On the infimum convolution inequality

Introduction and Notation

In the seminal paper , B. Maurey introduced the so called property $(\tau)$ for a probability measure ${\mu}$ with a cost function $\varphi$ (see Definition 2.1 below) and established a very elegant and simple proof of Talagrand’s two level concentration for the product exponential distribution $\nu^{n}$ using $(\tau)$ for this distribution and an appropriate cost function $w$.

It is natural to ask what other pairs $(\mu,\varphi)$ have property $(\tau)$? As any $\mu$ satisfies $(\tau)$ with $\varphi\equiv 0$, one will rather ask how big a cost function can one take. In this paper we study the probability measures $\mu$ that have property $(\tau)$ with respect to the largest (up to a multiplicative factor) possible convex cost function $\Lambda^{\star}_{\mu}$. This bound comes from checking property $(\tau)$ for linear functions. We say a measure satisfies the infimum convolution inequality ($IC$ for short) if the pair $(\mu,\Lambda^{\star}_{\mu})$ satisfies $\tau$.

It turns out that such an optimal infimum convolution inequality has very strong consequences. It gives the best possible concentration behaviour, governed by the so–called $L_{p}$-centroid bodies (Corollary 3.10). This, in turn, implies in particular a weak–strong moment comparison (Proposition 3.12), the Central Limit Theorem of Klartag and the tail estimates estimates of Paouris (Proposition 3.15). We believe that $IC$ holds for any log–concave probability measure, which is the main motivation for this paper.

Organization of the paper. This section, apart from the above introduction, defines the notation used throughout the paper. The second section is devoted to studying the general properties of the inequality $IC$. In subsection 2.1 we recall the definition of property $(\tau)$ and its ties to concentration from . In subsection 2.2 we study the opposite implication — what additional assumptions one needs to infer $(\tau)$ from concentration inequalities. In subsection 2.3 we show that $\Lambda^{\star}_{\mu}$ is indeed the largest possible cost function and define the inequality $IC$. In subsection 2.4 we show that product log–concave measures satisfy $IC$.

In the third section we give more attention to the concentration inequalities tied to $IC$. In subsection 3.1 we show the connection to $\mathcal{Z}_{p}$ bodies. In subsection 3.2 we continue in this vein with the additional assumption our measure is $\alpha$–regular. In subsection 3.3 we show how $IC$ implies a comparison of weak and strong moments and the results of and .

In the fourth section we give a modification of the two–level concentration for the exponential measure, in which for sets lying far away from the origin only an enlargement by $tB_{1}^{n}$ is used. This will be used in the fifth section, which focuses on the uniform measure on the $B_{p}^{n}$ ball. In subsection 5.1 we define and study two rather standard transports of measure used further on. In subsection 5.2 we use these transports along with the concentration from section 4 and a Cheeger inequality from to give a proof of $IC$ for $p\leq 2$. In section 5.3 we show a proof of $IC$ for $p\geq 2$ and a proof of the log–Sobolev inequality for $p\geq 2$.

We conclude with a few possible extensions of the results of the paper in the sixth section.

The letters $c,C$ denote absolute numerical constants, which may change from line to line. By $c(p),C(p)$ we mean constants dependent on $p$ (or, formally, a family of absolute constants indexed by $p$), these also may change from line to line. Other letters, in particular greek letters, denote constants fixed for a given proof or section. For any sets of positive real numbers $a_{i}$ and $b_{i}$, $i\in I$, by $a_{i}\sim b_{i}$ we mean there exist absolute numerical constants $c,C>0$ such that $ca_{i}<b_{i}<Ca_{i}$ for any $i\in I$. Similarly for collections of sets $A_{i}$ and $B_{i}$ by $A_{i}\sim B_{i}$ we mean $cA_{i}\subset B_{i}\subset CA_{i}$ for any $i\in I$, where again $c,C>0$ are absolute numerical constants. By $\sim_{p}$ we mean the constants above can depend on $p$.

Infimum convolution inequality

denotes the infimum convolution of $f$ and $g$.

The following two easy observations are almost immediate (c.f. ):

If pairs $(\mu_{i},\varphi_{i})$, $i=1,\ldots,k$ have property $(\tau)$ and $\varphi(x_{1},\ldots,x_{k})=\varphi_{1}(x_{1})+\ldots+\varphi_{k}(x_{k})$, then the couple $(\otimes_{i=1}^{k}\mu_{i},\varphi)$ also has property $(\tau)$.

Then the pair $(\mu\circ T^{-1},\psi)$ has property $(\tau)$.

Maurey noticed that property $(\tau)$ implies $\mu(A+B_{\varphi}(t))\geq 1-\mu(A)^{-1}e^{-t},$ where

We will need a slight modification of this estimate.

