Proof Without loss of generality, we may assume that $R=1$ . Let us first show that the left-hand side of equality (2.1) is less or equal to $-1$ . Writing the integral in polar coordinates, we get

$$ \begin{align*}\limsup_{t\to\infty}\frac{\ln{\int_{S^{n-1}}\int_t^{\infty} e^{-\phi(r)} r^{n-1}drd\theta}}{\phi(t)} =\limsup_{t\to\infty}\frac{\ln{\int_t^{\infty} e^{-\phi(r)} r^{n-1}dr}}{\phi(t)}.\end{align*} $$

We remind that $\lim _{t \to \infty } \phi (t)=\infty .$ Let $\eta (t)=-(n-1) \ln {t}+\phi (t)$ . Using that $\phi $ is a convex and nonconstant function, we get that there exists $t_0 \ge 0$ such that $ \phi '(t_0)>0$ . Thus, $\phi '(t)>0$ for all $t>t_0$ , and there exists a constant $a>0$ such that $\eta '(t)>a$ for all $t>t_0$ . Thus, $\eta (r)\geq \eta (t)+ a(r-t)$ for $r>t>t_0$ , and

$$ \begin{align*} \limsup_{t\to\infty}\frac{\ln{\int_t^{\infty} e^{-\phi(r)} r^{n-1}dr}}{\phi(t)} & \leq \limsup_{t\to\infty}\frac{\ln{\int_t^{\infty} e^{-\eta(t)-a(r-t)} dr}}{\phi(t)} \\ &=\limsup_{t\to\infty}\frac{\ln{e^{-\eta(t)}}+\ln{\int_t^{\infty} e^{-a(r-t)} dr}}{\phi(t)} \\ &=\limsup_{t\to\infty}\frac{\ln(t^{n-1}e^{-\phi(t)})+\ln{\frac{1}{a}}}{\phi(t)} \\ & =- 1 +\limsup_{t\to\infty}\frac{\ln \frac{1}{a} }{\phi(t)}\\ &=-1. \end{align*} $$

Next, we will show that the right-hand side of equality (2.1) is greater or equal to $-1$ . Since $r>t$ , we have

$$ \begin{align*} \limsup_{t\to\infty}\frac{\ln{\int_t^{\infty} e^{-\phi(r)} r^{n-1}dr}}{\phi(t)} & \geq \limsup_{t\to\infty}\frac{\ln\left(t^{n-1}\int_t^{\infty} e^{-\phi(r)} dr\right)}{\phi(t)} \\& =\limsup_{t\to\infty}\frac{\ln{\int_t^{\infty} e^{-\phi(r)} dr}}{\phi(t)}. \end{align*} $$

To finish proving the lemma, we prove the following claim.

Proof of Claim 2.3.

Assume the result is not true. Then, there exists $\alpha>1$ such that

$$ \begin{align*} \limsup_{t\to\infty}\frac{\ln{\int_t^{\infty} e^{-\phi(r)} dr}}{\phi(t)}<{-\alpha}. \end{align*} $$

Thus, there exists $t_0>0$ such that for all $t>t_0$ , we have

(2.2) $$ \begin{align} \int_t^{\infty}e^{-\phi(r)}dr\leq e^{-\alpha\phi(t)}. \end{align} $$

Let $F(t)=\int _t ^{\infty }e^{-\phi (r)} dr$ , and note that $F^\prime (t)=-e^{-\phi (t)}$ , and thus, (2.2) is equivalent to $ F(t)^{\frac {1}{\alpha }}\le -F^\prime (t). $ Therefore, for $t>t_0$ , we have

$$ \begin{align*} 1\le-\frac{F^\prime (t)}{F(t)^{\frac{1}{\alpha}}}. \end{align*} $$

Integrating both sides of the above inequality over $t \in [t_0, \infty )$ , we get that $\frac {1}{1-\frac {1}{\alpha }} F(t)^{1-\frac {1}{\alpha }}$ is unbounded, which gives a contradiction, and the claim is proved. This finishes the proof of Lemma 2.2.

