by $\gamma_{\partial K}$ we mean the measure on $\partial K$ with density
$(2\pi)^{-n/2}e^{-\frac{|x|^{2}}{2}}$ with respect to $H_{n-1}$, the
$(n-1)$-dimensional Hausdorff measure.

We denote by $W^{k,2}(K,\gamma)$ the Sobolev space of all functions $u$ such
that $u$ has weak partial derivatives up to order $k$, and all those partial
derivatives (including $u$ itself) belong to $L^{2}(K,\gamma)$. We will use
the notation

$\|u\|^{2}_{W^{1,2}(K,\gamma)}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
u|^{2}d\gamma$

(even though strictly speaking this is not a norm on $W^{1,2}(K,\gamma)$,
since $\|u\|^{2}_{W^{1,2}(K,\gamma)}=0$ for constant functions).

Recall that the trace operator is a continuous linear operator

$\operatorname{TR}:W^{1,2}(K,\gamma)\rightarrow L^{2}(\partial K,\gamma)$

such that for every $u\in C^{1}(K),$ continuous up to the boundary, we have

$\operatorname{TR}(u)=u|_{\partial K}.$

We shall use informal notation $\int_{\partial K}ud\gamma$ to mean
$\int_{\partial K}\operatorname{TR}(u)d\gamma$. Similarly, we use notation
$\int_{\partial K}\langle\nabla u,n_{x}\rangle d\gamma$ to mean
$\int_{\partial K}\langle\operatorname{TR}(\nabla u),n_{x}\rangle d\gamma$,
where $\operatorname{TR}(\nabla u)$ is the vector formed by the trace
functions of the weak first partial derivatives of $u$. We shall also use
notation $\nabla$, $\Delta$ and so on to denote the appropriate quantities in
the sense of weak derivatives. We will also use the following notation for the
Gaussian perimeter:

$\gamma^{+}(\partial K)=\int_{\partial K}d\gamma.$

The following result appears in [23]; we sketch its proof for completeness.

######  Theorem 2.1 (Gaussian Trace Theorem for convex sets containing the
origin).

Let $K$ be a convex domain such that $rB^{n}_{2}\subset K$ for some $r>0$. Fix
$g\in W^{1,2}(K,\gamma)$. Then

$\int_{\partial K}g^{2}d\gamma_{\partial
K}\leq\frac{1}{r}\int_{K}(ng^{2}+|\nabla g|^{2})d\gamma.$

###### Proof.

We use the estimate $\langle x,n_{x}\rangle\geq r$, and incorporate the trick
similar to the ones from [14], [18] and use the divergence theorem. We get

$\int_{\partial K}g^{2}d\gamma_{\partial K}\leq\frac{1}{r}\int_{\partial
K}\langle g^{2}x,n_{x}\rangle d\gamma_{\partial
K}=\frac{1}{r}\int_{K}(\operatorname{div}(g^{2}x)-g^{2}|x|^{2})d\gamma.$

Note that

$\operatorname{div}(g^{2}x)=ng^{2}+2g\langle\nabla g,x\rangle\leq
ng^{2}+|\nabla g|^{2}+g^{2}|x|^{2}.$

Combining the above yields the result. ∎

##  3\. Estimates on the special functions which concern the rate of the
stability estimate

In what follows, $C,c,C_{1}$ etc denote positive absolute constants that do
not depend on the dimension and whose value may change from line to line.
Recall that the in-radius $r(K)$ of a convex set $K\subset\mathbb{R}^{n}$ is
the largest number $r>0$ such that $rB_{2}^{n}\subset K$. We also denote by

$\gamma^{+}\left(\partial K\right)=\int_{\partial
K}\left(2\pi\right)^{-n/2}e^{-\left|x\right|^{2}/2}\mathrm{d}H_{n-1}$

the Gaussian surface area of $K$.

The main goal of this section is to prove the following technical estimate:

######  Proposition 3.1.

Let $K$ be a symmetric convex body in $\mathbb{R}^{n}$ with in-radius
$r=r(K)>0$. Assume that for $\delta<c_{0}$ we have

$\frac{\gamma(K)}{{\it\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma}}+\frac{\gamma(K)}{r\gamma^{+}(\partial
K)}\geq\frac{1}{\delta}$

Then either $r\geq\sqrt{\log\frac{1}{\delta}}$ or $r\leq
C\sqrt{n}\delta^{\frac{1}{n+1}}$.

For the proof we need several lemmas. First, we will need the Gaussian
isoperimetric inequality: Let
$\Phi\left(x\right)=\gamma\left((-\infty,x]\right)$ denote the CDF of a
standard normal random variable, and let $\Phi^{-1}:[0,1]\to\mathbb{R}$ denote
the inverse function. Define the isoperimetric profile $I:[0,1]\to\mathbb{R}$
by $I(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{\Phi^{-1}(x)^{2}}{2}}$. Then the
Gaussian isoperimetric inequality ([8], [28]) states that for every
$K\subset\mathbb{R}^{n}$ we have $\gamma^{+}\left(\partial K\right)\geq
I\left(\gamma(K)\right)$. Note that $I$ is concave and symmetric around
$x=\frac{1}{2}$ where it attains its maximum.

We first give a lower bound on $\gamma^{+}\left(\partial K\right)$ in terms of
$r$ instead of the measure $\gamma(K)$:

######  Lemma 3.2.

Let $K$ be a symmetric convex body in $\mathbb{R}^{n}$ with in-radius $r>0$.

  1. (1)

If $\gamma(K)\geq\frac{1}{2}$ then $\gamma^{+}\left(\partial
K\right)\geq\frac{1}{\sqrt{2\pi}}e^{-r^{2}/2}.$

  2. (2)

If $\gamma(K)\leq\frac{1}{2}$ then $\gamma^{+}\left(\partial
K\right)\geq\left(\frac{cr}{\sqrt{n}}\right)^{n}e^{-r^{2}/2}.$

###### Proof.

$(1)$ By the definition of $r$ we know that $K$ is contained in a strip

$S=\left\\{x:\ \left|\left\langle x,\theta\right\rangle\right|\leq r\right\\}$

for some $\theta\in S^{n-1}$. We therefore also have $K\subset H$ where
$H=\left\\{x:\ \left\langle x,\theta\right\rangle\leq r\right\\}$, and
therefore $\frac{1}{2}\leq\gamma(K)\leq\gamma\left(H\right)=\Phi(r)$. Since
$I$ is decreasing on $\left[\frac{1}{2},1\right]$ it follows from the Gaussian
isoperimetric inequality that $\gamma^{+}\left(\partial K\right)\geq
I\left(\Phi(r)\right)=\frac{1}{\sqrt{2\pi}}e^{-r^{2}/2}$.

$(2)$ By the concavity of $I$ we have for all $0\leq x\leq\frac{1}{2}$

$I(x)=I\left(2x\cdot\frac{1}{2}+(1-2x)\cdot 0\right)\geq 2x\cdot
I\left(\frac{1}{2}\right)=\sqrt{\frac{2}{\pi}}x.$

Therefore, by the Gaussian isoperimetric inequality

$\gamma^{+}\left(\partial K\right)\geq
I\left(\gamma(K)\right)\geq\sqrt{\frac{2}{\pi}}\cdot\gamma(K)\geq\sqrt{\frac{2}{\pi}}\cdot\gamma\left(rB_{2}^{n}\right).$

Since the density of $\gamma$ on $rB_{2}^{n}$ is bounded from below by
$\frac{1}{\left(2\pi\right)^{n/2}}e^{-r^{2}/2}$ we have

$\gamma\left(rB_{2}^{n}\right)\geq\frac{1}{\left(2\pi\right)^{n/2}}e^{-r^{2}/2}\cdot
r^{n}\omega_{n},$

where $\omega_{n}$ denotes the volume of $B_{2}^{n}$. Since
$\omega_{n}\geq\left(\frac{c}{\sqrt{n}}\right)^{n}$ we get

$\gamma^{+}\left(\partial
K\right)\geq\left(\frac{cr}{\sqrt{n}}\right)^{n}e^{-r^{2}/2}.$

as claimed. ∎

Next, we need some rough estimates for Gaussian integrals over balls. Sharper
estimates are definitely known, but this lemma will suffice for our needs:

######  Lemma 3.3.

  1. (1)

We have $\gamma\left(2\sqrt{n}B_{2}^{n}\right)\geq\frac{3}{4}$ and
$\int_{2\sqrt{n}B_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\geq cn$.

  2. (2)

For every $r>0$ we have
$\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\geq\left(\frac{c}{\sqrt{n}}\right)^{n}r^{n+2}e^{-r^{2}/2}$.