Property $(\tau)$ for $(\varphi,\mu)$ implies for any Borel set $A$ and $t\geq 0$,

Thus from property $(\tau)$ for $f$ we have

from which, extracting the condition upon ${\mu}(A+B_{\varphi}(t))$ by direct calculation, we get (2).

Let $f_{t}(p):=e^{t}p/((e^{t}-1)p+1)$, notice that $f_{t}$ is increasing in $p$ and for $p\leq e^{-t/2}/2$,

hence $f_{t}(p)>\min(e^{t/2}p,1/2)$ and (3) follows. Moreover for $p\geq 1/2$

Let $F(x)=\nu(-\infty,x]$ and $g_{t}(p)=F(F^{-1}(p)+t)$. Previous calculations show that for $t,p>0$, $f_{t}(p)\geq g_{t/2}(p)$ if $F^{-1}(p)+t/2\leq 0$ or $F^{-1}(p)\geq 0$. Since $g_{t+s}=g_{t}\circ g_{s}$ and $f_{t+s}=f_{t}\circ f_{s}$, we get that $f_{t}(p)\geq g_{t/2}(p)$ for all $t,p>0$, hence (2) implies (5). ∎

The main theorem of states that $\nu$ satisfies ($\tau$) with a sufficiently chosen cost function.

Let $w(x)=\frac{1}{36}x^{2}$ for $|x|\leq 4$ and $w(x)=\frac{2}{9}(|x|-2)$ otherwise. Then the pair $(\nu^{n},\sum_{i=1}^{n}w(x_{i}))$ has property $(\tau)$.

Theorem 2.5 together with Proposition 2.4 immediately gives the following two-level concentration:

that was first established (with different universal, rather large constants) by Talagrand .

2 From concentration to property (τ)𝜏(\tau)

Proposition 2.4 shows that property $(\tau)$ implies concentration, the next result presents the first approach to the converse implication.

Then the pair $(\mu,\frac{1}{36}\varphi(\frac{\cdot}{\beta}))$ has property $(\tau)$. In particular if $\varphi$ is convex, symmetric and $\varphi(0)=0$ then (7) implies property $(\tau)$ for $(\mu,\varphi(\frac{\cdot}{36\beta}))$.

To finish the proof of the first assertion, by Theorem 2.5 it is enough to show that

Since the set $A(g\Box w,u)$ is a halfline, it is enough to prove that

Let us fix $x_{1}$ and $x_{2}$ with $g(x_{1})+w(x_{2})<u$ and take $s_{1}>g(x_{1})$ $s_{2}=w(x_{2})$ with $s_{1}+s_{2}<u$. Put $A:=A(f,s_{1})$, then $\mu(A)=\nu(A(g,s_{1}))\geq\nu(-\infty,x_{1}]$. By the definition of $w$ it easily follows that $x_{2}\leq\max\{6\sqrt{s_{2}},9s_{2}\}$, hence by (7), $\mu(A+\beta B_{\varphi}(36s_{2}))\geq\nu(-\infty,x_{1}+x_{2}].$ Since

The last part of the statement immediately follows since any symmetric convex function $\varphi$ is radius-wise nondecreasing and if additionally $\varphi(0)=0$, then $\varphi(x/36)\leq\varphi(x)/36$ for any $x$. ∎

The next proposition shows that inequalities (3) and (4) are strongly related.

The following two conditions are equivalent for any Borel set $K$ and $\gamma>1$,

and we get the contradiction with (10). ∎

Let us fix the set $A$ with $\mu(A)=\nu(-\infty,x]$. Notice that $A+2K=A+K+K\supset A+K$. If $x+t\leq 0$, then $\mu(A+K)>e^{t}\mu(A)=\nu(-\infty,x+t]$. If $x\geq 0$, Proposition 2.7 gives

Finally, if $x\leq 0\leq x+t$, we get $\mu(A+K)\geq 1/2=\nu(-\infty,0]$, hence by the previous case,

Corollary 2.8 shows that if the cost function $\varphi$ is symmetric and convex, condition (7) (with $2\beta$ instead of $\beta$) for $t\geq 1$ is implied by the following:

To treat the case $t\leq 1$ we will need Cheeger’s version of the Poincaré inequality.

It is not hard to check that Cheeger’s inequality (cf. [6, Theorem 2.1]) implies

Finally, we may summarize this section with the following statement.

Suppose that the cost function $\varphi$ is convex, symmetric with $\varphi(0)=0$ and $1\wedge\varphi(x)\leq(\alpha|x|)^{2}$ for all $x$. If the measure $\mu$ satisfies Cheeger’s inequality with the constant $\beta=1/\delta$ and the condition (11) is satisfied for all $t\geq 1$ and $C=\gamma$ then $(\mu,\varphi(\cdot/C))$ has property $(\tau)$ with the constant $C=36\min\{2\gamma,\alpha\delta\}$.

