数理解析研究所講究録
第 1862 巻 2013 年 52-62

52

Properties of -Gaussian measures related tothe isoperimetric and concentration profiles$q$

Asuka Takatsu*

Graduate School of Mathematics, Nagoya University

1

Introduction

This note is devoted to properties related to the isoperimetric profile and the concentration profile of a non-Gaussian probability measure, in particular -Gaussian measures,on . We always assume that any measure and any set are Borel.
describes the
On one hand, the isoperimetric profile of a probability measure on$\mu(A)$boundarymeasuretheandrelation between the volume$:= \{x\in \mathbb{R}^{n}|\inf_{a\in A}|x-a|<\epsilon\}$ denotes the-open neighborhood, whereofprecise,the isoperimetric.
beToof with respect to the standard Euclidean metricprofileis the function on $[0,1]$ defined by$q$$\mathbb{R}^{n}$$\mathbb{R}^{n}$$\mu$$\mu^{+}(A)$$A\subset \mathbb{R}^{n}$$A^{\epsilon}$$:=\varliminf_{\downarrow 0}\mu[A^{\epsilon}\backslash A]/\epsilon$$\epsilon$$A$$|\cdot|$$I[\mu]$$I[\mu](a)$$:= \inf\{\mu^{+}(A)|A\subset \mathbb{R}^{n}$with $\mu(A)=a\}.$only on $[0,1/2]$ since a given set and its complement may
We sometimes consider
have the same boundary measure under suitable conditions.
On the other hand, the concentration profile of a probability measure on
estimates the volume of the -open neighborhood of sets having measure 1/2. To be
defined as
precise, the concentmtion profile $C[\mu]$ is the function on$I[\mu]$$\mathbb{R}^{n}$

$\mu$

$r$

$[0, \infty)$$C[\mu](r)$ $:= \sup\{1-\mu(A^{r})|A\subset \mathbb{R}^{n}$with $\mu(A)\geq 1/2\}.$

Note that the both profiles can be defined for a probability measure on a metricspace since the definition of the both profiles depend on only a probability measure
and a distance function, where we do not take advantage of the Euclidean structure.
It is usually difficult to obtain the isoperimetric profile and the concentration profile
of a given probability measure, however the both profiles of the Gaussian measure arewithknown. Here the Gaussian measure is an absolutely continuous measure on
density$\mathbb{R}^{n}$$\gamma_{n}$$\frac{d\gamma_{n}}{dx}(x)=(2\pi)^{-n/2}\exp(-\frac{|x|^{2}}{2})$with respect to the Lebesgue measure.
$*$

Supported in part by the Grant-in-Aid for Young Scientists (B) 24740042.

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Theorem 1.1 ([3, Theorem 3.1], [11, Corollary 1]) It holds for any $a\in[O, 1]$ that
$I[\gamma_{n}](a)=I[\gamma_{1}](a)=G’(\Phi(a))$

where$\Phi$

is the inverse function of$G$

which is defined for,by$r\in \mathbb{R}$$G(r);= \int_{-\infty}^{r}(2\pi)^{-1/2}\exp(-\frac{S^{2}}{2})ds=\gamma_{1}(-\infty, r].$

Theorem 1.1 easily induces$C[ \gamma_{n}](r)=C[\gamma_{1}](r)=1-G(r)=l^{\infty}(2\pi)^{-1/2}\exp(-\frac{S^{2}}{2})ds\leq\exp(-\frac{r^{2}}{2})$.

Since the isoperimetric profile and the concentration profile ofare dimension free,
we denote $I:=I[\gamma_{n}]$ and $C:=C[\gamma_{n}].$We say that a probability measure verffies a Gaussian isoperimetric inequality if
there exists a positive constant such that$\gamma_{n}$

$\mu$

$c$

$I[\mu](a)\geq cI(a)$holds for any $a\in[0,1]$ . Similarly, we say that a probability measure verffies a Gaussianconcentmtion inequality if there exist positive constants and such that$c$$\lambda$$C[\mu](r)\leq c\exp(-\lambda r^{2}/2)$

holds for any $r\geq 0$ . If a probability measure verifies a Gaussian isoperimetric inequality, then the probability measure also verifies a Gaussian concentration inequality,
which follows from Proposition 1.2 below and the fact that there exists a positive
constant such that$c$