Proof Let $R=r(K, B_2^n)$ . Then, $(tK)^c\subset (tRB_2^n)^c$ . Using Lemma 2.2, we get

$$ \begin{align*}\limsup_{t\to\infty}\frac{\ln{\mu((tK)^c)}}{\phi(tR)}\leq \limsup_{t\to\infty}\frac{\ln{\mu((tRB_2^n)^c)}}{\phi(tR)}=-1.\end{align*} $$

To obtain the reverse inequality, we denote by P a plank of width $2R$ which contains K. More precisely, using the maximality of R, there exist at least two tangent points $y, -y \in R{\mathbb S}^{n-1} \cap \partial K$ . Thus, we may consider $P=\{x \in {\mathbb R}^n: |\langle x, y\rangle | \le R\}$ . Next,

$$ \begin{align*}\limsup_{t\to\infty}\frac{\ln{\mu((tK)^c)}}{\phi(tR)}\geq \limsup_{t\to\infty}\frac{\ln{\mu((tP)^c)}}{\phi(tR)}.\end{align*} $$

By the rotation invariant of $\mu $ , we may assume that $y=Re_n$ , and so

$$ \begin{align*} \mu((tP)^c)=2\int_{tR}^{\infty} \int_{\mathbb R^{n-1}}e^{-\phi (|ze_{n}+x|)} dx dz. \end{align*} $$

Using the triangle inequality and the polar coordinates, we get

$$ \begin{align*} \mu((tP)^c) &\geq 2\int_{tR}^{\infty} \int_{\mathbb R^{n-1}}e^{-\phi (z+|x|)} dx dz \\ &=2\int_{tR}^{\infty} \int_{\mathbb S^{n-2}} \int_0 ^{\infty} e^{-\phi (z+r)} r^{n-2} dr d\theta dz\\ &=2|{\mathbb S^{n-2}}| \int_0^{\infty}r^{n-2}\int_{tR}^{\infty}e^{-\phi (z+r)} dz dr\\&=2|{\mathbb S^{n-2}}| \int_0^{\infty}r^{n-2}\int_{tR+r}^{\infty}e^{-\phi (z)} dz dr\\ &=2|\mathbb S^{n-2}|\int_{tR}^{\infty}e^{-\phi(z)} \int_0^{z-tR} r^{n-2} dr dz\\&=2\frac{|\mathbb{S}^{n-2}|}{n-1} \int_{tR}^{\infty} (z-tR)^{n-1}e^{-\phi(z)} dz. \end{align*} $$

Now to finish the proof of Lemma 2, we need to prove the following claim.

Claim 2.7

$$ \begin{align*} \limsup_{t\to\infty} \frac{\ln{\int_{tR}^{\infty} (z-tR)^m e^{-\phi(z)} dz} }{\phi (Rt)} \geq -1, \end{align*} $$

for any nonnegative integer m.

Proof of Claim 2.7.

Making the change of variables, we get

$$ \begin{align*} \limsup_{t\to\infty} \frac{\ln{\int_{tR}^{\infty} (z-tR)^m e^{-\phi(z)} dz} }{\phi (Rt)} = \limsup_{t\to\infty} \frac{\ln{\int_{t}^{\infty} (r-t)^m e^{-\phi(r)} dr} }{\phi (t)}. \end{align*} $$

We will first prove the following inductive step: fix a nonnegative integer $m $ , and let

$$ \begin{align*}F_m(t)=\int_{t}^{\infty} (r-t)^{m}e^{-\phi(r)}dr.\end{align*} $$

Then,

(2.6) $$ \begin{align} \liminf_{t\to\infty} \frac{\ln F_m(t)}{\ln F_{m-1}(t)} =1, \mbox{ for all } m\in {\mathbb N}. \end{align} $$

We note that $F_m(t)\le 1$ , for t large enough, and thus, the denominator and numerator are negative. It is a bit easier to work with a fraction when both the denominator and numerator are nonnegative. So we will prove that $\liminf \limits _{t\to \infty } \frac {-\ln F_m(t)}{-\ln F_{m-1}(t)} =1.$ Using integration by parts, we get

$$ \begin{align*} F_{m-1}(t) =\frac{1}{m}\int_t^\infty (r-t)^m \phi'(r) e^{-\phi(r)} dr \ge \frac{1}{m}\phi'(t)\int_t^\infty (r-t)^m e^{-\phi(r)} dr, \end{align*} $$

where, again, we denote by $\phi '(t)$ the left derivative of $\phi $ . Thus,

$$ \begin{align*}-\ln F_{m-1}(t) \le -\ln(\phi'(t)/m)-\ln F_m(t), \end{align*} $$

and

$$ \begin{align*} \liminf_{t\to\infty} \frac{-\ln F_m(t) }{-\ln F_{m-1}(t) } \ge \liminf_{t\to\infty} \frac{\ln(\phi'(t)/m)-\ln F_{m-1}(t)}{ -\ln F_{m-1}(t)}. \end{align*} $$

