** Rearrangement Inequalities (cf Lieb–Loss, Ch 3) **

DefinitionGiven a Borel set of finite measure, we define , an open ball in , with chosen so that . Similarly, we denote the (radially) symmetric decreasing rearrangement of a Borel function as follows: Suppose has finite measure for every and define

Remark

- .
- As a positive linear combination of spherically symmetric functions that are decreasing in radius, so has both of these properties.
- and are equi-measurable, that is,
To see this, note that

So in particular

which is to say and are equi-measurable.

- From the above,

PropositionFix measurable, thenFor any convex increasing function with (e.g., )

Moreover, if is strictly radially decreasing (that is, for all there is such that — note that cannot be compactly supported, nor have plateaus) and is strictly convex then equality occurs in either case if and only if .

ExampleIf , then both and give the same answer as without .

*Proof:* Suppose obey

and . This is implied by

and integrating over . For example, consider or . Note that in the second case we get

We will show that when ,

Now

Note that for equality we must have for almost everywhere and so in our example for a.e. .

Theorem (Riesz Rearrangement Inequality)For measurable non-negative

Moreover, if is strictly radially decreasing and equality eoccurs, then and for some .

RemarkBy linear changes of variables, one can essentially permute and modulo possible additional reflections.

**Application** (Non-rotating) stars are spherically symmetric. Minimize

Passing from to preserves but will decrease (because of the second term, cf theorem above) unless is already spherically symmetric about some point .

*Proof:* (of the 1D case, without cases of equality) First write and similarly for to see that it suffices to treat the case where and are characteristic functions of measurable sets.

By standard approximation arguments, we can pretend that our our open sets are finite collections of open intervals, e.g., with mutually disjoint closures.

Now add a new `time’ parameter

and similarly for . Note that when , then . We claim that is a decreasing function of , at least up until the first collision of a pair of intervals, for which it suffices to treat one summand from each of and . In this way, we are claiming that

Combining this with the natural amalgamation on collision algorithm yields the claim.

Before going to the multi-dimensional case, we first concretize a notion.

Definition (Steiner symmetrization)Given and a direction we define, writing with ,

Note that this is actually measurable! Oops I missed a picture. By Fubini, the -dimensional measure of is a measurable function of , and this measurable function completely describes the set .

*Proof:* (of the 2D case, as a model for general dimension, without cases of equality) Let and denote the Steiner symmetrization in the and directions, respectively: .

Let denote rotation by some . As in the 1D cases, it suffices to treat the case where , and are characteristic functions of bounded measurable sets, say . Now define with and similarly for and . By the 1D Riesz inequality (and the fact that is linear and measure preserving)

We just need to show that say a.e. (and DCT applies because all of the ‘s live in the same big ball that contain ) or or others, and same for .

Oops there is another picture here. Now, letting be some measurable subset of , we have with decreasing (lower semicontinuous).

We have sets with decreasing and uniformly bounded (by the radius above) with support contained in a common, compact set (e.g., ). By the Cantor diagonal slash trick we can find a subsequence of our that converge at each rational . Because our functions are monotone, we obtain pointwise convergence everywhere except a countable set where the limit function has jumps (there are at most countably many of them). Thus, along this sequence, converge almost everywhere (and so also in ) to for some with decreasing. We need to show is actually ; by the Dominated Convergence Theorem, we obtain . So, it suffices to show is actually a ball. Fix , which is strictly radially decreasing (in the technical sense that every appears as a super-level set). By the -contracting property of rearrangements (which incidentally is equivalent to the enhancement-of-dot-products property, noting ), therefore is decreasing in and so converges (even without passing to our subsequence). In particular,

From the case of equality for our first arrangement inequalities,

By our first rearrangement inequalities, we have

By our previous computation, therefore equality holds.

So now let us recall when equality holds in the 1D case: as is already symmetrical, so must the other functions be symmetrical. Therefore, , and hence is symmetrical across the -axis. Undoing the rotation, we see that has two a.e. reflection axes that are at an angle . Thus, the group generated by these two reflections, which is dense in , leaves for all in this group. Note that acts on , indeed and consequently for all and so is almost every a ball.

Recall that if is strictly radially decreasing, then equality requires and . So, for the proof of cases of equality, we use induction on dimension. Moreover we just need to consider the case when and for and being sets of finite measure. Similarly, writing as an integral over the characteristic functions of its super-level sets, our problem reduces to the following: if for a.e. ,

then .

In the one-dimensional case, if , then for almost and and so and thus for a.e. . Sending within this set, we see that every pair of Lebesgue points of and are separated by no more than . From this and elementary geometry, we see that and are a.e. intervals with a common center.

For the higher dimensional case,

and similarly for the right hand side. By Riesz, we have inequality of inner integrals for all , so equality of requires equality for a.e. and . Thus for a.e. and ,

and similarly for the slices of , with the same translation parameter . Because does not depend on nor on , we see that is a.e. constant. Thus, there is an axis parallel to and passing through so that and both are symmetric about this axis. We can make the same argument for a second axis, say one with an angle , and so see that and are a.e. invariant for a dense set of rotations about the intersection of the axes (If the intersections do not meet, we could use successive rotations to march our set to infinite, contradicting the fact that is invariant and of finite measure.) To finish, we use the fact that is a continuous function on the group of Euclidean motions (affine group, or ).

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