Let $\lambda$ denote the Lebesgue-measure on $\mathbb{R}^n$, and let $C\subset\mathbb{R}^n$ be a convex region.


My question is about 
$$f(C):=\int\_{C} \lambda(C \cap (x + C) ) \mathrm{d} x.$$ 
How large can $f(C)$ be? Of course, there is the trivial bound $f(C)\leq \lambda(C)^2$ but I would expect more something like $f(C) = O( \lambda(C) )$ at least. This question has probably been answered somewhere, and I suspect that the following might be very well known: for regions $C$ of constant measure, the value of $f(C)$ is maximal if $C$ is an $n$-ball (or something similar). I couldn't find any useful references regarding this question. If someone could help me out with some reference, or even a simple solution, I would be very grateful.



Average measure of intersection of a convex region with its translate
The following proposition answers OP's question regarding the upper bound of $$\tau(C) \Doteq f(C)/\lambda^2(C).$$


Let $B\_n$ be the closed Euclidean unit ball of $\mathbb{R^n}$ centred at $0$, that is $$B\_n = \{ (x\_1,\dots, x\_n) \in \mathbb{R^n} \, \vert \,\, 
x\_1^2 + \cdots + x\_n^2 \le 1\},$$
and let $\tau\_n = \tau(B\_n)$.



> 
> **Proposition.**
> The Euclidean unit ball $B\_n \subset \mathbb{R}^n$ maximizes $\tau(C)$ among all convex subsets of finite and positive Lebesgue measure, i.e., $\tau(C) \le \tau\_n$ holds for any such $C$.
> 
> 
> 


Claim 3 below gives a way to compute efficiently $\tau\_n$ for all $n > 0$.


The proof of the proposition follows Fedja's idea. The first part consists in proving that $\tau$ doesn't decrease after [Steiner symmetrization][2]. This is Claim 4 below. The second part uses [1, Theorem 7.1.10] showing that some Euclidean ball centred at $0$ can be obtained as the limit of Steiner symmetrizations of $C$ for any compact convex $C$ of positive Lebesgue measure (equivalently, with non-empty interior).


When $n = 1$, the result is slightly stronger.



> 
> **Claim 1.** Let $C \subset \mathbb{R}$ be a convex subset of finite Lebesgue measure $\lambda(C) > 0$. Then $\tau(C) \le \frac{3}{4}$ and equality holds if and only if $C$ is centred at $0$.
> 
> 
> 


So, the above claim settles OP's question when $n = 1$. Unsurprisingly, the result generalizes to $\mathbb{R}^n$ in the following way.



> 
> **Claim 2.** Let $C \subset \mathbb{R}^n$ be a parallelotope of positive Lebesgue measure. Then $\tau(C) \le (\frac{3}{4})^n$ and equality holds if and only if $C$ is centred at $0$.
> 
> 
> 


Computing the maximum $\tau\_n$ of $\tau$ can be achieved via the formula



> 
> **Claim 3.** We have
> $$ \tau\_n = \frac{3}{2} \frac{\int^{\pi/3}\_0 \sin^n(\theta) d\theta}{\int^{\pi/2}\_0 \sin^n(\theta) d\theta}.$$
> In particular, $\tau\_1 = \frac{3}{4}, \, \tau\_2 = 1 - \frac{3 \sqrt{3}}{4 \pi}, \tau\_3 = \frac{15}{32}, \, \tau\_4 = 1 - \frac{9 \sqrt{3}}{8 \pi},
> \tau\_5 = \frac{159}{512}, \, \tau\_6 = 1 - \frac{27 \sqrt{3}}{20 \pi}
> $.
> 
> 
> 


The above formula can be rephrased as $\tau\_n = \frac{3}{2}\frac{W\_n'}{W\_n}$ where $W\_n$ is the $n$-th [Wallis' integral][3] and $W\_n'$ is recursively defined by $W'\_0 = \frac{\pi}{3}, \,W\_1' = \frac{1}{2}$ and
$W\_n' = - \frac{1}{2n}(\frac{\sqrt{3}}{2})^{n - 1} + \frac{n - 1}{n}W'\_{n - 2}$. The latter formula allows an efficient computation of $\tau\_n$ for any $n > 0$. As $W\_n' \le \frac{\pi}{3}(\frac{\sqrt{3}}{2})^n$ and $W\_n \sim \sqrt{\frac{\pi}{2n}}$, the sequence $(\tau\_n)\_n$ converges to $0$ exponentially fast (we have for instance $10^{-7} < \tau\_{91} < 10^{-6}$).