Now we may use that $\phi '(t)>a>0$ for t large enough and $\lim \limits _{t \to \infty }\ln F_{m-1}(t)=-\infty $ to claim that

(2.7) $$ \begin{align} \liminf_{t\to\infty} \frac{-\ln F_m(t) }{-\ln F_{m-1}(t) } \ge \liminf_{t\to\infty} \frac{\ln(a/m)-\ln F_{m-1}(t)}{ -\ln F_{m-1}(t)} \ge 1. \end{align} $$

To prove the reverse inequality, we note that

$$ \begin{align*}F_m'(t)=-m\int_{t}^{\infty} (r-t)^{m-1} e^{-\phi(r)} dr. \end{align*} $$

Assume that

$$ \begin{align*} \liminf_{t\to\infty} \frac{ -\ln{F_m(t)}}{-\ln(-\frac{1}{m}F_m'(t))}>\alpha>1, \end{align*} $$

but then, again there exists $t_0>0$ such that for all $t> t_0,$ we have

$$ \begin{align*} -\ln{F_m(t)}> -\alpha \ln(-\frac{1}{m}F_m'(t)), \end{align*} $$

and thus,

$$ \begin{align*} m < \frac{-F_m'(t)}{F_m(t)^{\frac{1}{\alpha}}}. \end{align*} $$

Take an integral over $t \in [x, \infty )$ from both sides to get $ F_m^{1-\frac {1}{\alpha }}(x) =\infty , $ which is a contradiction. This finishes the proof of the inductive step, but we actually need a bit stronger statement, which is similar to Remark 2.4. Indeed, consider any $m\in {\mathbb N}$ and $\alpha> 1$ . Let

$$ \begin{align*} E_{m,\alpha}=\{t: -\ln F_m(t)>- \alpha \ln F_{m-1}(t)\}. \end{align*} $$

Then, using the same ideas as in (2.3), we get $|E_{m,\alpha }|<\infty $ .

To complete our proof, let

$$ \begin{align*}X_i (t)=\frac{\ln F_i(t)}{ \ln F_{i-1}(t)} \mbox{ and } Y(t)=\frac{\ln F_0(t)}{\phi(t)}.\end{align*} $$

Using (2.6), we get $\liminf \limits _{t\to {\infty }} X_{i}(t)=1$ , and using (2.1), we get $\limsup \limits _{t\to \infty }Y(t)= -1.$ Now let $X(t)=\prod _{i=1}^{m}X_{i}(t)$ . Our goal is to prove that

$$ \begin{align*} \limsup_{t\to\infty}X(t)Y(t) \ge -1. \end{align*} $$

Assume that this is not true. Then, there exists $\alpha>1$ such that

$$ \begin{align*} \limsup_{t\to\infty}X(t)Y(t)<{-\alpha}<-1. \end{align*} $$

Therefore, there exists $t_0$ such that for all $t>t_0$ ,

(2.8) $$ \begin{align}X(t)Y(t)\le-\alpha. \end{align} $$

Using (2.7), we may also assume that $X_i(t)>0$ for all $t>t_0.$ Next, consider the set

$$ \begin{align*} A:=\left\{t>t_0: X(t)>\frac{\alpha+1}{2}\right\}.\end{align*} $$

We claim that $|A|<\infty $ . Note that

$$ \begin{align*}\left|\left\{t:X(t)>\frac{\alpha+1}{2}\right\}\right| \le \left|\left\{t: X_i(t) >\left(\frac{\alpha+1}{2}\right)^{\frac{1}{m}}\!\!\!\!\!\!, \quad \mbox{for some } i\in\{1,\dots,m\}\right\}\right|\end{align*} $$

$$ \begin{align*}< \sum_{i=1}^{m} \left|\left\{t:X_{i}(t)>\left(\frac{\alpha+1}{2}\right)^{\frac{1}{m}} \right\}\right|<\infty.\end{align*} $$