###### Proof.

$(1)$ We use simple probabilistic bounds. Let
$Z=\left(Z_{1},\ldots,Z_{n}\right)$ denote a standard normal random vector in
$\mathbb{R}^{n}$. Then

$\displaystyle\mathbb{E}\left|Z\right|^{2}$
$\displaystyle=n\mathbb{E}Z_{1}^{2}=n$
$\displaystyle\operatorname{Var}\left|Z\right|^{2}$
$\displaystyle=n\operatorname{Var}Z_{1}^{2}=2n.$

It follows by Markov inequality that

$\mathbb{P}\left(\left|Z\right|\geq
2\sqrt{n}\right)=\mathbb{P}\left(\left|Z\right|^{2}\geq
4n\right)\leq\frac{n}{4n}=\frac{1}{4},$

or $\gamma\left(2\sqrt{n}B_{2}^{n}\right)\geq\frac{3}{4}$ as claimed. By
Chebyshev inequality we have for $n\geq 15$

$\mathbb{P}\left(\left|Z\right|<\frac{\sqrt{n}}{2}\right)=\mathbb{P}\left(\left|Z\right|^{2}<\frac{n}{4}\right)\leq\mathbb{P}\left(\left|Z\right|^{2}<n-2\sqrt{2n}\right)\leq\frac{1}{4},$

and therefore

$\gamma\left(2\sqrt{n}B_{2}^{n}\setminus\frac{\sqrt{n}}{2}B_{2}^{n}\right)=\mathbb{P}\left(\left|Z\right|\leq
2\sqrt{n}\right)-\mathbb{P}\left(\left|Z\right|\leq\frac{\sqrt{n}}{2}\right)\geq\frac{3}{4}-\frac{1}{4}=\frac{1}{2}.$

It follows that

$\int_{2\sqrt{n}B_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\geq\int_{2\sqrt{n}B_{2}^{n}\setminus\frac{\sqrt{n}}{2}B_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\geq\frac{n}{4}\gamma\left(2\sqrt{n}B_{2}^{n}\setminus\frac{\sqrt{n}}{2}B_{2}^{n}\right)=\frac{n}{8},$

and we choose $c>0$ so that the estimate will also hold for $1\leq n\leq 14$.

$(2)$ The density of $\gamma$ on $rB_{2}^{n}$ is bounded from below by
$\frac{1}{\left(2\pi\right)^{n/2}}e^{-r^{2}/2}$ and so

$\displaystyle\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma$
$\displaystyle\geq\frac{1}{\left(2\pi\right)^{n/2}}e^{-r^{2}/2}\cdot\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}x=\frac{1}{\left(2\pi\right)^{n/2}}e^{-r^{2}/2}\cdot
n\omega_{n}\cdot\frac{r^{n+2}}{n+2}$

Again we use $\omega_{n}\geq\left(\frac{c}{\sqrt{n}}\right)^{n}$ to get
$\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\geq\left(\frac{c}{\sqrt{n}}\right)^{n}r^{n+2}e^{-r^{2}/2}$.
∎

Now we can prove Proposition 3.1

###### Proof of Proposition 3.1.

Under the assumption of the proposition we have either
$\frac{\gamma(K)}{{\it\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma}}\geq\frac{1}{2\delta}$
or $\frac{\gamma(K)}{r\gamma^{+}(\partial K)}\geq\frac{1}{2\delta}$.

Assume first that
$\frac{\gamma(K)}{{\it\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma}}\geq\frac{1}{2\delta}$.
For $\delta$ small enough we must have $r\leq 2\sqrt{n}$: If this is not the
case then by Lemma 3.3(1) we have

$2\delta\geq
2\delta\gamma(K)\geq\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\geq\int_{2\sqrt{n}B_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\geq
cn\geq c,$

which is clearly impossible for $\delta$ small enough.

Next, as in Lemma 3.2 we use the fact that $K$ is contained in a strip $S$ of
width $2r$ and hence $\gamma(K)\leq\gamma(S)\leq Cr.$ Using Lemma 3.3(2) we
obtain

$\left(\frac{c}{\sqrt{n}}\right)^{n}r^{n+2}e^{-r^{2}/2}\leq\int_{rB_{2}^{n}}\left|x\right|^{2}\mathrm{d}\gamma\leq
2\delta\cdot\gamma(K)\leq Cr\delta,$

and since $r\leq 2\sqrt{n}$ we get

$\left(\frac{c}{\sqrt{n}}\right)^{n}r^{n+1}e^{-2n}\leq C\delta,$

or $r\leq C\sqrt{n}\delta^{\frac{1}{n+1}}$ as we claimed.

Finally, assume that $\frac{\gamma(K)}{r\gamma^{+}(\partial
K)}\geq\frac{1}{2\delta}$. Using again the fact that $\gamma(K)\leq Cr$ we
obtain

$\gamma^{+}\left(\partial K\right)\leq\frac{2\delta}{r}\gamma(K)=C\delta.$

If $\gamma\left(K\right)\geq\frac{1}{2}$ we see from Lemma 3.2(1) that
$e^{-r^{2}/2}\leq C\delta\leq\sqrt{\delta}$ (assuming we take
$\delta\leq\frac{1}{C^{2}}$), and then $r\geq\sqrt{\log\frac{1}{\delta}}$. If
on the other hand $\gamma(K)\leq\frac{1}{2}$ we see from Lemma 3.2(2) that

$\left(\frac{cr}{\sqrt{n}}\right)^{n}e^{-r^{2}/2}\leq C\delta.$

However, again in this case it follows from Lemma 3.3(1) that $r\leq
2\sqrt{n}$, so we have $\left(\frac{cr}{\sqrt{n}}\right)^{n}e^{-2n}\leq
C\delta$ or $r\leq C\sqrt{n}\delta^{\frac{1}{n}}$. ∎

######  Remark 3.4.

It is an interesting question – what (symmetric) convex set in
$\mathbb{R}^{n}$ has the smallest Gaussian perimeter if the largest ball
centered at the origin which is contained in this set has radius $r$.

In the non-symmetric case, it is natural to conjecture that the answer is the
ball for smaller $r$ and the half-space for larger $r$. In fact, if
$\gamma(rB_{2}^{n})\geq\frac{1}{2}$ then indeed one can prove using the
Gaussian isoperimetric inequality ([8], [28]) that the half-space minimize the
perimeter for a fixed $r$.

In the symmetric case, it is natural to conjecture that the answer is the ball
for smaller $r$ and the symmetric strip for larger $r.$ Once again, for large
enough $r$ this follows from the result of Latała and Oleszkiewicz [20], who
showed that $r(K)\gamma^{+}(\partial K)$ is minimized for a symmetric convex
set when $K$ is a symmetric strip of the same Gaussian measure as $K$. In
other words,

$\gamma^{+}(\partial
K)\geq\frac{2J^{-1}_{0}(\gamma(K))e^{-\frac{J^{-1}_{0}(\gamma(K))^{2}}{2}}}{\sqrt{2\pi}r}$

where $J_{0}(a)=2\Phi(a)-1$ denotes the Gaussian measure of a symmetric strip
of width $2a$. Let $\tilde{R}_{n}$ be the radius of the ball whose Gaussian
measure is $J_{0}(1)$, and suppose that $r\geq\tilde{R}_{n}.$ Then
$J_{0}(1)\leq\gamma(K)\leq J_{0}(r)$, and noting that
$J^{-1}_{0}(a)e^{-\frac{J^{-1}_{0}(a)^{2}}{2}}$ is decreasing for $a\geq
J_{0}(1),$ we conclude that

$\gamma^{+}(\partial
K)\geq\frac{2re^{-\frac{r^{2}}{2}}}{\sqrt{2\pi}r}=\frac{2}{\sqrt{2\pi}}e^{-\frac{r^{2}}{2}},$

and the inequality is sharp when $K$ is a strip. We could have used this
estimate that improves Lemma 3.2, but this sharper result would only affect
our outcome in terms of the value of the absolute constants which we are not
tracking.

##  4\. Stability in the Poincaré inequality on the convex set

The Gaussian Poincaré inequality on a convex set (which follows e.g. from the
Theorem of Brascamp and Lieb [10]) states that

(4)
$\gamma(K)\int_{K}f^{2}d\gamma-\left(\int_{K}fd\gamma\right)^{2}\leq\gamma(K)\int_{K}|\nabla
f|^{2}d\gamma.$

The result below is similar to Theorem 1.5 in [23], although the $W^{1,2}$
norm is replaced there by $L^{1}$ norm. Using the $W^{1,2}$ norm is crucial
for us, and thus for completeness, we outline the full proof here:

######  Theorem 4.1 (the quantitative stability in the Gaussian Poincaré).

Suppose that for a convex set $K$ containing $rB^{n}_{2}$, a function $f\in
W^{1,2}(K)\cap C^{1}(K),$ and for $\epsilon>0,$ we have

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}f^{2}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}fd\gamma\right)^{2}\geq\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
f|^{2}d\gamma-\epsilon.$

Then there exists a vector $\theta\in\mathbb{R}^{n}$ (possibly zero), which
depends only on $K$ and $f$, such that

  * •

$\int_{\partial K}\langle\theta,n_{x}\rangle^{2}d\gamma_{\partial
K}\leq\frac{2(n+1)\gamma(K)\epsilon}{r};$

