# Christ’s result on near-equality in Riesz-Sobolev inequality

It’s finally time to address one of Christ’s papers I talked about in the previous two blogposts. As mentioned there, I’ve chosen to read the one about the near-equality in the Riesz-Sobolev inequality because it seems the more approachable, while still containing one very interesting idea: exploiting the additive structure lurking behind the inequality via Freiman’s theorem.

1. Elaborate an attack strategy

Everything is in dimension ${d=1}$ and some details of the proof are specific to this dimension and don’t extend to higher dimensions. I’ll stick to Christ’s notation.

Recall that the Riesz-Sobolev inequality is

$\displaystyle \boxed{\left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle \leq \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle} \ \ \ \ \ (1)$

and its extremizers – which exist under the hypothesis that the sizes are all comparable – are intervals, i.e. the intervals are the only sets that realize equality in (1). See previous post for further details. The aim of paper [ChRS] is to prove that whenever ${\left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle}$ is suitably close to ${\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle}$ (i.e. we nearly have equality) then the sets ${A,B,C}$ are nearly intervals themselves.

As explained in the previous post, there is one powerful tool to prove that a set must be close to an interval: the continuum version of Freiman’s theorem, which asserts that if ${A}$ is a measurable set in ${\mathbb{R}}$ s.t. ${|A+A| < 3|A|}$ then ${A}$ nearly coincides with an interval (and this is bounded in size). I’ll restate it for convenience:

Theorem 1 (continuum Freiman’s theorem) Let ${A\subset \mathbb{R}}$ be a measurable set with finite measure ${>0}$. If

$\displaystyle |A+A|< 3|A|,$

then there exists an interval ${I}$ s.t. ${A\subset I}$ [1] and

$\displaystyle |I| \leq |A+A|-|A|.$

The problem is that we have no other information about ${A,B,C}$ other than the fact that (1) is nearly an equality when applied to these three sets – and we definitely need some strong information about the additive behaviour of the sets to prove something like ${|A+A|< 3|A|}$ out of the blue. Thus we are forced to reduce to well behaved sets. The most natural direction to look for is the superlevel sets of ${\chi_A \ast \chi_B}$, defined by

$\displaystyle S_{A,B}(t):=\{x \in \mathbb{R}\,:\, \chi_A \ast \chi_B (x)>t\},$

and at this point the (naive) strategy to prove ${C}$ is close to an interval is the following:

Strategy draft \# 1

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(t)}$ for some ${t}$ that will likely depend on all ${A,B,C}$, by proving that ${C}$ nearly coincides with ${S_{A,B}(t)}$;
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(t)+S_{A,B}(t)|< 3|S_{A,B}(t)|}$.

This would suffice, because then ${S_{A,B}(t)}$ would nearly coincide with an interval by Freiman’s theorem 1 and so ${C}$ would nearly coincide with an interval itself. This is obviously too naive to be expected to work, but there is something already there: indeed, one can prove that if ${A,B,C}$ nearly realizes inequality then ${C}$ nearly coincides with ${S_{A,B}(\tau)}$, where ${\tau}$ is s.t.

$\displaystyle |C| = |A| + |B| - 2 \tau,$

i.e. ${\tau = (|A|+|B|-|C|)/2}$. Thus point (1.) isn’t a problem. I should also mention that ${A,B,S_{A,B}(\tau)}$ realizes near-equality better than ${A,B,C}$ (the difference between the RHS and LHS of (1) is smaller).

The strategy will be refined and updated in these notes. Turns out there is no direct relationship between ${S_{A,B}(t)+S_{A,B}(t)}$ and ${S_{A,B}(t)}$, but there is a relationship between ${S_{A,B}-S_{A,B}}$ and ${S_{A,-A}}$ of the form

$\displaystyle S_{A,B} (t) - S_{A,B} (t') \subset S_{A,-A}(t+t' - |B|). \ \ \ \ \ (2)$

Remark 1 I haven’t discussed this point in the related post, but discrete Freiman’s theorem conclusion (and thus the continuum one) applies as well to the case ${|A-A|< 3|A|-3}$, and this follows from two facts: first, that the quantitative lemma where one estimates ${|A+A|}$ from below by partitioning in accordance to the remainders modulus some ${N}$ holds as well for ${|A-A|}$; second, by the fact that Kneser’s theorem is stated for two general sets ${A,B}$. If you don’t know what I’m talking about, check the blogpost.