$I(a)\geq ca\sqrt{\log 1}/a$holds for $a\in[O, 1/2].$
Proposition 1.2 ([6, Proposition 1.7]) For a continuous function, let be the inverse function of$[0, \infty)$$\sigma$: [log2,$\alpha$$r \mapsto\int_{\log 2}^{r}\frac{1}{\tilde{\sigma}(s)}ds,$

If a pmbability measure

$\mu$

on

$\tilde{\sigma}(s)=\{\begin{array}{ll}\sigma(s) if s\geq\log 2,\sigma(-\log(1-e^{-s})) if s<\log 2.\end{array}$$\mathbb{R}^{n}ver’ifie\mathcal{S}$$I[\mu](a)\geq a\sigma(\log 1/a)$on $[0,1/2]$ , then it holds for $r\geq 0$ that
$C[\mu](r)\leq\exp(-\alpha(r))$.

$\infty$)$arrow$

54
More generally, we have the following implication from an isoperimetric inequality toa concentration inequality since the difference of the volumes between a set and its-open neighborhood is roughly considered as an integral of the boundary measures ofthe -open neighborhoods of the given set on $t\in(O, r)$ .
$r$

$t$

Proposition 1.3 ([4, Corollary 2.2]) Let be an absolutely continuous probability measure on
with respect to the Lebesgue measure. If there exists a strictly increasing,$\mu$$\mathbb{R}^{n}$differentiable function from an interval of to $[0,1]$ such that $I[\mu]\geq v’ou$ holds on$[0,1]$ , whereis the inverse function of , then it holds for every $r>0$ that
$C[\mu](r)\leq 1-v(u(1/2)+r)$ .
$\mathbb{R}$

$v$

$u$

$v$

We thus find that a probability measure verifies a Gaussian concentration inequality if
the probability measure verifies a Gaussian isoperimetric inequality.
There are several criteria for a probability measure to verify a Gaussian isoperimetric inequality. For example, given an absolutely continuous logarithmic concaveprobability measure on
with respect to the Lebesgue measure, namely there exists a convex function : $\mathbb{R}^{n}arrow(-\infty, \infty] such that d\mu(x)/dx=\exp(-V(x))$ holds on, the following equivalent condition is known.
$\mathbb{R}^{n}$

$\mu$

$V$

$x\in \mathbb{R}^{n}$

Theorem 1.4 ([1, Theorem 1.3]) For an absolutely continuous logarithmic
probability measure on
with respect to the Lebesgue measure, the follovnngs areequivalent to each other.

$\omega ncave$$\mathbb{R}^{n}$$\mu$$\cdot$$\bullet$$\bullet$

$\mu$

$\mu$

$c$

verifies a Gaussian isoperimetric inequality.

verifies a logarithmic Sobolev inequality, that is there exists a positive constantsuch that$\int_{\mathbb{R}^{n}}f^{2}\log(f^{2})d\mu-\int_{R^{n}}f^{2}d\mu\log(\int_{\mathbb{R}^{n}}f^{2}d\mu)\leq c\int_{R^{n}}|\nabla f|^{2}d\mu$holds for every locally Lipschitz function

$f$

on

$\mathbb{R}^{n}$with its distri bution$al$gmdient$\nabla f.$$\bullet$$\mu$verifies a Herbst necessary$\omega$ndition, that is there exists a positive constant$\epsilon$

satisfying$\int_{R^{n}}\exp(\epsilon|x|^{2})d\mu(x)<\infty.$

Moreover, for an absolutely continuous probability measure on
with respect to$of-\log(d\mu/dx)$
measure,uniformly
if
the Lebesgue
the Hessian
bounded below byis, then verifying a Gaussian isoperimetric inequality is also equivalent tosomeverifying a Gaussian concentration inequality. This was proved for a more generalprobability measure on a Riemannian manifold (see [7, Theorems 1.1, 1.2]), where thelower boundedness of the -Ricci curvature is used instead of the uniform logarithmic
concavity of a probability measure.
$\mathbb{R}^{n}$$\mu$$K\in \mathbb{R}$$\infty$55
Definition 1.5 Let $(M, g)$ be an -dimensional complete connected Riemannian manifold without boundary and fix an arbitrary measure$n$$\omega=e^{-f}vo1_{g}, f\in C^{\infty}(M)$where$[n, \infty]$,

denotes the Riemannian volume measure of $(M, g)$ . Given $N\in(-\infty, 0)\cup$, we define the $N$ -Ricci curvature of by
and$vo1_{g}$$K\in \mathbb{R}$$\omega$$Ric_{N}^{\omega}:=\{\begin{array}{ll}Ric+Hessf if N=\infty,Ric+Hessf-\frac{Df\otimes Df}{N-n} if N\in(-\infty, 0)\cup(n, \infty) ,Ric+Hessf-\infty\cdot(Df\otimes Df) if N=n,\end{array}$where by convention $\infty\cdot 0=0.$

