$\proof$
Draw a picture! Suppose not, than there exists an $\e>0$ such that for
all $t > a-\e$: $V_1(t) > V_2(t)$; hence
$$
\int_0^{V_2(a-\e)}V_1^{-1}(s)\,ds>\int_0^{V_2(a-\e)}V_2^{-1}(s)\,ds
$$
which contradicts the assumtion that $V_2\succ V_1$.
$\eofproof$
Using Theorem, Theorem and Lemma we can now give a simple proof of Petty's projection inequality.
$\proof$
Put $a=\vol{}(C)$, $b_1=\vol{}(C-C)$ and $b_2=\vol{}(2C)$, then both
$V_C:[0,a]\rar[0,b_1]$ and $V_{B_2^d}:[0,a]\rar[0,b_2]$ are
homeomorphisms.
Since $V_C(a)=V_{B_2^d}(a)=0$ it follows from Theorem and
Lemma, that there exists a sequence $a_n\uar a$ such that for all
$n\in\N$: $V_C(a_n)\leq V_{B_2^d}(a_n)$.
Now the inequality is an immediate consequence of Theorem.
$\eofproof$
Remark for the careful reader of [K]: This proof
is somehow the reversed order of arguments given there.
$\proof$
Put $a_1=\max\{I_C*I_C(x)\}$, $a_2=\vol{}(C)$ and $b=\vol{}(2C)$,
then again both
$\wt V_C:[0,a_1]\rar[0,b]$ and $V_{B_2^d}:[0,a_2]\rar[0,b]$ are
homeomorphisms and $\wt V_C(0)=V_{B_2^d}(0)=b$, hence Lemma implies
the existence of a sequence $\d_n\dar0$ such that for all $n\in\N$:
$\wt V_C(\d_n)\geq V_{B_2^d}(\d_n)$. Now we conclude by Theorem.
$\eofproof$
In general symmetrization techniques are used to prove that the
value of a certain expression decreases or increases when applying Steiner
symmetrizations. In the above context this means the following: Let $C$
be a convex body in $\R^d$ and denote by $C_1$ the convex body obtained by
symmetrizing $C$ about a hyperplane $H$. Let $F_C$ and $F_{C_1}$ be the
convolution squares of $C$ and $C_1$ respectively. Then the following
conjecture seems to be natural: $F_{C_1}\succ F_C$. An affirmative
answer to this conjecture (cf. the Appendix) would imply that
$$
\vol{}(\Pi^*C_1)\geq\vol{}(\Pi^*C)
\quad\mbox{and}\quad
\O(C_1)\geq\O(C)~.
$$
$\proof$
Putting $1-t^2=(1-s^2)(1-\e^2)$ we get for all $0<\e<1$:
\begin{eqnarray*}
\vol{}\left(B_2^d(\e)\right)&=&
\vol{d-1}\left(B_2^{d-1}\right)\int_\e^1(1-t^2)^{\frac{d-1}2}\,dt\\
&=&\vol{d-1}\left(B_2^{d-1}\right)(1-\e^2)^{\frac{d+1}2}
\int_0^1(1-s^2)^{\frac{d-1}2}\tfrac st\,ds~.
\end{eqnarray*}
Since $1\geq s/t\geq s$ the result follows.
$\eofproof$
$\proof$
Let $D$ be a multiple of the euclidean ball such that
$\vol{}(D)=\vol{}(B)$. Since by Lemma
\begin{equation}\label{eq2}\tag{eq2}
\vol{}(D\cap(x+D))
=2\vol{}\left(D\cap\left[x=\tfrac12\Vert x\Vert_2^2\right]\right)
\leq\vol{}(D)\exp\left(-\tfrac18d\Vert x\Vert_D^2\right)
\end{equation}
we get
$$
V_D(\d)\leq\vol{}(D)\left(
\tfrac8d\log\left(\tfrac{\vol{}(D)}{\d}\right)\right)^{\frac d2}~.
$$
Hence
$$
V_D^{-1}(\a)\leq\vol{}(D)\exp\left(-\tfrac18d\left(
\tfrac{\a}{\vol{}(D)}\right)^{\frac2d}\right)~.