Thus we have a new refined strategy:

Strategy draft \# 2

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(\tau)}$ for ${\tau}$ as above, by proving that ${C}$ nearly coincides with ${S_{A,B}(\tau)}$; ✓
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(\tau)-S_{A,B}(\tau)|< |S_{A,-A}(2\tau-|B|)|}$; ✓
3. prove that ${|S_{A,-A}(2\tau-|B|)| < 3|S_{A,B}(\tau)|}$ under the hypotheses made;

then the result would follow from Freiman’s theorem as above. There isn’t a direct obvious way to prove something like (3.): point (2.) is tackled with a set theoretic inclusion, which is geometric information after all, but no such inclusion holds for those sets. Moreover, by Brunn-Minkowski, we seem to be going in the opposite direction: since ${|S_{A,B}(\tau)-S_{A,B}(\tau)| \geq 2|S_{A,B}(\tau)|}$, we have

$\displaystyle 2|S_{A,B}(\tau)| \leq |S_{A,-A}(2\tau-|B|)|.$

Do we have to discard the strategy? no, maybe we’re rushing. Since ${S_{A,B}(\tau)}$ is about the same size as ${C}$, we can contempt ourselves with proving that ${|S_{A,-A}(2\tau-|B|)|}$ isn’t much larger than ${2|C|}$.

Let’s shorten ${2\tau-|B|}$ to ${\gamma}$ and notice that ${\gamma = |A|-|C|}$. A moment’s thought reminds us that the inclusion relation (2) holds for all values of ${t}$, not just for our ${\tau}$. It might be worth it to integrate instead, as the integral over the superlevel sets has a clear interpretation, and see if we can squeeze out some information this way: since ${2|S_{A,B}(t)| \leq |S_{A,-A}(2t-|B|)|}$, integrating from ${\tau}$ to ${\infty}$ one gets

$\displaystyle \int_{\gamma}^{\infty}{|S_{A,-A}(t)|}\,dt \geq 4 \int_{\tau}^{\infty}{|S_{A,B}(t)|}\,dt. \ \ \ \ \ (3)$

Now, for a generic function ${F}$ and ${t>0}$ it is

$\displaystyle \int{|F(x)|}\,dx =\int_{t}^{\infty}{\left|\left\{|F|>u\right\}\right|}\,du + t\left|\left\{|F|>t\right\}\right| + \int_{\left\{|F|\leq t\right\}}{|F(x)|}\,dx$

$\displaystyle = \int_{t}^{\infty}{\left|\left\{|F|>u\right\}\right|}\,du + \int{\min(|F(x)|, t)}\,dx,$

and since ${\int_{\mathbb{R}}{\chi_A \ast \chi_B}\,dx = |A||B|}$ we can write

$\displaystyle \int_{\tau}^{\infty}{|S_{A,B}(t)|}\,dt = |A||B| - \int_{\mathbb{R}}{\min(\chi_A \ast \chi_B (x), \tau)}\,dx \ \ \ \ \ (4)$

(and an analogous expression for ${S_{A,-A}}$). We’ve just been rewriting our expression, but doing so we’ve introduced a quantity that has actually been studied before. I’m talking here about the last integral I’ve written: one has the so-called KPRGT inequality [2]

$\displaystyle \int_{\mathbb{R}}{\min(\chi_A \ast \chi_B (x), t)}\,dx \geq t \left(|A|+ |B| - t\right).$

But, because of (4), this can be recast as

$\displaystyle \int_{t}^{\infty}{|S_{A,B}(u)|}\,du \leq (|A|-t)(|B|-t).$

If one denotes by ${\mathcal{D}' (A,B,t)}$ the difference

$\displaystyle \mathcal{D}' (A,B,t)= (|A|-t)(|B|-t)-\int_{t}^{\infty}{|S_{A,B}(u)|}\,du,$

then (3) can be rewritten as

$\displaystyle (|A|-\gamma)^2 - \mathcal{D}'(A,-A,\gamma) \geq 4(|A|-\tau)(|B|-\tau) - 4\mathcal{D}'(A,B,\tau),$

and since by the definitions it is ${|A|-\gamma = |C|}$ and ${4(|A|-\tau)(|B|-\tau) = |C|^2 - (|A|-|B|)^2}$, the above is equivalent to