We remark that the $N$ -Ricci curvature is originally defined only for $N\in[n, \infty]$ and
if $Ric_{N}^{\omega}(v, v)\geq Kg(v, v)$ holds for every tangent vector to $M$ and for some$N\in[n, \infty)$ thenbehaves like a Riemannian manifold with dimension bounded$N$above by and Ricci curvature bounded below by $K$ . We refer to [5],[10] and references
therein for the details, and to [9] for the case of $N\in(-\infty, 0)$ .
$K\in \mathbb{R},$$v$$(M,\omega)$

Probability measure on an admissible quadruple2

It is known that if the -Ricci curvature of is bounded below by some $K>0,$
then verffies a Gaussian isoperimetric inequality and hence a Gaussian concentration
inequality (for instance, see [8, Theorem 5]). It is then natural to ask what kind of
an isoperimetric inequality and a concentration inequality hold for a non-Gaussian
probability measure whose -Ricci curvature is not bounded from below. Moreover,under a suitable condition, are the two inequalities equivalent to each other? To discuss
this, we deal with the following condition (see [9, Definition 4.3], where the conditionis slightly different).
$\infty$$\omega$$\omega$$\infty$

Definition 2.1 We say that a quadruple
conditions hold:$\bullet$$\bullet$$M$$\varphi$$(M,\omega, \varphi, \Psi)$

is admissible if all the following

is an -dimensional complete connected Riemannian manifold with $n\geq 2.$$n$

is a non-decreasing, positive, continuous function on$(0, \infty)$

such that$\theta_{\varphi}:=\sup_{s>0}\{\frac{s}{\varphi(s)}\cdot\varlimsup_{\epsilon\downarrow 0}\frac{\varphi(s+\epsilon)-\varphi(s)}{\epsilon}\}\in(0, \frac{n+1}{n}]$and$\bullet$$\Psi$$\theta_{\varphi}\neq 1,3/2$

with $\varphi(1)=1.$

is a function on $M$ such that
$M_{\Psi}^{\varphi}:= \{x\in M \Psi(x)\in(-\int_{1}^{\infty}\frac{1}{\varphi(s)}ds, -\int^{0}\frac{1}{\varphi(s)}ds)\}\neq\emptyset$

56
and

$\Psi>-L_{\theta_{\varphi}}$hold, where we set$L_{\theta_{\varphi}};=\{\begin{array}{ll}(\theta_{\varphi}-1)^{-1} if \theta_{\varphi}>1,\infty if \theta_{\varphi}\leq 1.\end{array}$$\bullet$is a positive measure on $M$ satisfying
for every tangent vector to$\omega$$v$$Ric_{N}^{\omega}(v, v)\geq 0$