Notice that $\alpha B_{\varphi}(t)\supset\sqrt{t}B_{2}^{n}$ for all $t<1$, hence Cheeger’s inequality implies that condition (7) holds for $t<1$ with $C=\alpha\delta$. Therefore (7) holds for all $t\geq 0$ with $C=\min\{2\gamma,\alpha\delta\}$ and the assertion follows by Corollary 2.6. ∎

3 Optimal cost functions

A natural question arises: what other pairs $(\mu,\varphi)$ have property ($\tau$)? First we have to choose the right cost function. To do this let us recall the following definitions.

The Legendre transform of any function is a convex function. If $f$ is convex and lower semi-continuous, then $\mathcal{L}\mathcal{L}f=f$, and otherwise $\mathcal{L}\mathcal{L}f\leq f$. In general, if $f\geq g$, then $\mathcal{L}f\leq\mathcal{L}g$. The Legendre transform satisfies $\mathcal{L}(Cf)(x)=C\mathcal{L}f(x/C)$ and if $g(x)=f(x/C)$, then $\mathcal{L}g(x)=\mathcal{L}f(Cx)$. For this and other properties of $\mathcal{L}$, cf. . The Legendre transform has been previously used in the context of convex geometry, see for instance and .

The function $\Lambda^{\star}_{\mu}$ plays a crucial role in the theory of large deviations cf. .

Take $f(x)=\left\langle x,v\right\rangle$. Then

where the last equality uses the fact that ${\mu}$ is symmetric. Thus by taking the logarithm we get $\mathcal{L}\varphi(v)\geq 2\Lambda_{\mu}(v)$, and by applying the Legendre transform we obtain $\varphi(v)=\mathcal{L}\mathcal{L}\varphi(v)\leq 2\Lambda^{\star}_{\mu}(v/2).$ The inequality $2\Lambda^{\star}_{\mu}(v/2)\leq\Lambda^{\star}_{\mu}(v)$ follows by the convexity of $\Lambda_{\mu}^{\star}$. ∎

The above remark motivates the following definition.

which substituted into (13) gives the thesis. ∎

Direct calculation shows that $\Lambda_{\nu}(x)=-\ln(1-x^{2})$ for $|x|<1$ and

Since $a/2\leq a-\ln(1+a/2)\leq a$ for $a\geq 0$, we get $\frac{1}{2}(\sqrt{1+x^{2}}-1)\leq\Lambda_{\nu}^{\star}(x)\leq\sqrt{1+x^{2}}-1$. Finally

The last statement follows by Theorem 2.5, since $\min((x/9)^{2},|x|/9)\leq w(x)$. ∎

4 Logaritmically concave product measures

It is easy to check that for any measure ${\mu}$ with a full–dimensional support there exists a linear map $L$ such that ${\mu}\circ L^{-1}$ is isotropic.

The next theorem (with a different universal, but rather large constant) may be deduced from the results of Gozlan . We give the following, relatively short proof for the sake of completeness.

then $g(x)\leq e^{-x/a}g(0)$ for $x>a$ and $g(x)\geq e^{-x/a}g(0)$ for $x\in[0,a)$. Therefore

Notice that $T^{\prime}(0)=1/(2g(0))\leq 4$, thus by concavity of $T$, $Tx\leq 4x$ for $x\geq 0$. Moreover for $x\geq 0$, $h(Tx)=x+\ln 2$.

where $w(x)$ is as in Theorem 2.5. We have two cases. i) $Tx\leq 16$, then

We expect that in fact a more general fact holds.

Concentration inequalities.

Sets ${\cal Z}_{p}(\mu_{K})$ for $p\geq 1$, when $\mu_{K}$ is the uniform disribution on the convex body $K$ are called $L_{p}$-centroid bodies of $K$, their properties were investigated in .

Let us take $v\in{\cal Z}_{p}(\mu)$, we need to show that $\Lambda_{\mu}^{\star}(v/(2^{1/p}e))\leq p$, that is

i) $\beta\leq 2^{1/p}ep$. Then, since $\Lambda_{\mu}(u)\geq\int\langle u,x\rangle d\mu(x)=0$,

Hence $\Lambda_{\mu}(2^{1/p}epu/\beta)\geq p$ and $\Lambda_{\mu}(u)\geq\frac{\beta}{2^{1/p}ep}\Lambda_{\mu}(2^{1/p}epu/\beta)\geq\frac{\beta}{2^{1/p}e}$. Therefore

Using the symmetry and isotropicity of $\mu$, we get

where to get the last inequality we used $\ln(1+x)\leq x$ for $x>-1$. ∎

2 α𝛼\alpha-regular measures.