$\displaystyle (|A|-|B|)^2 + 4\mathcal{D}'(A,B,\tau) \geq \mathcal{D}'(A,-A,\gamma). \ \ \ \ \ (5)$

It’s important to notice that this inequality is tighter if ${|A|=|B|}$. Now, our hope is that we can extract information from this quantity ${\mathcal{D}'}$, and it is indeed the case. It’s easily seen to be related to the Riesz-Sobolev inequality in the following way: define analogously

$\displaystyle \mathcal{D}(A,B,C) :=\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle - \left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle,$

which is a non-negative quantity. Now, for ${C = S_{A,B}(t)}$, we have ${\left\langle \chi_{A} \ast \chi_{B}, \chi_{S_{A,B}(t)}\right\rangle = \int_{S_{A,B}(t)}{\chi_A \ast \chi_B}}$, a quantity we know how to relate to ${\int{\min(\chi_A \ast \chi_B, t)}}$; but what about ${\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{{S_{A,B}(t)}^\ast}\right\rangle}$? we need to work out what this is. A little algebra shows that, if we define ${\sigma}$ by ${|S_{A,B}(t)| = |A|+|B|-2\sigma}$ (and thus suppose ${\left||A|-|B|\right|\leq |S_{A,B}(t)| \leq |A|+|B|}$), we have

$\displaystyle \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{{S_{A,B}(t)}^\ast}\right\rangle = |A||B|-\sigma^2. \ \ \ \ \ (6)$

Then

$\displaystyle \mathcal{D}'(A,B,t) = (|A|-t)(|B|-t) - \int_{t}^{\infty}{|S_{A,B}(u)|}\,du$

$\displaystyle = |A||B| - t (|A|+|B|-t) - \int_{S_{A,B}(t)}{\chi_A \ast \chi_B}\,dx + t |S_{A,B}(t)|$

$\displaystyle = \mathcal{D}(A,B,S_{A,B}(t)) + \sigma^2 - t (|A|+|B|-t) + t |S_{A,B}(t)|$

$\displaystyle = \mathcal{D}(A,B,S_{A,B}(t)) + \sigma^2 - t (|S_{A,B}(t)|+2\sigma -t) + t |S_{A,B}(t)|,$

and therefore

$\displaystyle \mathcal{D}'(A,B,t) = \mathcal{D}(A,B,S_{A,B}(t)) + (\sigma - t)^2$

(notice ${\sigma=\sigma(t)}$). This tells us plenty of things: first of all,

$\displaystyle \mathcal{D}'(A,B,t) \geq \mathcal{D}(A,B,S_{A,B}(t)),$

but also that ${\mathcal{D}'(A,B,t) \sim \mathcal{D}(A,B,S_{A,B}(t))}$ if we have some control of the kind ${|\sigma - t| \lesssim \mathcal{D}(A,B,S_{A,B}(t))^{1/2}}$. Turns out that this is the case in our situation above. Indeed,

$\displaystyle (\sigma - \tau)^2 = \left(\frac{|A|+|B|-|S_{A,B}(\tau)|}{2} - \frac{|A|+|B|-|C|}{2} \right)^2 = \frac{1}{4}(|S_{A,B}(\tau)|- |C|)^2,$

and as said we’ve chosen ${\tau}$ so that the difference ${|S_{A,B}(\tau)|- |C|}$ is small, in particular (but I haven’t mentioned it before) one has ${||S_{A,B}(\tau)|- |C||\lesssim \mathcal{D}(A,B,C)}$ (which is assumed small because we’re assuming near-equality); therefore ${\mathcal{D}'(A,B,\tau) \lesssim \mathcal{D}(A,B,S_{A,B}(\tau)) + \mathcal{D}(A,B,C)}$. Luckily though, it’s easy to prove that ${\mathcal{D}(A,B,S_{A,B}(\tau)) \leq \mathcal{D}(A,B,C)}$ (as one would naively expect), so using all of the above we can write for some constant ${c_0>0}$