Before proving the claims, let us state some obvious, but important, remarks.


**Remark 1.** Let $A \in GL\_n(\mathbb{R})$ and $d \in \mathbb{R}^n$. Let $C$ be any subset of $\mathbb{R}^n$. Then we have
$$f(AC) = \vert \det(A) \vert^2f(C), \quad f(C + d) = \int\_C \lambda\left(C \cap (C + d + x) \right)d\lambda(x).$$


In particular the ratio $\tau(C)$ is $GL\_n(\mathbb{R})$-invariant but it is **not invariant under translation**. Indeed, if $C$ is a convex subset of finite positive Lebesgue measure whose interior contains $0$, then the function $f\_C(d): d \mapsto f(C + d)$ has a non-empty and bounded support.


The following tightly relates to OP's question.



> 
> **Question.** Let $C \subset \mathbb{R}^n $ be a convex subset of finite positive Lebesgue measure. For which $d$ is $f\_C(d)$ maximal?
> 
> 
> 


The following relates $\tau$ to Steiner symmetrization.



> 
> **Claim 4.** Let $C \subset \mathbb{R}^n $ be a convex subset of finite positive Lebesgue measure. Let $H =\{0\} \times \mathbb{R}^{n -1}$ and let $C' \Doteq \text{St}\_H(C)$ be the [Steiner symmetrization][2] of $C$ with respect to $H$. Then we have $\tau(C) \le \tau(C')$.
> 
> 
> 


We now turn to the proofs of the above claims.



> 
> *Proof of Claim 1.* Because of Remark 1, we can restrict, without loss of generality, to intervals of the form $[-1, 1] + d$ with $d \ge 0$.
> A direct computation yields
> $$
> f([-1, 1] + d) = \int\_{-1}^1 (2 - \vert d + x \vert)^{+} dx = 
> \left\{\begin{array}{cc} 
> 3 -d^2 & \text{ if } 0 \le d \le 1, \\
> \frac{1}{2}(3 - d)^2 & \text{ if } 1 \le d \le 3, \\
> 0 & \text{ if } d \ge 3.
> \end{array}\right.
> $$ where $g^{+} \Doteq \max(g ,0)$ for $g$ a real-valued function. The result follows.
> 
> 
> 



> 
> *Proof of Claim 2.*
> By Remark 1, we can restrict to the convex sets of the form $C = [-1, 1]^n + d$ with $d = (d\_1, \dots, d\_n) \in \mathbb{R}^n$.
> We have $f(C) = \int\_{[-1, 1]^n} (2 - \vert x\_1 + d\_1 \vert)^{+} \cdots (2 - \vert x\_n + d\_n\vert)^{+}dx\_1 \cdots dx\_n$. It follows from Claim 1, that $f(C) \le \int\_{[-1, 1]^n} (2 - \vert x\_1 \vert)^{+} \cdots (2 - \vert x\_n \vert)^{+}dx\_1 \cdots dx\_n = 3^n$ and the inequality is strict whenever one of the $d\_i$'s is not zero. The result follows.
> 
> 
> 


We denote by $V\_n(r)$ the Lebesgue measure of the Euclidean ball of radius $r \ge 0$, that is, $V\_n(r) = r^n\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2} + 1)}$, and let $\nu\_n = V\_n(1) = \lambda(B\_n)$.
Given a real number $h$ such that $-1 \le h \le 1$, we denote by $B\_n^{\text{cap}}(h)$ the *hyperspherical cap of height $h$ along $x\_n$* , that is the set of points $x = (x\_1, \dots, x\_n) \in B\_n$ such that $x\_n \ge h$. We also set $V\_n^{\text{cap}}(h) = \lambda(B\_n^{\text{cap}}(h))$.