  * •

$\|f-\langle
x,\theta\rangle-\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}fd\gamma\|^{2}_{W^{1,2}(K,\gamma)}\leq
4\epsilon.$

###### Proof.

We first assume that the boundary of $K$ is of class $C^{\infty}$. Without
loss of generality we may also assume that $\int_{K}fd\gamma=0$. Consider the
function $u\in W^{2,2}(K)\cap C^{2}(K)$ such that $Lu=f$ and $\langle\nabla
u,n_{x}\rangle=0$ on $\partial K$ (see e.g. [23] for existence and
regularity). By Bochner’s formula (see e.g. [17]), we write

(5) $\displaystyle\int_{K}f^{2}d\gamma=$
$\displaystyle-\int_{K}\left(2\langle\nabla f,\nabla u\rangle+|\nabla
u|^{2}\right)d\gamma$
$\displaystyle-\int_{K}\|\nabla^{2}u\|^{2}d\gamma-\int_{\partial
K}\langle\mbox{\rm{II}}\nabla_{\partial K}u,\nabla_{\partial
K}u\rangle\,d\gamma_{\partial K}(x).$

Combining (5) with the Cauchy’s inequality

(6) $\int_{K}\left(2\langle\nabla f,\nabla u\rangle+|\nabla
u|^{2}\right)d\gamma\geq-\int_{K}|\nabla f|^{2}d\gamma,$

and an application of convexity of $K$

$\int_{\partial K}\langle\mbox{\rm{II}}\nabla_{\partial K}u,\nabla_{\partial
K}u\rangle)\,d\gamma_{\partial K}\geq 0,$

we get

(7)
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}f^{2}d\gamma\leq\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
f|^{2}d\gamma-\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\|\nabla^{2}u\|^{2}d\gamma.$

Under the assumption of the theorem, this yields

(8)
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\|\nabla^{2}u\|^{2}d\gamma\leq\epsilon.$

Therefore, there exists a vector $\theta\in\mathbb{R}^{n}$ such that

$u=-\langle x,\theta\rangle+v$

with
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\nabla
vd\gamma=0$ and
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\|\nabla^{2}v\|^{2}d\gamma\leq\epsilon.$
By the Poincaré inequality (4), this implies that
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
v|^{2}d\gamma\leq\epsilon$. By our choice of $u,$ we have

$\langle\nabla v,n_{x}\rangle=\langle\nabla
u,n_{x}\rangle+\langle\theta,n_{x}\rangle=\langle\theta,n_{x}\rangle.$

In order to show the first assertion, we apply the Trace Theorem 2.1 to all
the partial derivatives of $v$ and sum it up:

$\frac{1}{\gamma(K)}\int_{\partial K}|\langle\nabla
v,n_{x}\rangle|^{2}d\gamma_{\partial K}\leq\frac{1}{\gamma(K)}\int_{\partial
K}|\nabla v|^{2}d\gamma_{\partial K}\leq$
$\frac{n+1}{r}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}(|\nabla
v|^{2}+\|\nabla^{2}v\|^{2})d\gamma\leq\frac{2(n+1)}{r}\epsilon.$

The desired inequality follows if we remember that
$\langle\theta,n_{x}\rangle=\langle\nabla v,n_{x}\rangle$, which completes the
proof when $K$ is smooth.

In order to show the second assertion, note that we used (6) while proving the
Poincaré inequality, and therefore, the assumption of the theorem gives

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left(2\langle\nabla
f,\nabla u\rangle+|\nabla
u|^{2}\right)d\gamma-\epsilon\leq-\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
f|^{2}d\gamma,$

which amounts to

(9)
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
f+\nabla u|^{2}d\gamma\leq\epsilon.$

We write

$\|f-\langle
x,\theta\rangle\|_{W^{1}(K,\gamma)}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
f-\theta|^{2}d\gamma\leq$
$2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
f+\nabla
u|^{2}d\gamma+2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
v|^{2}d\gamma\leq 4\epsilon,$

where we used the properties of $v$ together with (9), and the conclusion
follows.

Next, assume $K$ is a general compact convex body. Choose a sequence
$\left\\{K_{i}\right\\}$ of $C^{\infty}$-smooth convex bodies such that
$K_{i}\subset K$ and $K_{i}\rightarrow K$ in the Hausdorff distance. Clearly

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K_{i}}f^{2}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K_{i}}fd\gamma\right)^{2}\geq\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K_{i}}|\nabla
f|^{2}d\gamma-\epsilon_{i}$

for $\epsilon_{i}\to\epsilon$, and $K_{i}\supseteq r_{i}B_{2}^{n}$ for
$r_{i}\to r$. Let $\theta_{i}\in\mathbb{R}^{n}$ be the vector constructed in
the proof for the body $K_{i}$. Since

$\displaystyle\left|\theta_{i}\right|^{2}$ $\displaystyle\leq
2\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K_{i}}\left|\nabla
f-\theta_{i}\right|^{2}d\gamma+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K_{i}}\left|\nabla
f\right|^{2}d\gamma\right)$ $\displaystyle\leq
8\epsilon_{i}+2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K_{i}}\left|\nabla
f\right|^{2}d\gamma\to
8\epsilon+2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\nabla
f\right|^{2}d\gamma$

it follows that $\left\\{\theta_{i}\right\\}$ is bounded. Therefore by passing
to a subsequence we may assume without loss of generality that
$\theta_{i}\to\theta\in\mathbb{R}^{n}$. The conclusion

$\left\|f-\langle
x,\theta\rangle-\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}fd\gamma\right\|_{W^{1,2}(K,\gamma)}^{2}\leq
4\epsilon.$

now follows by continuity. For the second conclusion, we note that for a fixed
$\eta\in\mathbb{R}^{n}$ the convergence

$\int_{\partial
K_{i}}\left\langle\eta,n_{K_{i},x}\right\rangle^{2}\mathrm{d}\gamma_{\partial
K_{i}}\to\int_{\partial
K}\left\langle\eta,n_{K,x}\right\rangle^{2}\mathrm{d}\gamma_{\partial K}$

follows e.g. from Proposition A.3 of [22]. Using this fact it is now
straightforward to deduce that

$\int_{\partial
K}\left\langle\theta,n_{K,x}\right\rangle^{2}\mathrm{d}\gamma_{\partial
K}=\lim_{i\to\infty}\int_{\partial
K_{i}}\left\langle\theta_{i},n_{K_{i},x}\right\rangle^{2}\mathrm{d}\gamma_{\partial
K_{i}}\leq\frac{2(n+1)\gamma(K)\epsilon}{r}$

as claimed.

Finally, if $K$ is not compact we approximate it by the bodies
$K_{m}=K\cap\left(mB_{2}^{n}\right)$ for $m=1,2,3,...$. This time the
convergence

$\int_{\partial
K_{m}}\left\langle\eta,n_{K_{m},x}\right\rangle^{2}\mathrm{d}\gamma_{\partial
K_{m}}\to\int_{\partial
K}\left\langle\eta,n_{K,x}\right\rangle^{2}\mathrm{d}\gamma_{\partial K}$

follows from the fact that on $\partial K_{m}\cap\partial K$ we have
$n_{K_{m},x}=n_{K,x}$ almost everywhere, and the contribution of the integral
on $\partial K_{m}\setminus\partial K\subset m\SS^{n-1}$ tends to zero as
$m\to\infty$. The rest of the argument is the same as before. ∎

##  5\. Stability in the “symmetric” Gaussian Poincaré inequality for
quadratic functions

The main result of this section is stability (in some partial cases) of the
“symmetric” Gaussian Poincaré inequality due to Cordero-Erasquin, Fradelizi
and Maurey: if $f$ is an even function and $K$ is a symmetric convex set, then

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}f^{2}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}fd\gamma\right)^{2}\leq\frac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|\nabla
f|^{2}d\gamma.$

######  Lemma 5.1.

Let $K$ be a symmetric convex body such that $K\supseteq rB_{2}^{n}$. Assume
an odd function $f:\mathbb{R}^{n}\to\mathbb{R}$ satisfies

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}f^{2}\mathrm{d}\gamma\geq\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\nabla
f\right|^{2}\mathrm{d}\gamma-\epsilon$

as well as

$\left\|f-\left\langle
x,\eta\right\rangle\right\|^{2}_{L^{2}\left(\gamma_{K}\right)}<\epsilon$

for some $\eta\in\mathbb{R}^{n}$ and $\epsilon>0$. Then

$\int_{\partial
K}\left\langle\eta,n_{x}\right\rangle^{2}\mathrm{d}\gamma_{\partial K}\leq
C\left(\frac{\gamma^{+}(\partial
K)}{\int_{rB_{2}^{n}}x^{2}\mathrm{d}\gamma}+\frac{1}{r}\right)n\epsilon\gamma(K).$