$\displaystyle (|A|-|B|)^2 + c_0 \mathcal{D}(A,B,C) \geq \mathcal{D}'(A,-A,\gamma) \geq (\nu -\gamma )^2,$

where ${\nu}$ is defined by [3]

$\displaystyle |S_{A,-A}(\gamma)| = |A| + |-A| - 2\nu.$

Then

$\displaystyle \nu - \gamma = \frac{2|A| - |S_{A,-A}(\gamma)| }{2} - (|A|-|C|) = \frac{1}{2}(2|C|-|S_{A,-A}(\gamma)| ),$

and therefore

$\displaystyle (|A|-|B|)^2 + c_0 \mathcal{D}(A,B,C) \gtrsim \left|2|C| - |S_{A,-A}(\gamma)|\right|.$

If ${|A|=|B|}$ this is what we wanted: ${\mathcal{D}(A,B,C)}$ is assumed to be small, and in particular is smaller than ${c_0^{-1}|C|}$, and therefore adjusting the constants we can have ${|S_{A,-A}(\gamma)| < 3|S_{A,B}(\tau)|}$, which was the original goal! On the other hand, if ${|A|\neq |B|}$ this miserably fails, and ${|A|=|B|}$ is an incredibly strong condition to ask for.

We revise the strategy as follows:

Strategy draft \# 3

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(\tau)}$ for ${\tau}$ as above, by proving that ${C}$ nearly coincides with ${S_{A,B}(\tau)}$; ✓
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(\tau)-S_{A,B}(\tau)|< |S_{A,-A}(2\tau-|B|)|}$; ✓
3. prove that ${|S_{A,-A}(2\tau-|B|)| < 3|S_{A,B}(\tau)|}$ under the hypotheses made and the further assumption that ${|A|=|B|}$; ✓
4. find a way to remove the assumption that ${|A|=|B|}$.

To deal with point (4.), Christ resorted to the machinery of truncations. Truncations are defined as follows: a truncation of a set ${E \subset \mathbb{R}}$ of positive measure with parameters ${\xi, \eta > 0}$ that satisfy ${\xi + \eta < |E|}$, is the subset ${E_{\xi, \eta} \subset E}$ given by ${E_{\xi, \eta} = E \cap [a,b]}$, where ${a,b}$ are such that

$\displaystyle \left|(-\infty, a) \,\cap\, E\right| = \xi$

$\displaystyle \left|E \, \cap\, (b, +\infty)\right| = \eta.$

In other words, ${E_{\xi, \eta}}$ is the middle portion of a trisection of ${E}$ s.t. the left section has measure ${\xi}$ and the right section has measure ${\eta}$ (and therefore the ${E_{\xi,\eta}}$ has measure ${|E|-\xi - \eta}$). This device allows to chop off portions of the sets ${A,B,C}$ in such a way that two of them are reduced to having the same size, but the triplet is still a near-extremizer of Riesz-Sobolev inequality. This is natural but not entirely obvious, and I don’t think I can sketch this part as I did for the previous three. Perhaps is time to move on to the rigourous proof. But first let me write the final

Strategy

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(\tau)}$ for ${\tau}$ as above, by proving that ${C}$ nearly coincides with ${S_{A,B}(\tau)}$;
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(\tau)-S_{A,B}(\tau)|< |S_{A,-A}(2\tau-|B|)|}$;
3. prove that ${|S_{A,-A}(2\tau-|B|)| < 3|S_{A,B}(\tau)|}$ under the hypotheses made and the further assumption that ${|A|=|B|}$;
4. Remove the assumption that ${|A|=|B|}$ by using truncations.

2. The actual proof

I will first of all state the theorem of Christ, but that requires a definition. We introduce a parameter to control how the sizes of the sets are comparable:

Definition 2 A triplet of sets ${A_1, A_2, A_3}$ is said to be ${\eta}$-strictly admissible for some ${0<\eta<1}$ if

$\displaystyle |A_i|+|A_j|> |A_k| + \eta \max(|A_1|,|A_2|,|A_3|)$

holds for all ${(i,j,k)}$ that are permutations of ${(1,2,3)}$.