> 
> *Proof of Claim 3.* By definition, we have $f(B\_n) = \int\_{B\_n} \lambda(B\_n \cap (B\_n + x)) d\lambda(x) = 2 \int\_{B\_n} V\_n^{\text{cap}}(\frac{\vert x \vert}{2}) d\lambda(x)$ where $\vert \cdot \vert$ stands for the Euclidean norm of $\mathbb{R}^n$. Changing to [$n$-dimensional spherical coordinates][4], we get
> $$f(B\_n) = 2 \int\_{D\_n} V\_n^{\text{cap}}(\frac{r}{2})r^{n - 1} \sin^{n - 2}(\phi\_1)\sin^{n - 3}(\phi\_2) \cdots \sin(\phi\_{n - 1}) dr d\phi\_1 d\phi\_2 \cdots d \phi\_{n - 1}$$ where $D\_n = [0,2\pi] \times [0, \pi] \times \cdots \times [0, \pi] \times [0, 1]$.
> Using the classical computation of $\nu\_n = \lambda(B\_n)$, we deduce that
> $$f(B\_n) = 2n\nu\_n \int\_{[0, 1]} V\_n^{\text{cap}}(\frac{r}{2})r^{n - 1}dr
> = 2^{n + 1}n\nu\_n \int\_{[0, \frac{1}{2}]} V\_n^{\text{cap}}(s)s^{n - 1}ds
> .$$ Since $V\_n^{\text{cap}}(s) = \int\_s^1 V\_{n - 1}(\sqrt{1 - h^2})dh = 
> \nu\_{n - 1} \int\_0^{\text{arcos}(s)} \sin^n(\theta)d \theta$, we obtain $$f(B\_n) = 2^{n + 1}n\nu\_n \nu\_{n - 1} \int\_{[0, \frac{1}{2}]} J\_n(\text{arcos}(s))s^{n - 1}ds$$ where $J\_n(\phi) = \int\_0^{\phi}\sin^n(\theta)d\theta$. Using the substitution $\theta = \text{arcos}(s)$, we get
> $$
> \begin{eqnarray}
> \frac{f(B\_n)}{2^{n + 1}n\nu\_n \nu\_{n - 1}} &=& \int\_{[\frac{\pi}{3}, \frac{\pi}{2}]} J\_n(\theta)\cos(\theta)^{n - 1} \sin(\theta)d\theta \\ 
> &=& \frac{[J\_n(\theta) \cos^n(\theta)]^{\frac{\pi}{3}}\_{\frac{\pi}{2}} + \int^{\frac{\pi}{2}}\_{\frac{\pi}{3}} (\cos(\theta)\sin(\theta))^n d\theta}{n} \\
> &=& \frac{2J\_n(\frac{\pi}{3}) + \int^{\pi}\_{\frac{2\pi}{3}} \sin(\theta)^n d\theta}{2^{n + 1} n} \\
> &=& \frac{3J\_n(\frac{\pi}{3})}{2^{n + 1} n}.
> \end{eqnarray}
> $$
> Now the result follows from the identity $\frac{\nu\_{n - 1}}{\nu\_n} = \frac{1}{2 W\_n}$.
> 
> 
> 


Claim 4 is a specialization of



> 
> **Fedja's Lemma.** Let
> $$\mathcal{I}(C, D, E) \Doteq \int\_C \lambda(D \cap (E + x)) d\lambda(x)$$ where $C$, $D$ and $E$ are convex subsets of $\mathbb{R}^n$ of positive finite Lebesgue measure. Then $\mathcal{I}(C, D, E) \le \mathcal{I}(C', D', E')$, where $X'$ denotes the Steiner symmetrization of $X$ with respect to $H = \{0\} \times \mathbb{R}^{n - 1}$.
> 
> 
> 


The proof of Fedja's Lemma can be reduced to the case $n = 1$. The proof of the case $n = 1$ is eased off by



> 
> **Lemma 1.** Let $g: \mathbb{R} \rightarrow \mathbb{R}^{+} \cup \{0\}$ be an even non-negative function which is concave on its support and non-decreasing on the negative real numbers. Then we have
> $$\int\_I g d\lambda \le \int\_{I'} g d\lambda $$ for every convex subset $I$ of $\mathbb{R}$ where $I'$ is the closed interval of measure $\lambda(I)$ centred at $0$.
> 
> 
> 



> 
> *Proof of Lemma 1.* The non-negative number $\int\_I g d\lambda$ is the area of convex subset $C(I) \subset \mathbb{R}^2$ located between the $x$-axis and the graph of $g$. By hypothesis, we have $C(I)' \subset C(I') = \int\_{I'} g d\lambda$, where $C(I)'$ is Steiner symmetrization of $C(I)$ with respect to $H = \{0\} \times \mathbb{R}$. As $C(I)$ and $C(I)'$ have the same area, the result follows.
> 
> 
> 


We are now in position to prove Fedja's Lemma.