$$
For all $\l\leq1$ we have by \eqref{eq1} and \eqref{eq2}
\begin{eqnarray*}
V_B^{-1}\left(\vol{}(D)t^d\right)&\leq&
\frac1{\vol{}(D)t^d}\int_0^{\vol{}(D)t^d}V_D^{-1}(\a)\,d\a\\
&\leq&
\frac{d\vol{}(D)}{t^d}\int_0^t s^{d-1}e^{-\frac18ds^2}\,ds\\
&\leq&\frac{d\vol{}(D)}{t^d}\left(\frac{\l^dt^d}d+
\frac{t^d(1-\l^d)}d\,e^{-\frac18d\l^2t^2}\right)\\
&\leq&\vol{}(D)\left(\l^d+e^{-\frac18d\l^2t^2}\right)
\end{eqnarray*}
Now we choose $\l$ in such a way that
$\l^d\leq e^{-\frac12d\l^2}$
and put $c=8\l^{-2}$, then for all $t\leq2$ (which is the possible
range of $t$):
$$
V_B^{-1}\left(\vol{}(D)t^d\right)
\leq2\vol{}(D)\exp\left(-\tfrac dct^2\right)~.
$$
This implies:
$$
V_B(\d)^{\frac1d}\leq\vol{}(D)^{\frac1d}\sqrt{\tfrac cd
\log\left(\tfrac{2\vol{}(D)}{\d}\right)}~.
$$
For $\l=\frac34$ we have $\l^2\leq-2\log\l$ and $c=\frac{128}9$.
$\eofproof$
Remark: In the nonsymmetric case we have to choose $\l$ in such
a way, that $\l^2\leq8t_0^{-2}\log\l$ where
$$
t_0=\left(\frac{\vol{}(C-C)}{\vol{}(C)}\right)^{\frac1d}~.
$$
By a theorem of Rogers and Shephard (cf. [RS]) this is bounded by $8$. Hence the upper bound of Proposition holds in the non symmetric case with a worse constant.
We now turn to estimates for the convolution square itself. The
following simple lemma is in constant use in the local theory of normed
spaces (cf. e.g. [LM]).
$\proof$
The formula
$$
\frac{\vol{}(B)}{\vol{}(B_2^d)}
=\int_{S^{d-1}}\Vert x\Vert_B^{-d}\,\s(dx)
$$
and Chebyshev's inequality imply: $\s(\Vert.\Vert_B < 1) < t^d$.
$\eofproof$
Before stating the next result we recall that the John ellipsoid of a
convex body $C$ is the ellipsoid of maximal volume contained in $C$. If
${\cal E}_C$ denotes the John ellipsoid of a $d$ dimensional convex body
$C$, then a famous result of F. John [J] states that $C\sbe d{\cal
E}_C$. If in addition $C$ is symmetric than $C\sbe\sqrt{d}{\cal
E}_C$.
$\proof$
W.l.o.g. we may assume $\vol{}(B)=1$. Choose $c_d>0$ such that
$\vol{}(c_d\sqrt{d}B_2^d)=1$; clearly: $c_d\leq1/2$. The
assumption on the John ellipsoid of $B$ now implies: $B\sbe\frac12d
B_2^d$.
For given $\d>0$ we choose $r(\d)>0$ such that
\begin{eqnarray}\label{eq3}
\vol{}(\tfrac12r(\d)B_2^d)^{\frac1d}&=&
\tfrac83\sqrt{\tfrac2d\log\left(\tfrac{2}{\d}\right)}\nonumber\\
\mbox{i.e.}\quad
r(\d)&=&\tfrac{16c_d}3\sqrt{2\log\left(\tfrac2{\d}\right)}~.
\end{eqnarray}
Let us denote by $B(\d)$ the set $[F_B>\d]$, then
from Proposition we conclude:
$$
\tfrac12\vol{}(r(\d)B_2^d)^{\frac1d}\geq\vol{}(B(\d))^{\frac1d}
$$
and from Lemma it follows that the set
$$
A_\d\colon=\{x\in S^{d-1}:\,\Vert x\Vert_{B(\d)}\geq r(\d)^{-1}\}
$$
is large i.e. $\s(A_\d)\geq1-2^{-d}$. For $n=1,\ldots,d$ define
$\d_n$ such that $r(\d_n)=n$ and $A\colon=\bigcap_nA_{\d_n}$. Then
$\s(A)\geq1-d2^{-d}$ and for all $x\in A$ and all $n=1,\ldots,d$ we
have: $$
F_B(nx)\leq\d_n=
2\exp\left(-\tfrac12\left(\tfrac{3n}{16c_d}\right)^2\right)~.
$$
Since for all $x\in S^{d-1}$ the function $t\mapsto F_B(tx)$ is
decreasing and vanishes for $t>d$, we get the assertion of the
corollary.