This has as a consequence that the sets are all comparable in sizes, and in particular

$\displaystyle \min(|A_1|,|A_2|,|A_3|) \geq \eta \max(|A_1|,|A_2|,|A_3|),$

because suppose ${|A_1|\geq|A_2|\geq |A_3|}$, then

$\displaystyle |A_3| + |A_2| > |A_1| + \eta |A_1|>|A_2| + \eta |A_1|.$

Theorem 3 (Christ, [ChRS]) Let ${A,B,C}$ be a ${\eta}$-strictly admissible triplet of measurable sets for some ${0<\eta <1}$; if

$\displaystyle \mathcal{D}(A,B,C)^{1/2} \lesssim \eta^3 \max(|A|,|B|,|C|),$

then there exist intervals ${I,J,K}$ s.t.

$\displaystyle |A\;\Delta \;I|,|B\;\Delta \;J|,|C\;\Delta \;K| \lesssim \eta^{-1} \mathcal{D}^{1/2}.$

Before we start, we prove (6):

Lemma 4 For ${||A|-|B||<|C|<|A|+|B|}$,

$\displaystyle \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle = |A||B|-\tau^2,$

where ${\tau}$ is defined by ${|C| = |A|+|B|-2\tau}$.

Proof: We write ${a}$ for ${|A|}$ and ${b,c}$ similarly. Assume without loss of generality ${b\leq a}$. The rearranged sets are intervals centered in 0. Write ${\chi_a}$ for ${\chi_{A^\ast} = \chi_{[-a/2,a/2]}}$, and similarly for ${\chi_b}$, ${\chi_c}$. Then

$\displaystyle \chi_a \ast \chi_b (x) = \int_{[-a/2,a/2]}{\chi_b (x+y)}\,dy = |A^\ast \cap (x- B^\ast)|=\begin{cases} b \qquad \text{ if } 2|x| < a-b, \\ \frac{a+b}{2}-|x| \qquad \text{ if } a-b \leq 2|x| \leq a+b,\\ 0 \qquad \text{ if } 2|x| > a+b. \end{cases}$

Integrating from ${-c/2}$ to ${c/2}$,

$\displaystyle \left\langle \chi_a \ast \chi_b , \chi_c\right\rangle = b (a-b) + 2 \int_{\frac{a-b}{2}}^{c/2}{\left(\frac{a+b}{2}-|x|\right)}\,dx$

$\displaystyle = b (a-b) + \frac{1}{2} \int_{a-b}^{c}{\left(a+b-x\right)}\,dx$

$\displaystyle = b (a-b) + (a+b)\frac{c-(a-b)}{2} - \frac{c^2}{4} + \frac{(a-b)^2}{4},$

and doing the algebra this is

$\displaystyle = a b - \left(\frac{a+b-c}{2}\right)^2 = |A||B| - \tau^2.$

$\Box$

Very nice formula indeed. Let me expand it for future reference:

$\displaystyle \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle = |A||B|-\left(\frac{|A|+|B|-|C|}{2}\right)^2.$

2.1. Reduction to superlevel sets

Remember that we defined

$\displaystyle \mathcal{D}(A,B,C) :=\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle - \left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle$

$\displaystyle = |A||B| - \tau^2 - \left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle.$

In the following, let ${\mathcal{D} = \mathcal{D}(A,B,C)}$ for shortness. Our parameter for “small” as used in the previous section will typically be ${\mathcal{D}^{1/2}}$. The proposition addresses point (1.) of the strategy directly.

Proposition 5 If ${A,B,C}$ are such that

$\displaystyle ||A| -|B|| + 2\mathcal{D}^{1/2} < |C|< |A|+|B| - 2\mathcal{D}^{1/2},$

then for ${\tau}$ as before (i.e. ${|C| =|A|+|B|-2\tau}$) it is

$\displaystyle |S_{A,B}(\tau) \; \Delta \; C| \lesssim \mathcal{D}^{1/2}.$

It obviously follows that ${||S_{A,B}(\tau)|- |C|| \lesssim \mathcal{D}^{1/2}}$ as well.