> 
> *Proof of Fedja's Lemma.*
> Let us assume first that $n = 1$.
> As $\mathcal{I}(C + x, D, E) = \mathcal{I}(C, D, E + x)$ and $\mathcal{I}(C, D + x, E + x) = \mathcal{I}(C, D, E)$, we can assume, without loss of generality, that $C$ and $D$ are centred at zero.
> Let us write $E = E’ + \eta$ and let $g(x) \Doteq \lambda(D \cap (E' + x))$.
> Then we have
> $$
> g(x) = \left\lbrace \begin{array}{ccc}
> \min(d, e) & \text{ if } \vert x \vert \le \frac{ \vert d - e \vert }{2}, \\
> \frac{d + e}{2} - \vert x \vert & \text{ if } \frac{ \vert d - e \vert }{2} \le \vert x \vert \le \frac{d + e}{2}, \\
> 0 & \text{ if } \vert x \vert \ge \frac{d + e}{2},
> \end{array}\right.
> $$
> where $d = \lambda(D) > 0$ and $e = \lambda(E) > 0$.
> The function $g$ satisfies the hypothesis of Lemma $1$, thus $$
> \mathcal{I}(C, D, E) = \mathcal{I}(C + \eta, D, E’) = \int\_{C + \eta} g(x) d\lambda(x) \le \int\_{C} g(x) d\lambda(x) = \mathcal{I}(C, D, E’)$$ which proves the result for $n = 1$.
> Assume now that $n > 1$. We denote by $\lambda\_{n - 1}$ the Lebesgue measure on $\mathbb{R}^{n - 1}$ and we identify $H$ with $\mathbb{R}^{n - 1}$.
> For $\xi \in \mathbb{R^n}$, we write $\xi = (\xi\_1, \zeta)$, with $\xi\_1 \in \mathbb{R}$ and $\zeta \in H$.
> Given $A \subset \mathbb{R}^n$ and $\zeta \in H$, we set $A^{\zeta} \Doteq \{ \xi\_1 \, \vert \, (\xi\_1, \zeta) \in A\} \subset \mathbb{R}$. Let $x = (x\_1, z) \in \mathbb{R} \times H$.
> Observe that $(A^{\zeta})’ = (A’)^{\zeta}$ and $(A + x)^{\zeta} = A^{\zeta - z} + x\_1$.
> By the Fubini-Tonelli Theorem, we have
> $\lambda\_n(D \cap (E + x)) = \int\_H \lambda\_1(D^{\zeta} \cap (E^{\zeta - z} + x\_1)) d \lambda\_{n - 1}(\zeta)$.
> We shall use the abbreviations $dz = d \lambda\_{n - 1}(z)$ and $d\zeta = d \lambda\_{n - 1}(\zeta)$.
> Using Fubini-Tonelli’s Theorem and the case $n = 1$, we obtain
> $$
> \begin{eqnarray}
> \mathcal{I}(C, D, E) & = & \int\_{H^2}
> \mathcal{I}(C^{z}, D^{\zeta}, E^{\zeta - z})d\zeta dz \\
> & \le & \int\_{H^2} \mathcal{I}((C’)^{z}, (D’)^{\zeta}, (E’)^{\zeta - z}) d\zeta dz \\
> & = & \mathcal{I}(C’, D’, E’). 
> \end{eqnarray}
> $$
> 
> 
> 


Eventually, we can prove the proposition.



> 
> *Proposition's proof*. By [1, Theorem 7.1.10], there is a Euclidean ball $B$ centred at zero which is the limit in the Hausdorff distance topology of Steiner symmetrizations of $C$ with respect to linear hyperplanes. As $\lambda$, $f$ and hence $\tau$ are continuous with respect to the Hausdorff distance topology (we use that $C$ has non-empty interior), we have $\lambda(B) = \lambda(C) > 0$ and we infer from Claim 4 that $\tau(C) \le \tau(B)$. By Remark 1, we have $\tau(B) = \tau\_n$, which completes the proof.
> 
> 
> 




---


[1] S. G. Krantz, H. R. Parks, "The Geometry of Domains in Space", 1999.  

[2]: <https://en.wikipedia.org/wiki/Symmetrization_methods>
[3]: <https://en.wikipedia.org/wiki/Wallis%27_integrals>
[4]: <https://en.wikipedia.org/wiki/N-sphere>


