Measuring the distance of a convex set from a ball (Nikodym distance)

Question

Assume we have a centrally symmetric convex set $K \subset \mathbb{R}^n$ such that Vol(K)=1. In addition, assume that for every direction $u$ we know that $Vol(K \Delta R_u(K)) < \epsilon$, where $A \Delta B$ is the symmetric difference and $R_u(K)$ denotes the reflection of $K$ with respect to $u^\perp$. Does this imply that $K$ is close (in terms of $\epsilon$) to a Euclidean ball (in the same metric)?

Answer 1

Let $r$ be the radius of the maximal ball contained in $K$ and $R$ the radius of the minimal ball containing $K$. I claim that $\frac rR>1-2\epsilon^{1/n}$. It follows that the Hausdorff distance from $K$ to a ball is bounded by $C_n\epsilon^{1/n}$. The argument does not use symmetry (although the constant can be improved in the symmetric case).

Let $v$ and $v'$ be unit vectors such that $Rv$ and $rv'$ belong to the boundary of $K$. Apply a reflection which sends $v'$ to $v$ and let $K'$ be the resulting body. Let $h$ be the homothety centered at $Rv$ with ratio $\frac{R-r}{2R}$. Then $h(K)$ does not intersect the interior of $K'$ because they are separated by the hyperplane through $rv$ orthogonal to $v$. On the other hand, $h(K)\subset K$ due to convexity. Thus $h(K)\subset K\setminus int(K')$, hence $Vol(h(K))\le Vol(K\Delta K')<\epsilon$. But $Vol(h(K))=\left(\frac{R-r}{2R}\right)^n=\frac1{2^n}\left(1-\frac rR\right)^n$. Hence $1-\frac rR<2\epsilon^{1/n}$ as claimed.

Added later.

Now let us show that the Nikodym distance from $K$ to a ball is bounded by $O(\epsilon)$. Let $f:S^{n-1}\to\mathbb R_+$ be the radial function of $K$. Then $$ Vol(K\Delta R_u(K)) =\frac1n \int_{S^{n-1}} |f(x)^n-f(R_ux)^n| dx \sim c(n) \int_{S^{n-1}} |f(x)-f(R_ux)| dx . $$ The last equivalence holds because we already know that $f$ is uniformly close to a known constant. Thus $f$ lies within $L^1$ distance $O(\epsilon)$ from every its reflection and hence from any rotation. Averaging over all rotations yields that $$ \int_{S^{n-1}}\int_{S^{n-1}} |f(x)-f(y)| dxdy \le C\epsilon $$ (where $C$ depends on $n$). This easily implies that $f$ lies within $L^1$ distance $O(\epsilon)$ from a constant. Indeed there is $r_0$ such that the volumes of both sets $A=\{x:f(x)\le r_0\}$ and $B=\{x:f(x)\ge r_0\}$ are at least half the total volume. Restricting the above integral to $x\in A$ and $y\in B$ shows that the integral mean of $|f(x)-f(y)|$ is at least 1/4 of the integral mean of $|f(x)-r_0|$ over $x\in S^{n-1}$. Thus $$ \int_{S^{n-1}} |f(x)-r_0| dx \le C_1\epsilon $$ for some $C_1=C_1(n)$. This means that $K$ lies within Nikodym distance $C_2\epsilon$ from the ball of radius $r_0$ and hence within Nikodym distance $C_3\epsilon$ from the ball of volume 1.

Answer 2

For convex closed sets with uniform upper-diameter and lower-volume bounds, the Nikodym distance is equivalent to Hausdorff distance.

Your condition for the Hausdorff distance, implies that $K$ is close to any ball which does not contain $K$ and is not contained in $K$. Hence the statement follows.

Answer 3

Isn't the diameter bounded by square root of n?