Proof: We prove it separately for ${|S_{A,B}(\tau) \backslash C|}$ and ${|C \backslash S_{A,B}(\tau)|}$. Write ${S}$ for ${S_{A,B}(\tau)}$ for shortness. Suppose ${|C \backslash S| \geq 2\mathcal{D}^{1/2}}$; since ${|C| > ||A|-|B||+2\mathcal{D}^{1/2}}$, we can find a measurable set ${T}$ s.t. ${C \cap S \subseteq T \subset C}$, with size ${|T|\geq ||A|-|B||}$ but ${|C\backslash T|>2\mathcal{D}^{1/2}}$ (e.g. ${T=C\cap S}$ if ${|C\cap S|> ||A|-|B||}$ too). Then, in ${C\backslash T}$ it is ${\chi_A \ast \chi_B \leq \tau}$ by definition of ${S}$; moreover, by the assumption on the size of ${T}$ we can apply the formula in Lemma 4, therefore

$\displaystyle \left\langle\chi_A \ast \chi_B, \chi_C \right\rangle = \left\langle\chi_A \ast \chi_B, \chi_T \right\rangle + \left\langle\chi_A \ast \chi_B, \chi_{C\backslash T} \right\rangle \leq \left\langle\chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{T^\ast} \right\rangle + \tau |C\backslash T|$

$\displaystyle = |A||B| - \left(\frac{|A|+|B|-|T|}{2}\right)^2 + \tau |C\backslash T|= |A||B| - \left(\frac{|C| + 2\tau - |T|}{2}\right)^2 + \tau |C\backslash T|$

$\displaystyle = |A||B| - \left(\tau + \frac{|C\backslash T|}{2}\right)^2 + \tau |C\backslash T|=|A||B| - \tau^2 - \frac{1}{4}\left(|C\backslash T|\right)^2 - \tau |C\backslash T| + \tau |C\backslash T|$

$\displaystyle = |A||B| - \tau^2 - \frac{1}{4}\left(|C\backslash T|\right)^2,$

and thus

$\displaystyle |C\backslash T|^2 \leq 4\mathcal{D},$

The proof for ${|S\backslash C|}$ is similar: one takes any measurable ${T}$ s.t. ${C \subset T \subset C \cup S}$ and s.t. ${|T| \leq |A|+|B|}$, then for all of them proves ${|T \backslash C| \leq 2\mathcal{D}^{1/2}}$ as above, and the result follows for ${S}$ as well. $\Box$

Thus we’ve proved that ${C}$ is nearly a superlevel set, with error of size ${O(\mathcal{D}^{1/2})}$.

Since we’ll need it in the following (it was mentioned above), we also prove

Proposition 6 Under the assumptions above,

$\displaystyle \mathcal{D}(A,B,S_{A,B}(\tau)) \leq \mathcal{D}(A,B,C).$

In other words, ${(A,B,S_{A,B}(\tau))}$ is a tighter near-extremizer than ${(A,B,C)}$.

Proof: One has ${\chi_A \ast \chi_B > \tau}$ on ${S\backslash C}$ and viceversa ${\chi_A \ast \chi_B \leq \tau}$ on ${C \backslash S}$. Therefore

$\displaystyle \left\langle\chi_A \ast \chi_B, \chi_S \right\rangle = \left\langle\chi_A \ast \chi_B, \chi_C + \chi_{S \backslash C} - \chi_{C\backslash S} \right\rangle \geq \left\langle\chi_A \ast \chi_B, \chi_C \right\rangle + \tau |S \backslash C| - \tau |C\backslash S|$

$\displaystyle = |A||B| - \tau^2 -\mathcal{D}(A,B,C) + \tau (|S| - |C|).$

Define ${\sigma}$ by ${|S_{A,B}(\tau)|=|A|+|B|-2\sigma}$ (the analogous of ${\tau}$ for the new triplet), and the last line can be rewritten as

$\displaystyle |A||B| - \tau^2 - \mathcal{D}(A,B,C) + \tau (-2\sigma + 2 \tau)$

$\displaystyle = |A||B| + (\sigma - \tau)^2 - \mathcal{D}(A,B,C) - \sigma^2$

$\displaystyle \geq |A||B| - \sigma^2 - \mathcal{D}(A,B,C),$

and rearranging we get exactly the conclusion. $\Box$

2.2. Additive structure in superlevel sets

This section addresses point (2.) of the strategy.