To establish inlusions opposite to those in the previous subsection, we introduce the following property:

If $\mu$ is $\alpha$-regular for some $\alpha\geq 1$, then for any $p\geq 2$,

Take any $v\notin 4e\alpha{\cal Z}_{p}(\mu)$, then we may find $u\in{\cal M}_{p}(\mu)$ such that $\left\langle v,u\right\rangle>4e\alpha$ and obtain

If $\mu$ is symmetric, isotropic $\alpha$-regular for some $\alpha\geq 1$, then

We have by the symmetry, isotropicity and regularity of $\mu$,

so $\Lambda_{\mu}(u)\leq\alpha^{2}e^{2}|u|^{2}/2$ for $\alpha e|u|\leq 1$. Thus $\Lambda_{\mu}^{*}(u)\geq\min\{\frac{|u|}{2\alpha e},\frac{|u|^{2}}{2\alpha^{2}e^{2}}\}$ for all $u$. ∎

We always have for $p\geq q$, ${\cal M}_{p}(\mu)\subset{\cal M}_{q}(\mu)$ and ${\cal Z}_{q}(\mu)\subset{\cal Z}_{p}(\mu)$. If the measure $\mu$ is $\alpha$-regular, then ${\cal M}_{q}(\mu)\subset\frac{\alpha p}{q}{\cal M}_{p}(\mu)$ and ${\cal Z}_{p}(\mu)\subset\frac{\alpha p}{q}{\cal Z}_{q}(\mu)$ for $p\geq q\geq 2$. Moreover for any symmetric measure $\mu$, $\Lambda_{\mu}^{*}(0)=0$, hence by the convexity of $\Lambda_{\mu}^{*}$, $B_{q}(\mu)\subset B_{p}(\mu)\subset\frac{p}{q}B_{q}(\mu)$ for all $p\geq q>0$.

Symmetric log–concave measures are 1-regular.

so it is enough to show that the function $f(x):=\frac{1}{x}(\Gamma(x+1))^{1/x}$ is nonincreasing on $[2,\infty)$. Binet’s form of the Stirling formula (cf. [1, Theorem 1.6.3]) gives

where $\mu(x)=\int_{0}^{\infty}{\rm arctg}(t/x)(e^{2\pi t}-1)^{-1}dt$ is decreasing function. Thus

is indeed nonincreasing on $[2,\infty)$. ∎

This definition is motivated by the following Corollary:

By Remark 3.7, Proposition 2.4 and the definition of $B_{p}(\mu)$ we have

so the first part of the statement immediately follows by Proposition 3.5.

By Proposition 2.7 this implies property (11). Additionally $\Lambda^{\star}_{{\mu}}$ is convex, symmetric and $\Lambda^{\star}_{\mu}(0)=0$. Finally, from Proposition 3.3 we have $\min\{1,\Lambda^{\star}_{\mu}(u)\}\leq|u|^{2}$. Thus, from Proposition 2.9 we get the second part of the statement.

By Proposition 2.7 in the definition 3.9 we could use the equivalent condition $\mu(A+\beta{\cal Z}_{p}(\mu))\geq\min\{e^{p}\mu(A),1/2\}$. The next proposition shows that for log-concave measures these conditions are satisfied for large $p$ and for small sets.

Using a standard volumetric estimate for any $r>0$ we may choose $S\subset{\cal M}_{r}(\mu)$ with $\#S\leq 5^{n}$ such that ${\cal M}_{r}(\mu)\subset\bigcup_{u\in S}(u+\frac{1}{2}{\cal M}_{r}(\mu))$. Then for $t>0$,

Let $\mu(A)=e^{-q}$, we will consider two cases.

i) $p\geq\max\{q,cn\}$. Then by Remark 3.7,

so $A\cap 30c^{-1}{\cal Z}_{p}(\mu)\neq\emptyset$, hence $0\in A+30c^{-1}{\cal Z}_{p}(\mu)$ and

The previous facts motivate the following.

Proposition 3.10 shows that Conjecture 1 implies Conjecture 2. Both hypotheses would be equivalent provided that the following conjecture of Kannan, Lovász and Simonovits holds.

There exists an absolute constant $C$ such that any symmetric isotropic log–concave probability measure satisfies Cheeger’s inequality with constant $1/C$.

3 Comparison of weak and strong moments

where $\|\cdot\|_{*}$ denotes the norm dual to $\|\cdot\|$.

hence $\|x\|\leq M+tm_{p}$ for $x\in A+t{\cal Z}_{p}(\mu)$. Thus for $t\geq p$,

Under the assumptions of Proposition 3.12 by the triangle inequality we get for $\gamma=4\alpha\beta$,

Notice that $\int\|x\|_{2}^{2}d\mu=n$ and $\|u\|_{2}^{*}=\|u\|_{2}$. Hence i) follows directly from (19) with $p=q=2$. Moreover (19) with $q=2$ implies

by the $\alpha$-regularity and isotropicity of $\mu$. ∎

Property i) plays the crucial role in the Klartag proof of the central limit theorem for convex bodies . Paouris showed that moments of the Euclidean norm for symmetric isotropic log-concave measures are bounded by $C(p+\sqrt{n})$. Thus Conjecture 4 would imply both Klartag CLT (with the optimal speed of convergence) and Paouris concentration.