###### Proof.

By Theorem 4.1 there exists $\theta\in\mathbb{R}^{n}$ such that the
conclusions (1) and (2) hold. Since
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}f\mathrm{d}\gamma=0$
we have $\left\|f-\left\langle
x,\theta\right\rangle\right\|_{L^{2}\left(K,\gamma\right)}^{2}\leq 4\epsilon$.
Also, by our assumption we have $\left\|f-\left\langle
x,\eta\right\rangle\right\|_{L^{2}\left(K,\gamma\right)}^{2}\leq\epsilon$.
Thus by the triangle inequality, we conclude

$\frac{1}{\gamma(K)}\int_{K}\left\langle
x,\theta-\eta\right\rangle^{2}\mathrm{d}\gamma=\left\|\left\langle
x,\theta\right\rangle-\left\langle
x,\eta\right\rangle\right\|_{L^{2}\left(K,\gamma\right)}^{2}\leq$ (10)
$\left\|\left\langle
x,\theta\right\rangle-f\right\|_{L^{2}\left(K,\gamma\right)}^{2}+\left\|f-\left\langle
x,\eta\right\rangle\right\|_{L^{2}\left(K,\gamma\right)}^{2}\leq(2\cdot
4+2)\epsilon=10\epsilon.$

Using the fact that $K\supseteq rB_{2}^{n}$ we have from (10):

$\left|\theta-\eta\right|^{2}\cdot\frac{1}{n}\int_{rB^{n}_{2}}|x|^{2}d\gamma=\int_{rB_{2}^{n}}\left\langle
x,\theta-\eta\right\rangle^{2}\mathrm{d}\gamma\leq\int_{K}\left\langle
x,\theta-\eta\right\rangle^{2}\mathrm{d}\gamma\leq 10\gamma(K)\epsilon,$

and thus

(11) $\left|\theta-\eta\right|^{2}\leq
10\frac{n\gamma(K)}{\int_{rB^{n}_{2}}|x|^{2}d\gamma}\epsilon.$

It follows from (11) and the conclusion (2) of Theorem 4.1:

$\displaystyle\int_{\partial
K}\left\langle\eta,n_{x}\right\rangle^{2}\mathrm{d}\gamma_{\partial K}$
$\displaystyle\leq 2\left(\int_{\partial
K}\left\langle\eta-\theta,n_{x}\right\rangle^{2}\mathrm{d}\gamma_{\partial
K}+\int_{\partial
K}\left\langle\theta,n_{x}\right\rangle^{2}\mathrm{d}\gamma_{\partial
K}\right)$ $\displaystyle\leq
2\cdot\left(\frac{10n\gamma(K)}{\int_{rB^{n}_{2}}|x|^{2}d\gamma}\epsilon\cdot\gamma^{+}\left(\partial
K\right)+\frac{2(n+1)}{r}\gamma(K)\epsilon\right)$
$\displaystyle\leq\left(\frac{20\gamma^{+}(\partial
K)}{\int_{rB^{n}_{2}}|x|^{2}d\gamma}+\frac{8}{r}\right)\gamma(K)n\epsilon$

∎

Lemma 5.1 allows us to deduce:

######  Theorem 5.2.

Let $K$ be a symmetric convex set with the in-radius $r$. Let $T$ be a
positive definite matrix with columns $t_{i}=Te_{i},$ $i=1,...,n,$ and let
$s\geq 0$ denote the smallest eigenvalue of $T$. Assume that

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left\langle
Tx,x\right\rangle^{2}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left\langle
Tx,x\right\rangle d\gamma\right)^{2}\geq
2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|Tx\right|^{2}\mathrm{d}\gamma-\epsilon$

for small enough $\epsilon>0$. Then for every $i=1,...,n,$ we have

$\int_{\partial K}\left\langle
t_{i},n_{x}\right\rangle^{2}\mathrm{d}\gamma_{\partial K}\leq
C\left(\frac{\gamma^{+}(K)}{\int_{rB^{n}_{2}}|x|^{2}d\gamma}+\frac{1}{r}\right)n^{2}\epsilon\gamma(K).$

Therefore, if $s>0$ and $\epsilon<c(\frac{s}{n})^{2}$ then either
$r\geq\sqrt{\log\frac{cs^{2}}{n^{2}\epsilon}},$ or $r\leq
C\sqrt{n}\left(\frac{\epsilon}{s^{2}}\right)^{\frac{1}{n+1}}.$

###### Proof.

Using the same approximation argument as in Theorem 4.1, we may assume
$\partial K$ is $C^{\infty}$-smooth. Write $f=\left\langle
Tx,x\right\rangle+c$ such that
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}f\mathrm{d}\gamma=0$.
Consider the function $u\in W^{2,2}(K)\cap C^{2}(K)$ such that $Lu=f$ and
$\langle\nabla u,n_{x}\rangle=0$ on $\partial K$ (again see e.g. [23] for
existence and regularity). By Bochner’s formula (again see e.g. [17]), we
write

$\displaystyle\int_{K}f^{2}\mathrm{d}\gamma$
$\displaystyle=-\int_{K}\left(\left\|\nabla^{2}u\right\|^{2}+\left|\nabla
u\right|^{2}+2\left\langle\nabla u,\nabla
f\right\rangle\right)\mathrm{d}\gamma-\int_{\partial
K}\left\langle\operatorname{II}\nabla_{\partial K}u,\nabla_{\partial
K}u\right\rangle\mathrm{d}\gamma_{\partial K}.$
$\displaystyle\leq-\int_{K}\left(\left\|\nabla^{2}u\right\|^{2}+\left|\nabla
u\right|^{2}+2\left\langle\nabla u,\nabla
f\right\rangle\right)\mathrm{d}\gamma,$

where in the last passage we used that $\rm{II}\geq 0$ since $K$ is convex.
Since $\nabla u$ is odd we have $\int\nabla u\mathrm{d}\gamma_{K}=0$, so by
the Poincaré inequality we have

(12) $\delta_{1}=\int_{K}\left(\left\|\nabla^{2}u\right\|^{2}-\left|\nabla
u\right|^{2}\right)\mathrm{d}\gamma\geq 0,$

and thus

(13) $\displaystyle\int_{K}f^{2}\mathrm{d}\gamma$
$\displaystyle\leq-\delta_{1}-\int_{K}\left(2\left|\nabla
u\right|^{2}+2\left\langle\nabla u,\nabla
f\right\rangle\right)\mathrm{d}\gamma$
$\displaystyle=-\delta_{1}-\int_{K}\left(\left|\sqrt{2}\nabla
u+\frac{1}{\sqrt{2}}\nabla f\right|^{2}-\frac{1}{2}\left|\nabla
f\right|^{2}\right)\mathrm{d}\gamma.$

It follows from (13) that

$\displaystyle-\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left\langle
Tx,x\right\rangle^{2}d\gamma+\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left\langle
Tx,x\right\rangle
d\gamma\right)^{2}+2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|Tx\right|^{2}\mathrm{d}\gamma$
$\displaystyle=\frac{1}{\gamma(K)}\int_{K}\left(\frac{1}{2}\left|\nabla
f\right|^{2}-f^{2}\right)\mathrm{d}\gamma$
$\displaystyle\geq\frac{\delta_{1}}{\gamma(K)}+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\sqrt{2}\nabla
u+\frac{1}{\sqrt{2}}\nabla f\right|^{2}\mathrm{d}\gamma\geq 0.$

Therefore our assumption implies that

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\sqrt{2}\nabla
u+\sqrt{2}Tx\right|^{2}\mathrm{d}\gamma=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\sqrt{2}\nabla
u+\frac{1}{\sqrt{2}}\nabla f\right|^{2}\mathrm{d}\gamma\leq\epsilon,$

and so
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\nabla
u+Tx\right|^{2}\mathrm{d}\gamma\leq\frac{\epsilon}{2}$. In particular, for
every $i$ we have

(14)
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\partial_{i}u+\left\langle
x,t_{i}\right\rangle\right|^{2}\mathrm{d}\gamma\leq\epsilon;$

recall that $t_{i}=Te_{i}\in\mathbb{R}^{n}$ denotes the $i$’th row of $T$.