$\eofproof$
In order to obtain an affinely invariant version of this corollary,
we need to recall the classical notion of the volume ratio: Let $C$
be convex body in $\R^d$ and let ${\cal E}_C$ be the John ellipsoid
of $C$, then the quantity $vr(C)\colon=(\vol{}(C)/\vol{}({\cal
E}_C))^{\frac1d}$ is called the volume ratio of $C$.
Remark: There is no chance to prove the above result for all
directions $x$: take the unit ball $B_1^d$ of $\ell_1^d$ and choose for
$x$ one of the basis vectors. Another even more symmetric
counterexample is the set
$$
\left\{(y,t)\in\R^{d-1}\times\R:\norm
y_2+\tfrac{|t|}{\sqrt{d-1}}\leq1\right\}
$$
and the direction $x=(0,\ldots,0,1)$. Corollary has to be seen in connection with the following
conjecture of V.D. Milman:
An affirmative solution of this 'deviation inequality', would
solve the hyperplane problem (cf. e.g. [MP]): For all convex symmetric
bodies $B$ in $\R^d$ there exists a hyperplane $H$ such that
$\vol{d-1}(H\cap B)\geq c\vol{}(B)^{\frac{d-1}d}$, where $c$ denotes
an absolute constant.
A lower bound for $V_B$ can be obtained in a
similar way: Lemma implies:
$$
V_B^{-1}(\a)\geq\tfrac c{\sqrt d}\vol{}(D)
\left(1-\left(\tfrac{\a}{2^d\vol{}(D)}\right)^{\frac 2d}
\right)^{\frac{d+1}2}
$$
where, as before, $D$ denotes a multiple of the euclidean ball with
the same volume as $B$. Now we conclude again by \eqref{eq1}:
\begin{eqnarray*}
V_B^{-1}\left(\vol{}(D)t^d\right)&\geq&
\tfrac1{(2^d-t^d)\vol{}(D)}
\int_{t^d\vol{}(D)}^{2^d\vol{}(D)}V_d^{-1}(\a)\,d\a\\
&\geq&\tfrac c{\sqrt d}\vol{}(B)
\int_{\frac{t^2}4}^1(1-s)^{\frac{d+1}2}\tfrac d2\tfrac{s^{\frac d2-1}}
{1-(t^2/4)^{\frac d2}}\,ds\\
&=&\tfrac c{\sqrt d}\vol{}(B)H(t^2/4)
\end{eqnarray*}
where $H$ denotes the function
$$
H(x)=\tfrac d2\b\left(1-x,\tfrac{d+3}2,\tfrac d2\right)\left(1-x^{\frac d2}
\right)^{-1}
$$
and $\b(x,p,q)$ the incomplete beta function. Taking the inverse we abtain:
$$
V_B(\d)^{\frac1d}\geq 2\vol{}(B)^{\frac1d}
\sqrt{H^{-1}\left(\tfrac{\d\sqrt{d}}{\vol{}(B)}\right)}~.
$$
If $x$ is close to $0$ then $H^{-1}(x)\sim1-x^{\frac2{d+1}}$, actually:
$$
\lim_{x\dar0}\tfrac{1-H^{-1}(x)}{x^{\frac2{d+1}}}=1~.
$$
Finally, let $B$ be a ball in $\R^d$. For $0\leq\d\leq\frac12\vol{}(B)$ let
$B_\d$ be the floating body of $B$ i.e. $B_\d$ is a subset of $B$ such that
every tangent hyperplane of $B_\d$ cuts off a cap of volume
$\d$ from $B$. K. Ball [B] and independently M. Meyer and S. Reisner [MR]
proved that $B_\d$ is a convex symmetric body. This is not true in the
general case. C. Schütt and E. Werner [SW] introduced the notion of
the convex floating body, which coincides with the floating body provided
that the latter is convex. Since
$$
B_\d\spe\tfrac12[I_B*I_B>2\d] \quad\mbox{i.e.}\quad
\vol{}(B_\d)\geq2^{-d}V_B(2\d)
$$
a lower bound for $V_B$ implies a lower bound for $\vol{}(B_\d)$.
We conclude this paper with two additional results on the distribution
function. The first one being a characterization of ellipsoids
$\proof$
If $H$ is a hyperplane supporting $B_\d$ at $x_0$, then
$$
H^+\cap B\cap(B+2x_0)=H^+\cap B
$$
If this were false, we would have
\begin{eqnarray*}
\d=\vol{}(H^+\cap B)&=&\vol{}(H^+\cap B\cap(B+2x_0))\\
&&+\vol{}(H^+\cap B\cap(B+2x_0)^c)\\
&>&\tfrac12(2\d)=\d~.