for

and

$N=(\theta_{\varphi}-1)^{-1}$$M_{\Psi}^{\varphi}.$Note that if is differentiable, thenis the upper bound of the differentiable coefficient of . We denote by
the quantity corresponding to the lower bound of thedifferentiable coefficient of , that is,$\theta_{\varphi}$$\varphi$$\delta_{\varphi}$$\varphi$$\varphi$$\delta_{\varphi}:=\inf_{s>0}\{\frac{s}{\varphi(s)}\cdot\varlimsup_{\epsilon\downarrow 0}\frac{\varphi(s+\epsilon)-\varphi(s)}{\epsilon}\}.$We also define the -exponential function by$\varphi$$\exp_{\varphi}(\tau) :=\sup\{t>0 l^{t}\frac{1}{\varphi(s)}ds\leq\tau\},$$:=0$ forwhere we set$q\neq1$, then we havewith$\tau\leq\int_{1}^{0}1/\varphi(s)ds$$\exp_{\varphi}(\tau)$by convention. Take for example, if$\varphi_{q}(s)=s^{q}$$\exp_{q}(\tau) :=\exp_{\varphi_{q}}(\tau)=(1+(1-q)\tau)_{+}^{1/(1-q)},$$:=\infty$ for $a<0$ .
Since
where we set $[ \tau]_{+};=\max\{\tau, 0\}$ and by convention$:=\exp(\tau)$$qarrow1$.
, we set
recovers the usual exponential function when
, then we may assume
We remark that if is $K$-convex for some $K>0$ onthat the measure
on an admissible quadruple
is a probability
case,
[9,
(see
In
4.5]).
generality
thisthe probability
Lemmameasure without loss ofmeasureverifies a non-Gaussian concentration inequality. Here the $K$
convexity of a function is roughly equivalent to that the Hessian of a function along
any geodesic is bounded below by $K$ (see [9, Definition 4.1] for the precise definition).
$0^{a}$$\exp_{q}$$\exp_{1}(\tau)$$\Psi$$M_{\Psi}^{\varphi}$$(M,\omega, \varphi, \Psi)$$\exp_{\varphi}(-\Psi)\omega$$\exp_{\varphi}(-\Psi)\omega$-, we set
Proposition 2.2 ([9, Theorem 7.9]) For an admissible quadruple$K$. Suppose the -convexity of forand,some $K>0$ and $\mu[M]=1.$$(M,\omega, \varphi, \Psi)$(i) If$\theta_{\varphi}$$\theta_{\varphi}<1$

and$\delta_{\varphi}$, then there exists a positive constantandsuch that we have for any $r>0$$\delta_{\varphi}>0$$C[ \mu](r)\leq c_{1}/\exp_{\delta_{\varphi}}(\frac{K}{4},0^{\varphi}\delta-1r^{2})$(ii) If
constants$\Psi$$0:= \max\{1, \Vert\exp_{\varphi}(-\Psi)\Vert_{\infty}\}$$\mu:=\exp_{\varphi}(-\Psi)\omega$$c_{3}$

depending only on.

positiveand if $\omega[M]<\infty$ , then there
such that we have for any $r>0$
depending only on
and$\theta_{\varphi}\in(1,3/2),$$c_{2},$$c_{1}$$e\dot{m}t$$\delta_{\varphi}>3(\theta_{\varphi}-1)$$\theta_{\varphi}$$\delta_{\varphi}$$C[p](r)\leq c_{2}\exp_{2\theta_{\varphi}-\delta_{\varphi}}(-c_{3_{0}^{\frac{K}{2},\omega[M]^{1-\theta_{\varphi}}r^{2})}}^{\delta_{\varphi}-\theta_{\varphi}}.$

57
Moreover, whenanda Gaussian concentmtion inequality.
$\varphi(s)=s^{q}$$\theta_{\varphi}=\delta_{\varphi}=qarrow 1$, the two inequalities above recover

A fundamental and important example of an admissible quadruple isequipped with the Lebesgue measure andwith $q\in(0, (n+1)/n]$ and$q\neq 1,3/2,$ $\Psi(x)=|x|^{2}/2$ . In this case, there exists a constant $c(n, q)$ such that$1+(1-q)c(n, q)>0$ and$\mathbb{R}^{n}(n\geq 2)$$\varphi_{q}(s)=s^{q}$$\int_{\mathbb{R}^{n}}\exp_{q}(-\frac{|x|^{2}}{2}+c(n, q))dx=1$$(see [12] and$ Section$3$

below $for the$ explicit value $of c(n, q)$ ). In addition,$B_{q}^{n}:= \{x\in \mathbb{R}^{n} \exp_{q}(-\frac{|x|^{2}}{2}+c(n, q))>0\}$contains the origin and is bounded (resp. unbounded) if $q<1$ (resp. $q>1$ ). An
absolutely continuous probability measureonwith the density$\mathbb{R}^{n}$$\gamma_{n}^{q}$$\frac{d\gamma_{n}^{q}}{dx}=\exp_{q}(-\frac{|x|^{2}}{2}+c(n, q))$with respect to the Lebesgue measure is called the -Gaussian measure. According
to [9, Theorem 5.7], the -Gaussian measure can be regarded as an extremal element among all the probability measureson an admissible quadruple
as well as the Gaussian measure among all the probability measures on a
Riemannian manifold whose -Ricci curvature is bounded from below.
In this way, it turns out that a probability measure
on an admissible
quadruple
with certain conditions verifies a non-Gaussian isoperimetricinequality characterized by, wheredepends on
and . In particular, if$\varphi(s)=s^{q}$ , then $q(\varphi)=q$
holds. However, as far as the author knows, the isoperimetricinequality for such a probability measure is not available in the literature, even for thecase of the -Gaussian measure.
$q$