However, our assumption also implies that

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left(\left\|\nabla^{2}u\right\|^{2}-\left|\nabla
u\right|^{2}\right)\mathrm{d}\gamma=\frac{\delta_{1}}{\gamma(K)}\leq\epsilon,$

so in particular for all $i$ we have

(15)
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left(\partial_{i}u\right)^{2}\mathrm{d}\gamma\geq\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}\left|\nabla\partial_{i}u\right|^{2}\mathrm{d}\gamma-\frac{\epsilon}{2}.$

The first conclusion now follows from (14) and (15), and Lemma 5.1:

$\int_{\partial K}\left\langle
t_{i},n_{x}\right\rangle^{2}\mathrm{d}\gamma_{\partial K}\leq
C\left(\frac{\gamma^{+}(K)}{\int_{rB^{n}_{2}}|x|^{2}d\gamma}+\frac{1}{r}\right)n\epsilon\gamma(K).$

Summing over all $i$, and using the bound $|Tn_{x}|\geq s,$ we obtain the
second conclusion:

$s^{2}\gamma^{+}\left(\partial K\right)\leq\int_{\partial
K}\left|Tn_{x}\right|^{2}\mathrm{d}\gamma_{\partial K}\leq
C\left(\frac{\gamma^{+}(\partial
K)}{\int_{rB^{n}_{2}}|x|^{2}d\gamma}+\frac{1}{r}\right)n^{2}\epsilon\gamma(K),$

or

$\frac{\gamma(K)}{\int_{rB^{n}_{2}}|x|^{2}d\gamma}+\frac{\gamma(K)}{r\gamma^{+}\left(\partial
K\right)}\geq\frac{cs^{2}}{n^{2}\epsilon}.$

Applying Proposition 3.1 with $\delta=\frac{n^{2}\epsilon}{cs^{2}},$ we get
the second conclusion. ∎

From Theorem 5.2 we deduce

######  Corollary 5.3.

Let $K$ be a symmetric convex set with the in-radius $r$. Assume that

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|x|^{4}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|x|^{2}d\gamma\right)^{2}\geq
2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|x|^{2}\mathrm{d}\gamma-\epsilon$

for $\epsilon<\frac{c}{n^{2}}$. Then either
$r\geq\sqrt{\log\frac{c}{n^{2}\epsilon}},$ or $r\leq
C\sqrt{n}\epsilon^{\frac{1}{n+1}}.$

##  6\. Proofs of the main results.

We first point out the following very nice fact:

######  Lemma 6.1.

Suppose $V\in C^{2}(\mathbb{R}^{n})$. Then

(16)
$V\left(\frac{z_{1}+z_{2}}{2}\right)+\beta(z_{1},z_{2})=\frac{V(z_{1})+V(z_{2})}{2},$

where, letting $z(t)=\frac{(1-t)z_{1}+(1+t)z_{2}}{2},$ we have

(17)
$\beta(z_{1},z_{2})=\frac{1}{8}\cdot\int_{-1}^{1}(1-|t|)\langle\nabla^{2}V(z(t))(z_{1}-z_{2}),z_{1}-z_{2}\rangle
dt.$

###### Proof.

Note that for $b\in C^{2}[-1,1],$

$\int_{-1}^{1}(1-|t|)b^{\prime\prime}(t)dt=\int_{0}^{1}(1-t)b^{\prime\prime}(t)dt+\int_{-1}^{0}(1+t)b^{\prime\prime}(t)dt=$
$(1-t)b^{\prime}(t)|^{1}_{0}+\int_{0}^{1}b^{\prime}(t)dt+(1+t)b^{\prime}(t)|^{0}_{-1}-\int_{-1}^{0}b^{\prime}(t)dt=$
(18) $b(1)+b(-1)-2b(0).$

Let $b(t)=V(z(t))$. Then $b(-1)=V(z_{1}),$ $b(1)=V(z_{2}),$
$b(0)=V(\frac{z_{1}+z_{2}}{2}),$ and

(19)
$b^{\prime\prime}(t)=\frac{1}{4}\langle\nabla^{2}V(z(t))(z_{1}-z_{2}),z_{1}-z_{2}\rangle.$

Combining (18) and (19) finishes the proof. ∎

Proof of the Theorem 1.1. Let $F(t)=\frac{\gamma(e^{t}K)}{\gamma(K)}.$ Then
$F^{\prime}(0)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int(n-|x|^{2})d\gamma,$
and

$F^{\prime\prime}(0)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}(n-|x|^{2})^{2}d\gamma-2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}|x|^{2}d\gamma.$

Next, let $V(t)=\log F(t)$. In view of the fact that $F(0)=1,$ we have
$V^{\prime\prime}(0)=F^{\prime\prime}(0)-F^{\prime}(0)^{2}$, and thus

$V^{\prime\prime}(t)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{t}K}(n-|x|^{2})^{2}d\gamma-2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{t}K}|x|^{2}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{t}K}(n-|x|^{2})d\gamma\right)^{2}.$

On the other hand, by the assumption, letting $\alpha=\log a$ and $\beta=\log
b,$ we have

(20)
$V\left(\frac{\alpha+\beta}{2}\right)-\frac{1}{2}V(\alpha)-\frac{1}{2}V(\beta)\leq\log(1+\epsilon)\leq\epsilon.$

By Lemma 6.1, (20) implies that there exists a $t\in[\log a,\log b]$ such that

$V^{\prime\prime}(t)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{t}K}|x|^{4}d\gamma-2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{t}K}|x|^{2}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{t}K}|x|^{2}d\gamma\right)^{2}\geq-\frac{8\epsilon}{(\log
b/a)^{2}}.$

An application of the Corollary 5.3 with $e^{t}K$ in place of $K,$ and with
$\frac{8\epsilon}{(\log b/a)^{2}}$ in place of $\epsilon$ finishes the proof.
$\square$

Proof of the Theorem 1.4. This time, we let
$F_{i}(t)=\frac{\gamma(e^{te_{i}}K)}{\gamma(K)}.$ Then
$F_{i}^{\prime}(0)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int(1-x_{i}^{2})d\gamma,$
and

$F_{i}^{\prime\prime}(0)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}(1-x_{i}^{2})^{2}d\gamma-2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{K}x_{i}^{2}d\gamma.$

Next, for $x\in\mathbb{R}^{n},$ let
$V(x)=\log\frac{\gamma(e^{x}K)}{\gamma(K)}$. We have $\frac{\partial
V}{\partial x_{i}\partial x_{j}}=0$ whenever $i\neq j,$ and for all
$i=1,...,n,$

(21) $\frac{\partial^{2}V}{\partial
x^{2}_{i}}(0)=F_{i}^{\prime\prime}(0)-F_{i}^{\prime}(0)^{2}.$

On the other hand, by our assumption,

(22)
$V\left(\frac{x+y}{2}\right)-\frac{1}{2}V(x)-\frac{1}{2}V(y)\leq\log(1+\epsilon)\leq\epsilon.$

By Lemma 6.1, and in view of (21), (22) implies that there exists a
$z\in[x,y]$ such that for all $i=1,..,n$

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{z}K}x_{i}^{4}d\gamma-2\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{z}K}x_{i}^{2}d\gamma-\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{e^{z}K}x_{i}^{2}d\gamma\right)^{2}\geq-\frac{8\epsilon}{(x_{i}-y_{i})^{2}}.$

By Theorem 5.2, we get, denoting $\tilde{K}=e^{z}K$ and
$\tilde{r}=|e^{z}|r(K)$, and summing up:

(23)
$\int_{\partial\tilde{K}}\sum_{i=1}^{n}(x_{i}-y_{i})^{2}(n^{i}_{x})^{2}d\gamma_{\partial\tilde{K}}\leq
C\epsilon
n^{2}\gamma(\tilde{K})\left(\frac{1}{\tilde{r}}+\frac{\gamma^{+}(\partial\tilde{K})}{\int_{\tilde{r}B^{n}_{2}}|x|^{2}d\gamma}\right).$

On the other hand, recalling the notation from the statement of the theorem,
we have

(24)
$\int_{\partial\tilde{K}}\sum_{i=1}^{n}(x_{i}-y_{i})^{2}(n^{i}_{x})^{2}d\gamma_{\partial\tilde{K}}\geq\delta^{2}\int_{\partial\tilde{K}}\sum_{i\in\sigma^{\delta}}(n^{i}_{x})^{2}d\gamma_{\partial\tilde{K}}\geq\delta^{2}\alpha\gamma^{+}(\tilde{\Omega}_{\delta,\alpha}),$

where $\tilde{\Omega}_{\delta,\alpha}=e^{z}\Omega_{\delta,\alpha}.$

Combining (23), (24) and Proposition 3.1 (applied with $\tilde{K}$,
$\tilde{r}$ and $\delta^{2}\epsilon\alpha\beta$), and recalling that
$r(K)\in[|e^{x}|\tilde{r},|e^{y}|\tilde{r}],$ we get the conclusion. $\square$

Finally, we are moving to proving Corollary 1.5. First, we shall need the
following

######  Lemma 6.2.

Let $H\subset\mathbb{R}^{n}$ be a subspace. Let $K\subset\mathbb{R}^{n}$ be a
closed, convex set with non-empty interior. Assume that at almost every point
$x\in\partial K$ a unique normal $n_{x}$ exists and $n_{x}\in H$ (Here ”almost
everywhere” is with respect to the $(n-1)$-dimensional Hausdorff measure on
$\partial K$). Then there exists a closed convex set $K_{0}\subset H$ such
that $K=K_{0}\times H^{\perp}$.

###### Proof.

By translating $K$ we may assume without loss of generality that $0$ is an
interior point of $K$. Define $K_{0}$ to be the orthogonal projection
$K_{0}=\operatorname{Proj}_{H}K$. Then we clearly have $K\subset K_{0}\times
H^{\perp}$, and $K_{0}$ also has $0$ in its (relative) interior.