We conclude this section with the estimate that shows comparison of weak and strong moments for any probability measure and $p>n/C$.

As in the proof of Proposition 3.11 we can find $u_{1},\ldots,u_{N}$ with $\|u_{i}\|_{*}\leq 1$, $N\leq 5^{n}$ such that $\|x\|\leq 2\max_{i\leq N}\left\langle u_{i},x\right\rangle$ for all $x$. Then

Modified Talagrand concentration for exponential measure

In this section we show that for a set lying far from the origin Talagrand’s two level concentration for the exponential measure may be somewhat improved, namely (for sufficiently large $t$) it is enough to enlarge the set by $tB_{1}^{n}$ instead of $tB_{1}^{n}+\sqrt{t}B_{2}^{n}$.

If $u\geq t>0$ then for any $i\in\{1,\ldots,n\}$ we have

Obviously we may assume that $i=1$ and $u\leq n$. Let $A_{1}:=A\cap nB_{1}^{n}\cap\{x\colon x_{1}\geq u\}$ and $B:=\{x\in B_{1}^{n}:x_{1}\geq\sum_{i\geq 2}|x_{i}|\}$. From the definition of $B$ and $A_{1}$ we have $A_{1}-tB\subset nB_{1}^{n}$. On the other hand $B=\{x:|x_{1}-1/2|+\sum_{i\geq 2}|x_{i}|\leq 1/2\}$, so $|B|=2^{-n}|B_{1}^{n}|=(2r_{1,n})^{-n}$. Thus

Then we easily check that $|tB/(1-s)|=|A_{1}/s|$. Since $A_{1}\subset\{x\in nB_{1}^{n}\colon x_{1}\geq t\}$ we get $|A_{1}|^{1/n}\leq(n-t)/r_{1,n}$ and $s\leq 2(n-t)/(2n-t)$. Now we can use the Brunn-Minkowski inequality to get

Notice that $A_{1}+tB_{1}^{n}\subset\{x\colon x_{1}\geq u-t\}$, so we obtain

A similar result (although with a constant multiplicative factor) can be obtained using the same technique and more calculations for $n^{1/p}B_{p}^{n}$ instead of $nB_{1}^{n}$ for $p\in$.

If $u\geq t>0$ then for any $i\in\{1,\ldots,n\}$ we have

and also $P^{-1}(A)+B_{1}^{n+k}\subset P^{-1}(A+B_{1}^{n})$. From Lemma 4.1 we have

Let $A_{t}=A+tB_{1}^{n}$. By Lemma 4.3 we get for any $s\geq 0$ and any $i$:

To get the assertion it is enough to take the sum over all $i$ and notice that the function $f(y):=(\sqrt{y}-t)_{+}^{2}$ is convex on $[0,\infty)$, hence

Then $A_{k}+tB_{1}^{n}\subset\{x\colon 5t\sqrt{n}+t(2k-3)\leq|x|<5t\sqrt{n}+t(2k+1)\}$, hence

From Proposition 4.4 applied for $A_{k}$ we have

In particular $(\ref{l1_enl})$ holds if $A\cap(50\sqrt{n}B_{2}^{n}+tB_{1}^{n})=\emptyset$.

Let $A_{k}$ denote $A+10kB_{1}^{n}$ for $k=0,1,\ldots$. If for any $0\leq k\leq t/10$ we have $\nu^{n}(A_{k}\cap 50\sqrt{n}B_{2}^{n})\geq\nu^{n}(A)/2$, the thesis is proved. Thus we assume otherwise. Let $A_{k}^{\prime}:=A_{k}\setminus 50\sqrt{n}B_{2}^{n}$. From Lemma 4.5 we have

By a simple induction we get $\nu^{n}(A_{k})\geq e^{2k}\nu^{n}(A)$ for any $k\leq t/10$. Thus we get

where the last part follows from Stirling’s formula. Thus ${r_{p,n}}\sim n^{1/p}$.

where $f_{p}(t)=t^{2}$ for $t<1$ and $f_{p}(t)=t^{p}$ for $t\geq 1$.