We also note that $\frac {2\alpha }{\alpha +1}>1$ , and thus,

$$ \begin{align*} \left|\left\{t: Y(t) < -\frac{2 \alpha}{\alpha+1}\right\}\right|<\infty. \end{align*} $$

Finally,

$$ \begin{align*} \left|\left\{t> t_0: X(t)Y(t) < -\alpha\right\}\right| =\left|\left\{t> t_0: Y(t) < -\frac{\alpha}{X(t)}\right\}\right|\end{align*} $$

$$ \begin{align*}\le |A|+ \left|\left\{t> t_0: Y(t) < -\frac{\alpha}{X(t)} \mbox{ and } X(t)<\frac{\alpha+1}{2}\right\}\right|\end{align*} $$

$$ \begin{align*} \le |A|+ \left|\left\{t: Y(t) < -\frac{2 \alpha}{\alpha+1}\right\}\right|<\infty, \end{align*} $$

which contradicts with (2.8). The claim is proved, and this finishes the proof of Lemma 2.6.

Proof Let $K\subset {\mathbb R}^n $ be a convex, symmetric body such that $\mu (tRB_2^n) \le \mu (tK)$ holds for for some fixed $R>0$ and every t large enough, but $RB_2^n \not \subset K.$ Thus, the maximal Euclidean ball in K has radius $rR$ , with $r\in (0,1)$ . From the assumption, it follows that

$$ \begin{align*}\mu((tRB_2^n)^c)\ge\mu((tK)^c),\end{align*} $$

which implies that

$$ \begin{align*} \frac{\ln{\mu((tRB_2^n)^c)}}{\phi(tR)}\ge\frac{\ln{\mu((tK)^c)}}{\phi(trR)}\frac{\phi(trR)}{\phi(tR)}. \end{align*} $$

From the convexity of $\phi $ and $r \in (0,1)$ , we get that

$$ \begin{align*}\phi(trR)=\phi(trR+(1-r)0)\le r\phi(tR)+(1-r)\phi(0).\end{align*} $$

Using that $\phi (tR) \to \infty $ , we get that there exists $r'\in (0,1)$ and $t_0>0$ such that $\frac {\phi (trR)}{\phi (tR)}\le r'$ for all $t>t_0$ . Thus,

(2.9) $$ \begin{align} \frac{\ln{\mu((tRB_2^n)^c)}}{\phi(tR)}\ge r'\frac{\ln{\mu((tK)^c)}}{\phi(trR)} \end{align} $$

for all $t>t_0$ . Taking the limit superior, as $t\to \infty $ , from both sides of the inequality (2.9), we obtain $-1\ge -r'.$ But this contradicts the fact that $r'$ is less than $1$ . Therefore, our assumption that $RB_2^n \not \subset K$ must be false.

Proof We have to check Lemma 2.2 and Lemma 2.6 (i.e., to prove yet another generalization of the classical large deviation principle (see (3.2) below)).

We claim that for any $R>0$ ,

$$ \begin{align*} \limsup_{t\to \infty}\frac{\ln{\mu((tRL)^c)}}{\phi(tR)}=-1. \end{align*} $$

We can assume $R=1$ . Moreover, as before, using convexity of $\phi $ , we may assume that $\phi (t)$ is a strictly increasing function for large enough t. Thus,

(3.1) $$ \begin{align} \mu((tL)^c) &= \int_{(tL)^c} e^{-\phi (\|x\|_L)} dx = \int_{(tL)^c} \int_{\phi(\|x\|_L) }^{\infty} e^{-u} du dx \nonumber \\&= \int_{\mathbb{R}^n} \int_{\phi(\|x\|_L)}^{\infty} \!\!\!\chi_{(tL)^c} (x) e^{-u} du dx= \int_{0}^{\infty} \int_{\{x: \phi(\|x\|_L)<u\}}\!\!\! \chi_{(tL)^c} (x) e^{-u} dx du \nonumber \\&= \!\int_{0}^{\infty}\!\!\!e^{-u} |\{x\!\!: \|x\|_L \in [t, \phi^{-1}(u)]\}| du = |L| \!\int_{\phi(t)}^{\infty} ((\phi^{-1}(u))^n-t^n) e^{-u}du \nonumber\\&= |L| \int_{t}^{\infty} (v^n-t^n) \phi^{\prime}(v) e^{-\phi(v)} dv = -|L| \int_{t}^{\infty} (v^n-t^n) d e^{-\phi(v)}\nonumber\\& = n|L| \int_{t}^{\infty} v^{n-1} e^{-\phi(v)} dv. \end{align} $$