We first show that $\partial K\subset\partial K_{0}\times H^{\perp}$. Indeed,
fix $x\in\partial K$. If we define

$\Omega=\left\\{y\in K:\ n_{y}\text{ is unique and }n_{y}\in H\right\\}$

then by assumption $K\setminus\Omega$ has measure $0$, so in particular
$\Omega$ is dense in $\partial K$. Choose a sequence $x_{k}\in\Omega$ such
that $x_{k}\to x$. By compactness we may pass to a subsequence and assume
without loss of generality that $n_{x_{k}}\xrightarrow{k\to\infty}v\in H$.

We know that $x_{k}+n_{x_{k}}^{\perp}$ is a supporting hyperplane for $K$ for
all $k$. Therefore for every $z\in K$ we have

$\left\langle z,n_{x_{k}}\right\rangle\leq\left\langle
x_{k},n_{x_{k}}\right\rangle.$

Taking the limit as $k\to\infty$ we see that for every $z\in K$ we have
$\left\langle z,v\right\rangle\leq\left\langle x,v\right\rangle$. In
particular, every point $z_{0}\in K_{0}$ may we written as
$z_{0}=\operatorname{Proj}_{H}\left(z\right)$ for some $z\in K$, and since
$v\in H$ we have

$\left\langle z_{0},v\right\rangle=\left\langle
z,v\right\rangle\leq\left\langle
x,v\right\rangle=\left\langle\operatorname{Proj}_{H}x,v\right\rangle.$

For every $\epsilon>0$ the point
$z_{\epsilon}=\operatorname{Proj}_{H}x+\epsilon v$ satisfies

$\left\langle
z_{\epsilon},v\right\rangle=\left\langle\operatorname{Proj}_{H}x,v\right\rangle+\epsilon\left|v\right|^{2}>\left\langle\operatorname{Proj}_{H}x,v\right\rangle,$

so $z_{\epsilon}\notin K_{0}$. Therefore $\operatorname{Proj}_{H}x\in\partial
K_{0}$, so indeed $x\in\partial K_{0}\times H^{\perp}$ as claimed.

Now we prove that $K_{0}\times H^{\perp}\subset K$, finishing the proof.
Indeed, for every $x\in K_{0}\times H^{\perp}$ consider

$A=\left\\{t\in[0,1]:\ tx\in K\right\\}.$

We know that $0\in A$ and $A$ is closed. Define $t_{0}=\max A$. If $t_{0}<1$
it follows that $t_{0}x\in\partial K\subset\partial\left(K_{0}\times
H^{\perp}\right)$, which is impossible since $x\in K_{0}\times H^{\perp}$.
Hence $t_{0}=1$ and $x\in K$ as claimed. ∎

Proof of Corollary 1.5. We may approximate $K$ with a convex set whose
boundary is $C^{2}$ arbitrarily closely. We let $\sigma$ so that

$\sigma^{c}=\\{i\in[n]:\,x_{i}=y_{i}\\},$
$H=\operatorname{span}\\{e_{i}:\,i\in\sigma^{c}\\},$

and

$\Omega=\\{x\in\partial K:\,n_{x}\not\in H^{\perp}\\}.$

We see that the assumption of Theorem 1.4 is satisfied for an arbitrarily
small $\epsilon>0.$ Pick $\alpha=\beta=\delta=\epsilon^{\frac{1}{10}}$ and let
$\epsilon\rightarrow 0.$ Note that as $\epsilon\rightarrow 0,$ we might get
different $z=z(\epsilon)$ in the conclusion of Theorem 3, but as $z\in[x,y]$,
then by compactness we may select a convergent sub-sequence of $z(\epsilon)$
to some point $z_{0}.$ We conclude that either $r(e^{z_{0}}K)=0$ (in which
case $K$ has an empty interior), or $r(e^{z_{0}}K)=\infty$ (in which case
$K=\mathbb{R}^{n}$), or $\gamma^{+}(\Omega)=0$. Since $\gamma_{\partial K}$ is
absolutely continuous with respect to the Hausdorff measure on $\partial K,$
we conclude that in this case, for almost every $x\in\partial K,$ we have
$n_{x}\in H^{\perp}.$ The conclusion follows from Lemma 6.2. $\square$

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# An Empirical Study on Leveraging Position Embeddings for Target-oriented
Opinion Words Extraction

Samuel Mensah
Computer Science Department
University of Sheffield, UK
<EMAIL_ADDRESS>
&Kai Sun
BDBC and SKLSDE
Beihang University, China
<EMAIL_ADDRESS>
&Nikolaos Aletras
Computer Science Department
University of Sheffield, UK
<EMAIL_ADDRESS>

###### Abstract

Target-oriented opinion words extraction (TOWE) Fan et al. (2019b) is a new
subtask of target-oriented sentiment analysis that aims to extract opinion
words for a given aspect in text. Current state-of-the-art methods leverage
position embeddings to capture the relative position of a word to the target.
However, the performance of these methods depends on the ability to
incorporate this information into word representations. In this paper, we
explore a variety of text encoders based on pretrained word embeddings or
language models that leverage part-of-speech and position embeddings, aiming
to examine the actual contribution of each component in TOWE. We also adapt a
graph convolutional network (GCN) to enhance word representations by
incorporating syntactic information. Our experimental results demonstrate that
BiLSTM-based models can effectively encode position information into word
representations while using a GCN only achieves marginal gains. Interestingly,
our simple methods outperform several state-of-the-art complex neural
structures.

##  1 Introduction

Target-oriented opinion words extraction (TOWE) Fan et al. (2019b) is a fine-
grained task of target-oriented sentiment analysis Liu (2012) aiming to
extract opinion words with respect to an opinion target (or aspect) in text.
Given the sentence “The food is good but the service is extremely slow”, TOWE
attempts to identify the opinion words “good” and “extremely slow”
corresponding respectively to the targets “food” and “service”. TOWE is
usually treated as a sequence labeling problem using the BIO tagging scheme
Ramshaw and Marcus (1999) to distinguish the Beginning, Inside and Outside of
a span of opinion words. Table 1 shows an example of applying the BIO tagging
scheme for TOWE.

Sentence:
---
The food is good but the service is extremely slow.
True Labels for target ‘food’:
The/O food/O is/O good/B but/O the/O service/O
is/O extremely/O slow/O.
True Labels for target ‘service’:
The/O food/O is/O good/O but/O the/O service/O
is/O extremely/B slow/I.
TOWE Extraction Results:
{(food, good), (service, extremely slow)}
Table 1: Identifying target-oriented opinion words in a sentence. Underlined
words are opinion targets. Spans tagged B and I are considered as opinion
words.

Learning effective word representations is a critical step towards tackling
TOWE. Traditional work Zhuang et al. (2006a); Hu and Liu (2004a); Qiu et al.
(2011) has used hand-crafted features to represent words which do not often
generalize easily. More recent work Liu et al. (2015); Fan et al. (2019b); Wu
et al. (2020a); Veyseh et al. (2020) has explored neural networks to learn
word representations automatically.

Previous neural-based methods Liu et al. (2015); Fan et al. (2019b) has used
word embeddings Collobert and Weston (2008); Mikolov et al. (2013); Pennington
et al. (2014) to represent the input. However, TOWE is a complex task that
requires a model to know the relative position of each word to the aspect in
text. Words that are relatively closer to the target usually express the
sentiment towards that aspect Zhou et al. (2020).

Fan et al. (2019b) employ Long Short-Term Memory (LSTM) networks Hochreiter
and Schmidhuber (1997) to encode the target position information in word
embeddings. Wu et al. (2020a) transfer latent opinion knowledge into a
Bidirectional LSTM (BiLSTM) network that leverages word and position
embeddings Zeng et al. (2014). Recently, Veyseh et al. (2020) have proposed
ONG, a method that combines BERT (Bidirectional Encoder Representations from
Transformers) Devlin et al. (2018), position embeddings, Ordered Neurons LSTM
(ON-LSTM) Shen et al. (2018),111An LSTM variant with an inductive bias toward
learning latent tree structures in sequences. and a graph convolutional
network (GCN) Kipf and Welling (2016) to introduce syntactic information into
word representations. While this model achieves state-of-the-art results,
previous studies have shown that the ON-LSTM does not actually perform much
better than LSTMs in recovering latent tree structures Dyer et al. (2019).
Besides, ON-LSTMs perform worse than LSTMs in capturing short-term
dependencies Shen et al. (2018). Since opinion words are usually close to
targets in text, ON-LSTM risks missing the relationship between the aspect and
any information (e.g. position) relating to the opinion words.

In this paper, we empirically evaluate a battery of popular text encoders
which apart from words, take positional and part-of-speech information into
account. Surprisingly, we show that methods based on BiLSTMs can effectively
leverage position embeddings to achieve competitive if not better results than
more complex methods such as ONG on standard TOWE datasets. Interestingly,
combining a BiLSTM encoder with a GCN to explicitly capture syntactic
information achieves only minor gains. This empirically highlights that
BiLSTM-based methods have an inductive bias appropriate for the TOWE task,
making a GCN less important.