\end{eqnarray*}
Hence for every hyperplane $H$ supporting $B_\d$ at $x_0$
$$
H\cap B\cap(B+2x_0)=H\cap B
$$
i.e. $x_0$ is a center of symmetry for $H\cap B$. Since every hyperplane
intersecting $B$ is supporting some $B_\d$ we conclude, that $H\cap B$ has
a center of symmetry for all hyperplanes $H$. Hence, by a theorem of Brunn
(cf. [A] p 123), $B$ is an ellipsoid.
$\eofproof$
Remark: Using $\wt V_C$ instead of $V_C$ it is possible to get a
characterization of ellipsoids as the only convex bodies $C$ satisfying
$\vol{}(C_\d)=2^{-d}\wt{V}_C(2\d)$, where $C_\d$ denotes the convex
floating body.
$\proof$
Let $B(\d)$ be the set $[F_B>\d]$. For $h>0$ we have
\begin{eqnarray*}
\tfrac1h(V_B(\d+h)-V_B(\d))&=&-\tfrac1{dh}\int_{S^{d-1}}
\Vert x\Vert^{-d}_{B(\d)}
-\Vert x\Vert^{-d}_{B(\d+h)}\,dx\\
&=&-\tfrac1{dh}\int_{S^{d-1}}
\Vert x\Vert^{-d}_{B(\d+h)}\sum_{k=1}^d\mbox{${d\choose k}$}\left(\D r
\Vert x\Vert_{B(\d+h)}\right)^k\,dx
\end{eqnarray*}
where $\D r=\Vert x\Vert^{-1}_{B(\d)}-\Vert x\Vert^{-1}_{B(\d+h)}$. From
[Sm] Lemma 4 it follows that
$$
\tfrac1{\s_\d(x)}\leq\tfrac{\D r}h\leq\tfrac{1+\frac{\d}{h}}{\s_\d(x)}
\log\left(1+\tfrac h\d\right)~.
$$
where $\s_\d(x)$ denotes the $d-1$ dimensional volume of the orthogonal
projection of the set
$$
B\cap\left(B+\tfrac x{\Vert x\Vert_{B(\d)}}\right)
$$
on the hyperplane orthogonal to $x$. Hence
$$
\lim_{h\to0}\tfrac1h(V(\d+h)-V(\d))
=-\int_{S^{d-1}}\Vert x\Vert^{-n}_{B(\d)}\,\tfrac{
\Vert x\Vert_{B(\d)}}{\s_\d(x)}\,dx~.
$$
Since $\s_\d(x)$ is a decreasing function of $\d$ the result follows.
$\eofproof$
$\proof$
Let $U$ be any measurable subset of $\R$ of volume $2a$ and put
$V=-B(x,\d)\times-U$. We then have by the Riesz-Sobolev inequality
$$
F*I_V(0)=I_A*I_B*I_V(0)\leq I_{A^\prime}*I_{B^\prime}*I_{V^\prime}
=G*I_{V^\prime}~.
$$
Now, since
$$
F*I_V(0)=\int F(y)I_{-V}(y)\,dy=
\int_{B(x,\d)}\int_U F(z,t)\,dt\,dz
$$
the above inequality implies:
$$
\frac1{\vol{}(B(x,\d))}\int_{B(x,\d)}\int_U F(z,t)\,dt\,dz\leq
\frac1{\vol{}(B(x,\d))}\int_{B(x,\d)}\int_{-a}^a G(z,t)\,dt\,dz
$$
By taking the limit as $\d\rar 0$ we get:
$$
\int_U F(x,t)\,dt\leq\int_{-a}^a G(x,t)\,dt
$$
i.e. for all $x\in\R^{d-1}$ the function $F(x,.)$ is dominated by
$G(x,.)$ in the sense of Hardy and Polya. Therefore for any
increasing convex function $\vp:\R_0^+\rar\R_0^+$:
$$
\int_\R\vp\circ F(x,t)\,dt\leq\int_\R\vp\circ G(x,t)\,dt
$$
Integrating with respect to $x\in\R^{d-1}$ we obtain:
$$
\int_{\R^d}\vp\circ F(y)\,dy\leq\int_{\R^d}\vp\circ G(y)\,dy
$$
which is equivalent to Hardy Polya domination of $F$ by $G$.
$\eofproof$