We shall use the facts proved in Section 3 to approximate $B_{t}({\nu_{p}})$. Note that ${\nu_{p}}$ is log-concave (as its density is log-concave) and symmetric. It is 1–regular from Proposition 3.8. Also

Thus $Z_{t}({\nu_{p}})\sim[-t^{1/p},t^{1/p}]$ for $|t|\geq 1$, so by Propositions 3.2 and 3.5, $B_{t}({\nu_{p}})\sim[-t^{1/p},t^{1/p}]$. Hence, for all $t\geq 0$ we have $\{x\colon f_{p}(|x|)\leq t\}\sim\{x:\Lambda^{\star}_{\nu_{p}}(x)\leq t\}$, so $\Lambda^{\star}_{\nu_{p}}(t/C)\leq f_{p}(t)\leq\Lambda^{\star}_{{\nu_{p}}}(Ct)$. As $\Lambda^{\star}_{\nu_{p}}$ is symmetric, the proof is finished. ∎

For $t<1$ we use Propositions 3.3 and 3.6. Both ${\mu_{p,n}}$ and ${\nu_{p}^{n}}$ are symmetric, log–concave measures, and both can be rescaled as in Proposition 5.1 to be isotropic, thus $B_{t}({\mu_{p,n}})\sim\sqrt{t}B_{2}^{n}\sim B_{t}({\nu_{p}^{n}})$.

Lemma 6 from gives (after rescaling by ${r_{p,n}}$),

It is not hard to verify that $B_{t}({\mu_{p,n}})\sim{r_{p,n}}B_{p}^{n}$ for $t\geq n$.

We are now going to investigate two transports of measure. They will combine to transport a measure with known concentration properties ($\nu^{n}$ or $\nu_{2}^{n}$, that is the exponential or Gaussian measure) to the uniform measure ${\mu}_{p,n}$. We will investigate the contractive properties of these transports with respect to various norms. Our motivation is the following:

Let us prove the first statement, the second proof is almost identical. Suppose $U(x)\in U(A)+\delta^{1/p}t^{1/p}B_{p}^{n}$. Then there exists $y\in A$ such that $\|U(x)-U(y)\|_{p}^{p}\leq\delta t.$ From the assumption we have $t\geq\|x-y\|_{q}^{q}$, which means $x\in A+t^{1/q}B_{q}^{n}$, and $U(x)\in U(A+t^{1/q}B_{q}^{n})$. ∎

Let us first show the following simple estimate.

Then $f(0)=0$ and $f^{\prime}(u)=e^{-u}u^{q}(1-2u/q+2/q)\geq 0$ for $0\leq u\leq q/2$. ∎

Now we are ready to state the basic properties of $T_{p,n}$.

i) The map $T_{p,n}$ transports the probability measure $\nu_{p}^{n}$ onto the measure ${\mu}_{p,n}$. ii) For all $t>0$ we have $e^{-t^{p}/n}t\leq 2\gamma_{p}f_{p,n}(t)\leq t$ and $f_{p,n}^{\prime}(t)\leq(2\gamma_{p})^{-1}\leq 1$. iii) For any $t>0$, $0\leq f_{p,n}(t)/t-f_{p,n}^{\prime}(t)\leq\min\{1,2pt^{p}/n\}$. iv) The function $t\mapsto f_{p,n}(t)/t$ is decreasing on $(0,\infty)$ and for any $s,t>0$,

The definition of $T_{p,n}$ directly implies i). Differentiation of (22) gives

which, when the $n$-th root is taken, give the first part of ii).

For the second part of ii) we use (23) and the estimate above to get

To show iii) first notice that by (23) and ii),

thus $f_{p,n}(t)/t-f_{p,n}^{\prime}(t)\geq 0$. Moreover by ii), $f_{p,n}(t)/t-f_{p,n}^{\prime}(t)\leq f_{p,n}(t)/t\leq 1$, so we may assume that $2pt^{p}/n\leq 1$. By (22) and Lemma 5.7 we obtain

Thus using again (23) and part ii) we get

By iii) we get $(f_{p,n}(t)/t)^{\prime}\leq 0$, which proves the first part of iv). For the second part suppose that $s>t>0$, then

The next Proposition may be also deduced (with different constant) from the more general fact proved in .

Assume $s:=\|x\|_{p}\geq t:=\|y\|_{p}$, we apply Proposition 5.8 and get

Let $s=\|x\|_{p}$ and $t=\|y\|_{p}$, we use Proposition 5.8 as in the proof of Proposition 5.9, and the Hölder inequality,

The second transport we will use is a simple product transport which transports the measure $\nu_{p}^{n}$ onto $\nu_{q}^{n}$. We shall be particularly interested in the cases $p=1$ and $p=2$, but most of the results can be stated in the more general setting.

Note that $w_{p,q}^{-1}=w_{q,p}$ and $(W_{p,q}^{n})^{-1}=W_{q,p}^{n}$. Differentiating equality (24) we get

We will prove that $w_{p,q}$ behaves very much like $x^{p/q}$ for large $x$, and is more or less linear for small $x$. We begin with the bound for $q=1$.