$q$

$\exp_{\varphi}(-\Psi)\omega$$(M, \omega, \varphi, \Psi)$$\infty$$\exp_{\varphi}(-\Psi)\omega$$(M,\omega, \varphi, \Psi)$$\exp_{q(\varphi)}$$q(\varphi)$$\theta_{\varphi}$$\delta_{\varphi}$$q$3Properties of -Gaussian measure$\varphi$

In this section, we provide some properties of the -Gaussian measure, which are related
to the concentration profile and may be useful to investigate the isoperimetric profile.
We first discuss the logarithmic concavity of the -Gaussian measure.
$q$

$q$

Proposition 3.1 For any
function on the open set$n\in \mathbb{N}$and any $q\in(0, (n+1)/n]$ with $q\neq 3/2$ , define the$V_{q}$$B_{q}^{n}= \{x\in \mathbb{R}^{n} \exp_{q}(-\frac{|x|^{2}}{2}+c(n, q))>0\}$

58
$by$

$V_{q}(x):=- \log(\frac{d\gamma_{n}^{q}(x)}{dx})=-\log(\exp_{q}(-\frac{|x|^{2}}{2}+c(n, q)))$.

We moreover set $\lambda_{q}(n):=1+(1-q)c(n, q)>0$ . Then for the smallest eigenvalue
satisfiesof the Hessian $mat\dot{m}$ of at$V_{q}$$\lambda(x)$$x\in B_{q}^{n}$(3.1)$\lambda(x)\geq\{\begin{array}{ll}\frac{1}{\lambda_{q}(n)} if q\leq 1,-\frac{1}{8\lambda_{q}(n)} if q>1.\end{array}$Pmof. Consider the function on$B_{q}^{n}$of the form

$f_{q}(x):=1+(1-q)(- \frac{|x|^{2}}{2}+c(n, q))>0.$

We compute $f_{q}(0)=\lambda_{q}(n)$ and $\nabla f_{q}(x)=-(1-q)x$ . It follows from the relation$V_{q}=-\log(f_{q})/(1-q)$ that$\nabla V_{q}(x)=x/f_{q}(x)$,moreover that the $(i,j)$ -component of the Hessian matrix of$V_{q}$

at$x$

is given by$(1-q) \frac{x_{i}x_{j}}{f_{q}(x)^{2}}+\frac{\delta_{ij}}{f_{q}(x)},$

if $i\neq j$ . It is easy to check that all the eigenvalue of
and$1/f_{q}(0)=1/\lambda_{q}(n)$. In the case of $x\neq 0$ , letbe an orthogare$n,$is the eigenvector ofwith $v_{1}=x/|x|$ . Then, for $k=1,$
onal basis ofwhose eigenvalue is
where$\delta_{ii}=1$$\delta_{ij}=0$$(H_{ij}(0))_{1\leq i,j\leq n}$$\{v_{k}\}_{k=1}^{n}$$\mathbb{R}^{n}$$\ldots,$$v_{k}$$(H_{ij}(x))_{1\leq i,j\leq n}$$(1-q) \frac{|x|^{2}\delta_{1k}}{f_{q}(x)^{2}}+\frac{1}{f_{q}(x)}$In the case of $q\leq 1$ , it follows from$f_{q}\in(0, \lambda_{q}(n)]$

.

(3.2)

that

$(1-q) \frac{|x|^{2}}{f_{q}(x)^{2}}+\frac{1}{f_{q}(x)}\geq\frac{1}{f_{q}(x)}\geq\frac{1}{\lambda_{q}(n)}.$

For $q>1$ , we have$f_{q}\in[\lambda_{q}(n), \infty)$

and$\frac{1}{f_{q}(x)}\geq(1-q)\frac{|x|^{2}}{f_{q}(x)^{2}}+\frac{1}{f_{q}(x)}=\frac{\lambda_{q}(n)+(1-q)|x|^{2}/2}{(\lambda_{q}(n)-(1-q)|x|^{2}/2)^{2}}\geq-\frac{1}{8\lambda_{q}(n)}.$This completes the proof of the proposition.