##  2 Methodology

Given sentence $s=\\{w_{1},\ldots,w_{n}\\}$ with aspect $w_{t}\in s$, our
approach consists of a text encoder that takes as input a combination of
words, part-of-speech and position information for TOWE. We further explore
enhancing text encoding by incorporating information from a syntactic parse of
the sentence through a GCN encoder.

###  2.1 Input Representation

#### Word Embeddings:

We experiment with Glove word vectors Pennington et al. (2014) as well as
BERT-based representations, extracted from the last layer of a BERT base model
Devlin et al. (2018) fine-tuned on TOWE.

#### Position Embeddings (POSN):

We compute the relative distance $d_{i}$ from $w_{i}$ to $w_{t}$ (i.e.,
$d_{i}=i-t$), and lookup their embedding in a randomly initialized position
embedding table.

#### Par-of-Speech Tag Embeddings (POST):

We assign part-of-speech tags to each word token using the Stanford
parser,222https://stanfordnlp.github.io/CoreNLP/ and lookup their embedding in
a randomly initialized POST embedding table.

#### Combined Input:

We consider two types of input representations:

  1. 1.

Glove Input (G): Constructed from concatenating Glove word embeddings, POST
and POSN embeddings for each token.

  2. 2.

BERT Input (B): Constructed from concatenating BERT vectors with POSN
embeddings for each word token following a similar approach as Veyseh et al.
(2020).333Early experimentation with RoBERTa Liu et al. (2019) yielded lower
performance. We ignore POST embeddings since BERT is efficient in modeling
such information Tenney et al. (2019).

###  2.2 Text Encoders

We experiment with the following neural encoders that take word vector
representations as input:

#### CNN:

A single layer convolutional neural network LeCun et al. (1990). Given a word
$w_{i}\in s$, the CNN takes a fixed window of words around it and applies a
filter on their representation to extract a feature vector for $w_{i}$. We
concatenate the feature vectors corresponding to different filters for $w_{i}$
to compute word representations.

#### Transformer:

A Transformer encoder Vaswani et al. (2017) that takes a linear transformation
of the input words to learn contextualized representations.

#### BiLSTM:

A bi-directional LSTM that takes the input representation and models the
context in a forward and backward direction.

#### ON-LSTM:

A variant of the LSTM neural network proposed by Shen et al. (2018) which has
an inductive bias toward learning latent tree structures.

###  2.3 GCN Encoder

First, we interpret the syntactic parse tree as an adjacency binary matrix
$A^{n\times n}$ ($n$ is the sentence length) with entries $A_{ij}=1$ if there
is a connection between nodes $i$ and $j$, and $A_{ij}=0$ otherwise. To apply
a GCN on $A$, we consider the tree with self-loops at each node (i.e.,
$A_{ii}=1$), ensuring nodes are informed by their corresponding
representations at previous layers. Formally, let $H^{(k)}$ be the output at
the $k$-th GCN layer, $H^{(k)}$ is given by:

$H^{(k)}={\rm ReLU}(AH^{(k-1)}W^{(k)})+H^{(k-1)}$ (1)

where $k=1,\ldots,K$, $W^{(k)}$ is a parameter matrix at layer $k$. $RELU$ is
used as the activation function. $H^{(0)}$ corresponds to the set of word
representations extracted by the text encoder. The second term in (1) induces
a residual connection that retains the contextual information of $H^{(0)}$
during the propagation process Sun et al. (2020).

###  2.4 Classification and Optimization

Our model uses the representation $H^{(l)}$ (where $l\geq 0$), applies a
linear layer and then normalize it with a softmax function to output a
probability distribution over the set {B,I,O} for each word in the input.
During training, we minimize the cross-entropy function for each word in text
for the entire training set.

##  3 Experiments and Results

###  3.1 Baselines

We compare our methods with Distance-rule Hu and Liu (2004b); Dependency-rule
Zhuang et al. (2006b); LSTMword and BiLSTMword Liu et al. (2015); Pipeline Fan
et al. (2019b); TC-BiLSTM Fan et al. (2019b); IOG Fan et al. (2019b); LOTN Wu
et al. (2020a); and ONG Veyseh et al. (2020).444Note that LSTMword/BiLSTMword
only use word embeddings as input.

###  3.2 Data

Following Wu et al. (2020b), we use four benchmark datasets including
restaurant (Res14, Res15, Res16) and laptop (Lap14) reviews from Semeval
Pontiki et al. (2014, 2015, 2016). We use the preprocessed data provided by
Fan et al. (2019a). Table 2 shows the dataset statistics.

Dataset | #Sent. | #ASL | #AT | #OT | #D.Dist. | #S.Dist.
---|---|---|---|---|---|---
Lap14 (Train) | 1151 | 20.78 | 1632 | 1877 | 2.40 | 4.25
Lap14 (Test) | 343 | 17.33 | 482 | 567 | 2.03 | 4.00
Res14 (Train) | 1625 | 19.11 | 2636 | 3057 | 2.11 | 3.68
Res14 (Test) | 500 | 19.22 | 862 | 1028 | 2.01 | 3.97
Rest15 (Train) | 754 | 16.50 | 1076 | 1277 | 1.97 | 3.62
Rest15 (Test) | 325 | 17.47 | 436 | 493 | 2.13 | 3.53
Rest16 (Train) | 1079 | 16.78 | 1512 | 1770 | 2.01 | 3.59
Rest16 (Test) | 328 | 16.54 | 456 | 524 | 1.93 | 3.43
Table 2: Dataset Statistics. No. of sentences (#Sent), Avg. sentence length
(#ASL), No. of aspect terms (#AT), No. of opinion words (#OT), Avg. dependency
distance (#D.Dist) and Avg. sequential distance (#S.Dist) between aspect and
opinion.

###  3.3 Implementation Details

Hyper-parameters are tuned on 20$\%$ of samples randomly selected from the
train set since there is no development set.555We use 300-dim Glove word
vectors Pennington et al. (2014) and apply a dropout of 0.8. Dimensions of
part-of-speech and position embeddings are set to 30, but dimensions of
position embeddings for pretrained models are set to 100. The CNN uses three
filters with sizes 3, 4 and 5 and has a hidden dimension of 300. All other
models have a hidden dimension of 200. The number of GCN layers is set over
$K\in\\{1,\ldots,5\\}$. Experiments are performed on NVIDIA Tesla V100. We use
the Adam optimizer to train all models. Models that use Glove word vectors are
optimized with learning rate $1e^{-3}$ and trained for 100 epochs with batch
size 16. Models that use BERT hidden vectors are optimized with learning rate
$1e^{-5}$ and trained with batch size 6. Our source code is publicly
available.666https://github.com/samensah/Encoders_TOWE_EMNLP2021