One has the following additive relation amongst superlevel sets of convolutions of characteristic functions

Proposition 7 Let ${U,V}$ be measurable sets, then

$\displaystyle S_{U,V}(\alpha) - S_{U,V}(\beta) \subset S_{U,-U}(\alpha+\beta - |V|).$

Proof: It will follow from the fact that

$\displaystyle S_{U,V}(\alpha) =\{ x \,:\, \|\chi_{U-x} - \chi_{-V}\|_{L^1} < |U|+|V| - 2\alpha\}.$

To see this, notice that ${\chi_U \ast \chi_V (x) = |(U-x) \cap (-V)|}$, and therefore ${\chi_U \ast \chi_V (x) > \alpha}$ is equivalent to

$\displaystyle 2\alpha < 2|(U-x) \cap (-V)| = |U| + |V| - |(U-x)\;\Delta\; (-V)| = |U|+|V| - \|\chi_{U-x} - \chi_{-V}\|_{L^1}.$

Now, take ${x \in S_{U,V}(\alpha)}$, ${y \in S_{U,V} (\beta)}$ and ${z= x-y}$. Notice the superlevel sets are open, so the points lie in the inside. We want to prove ${z \in S_{U,-U}(\alpha + \beta -|V|)}$. By the above

$\displaystyle \|\chi_{U-x} - \chi_{-V}\|_{L^1} < |U|+|V| - 2\alpha$

and analogously for ${y}$, thus

$\displaystyle \|\chi_{U-z} - \chi_{U}\|_{L^1} \leq \|\chi_{U-z} - \chi_{-V+y}\|_{L^1} +\|\chi_{U} - \chi_{-V+y}\|_{L^1}$

$\displaystyle =\|\chi_{U-x} - \chi_{-V}\|_{L^1} +\|\chi_{U-y} - \chi_{-V}\|_{L^1}< 2|U|+2|V| - 2\alpha - 2\beta = |U|+|-U| - 2(\alpha + \beta -|V|),$

which is equivalent to ${z \in S_{U,-U}(\alpha + \beta -|V|)}$. $\Box$

As we’ve seen before, then, we have

$\displaystyle S_{A,B} (\tau) - S_{A,B} (\tau) \subset S_{A,-A}(2\tau - |B|) = S_{A,-A}(|A|-|C|).$

The next step is then to estimate ${|S_{A,-A}(|A|-|C|)|}$.

2.3. The KPRGT inequality

As mentioned in the first section, we’re gonna need the KPRGT inequality, which I recall:

$\displaystyle \int_{\mathbb{R}}{\min(\chi_A\ast \chi_B (x) , t)}\,dx \geq t (|A|+|B| - t).$

As seen before, it can be recast as

$\displaystyle \int_{t}^{\infty}{|S_{A,B}(u)|}\,du \leq (|A|-t)(|B|-t);$

moreover, if we define the difference

$\displaystyle \mathcal{D}' (A,B,t)= (|A|-t)(|B|-t)-\int_{t}^{\infty}{|S_{A,B}(u)|}\,du,$

then we’ve seen that we have the relationship

$\displaystyle \mathcal{D}'(A,B,t) = \mathcal{D}(A,B,S_{A,B}(t)) + (\sigma - t)^2, \ \ \ \ \ (7)$

where ${|S_{A,B}(t)|=|A|+|B|-2\sigma}$, under the assumption that ${||A|-|B||< |S_{A,B}(t)|< |A|+|B|}$. For ${t = \tau}$ this requirement is satisfied. Notice that (7) is essentially a sharpened KPRGT inequality, since it means ${\mathcal{D}'(A,B,t) - (\sigma - t)^2 \geq 0}$, or

$\displaystyle \int_{t}^{\infty}{|S_{A,B}(u)|}\,du + (\sigma - t)^2 \leq (|A|-t)(|B|-t).$

2.4. Case ${|A|=|B|}$

This section addresses point (3.) of the strategy, under the assumption that ${|A|=|B|}$, assumption that will be removed in the next subsection. Remember that ${\mathcal{D}}$ is short for ${\mathcal{D}(A,B,C)}$.