For $p\geq 1$ we have i) $v_{p}(x)\geq x^{p}+\ln(p\gamma_{p}x^{p-1})$ and $v_{p}^{\prime}(x)\geq px^{p-1}$ for $x\geq 0$, ii) $v_{p}(x)\leq e+x^{p}+\ln(p\gamma_{p}x^{p-1})$ and $v_{p}^{\prime}(x)\leq e^{e}px^{p-1}$ for $x\geq 1$, iii) $|v_{p}(x)-v_{p}(y)|\geq 2^{1-p}|x-y|^{p}$.

Note that $\gamma_{1}=1$. We have for $x\geq 0$,

and for $x\geq 1$, since $(1+r/p)^{p}\leq e^{r}\leq 1+er$ for $r\in$, we get

Notice that by (25), $v_{p}^{\prime}(x)=e^{-x^{p}+v_{p}(x)}/\gamma_{p}$, hence we may estimate $v_{p}^{\prime}$ using the just derived bounds on $v_{p}$.

The lower bound on $v_{p}^{\prime}$ yields $|v_{p}(x)-v_{p}(y)|\geq|x-y|^{p}$ for $x,y\geq 0$. The same estimate holds for $x,y\leq 0$, since $v_{p}$ is odd. Finally for $x\geq 0\geq y$ we have

i) For $p\geq q\geq 1$, $|w_{p,q}(x)|\geq|x|^{p/q}$ and $w_{p,q}^{\prime}(x)\geq\frac{\gamma_{q}}{\gamma_{p}}\geq\frac{1}{2}$. ii) For $p\geq 2$, $w_{p,2}^{\prime}(x)\geq\frac{1}{8}\sqrt{p}|x|^{p/2-1}.$

Since the function $w_{p,q}$ is odd, we may and will assume that $x\geq 0$.

i) We have by the monotonicity of $u^{p/q-1}$ on $[0,\infty)$,

thus $w_{p,q}(x)^{q/p}\geq x$ and $w_{p,q}(x)\geq x^{p/q}$. Formula (25) gives $w_{p,q}^{\prime}(x)\geq\gamma_{q}/\gamma_{p}\geq 1/2$.

ii) We begin by the following Gaussian tail estimate for $z>0$:

We have equality when $z{\rightarrow}\infty$, and direct calculation shows the derivative of the left–hand–side is no larger than the derivative of the right–hand–side.

Let $\kappa:=4\sqrt{\pi}$, we will now show that for all $x>0$ and $p\geq 2$,

Suppose on the contrary that $w_{p,2}(x)<u_{p}(x)$ for some $p\geq 2$ and $x>0$. Note that by i) we have $w_{p,2}^{\prime}\geq\gamma_{2}/\gamma_{p}\geq\gamma_{2}=\sqrt{\pi}/2$. Thus $u_{p}(x)$ is equal to the second part of the maximum. This in particular implies that $x\geq 2/3$, since for $x<2/3$ we have

Therefore $u_{p}(x)\geq\sqrt{\pi}x/2\geq 1/\sqrt{3}$. Now by (26), (24) and (27),

After simplifying this gives $u_{p}(x)>\sqrt{p}x^{p/2}$. Hence

which is impossible. This condratiction shows that (28) holds.

Thus we have $w_{p,2}(x)\geq u_{p}(x)$ and by (25) we obtain

By taking $u_{p}(x)=\max\{\sqrt{\pi}x/2,\sqrt{(x^{p}+\ln(px^{p/2-1}/(\kappa\ln p)))_{+}}\}$ for sufficiently large $\kappa$ and estimating carefully one may arrive at the bound $w_{p,2}^{\prime}(x)\geq C^{-1}px^{p/2-1}/\ln p$. One cannot, however, receive a bound of the order of $px^{p/2-1}$.

Property i) follows from the definition of $w_{q,p}$ and $W_{q,p}^{n}$. Since $w_{q,p}=w_{p,q}^{-1}$ we get ii) by Lemma 5.13 i). Property iii) is a direct consequence of ii).

The above inequality together with ii) gives iv) and iv) yields v). ∎

Now we define a transport from the exponential measure $\nu^{n}$ to ${\mu_{p,n}}$ for $p\geq 2$:

This transport satisfies the following bound:

Let $s=\|W_{1,p}^{n}(x)\|_{p}$. By direct calculation we get

Since $w_{1,p}=w_{p,1}^{-1}$, Proposition 5.13 i) implies $|w_{1,p}^{\prime}(x_{j})|\leq 2$, while by Proposition 5.8 we have $f_{p,n}(s)/s\leq 1$. Thus the first summand can be bounded by $2$.