| Lap14 | Res14 | Res15 | Res16 |
---|---|---|---|---|---
Model | Prec | Rec | F1 | Prec | Rec | F1 | Prec | Rec | F1 | Prec | Rec | F1 | Avg.F1
Distance-rule | 50.13 | 33.86 | 40.42 | 58.39 | 43.59 | 49.92 | 54.12 | 39.96 | 45.97 | 61.90 | 44.57 | 51.83 | 47.04
Dependency-rule | 45.09 | 31.57 | 37.14 | 64.57 | 52.72 | 58.04 | 65.49 | 48.88 | 55.98 | 76.03 | 56.19 | 64.62 | 53.95
LSTMword | 55.71 | 57.53 | 56.52 | 52.64 | 65.47 | 58.34 | 57.27 | 60.69 | 58.93 | 62.46 | 68.72 | 65.33 | 59.78
BiLSTMword | 64.52 | 61.45 | 62.71 | 58.34 | 61.73 | 59.95 | 60.46 | 63.65 | 62.00 | 68.68 | 70.51 | 69.57 | 63.56
Pipeline | 72.58 | 56.97 | 63.83 | 77.72 | 62.33 | 69.18 | 74.75 | 60.65 | 66.97 | 81.46 | 67.81 | 74.01 | 68.50
TC-BiLSTM | 62.45 | 60.14 | 61.21 | 67.65 | 67.67 | 67.61 | 66.06 | 60.16 | 62.94 | 73.46 | 72.88 | 73.10 | 66.22
IOG | 73.24 | 69.63 | 71.35 | 82.85 | 77.38 | 80.02 | 76.06 | 70.71 | 73.25 | 82.25 | 78.51 | 81.69 | 76.58
LOTN | 77.08 | 67.62 | 72.02 | 84.00 | 80.52 | 82.21 | 76.61 | 70.29 | 73.29 | 86.57 | 80.89 | 83.62 | 77.79
ONG | 73.87 | 77.78 | 75.77 | 83.23 | 81.46 | 82.33 | 76.63 | 81.14 | 78.81 | 87.72 | 84.38 | 86.01 | 80.73
Glove Input |
Transformer(G) | 68.33 | 61.91 | 64.91 | 71.77 | 70.29 | 70.98 | 78.90 | 59.07 | 67.41 | 83.59 | 70.57 | 76.49 | 69.94
CNN(G) | 64.81 | 73.83 | 69.00 | 75.86 | 78.83 | 77.29 | 68.21 | 73.87 | 70.91 | 76.93 | 84.77 | 80.64 | 74.46
ON-LSTM(G) | 69.27 | 69.70 | 69.47 | 83.01 | 76.98 | 79.87 | 76.19 | 74.24 | 75.20 | 84.17 | 82.90 | 83.52 | 77.02
BiLSTM(G) | 76.49 | 70.94 | 73.59 | 86.22 | 83.44 | 84.80 | 81.49 | 77.93 | 79.66 | 88.96 | 84.05 | 87.36 | 81.35
Transformer+GCN(G) | 66.32 | 70.83 | 68.45 | 82.98 | 75.14 | 78.82 | 76.80 | 69.45 | 72.88 | 84.71 | 79.92 | 82.25 | 75.60
CNN+GCN(G) | 66.88 | 74.88 | 70.65 | 82.45 | 80.12 | 81.24 | 75.32 | 73.75 | 74.51 | 82.17 | 84.89 | 83.48 | 77.47
ON-LSTM+GCN(G) | 71.63 | 74.04 | 72.75 | 87.06 | 80.97 | 83.90 | 80.18 | 77.53 | 78.83 | 89.89 | 83.97 | 86.82 | 80.58
BiLSTM+GCN(G) | 76.49 | 74.46 | 75.46 | 87.60 | 83.66 | 85.57 | 82.32 | 78.82 | 80.52 | 91.63 | 85.65 | 88.52 | 82.52
BERT Input |
Transformer(B) | 78.88 | 78.03 | 78.13 | 83.97 | 84.40 | 84.18 | 82.37 | 78.21 | 80.22 | 88.22 | 84.05 | 86.06 | 82.14
CNN(B) | 77.94 | 75.91 | 76.87 | 86.35 | 82.16 | 84.20 | 80.01 | 78.62 | 79.30 | 88.50 | 82.41 | 85.33 | 81.43
ON-LSTM(B) | 77.96 | 77.53 | 77.71 | 85.58 | 83.25 | 84.39 | 82.57 | 78.34 | 80.38 | 87.76 | 83.55 | 86.54 | 82.26
BiLSTM(B) | 78.38 | 78.27 | 78.25 | 86.38 | 84.82 | 85.60 | 82.17 | 78.78 | 80.41 | 89.94 | 84.16 | 86.94 | 82.80
Transformer+GCN(B) | 79.38 | 77.04 | 78.19 | 85.43 | 84.18 | 84.79 | 82.21 | 79.55 | 80.84 | 89.34 | 84.16 | 86.66 | 82.62
CNN+GCN(B) | 79.19 | 76.19 | 77.62 | 84.96 | 84.08 | 84.50 | 82.39 | 77.36 | 79.77 | 88.16 | 84.09 | 86.06 | 81.98
ON-LSTM+GCN(B) | 80.33 | 76.01 | 77.96 | 85.68 | 84.03 | 84.83 | 82.14 | 78.18 | 80.07 | 89.35 | 83.93 | 86.54 | 82.35
BiLSTM+GCN(B) | 79.72 | 78.06 | 78.82 | 86.45 | 85.06 | 85.74 | 83.37 | 77.93 | 80.54 | 88.98 | 85.80 | 87.35 | 83.11
Table 3: Results of experiments across baseline methods (across 5 runs). Results of compared models are retrieved from Veyseh et al. (2020). The best F1 performance is bold-typed. | Lap14 | Res14 | Res15 | Res16
---|---|---|---|---
Model | Prec | Rec | F1 | Prec | Rec | F1 | Prec | Rec | F1 | Prec | Rec | F1
BiLSTM+GCN(G) | 76.49 | 74.46 | 75.46 | 87.60 | 83.66 | 85.57 | 82.32 | 78.82 | 80.52 | 91.63 | 85.65 | 88.52
— GCN | 76.49 | 70.94 | 73.59 | 86.22 | 83.44 | 84.80 | 81.49 | 77.93 | 79.66 | 88.96 | 84.05 | 87.36
— GCN, POST | 75.38 | 70.12 | 72.63 | 86.83 | 82.94 | 84.83 | 82.45 | 75.58 | 78.85 | 88.71 | 84.01 | 86.29
— GCN, POST, POSN | 61.65 | 62.08 | 61.80 | 63.17 | 56.63 | 59.66 | 62.16 | 61.54 | 61.76 | 70.11 | 70.23 | 70.08
BiLSTM+GCN(B) | 79.72 | 78.06 | 78.82 | 86.45 | 85.06 | 85.74 | 83.37 | 77.93 | 80.54 | 88.98 | 85.80 | 87.35
— GCN | 78.38 | 78.27 | 78.25 | 86.38 | 84.82 | 85.60 | 82.17 | 78.78 | 80.41 | 89.94 | 84.16 | 86.94
— GCN, POSN | 62.92 | 72.17 | 67.21 | 60.84 | 64.42 | 62.54 | 63.88 | 64.42 | 63.97 | 69.59 | 71.45 | 70.39
Table 4: Precision, Recall and F1 scores of ablated models on the benchmark
datasets (across $5$ runs).

###  3.4 Performance Comparison

Table 3 presents the results of all methods. Our models that use Glove Input
(or BERT Input) are appended with “G”(or “B”) to distinguish them. We report
precision (Prec), recall (Rec), F1 score and average F1 score (Avg.F1) across
all datasets.

#### Comparison of Text Encoders:

We first observe that CNN(G) is adept at exploiting the information from
simpler word representations (Glove), outperforming the Transformer(G) by
+4.52 Avg.F1. We believe that this behavior is due to the fact that TOWE is a
short-sequence task (see #ASL in Table 2). This assumption lies well with
previous observations by Yin et al. (2021), which found that CNNs often
perform better than Transformers at short-sequence tasks. However, the
Transformer(B) is able to improve performance and even outperform CNN(B) by
+0.71 Avg.F1 by using BERT.

In addition, we find that ON-LSTM(G) and ON-LSTM(B) lag behind BiLSTM(G) and
BiLSTM(B) by 4.33 and 0.54 Avg.F1 respectively. ON-LSTM performs worse than
LSTMs on tasks that require tracking short-term dependencies Shen et al.
(2018). Since opinion words are usually close to the target in the sequence
(see #ASL vrs. #S.Dist. in Table 2), tracking short-term dependency
information is important in TOWE. This explains why BiLSTM(G)(or BiLSTM(B))
achieves a better performance over ON-LSTM(G)(or ON-LSTM(B)).

The performance of BiLSTM(G) over BiLSTMword suggests that the substantial
boost in performance comes from either part-of-speech or position embeddings.
We later perform an ablation experiment to examine which information is more
useful. Interestingly, BiLSTM(G) outperforms the current state-of-the-art ONG
by +0.62 Avg.F1 despite its simple architecture, demonstrating the importance
to first experiment with simpler methods before designing more complex
structures.

#### Comparison of Text+GCN Encoders:

Adding a GCN over any text encoder generally improves performance. This
happens because the GCN provides additional syntactic information that is
helpful for representation learning. We find that BiLSTM+GCN(G) achieves few
gains over BiLSTM(G) while other text encoders including Transformer+GCN(G)
and CNN+GCN(G) achieve relatively higher gains than their counterparts. This
suggest that BiLSTM(G) has an inductive bias appropriate for the TOWE task and
the performance mostly depends on the quality of the input representation. We
observe that when using BERT embeddings, there is a minimal performance
difference between using GCNs or not. We attribute this to the expressiveness
of BERT embeddings and its ability to capture syntactic dependencies Jawahar
et al. (2019). Overall results suggest that our proposed method outperforms
SOTA consistently across datasets.

###  3.5 Ablation Study

We perform ablation experiments on the two best performing models,
BiLSTM+GCN(G) and BiLSTM+GCN(B), to study the contribution of their different
components. The results are shown in Table 4. On BiLSTM+GCN(G), as we
consecutively remove the GCN and POST embeddings from the input
representation, we observe a slight drop in performance. The results indicate
that POST embeddings as well as the GCN are not critical components for
BiLSTM+GCN(G). Therefore, they can be ignored to reduce model complexity.
However, we observe a substantial drop in performance by removing the position
embedding from the input representation, obtaining an F1 score equivalent to
BiLSTMword across datasets. Similarly, removing the position embeddings in
BiLSTM+GCN(B) causes a substantial drop in performance. The results suggest
that leveraging position embeddings is crucial for TOWE performance.

##  4 Conclusion

We presented through extensive experiments that by employing a simple BiLSTM
architecture that uses input representations from pre-trained word embeddings
or language models, POST embeddings and position embeddings, we can obtain
competitive, if not better results than the more complex current state-of-the-
art methods Veyseh et al. (2020). The BiLSTM succeeds in exploiting position
embeddings to improve performance. By adapting a GCN to incorporate syntactic
information from the sentence we achieve further gains. In future work, we
will explore how to improve existing TOWE models by effectively leveraging
position embeddings.

## Acknowledgements

Samuel Mensah and Nikolaos Aletras are supported by a Leverhulme Trust
Research Project Grant.

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