Note that $\phi (t)$ may be a constant function on some interval $[0, t_0]$ and strictly increasing on $[t_0, \infty )$ . In such a case, we define $\phi ^{-1}(\phi (0))=t_0$ . So, we have

$$ \begin{align*} \limsup_{t\to\infty}\frac{\ln\mu((tL)^c)}{\phi(t)} &=\limsup_{t\to\infty}\frac{\ln ({n|L|\int_t^{\infty} e^{-\phi(v)} v^{n-1}dv})}{\phi(t)} \\&=\limsup_{t\to\infty}\frac{\ln {\int_t^{\infty} e^{-\phi(v)} v^{n-1}dv}}{\phi(t)},\\ &=-1, \end{align*} $$

where the last equality follows from the proof of Lemma 2.2.

To finish the proof, we must check Lemma 2.6. In particular, we want to show that

(3.2) $$ \begin{align}\limsup_{t\to\infty}\frac{\ln{\mu((tK)^c)}}{\phi(r(K,L)t)}=-1 \end{align} $$

for symmetric, convex bodies $K, L \subset {\mathbb R}^n$ , convex, increasing function $\phi :[0, \infty ) \to [0, \infty )$ and measure $\mu $ with density $e^{-\phi (\|x\|_L)}$ .

Let $R=r(K,L)$ . Then, we have $(tK)^c\subset (tRL)^c$ from the assumption. Thus, using Lemma 2.2, we get

$$ \begin{align*}\limsup_{t\to\infty}\frac{\ln{\mu((tK)^c)}}{\phi(tR)}\leq \limsup_{t\to\infty}\frac{\ln{\mu((tRL)^c)}}{\phi(tR)}=-1.\end{align*} $$

Using that $RL$ is the maximal dilate of L inside K, we get that there is a pair of points $v, -v \in \partial RL\cap \partial K$ . Let P be a plank created by tangent planes to $RL$ and K at v and $-v$ . Let $n_v$ be a normal vector to $\partial RL$ at v. Then, the width of the plank P is $2Rh_L(n_v)=2h_K(n_v)$ : $P=\{x \in {\mathbb R}^n: |\langle x, n_v\rangle | \le Rh_L(n_v)\}$ , where $h_L (x)=\sup \{\langle x, y \rangle : y\in L\}$ is the support function of L (see [Reference Schneider20] for basic definitions and properties). Next,

$$ \begin{align*}\limsup_{t\to\infty}\frac{\ln{\mu((tK)^c)}}{\phi(tR)}\geq \limsup_{t\to\infty}\frac{\ln{\mu((tP)^c)}}{\phi(tR)}.\end{align*} $$

Selecting a proper system of coordinates, we may assume that $n_v=e_n$ . Let $a=tRh_L(e_n).$ Then,

$$ \begin{align*} \mu((tP)^c) &= 2\int_{a}^{\infty} \int_{e_n^\perp}e^{-\phi (\|ze_n +x\|_L)} dx dz\\ &= 2\int_{a}^{\infty}\int_{e_n^\perp} \int^\infty_{\phi (\|z e_n + x\|_L)} e^{-u} du dx dz \\ &= 2\int_{a}^{\infty}\int_{0}^\infty \int_{\{x \in e_n^\perp: \phi (\|ze_n+ x\|_L)<u\}} e^{-u} dx du dz \\ &= 2\int_{a}^{\infty}\int_{0}^\infty e^{-u}\left|\left\{x \in e_n^\perp: \|ze_n + x\|_L\le \phi^{-1}(u)\right\}\right| du dz. \end{align*} $$

Now note that

$$ \begin{align*} \left|\left\{x \in e_n^\perp: \|ze_n + x\|_L\le \phi^{-1}(u)\right\}\right|= \left|\left\{x \in e_n^\perp: ze_n + x \in \phi^{-1}(u)L\right\} \right|. \end{align*} $$