For the second summand note that by Proposition 5.8 iii),

Moreover, $\|W_{1,p}^{n}(x)\|_{2}\leq n^{1/2-1/p}s$ by the Hölder inequality and

Let $u_{i}(t)=(y_{1},y_{2},\ldots,y_{i-1},t,z_{i+1},z_{i+2},\ldots,z_{n})$ for $i=1,\ldots,n$. Note that $u_{i}(y_{i})=u_{i+1}(z_{i+1})$, $u_{1}(z_{1})=z$ and $u_{n}(y_{n})=y$, hence

Let $s_{i}(t):=\|w_{1,p}(u_{i}(t))\|_{p}$. By vector–valued integration and (29) we get

As in the proof of Proposition 5.17 we show that

To deal with the sum of $a_{i}$’s we notice that, since $f_{p,n}(s)/s\leq 1$ and $w_{1,p}^{\prime}(x)\geq 0$,

If $x-y\in tB_{1}^{n}+t^{1/2}B_{2}^{n}$ for some $t>0$, then for all $p\geq 2$, $S_{p,n}x-S_{p,n}y\in 10(t^{1/2}B_{2}^{n}\cap t^{1/p}B_{p}^{n})$.

Let us fix $x,y$ with $x-y\in tB_{1}^{n}+t^{1/2}B_{2}^{n}$. By Proposition 5.15 iv),

By Hölder’s inequality $\|S_{p,n}x-S_{p,n}y\|_{2}\leq n^{1/2-1/p}\|S_{p,n}x-S_{p,n}y\|_{p}\leq 8t^{1/2}$ for $t\geq n$.

Assume now that $t\leq n$. Let $z$ be such that $x-z\in t^{1/2}B_{2}^{n}$ and $z-y\in tB_{1}^{n}$. Then $S_{p,n}x-S_{p,n}z\in 4t^{1/2}B_{2}^{n}$ by Proposition 5.17 and $\|W_{1,p}^{n}z-W_{1,p}^{n}y\|_{2}\leq 2\sqrt{t}$ by Proposition 5.15 v). Thus by Proposition 5.18,

Hence $S_{p,n}x-S_{p,n}y\in 10t^{1/2}B_{2}^{n}$. ∎

The last function we define transports the Gaussian measure $\nu_{2}^{n}$ to ${\mu_{p,n}}$ for $p\geq 2$.

The first summand is bounded by 2 as in the proof of Proposition 5.17. Since $w_{2,p}=w_{p,2}^{-1}$ we get by Lemma 5.13 ii)

We start with the version of Theorem 4.6 for $\nu_{p}$.

${\nu_{p}^{n}}(A+20t^{1/p}B_{p}^{n})\geq e^{t}{\nu_{p}^{n}}(A)$ or

{\nu_{p}^{n}}\big{(}(A+20t^{1/p}B_{p}^{n})\cap 100\sqrt{n}B_{2}^{n}\big{)}\geq\frac{1}{2}{\nu_{p}^{n}}(A).

We will use the transport $W_{1,p}^{n}$ from ${\nu^{n}}$ to ${\nu_{p}^{n}}$. Proposition 5.15 v) gives $\|W_{1,p}^{n}(x)-W_{1,p}^{n}(y)\|_{p}^{p}\leq 2^{p}\|x-y\|_{1}.$ By Remark 5.5 this means that $A+2(10t)^{1/p}B_{p}^{n}\supset W_{1,p}^{n}(W_{p,1}^{n}(A)+10tB_{1}^{n})$. Let us fix $t\geq 1$ and apply Theorem 4.6 to $W_{p,1}^{n}(A)$ and $10t$. If the second case occurs, we have

If the first case of Theorem 4.6 occurs, then due to Proposition 5.15 iii) we have $\|W_{1,p}^{n}(x)\|_{2}\leq 2\|x\|_{2}$, so $2\alpha B_{2}^{n}\supset W_{1,p}^{n}(\alpha B_{2}^{n})$ for any $\alpha>0$. Thus

Corollary 5.2 gives $B_{s}({\nu_{p}^{n}})\subset C(s^{1/p}B_{p}^{n}+s^{1/2}B_{2}^{n})$ for $s>0$. By Corollary 2.19, ${\nu_{p}^{n}}$ satisfies $IC(48)$, which, due to Proposition 2.4 implies $\nu_{p}^{n}(A+48B_{2t}({\nu_{p}^{n}}))\geq\min\{1/2,e^{t}{\nu_{p}^{n}}(A)\}$ for any Borel set $A$. Thus we have

where in the last step we use the Stirling approximation and $C$ as always denotes a universal constant. Thus it is enough to take $c(\alpha)<(C\alpha)^{-1}$. ∎

By Propositions 2.7, 3.11, 3.5 and 5.3 it is enough to show

for $1\leq t\leq n$ and $\mu_{p,n}(A)\geq e^{-n}$.