2
$\begingroup$

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)?

$\endgroup$
1
  • $\begingroup$ I gather you're mostly interested in the Nikodym distance of $K$ to a ball. Do you expect the Nikodym distance between $K$ and the ball to be at most $c_n \epsilon$, where $c_n$ depends only on the dimension of the space? $\endgroup$ May 6, 2013 at 19:20

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thank you! This was very helpful. Maybe I'm missing something here, but couldn't you apply the same argument for $f(x)^n$ without using the Hausdorff distance result to drop the power, and get that $$\int_{S^{n-1}} |f(x)^n - r_0| dx \leq \epsilon$$ ? This would give a constant independent of the dimension. $\endgroup$
    – alex
    May 10, 2013 at 10:07
  • $\begingroup$ @alex: I think you are right. This simplification did not occur to me. $\endgroup$ May 10, 2013 at 12:01
2
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ I think one just needs to add that every rotation is a composition of reflections and so the fact that the body $K$ or its gauge function is almost reflection invariant implies that is is almost invariant under rotations. $\endgroup$ May 6, 2013 at 18:14
  • $\begingroup$ Yes, I will make an update. $\endgroup$ May 6, 2013 at 18:25
  • $\begingroup$ You are correct that the statement follows, but with an estimate of order of $\epsilon^{1/n}$ for the distance from the ball. I was hoping that one can get a better estimate without using Hausdorff distance. $\endgroup$
    – alex
    May 6, 2013 at 18:34
  • 2
    $\begingroup$ It is not immediately clear that the diameter is bounded. $\endgroup$ May 6, 2013 at 18:38
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.