$\square$59
Remark 3.2 (1) Note thatas $qarrow 1$ , and $\lambda(x)=\lambda_{1}(n)=1$ on. On one hand, (3.1) recovers. On the other hand, whenas(3.1) does not recovers, however (3.2) recovers $\lambda(x)=1.$(2) Given any $q\in(0, (n+1)/n]$ with $q\neq 1,3/2$ , let $N_{q}\in(-\infty, 0)\cup(n, oo)$ satisfy$1-q\geq 1/(N_{q}-n)$ . It then holds for any
andthat$\lambda_{q}(n)arrow\lambda_{1}(n)=1$$\mathbb{R}^{n}$$\lambda(x)\geq 1$$q\nearrow 1$$q\searrow 1,$$\lambda(x)\geq 1$$v\in \mathbb{R}^{n}$$x\in B_{q}^{n}$$HessV_{q}(x)(v, v)-\frac{DV_{q}(x)\otimes DV_{q}(x)(v,v)}{N_{q}-n}=(1-q)\frac{\langle v,x\rangle^{2}}{f_{q}(x)^{2}}+\frac{|v|^{2}}{f_{q}(x)}-\frac{\langle v,x\rangle^{2}}{(N_{q}-n)f_{q}(x)^{2}}$$\geq\frac{|v|^{2}}{f_{q}(x)}.$This implies that, for $q>1$ (henceis negative), the-Ricci curvature ofonequipped with the standard Euclidean metric is non-negative on the whole of, however little is known conceming a measure having the non-negative $N$-Ricci
curvature for some negative $N$ . For example, although a Poincar\’e type inequalities forare proved in [2], the condition $\omega(M)<\infty$ in Proposition 2.2(ii) does not hold forequipped with the Lebesgue measure and thenmay not verify a concentration
inequality in terms of the -exponential function.
On the other hand, for $q<1$ , the $N$-Ricci curvature ofequipped withon
the standard Euclidean metric is bounded below by $K$ on
if $N\geq n+(1-q)^{-1}$and $K\leq 1/f_{q}(0)$ . There are many study about a measure whose $N$-Ricci curvature isbounded from below for some positive $N$ , however we usually assume the positivity ofa measure and the completeness of a metric space.
$N_{q}$$N_{q}$$\gamma_{n}^{q}$$\mathbb{R}^{n}$$\mathbb{R}^{n}$$\gamma_{n}^{q}$$\mathbb{R}^{n}$$\gamma_{q}^{n}$$q$$\mathbb{R}^{n}$$\gamma_{n}^{q}$$B_{q}^{n}$

We finally estimate the smallest Lipchitz constant $L_{q}(n)$ ofwhich pushes forward
to . The existence of such a map
is guaranteed for any $q\in(0,1)$ andby [13, Section 4]. To do this, set$T_{n,q}$$\gamma_{n}$$T_{n,q}$$\gamma_{n}^{q}$$n\in \mathbb{N}$$R_{q}(n):= \sup\{r\in \mathbb{R} \exp_{q}(-\frac{r^{2}}{2}+c(n, q))>0\}=(\frac{2\lambda_{q}(n)}{1-q})^{1/2}<\infty.$

Proposition 3.3 For any $q\in(0,1)$ and$n\in \mathbb{N}$, we have$R_{q}(n)^{n+2/(1-q)}= \pi^{-n/2}(\frac{2}{1-q})^{1/(1-q)}\Gamma(\frac{n}{2}+\frac{2-q}{1-q})/\Gamma(\frac{2-q}{1-q})$,$R_{q}(n)^{2} \cdot\frac{(1-q)}{(n+2)(1-q)+2}\leq L_{q}(n)^{2},$whereProof.

$\Gamma$stands for the Gamma function.