The above volume is zero if $z> \phi ^{-1}(u)h_L(e_n)$ (or $\phi (z/h_L(e_n))>u$ ). For $z \in [0, \phi ^{-1}(u)h_L(e_n)]$ , we note that $\phi ^{-1}(u) L$ is a convex body and thus contains inside a pyramid $\Delta $ with base $L\cap e_n^\perp $ and the height $\phi ^{-1}(u)h_L(e_n)$ (with apex $\phi ^{-1}(u)h_L(e_n)v/R$ ). Then,

$$ \begin{align*} &\left|\left\{x \in e_n^\perp: ze_n + x \in \phi^{-1}(u)L \right\}\right| \ge |\Delta\cap (e_n^\perp+ze_n)|\\&\qquad=(\phi^{-1}(u)h_L(e_n)-z)^{n-1}|L\cap e_n^\perp|.\end{align*} $$

Thus,

$$ \begin{align*} \mu((tP)^c) &\ge 2 |L\cap e_n^\perp|\int_{a}^{\infty}\int_{\phi(z/h_L(e_n))}^\infty e^{-u} (\phi^{-1}(u)h_L(e_n)-z)^{n-1} du dz \\ &= 2 |L\cap e_n^\perp|\int_{a}^{\infty}\int_{z/h_L(e_n)}^\infty e^{-\phi(u)}\phi'(u) (u h_L(e_n)-z)^{n-1} du dz\\ &= -2 |L\cap e_n^\perp|\int_{a}^{\infty}\int_{z/h_L(e_n)}^\infty (u h_L(e_n)-z)^{n-1} de^{-\phi(u)} dz\\ &= 2(n-1) |L\cap e_n^\perp|\int_{a}^{\infty}\int_{z/h_L(e_n)}^\infty e^{-\phi(u)} (u h_L(e_n)-z)^{n-2} du dz\\ &= 2(n-1) |L\cap e_n^\perp|\int_{a/h_L(e_n)}^{\infty}\int_{a}^{h_L(e_n)u} e^{-\phi(u)} (u h_L(e_n)-z)^{n-2} dz du\\ &= 2 |L\cap e_n^\perp|\int_{a/h_L(e_n)}^{\infty} e^{-\phi(u)} (u h_L(e_n)-a)^{n-1} du\\ &= 2 h_L^{n-1}(e_n)|L\cap e_n^\perp|\int_{tR}^{\infty} e^{-\phi(u)} (u -tR)^{n-1} du.\\ \end{align*} $$

So, we have

$$ \begin{align*} \limsup_{t\to\infty}\frac{\ln \mu((tP)^c)}{\phi(tR)} &\ge \limsup_{t\to\infty}\frac{\ln \left(2 h_L^{n-1}(e_n)|L\cap e_n^\perp|\int_{tR}^{\infty} e^{-\phi(u)} (u -tR)^{n-1} du\right)}{\phi(tR)} \\&=\limsup_{t\to\infty}\frac{\ln\int_{tR}^{\infty} e^{-\phi(u)} (u -tR)^{n-1}du }{\phi(tR)}. \end{align*} $$

By Claim 2, the above quantity is greater than or equal to $-1$ ; thus, Lemma 2.6 is applied here, which finishes the proof for our main result.

Fact If $d\mu = e^{-\phi (\|x\|_L)} dx$ is a $\log $ -concave measure and $K, L\subset \mathbb {R}^n$ are convex, symmetric bodies such that $|K|\le |RL|$ for some $R>0,$ then

$$ \begin{align*}\mu(K)\le \mu(RL).\end{align*} $$

Proof Using calculations similar to (3.1), we get

$$ \begin{align*} \mu(K) = \int_{0}^{\infty} e^{-u} |K\cap\phi^{-1}(u) L| du \end{align*} $$

and

$$ \begin{align*} \mu(RL)= \int_{0}^{\infty} e^{-u} |RL\cap\phi^{-1}(u) L| du. \end{align*} $$

To get $ \mu (K)\le \mu (RL)$ , we only need to check that $|K\cap \phi ^{-1}(u) L|\le |RL\cap \phi ^{-1}(u) L|.$ Indeed, if $R\le \phi ^{-1}(u),$ then we have

$$ \begin{align*}|K\cap\phi^{-1}(u) L|\le |K|\le|RL|= |RL\cap\phi^{-1}(u) L|,\end{align*} $$

and if $R\ge \phi ^{-1}(u),$ then we get

$$ \begin{align*}|K\cap\phi^{-1}(u) L|\le |\phi^{-1}(u) L|=|RL\cap\phi^{-1}(u) L|.\end{align*} $$