The direct calculation gives$1= \int_{\mathbb{R}^{n}}d\gamma_{n}^{q}(x)=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{0}^{R_{q}(n)}\exp_{q}(-\frac{r^{2}}{2}+c(n, q))r^{n-1}dr$$= \frac{2\pi^{n/2}}{\Gamma(n/2)}\lambda_{q}(n)^{1/(1-q)}R_{q}(n)^{n}\int_{0}^{1}(1-\mathcal{S}^{2})^{1/(1-q)}s^{n-1}d_{\mathcal{S}}$$= \pi^{n/2}\lambda_{q}(n)^{1/(1-q)}R_{q}(n)^{n}\Gamma(\frac{2-q}{1-q})/\Gamma(\frac{n}{2}+\frac{2-q}{1-q})$,

60
which implies the first equality. Similarly, we compute$\int_{R^{n}}|x|^{2}d\gamma_{n}^{q}(x)=\frac{n\pi^{n/2}}{2}\lambda_{q}(n)^{1/(1-q)}R_{q}(n)^{n+2}\Gamma(\frac{2-q}{1-q})/\Gamma(\frac{n}{2}+\frac{2-q}{1-q}+1)$$=R_{q}(n)^{2} \cdot\frac{n(1-q)}{(n+2)(1-q)+2}$

On the other hand, by the definition of the push-forward measure, we have$\int_{R^{n}}|x|^{2}d\gamma_{n}^{q}(x)=\int_{R^{n}}|T_{n,q}(x)|^{2}d\gamma_{n}(x)\leq\int_{R^{n}}L_{q}(n)^{2}|x|^{2}d\gamma_{n}(x)=nL_{q}(n)^{2}.$Combining the these implies$R_{q}(n)^{2} \cdot\frac{(1-q)}{(n+2)(1-q)+2}\leq L_{q}(n)^{2}.$$\square$From [13, Theorem 1.2] we deduce the another estimate of $L_{q}(n)$$(2 \pi)^{1/2}L_{q}(n)\geq\lambda_{q}(n)^{-1/n(1-q)}=(\frac{1-q}{2}R_{q}(n)^{2})^{-1/n(1-q)}$$= \pi^{1/2}R_{q}(n)[\Gamma(\frac{2-q}{1-q})/\Gamma(\frac{n}{2}+\frac{2-q}{1-q})]^{1/n}$

where the equalities follow from the equality in Proposition 3.3. This estimate is better
than the estimate in Proposition 3.3. For simplicity, let us consider the case of $n=2k.$
We then have$(k+1+ \frac{1}{1-q})^{k}\geq\prod_{j=1}^{k}(k+1-j+\frac{1}{1-q})=\Gamma(k+\frac{2-q}{1-q})/\Gamma(\frac{2-q}{1-q})$,which implies$\frac{R_{q}(2k)^{2}1-q}{2(k+1)(1-q)+1}\leq\frac{R_{q}(2k)^{2}}{2}[\Gamma(\frac{2-q}{1-q})/\Gamma(k+\frac{2-q}{1-q})]^{1/k}$

The asymptotic behavior of $L_{q}(2k)$ as $karrow\infty$ is unknown, however we have$(2 \pi)^{1/2}L_{q}(2k)\geq(\frac{1-q}{2}R_{q}(2k)^{2})^{-1/2k(1-q)}=\pi^{1/a_{k}}(\frac{2}{1-q})^{1/a_{k}}P_{k}^{-1/a_{k}}arrow 1$

as$karrow\infty$, where we set$P_{k}:=[ \prod_{=1}^{k}(k-j+\frac{2-q}{1-q})]^{1/k}\in[1+\frac{1}{1-q},$ $\frac{a_{k}}{2(1-q)}],$$a_{k}:=2(k(1-q)+1)$ .

61
It thus is enough to show
follows from the observation that

$P_{k}^{-1/a_{k}}arrow 1$, or equivalently$\log P_{k}^{-1/a_{k}}arrow 0$, as $karrow\infty$ . This

$0= \lim_{karrow\infty}\frac{-1}{a_{k}}\log\frac{a_{k}}{2(1-q)}\leq\lim_{karrow\infty}\log P_{k}^{-1/a_{k}}\leq\lim_{karrow\infty}\frac{-1}{a_{k}}\log(1+\frac{1}{1-q})=0.$

This suggests that, for $q\in(O, 1)$ , the familyof the -Gaussian measures may
not have the L\’evy property (for instance, see [4, Section 3.3] about the definition of
the L\’evy property) and then suggests how difficult and interesting to investigate theasymptotic behavior of the concentration profiles of$\{\gamma_{n}^{q}\}_{n\in \mathbb{N}}$$q$$\{\gamma_{n}^{q}\}_{n\in \mathbb{N}}.$

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Asuka TAKATSU
Graduate School of Mathematics, Nagoya University
Nagoya 464-8602

JAPANE–mail address: takatsu@math.nagoya-u.ac.jp