Hence, $ \mu (K)\le \mu (RL)$ for any $R>0.$

Proof Let us assume, toward the contradiction, that Question 3 has an affirmative answer in $\mathbb {R}^n$ for some fixed $n\ge 5.$ So, for any pair of convex symmetric bodies $K, L$ that satisfy (4.1), we would get $\mu (K)\le \mu (L).$ The condition on sections (4.1) will be also satisfied for the dilated bodies $tK$ and $tL$ , for all $t>0.$ Therefore, we have

(4.2) $$ \begin{align} \mu(tK) \le \mu (tL), \hspace{.5cm} \forall t>0, \end{align} $$

which, by definition of $\mu $ , means

$$ \begin{align*}\int_{tK}{e^{-\phi(||x||_L)} dx} \le \int_{tL}{e^{-\phi(||x||_L)} dx},\end{align*} $$

or equivalently, applying the change of variables $x = tx$ , we have

$$ \begin{align*}\int_{K}{e^{-\phi(t||x||_L)} dx} \le \int_{L}{e^{-\phi(t||x||_L)} dx}.\end{align*} $$

Using the continuity of $\phi $ and compactness of K and L, we can take the limit for the above inequality as $t \to 0^+$ to obtain

$$ \begin{align*}|K|\le|L|.\end{align*} $$

Therefore, we have a relation between the dilation problem for a $\log $ -concave probability measure, with the Busemann-Petty problem for volume measure, which is if

$$ \begin{align*}\mu (rK\cap \xi^\perp)\le \mu (rL\cap \xi^\perp), \hspace{0.5cm} \forall \xi\in\mathbb{S}^{n-1}, \hspace{0.5cm}\forall r>0,\end{align*} $$

then $|K|\le |L|$ .

A number of very interesting counterexamples to the Busemann-Petty problem were shown by Papadimitrakis [Reference Papadimitrakis17]; Gardner [Reference Gardner7]; Gardner, Koldobsky, and Schlumprecht [Reference Gardner, Koldobsky and Schlumprecht10]: there are convex symmetric bodies $K, L$ in $\mathbb {R}^n$ for $n\ge 5$ such that

(4.3) $$ \begin{align} |K \cap\xi ^\perp|\le|L \cap \xi ^\perp|, \hspace{.5cm} \forall \xi\in\mathbb{S}^{n-1}, \end{align} $$

but

(4.4) $$ \begin{align} |K|> |L|. \end{align} $$

Note that because the volume measure is homogeneous, the condition on sections (4.3) is also true for dilates of K and L, so we have

(4.5) $$ \begin{align} |rK \cap\xi ^\perp|\le|rL \cap \xi ^\perp|, \hspace{.5cm} \forall \xi\in\mathbb{S}^{n-1},\hspace{0.5cm}\forall r>0. \end{align} $$

Now, applying the fact to (4.5), we get that

$$ \begin{align*}\mu (rK\cap \xi^\perp)\le \mu (rL\cap \xi^\perp), \hspace{0.5cm} \forall \xi\in\mathbb{S}^{n-1}, \hspace{0.5cm}\forall r>0.\end{align*} $$

Thus, using (4.2), we have

$$ \begin{align*}\mu(tK) \le \mu (tL), \hspace{.5cm} \forall t>0.\end{align*} $$

Dividing by $t^n$ and taking the limit of the above inequality as $t \to 0^+$ , we get

$$ \begin{align*}|K| \le |L|,\end{align*} $$

and this contradicts (4.4).

Proof Giannopoulos [Reference Giannopoulos11] and Bourgain [Reference Bourgain4] constructed an example in $\mathbb {R}^n$ for $n\ge 7$ of convex body $K\subset \mathbb {R}^n$ that satisfies

$$ \begin{align*}|K \cap\xi ^\perp|\le|B_2^n \cap \xi ^\perp|, \hspace{.5cm} \forall \xi\in\mathbb{S}^{n-1},\end{align*} $$

but $|K|> |B_2^n|$ . To prove Theorem 4.2, one may take the same convex body K and $B_2^n$ as provided in [Reference Giannopoulos11, Reference Bourgain4] and repeat the proof of Theorem 4.1.