We want to prove that ${|S_{A,-A} (\gamma)| < 3|S_{A,B}(\tau)|}$, where remember ${\gamma = |A|-|C|}$. We need to be rigorous, so we state:

Proposition 8 Let ${A,B,C}$ be ${\eta}$-strictly admissible, s.t.

• ${|A|=|B|}$;
• ${|A|-|C| \geq 4 \mathcal{D}^{1/2}}$;
• ${4\mathcal{D}^{1/2} < \eta |A|}$.

Then

$\displaystyle ||S_{A,-A} (\gamma)| - 2|C|| \lesssim \mathcal{D}^{1/2}.$

The hypotheses made here are so that the hypothesis in Proposition 5 applies. Indeed ${|C|\leq |A|\leq |A|+|A| - 2\mathcal{D}^{1/2}}$ (since ${\eta\leq 1}$), and ${|C|\geq \eta |A| > 4\mathcal{D}^{1/2}> 2 \mathcal{D}^{1/2}}$ by strict-admissibility. Thus ${||S_{A,B}(\tau)|-|C||\leq 4 \mathcal{D}^{1/2}}$ (check the proof for the factor ${4}$).

Proof: As seen above: the additive relation in Proposition 7 gives us (by Brunn-Minkowski)

$\displaystyle 2|S_{A,B}(t)|\leq |S_{A,B}(t) -S_{A,B}(t)| \leq |S_{A,-A}(2t -|B|)|,$

which when integrated from ${\tau}$ to ${+ \infty}$ becomes

$\displaystyle 4 \int_{\tau}^{\infty}{|S_{A,B}(t)|}\,dt \leq \int_{\gamma}^{\infty}{|S_{A,-A}(t)|}\,dt,$

which by the same algebraic manipulations done before can be equivalently rewritten as (5), that is

$\displaystyle \mathcal{D}'(A,-A,\gamma)\leq 4 \mathcal{D}'(A,B,\tau).$

Now, we want to prove ${\mathcal{D}'(A,B,\tau)\lesssim \mathcal{D}}$ by using (7), but to use that we need to show that ${0=||A|-|B||<|S_{A,B}(\tau)|<|A|+|B| = 2|A|}$. Indeed, as seen just before this proof, ${|S_{A,B}(\tau)|\geq |C| - 4 \mathcal{D}^{1/2}>0}$, and on the other hand ${|S_{A,B}(\tau)| \leq |C| + 4\mathcal{D}^{1/2} \leq |A|+|A|}$. So we can use (7) safely, and doing the calculations (see above)

$\displaystyle \mathcal{D}'(A,B,\tau) \leq \mathcal{D}(A,B,S_{A,B}(\tau)) + \frac{1}{4}\left(|S_{A,B}(\tau)|-|C|\right)^2 \leq \mathcal{D} + \mathcal{D} = 2\mathcal{D}.$

Therefore we have

$\displaystyle \mathcal{D}'(A,-A,\gamma)\leq 8\mathcal{D} \ \ \ \ \ (8)$

so far. We want to invoke (7) again, so that we can say ${\mathcal{D}'(A,-A,\gamma) \geq (\nu - \gamma)^2}$ with ${|S_{A,-A}(\gamma)|=2|A|-2\nu}$, and therefore we have to verify the hypothesis again: ${|S_{A,-A}(\gamma)|>0}$ follows from Brunn-Minkowski since ${|S_{A,B}(\tau)|>0}$, as for the other one suppose the contrary, i.e. that ${|S_{A,-A}(\gamma)|> 2|A|}$. By (8) expanded,

$\displaystyle (|A|-\gamma)^2 - \int_{\gamma}^{+\infty}{|S_{A,-A}(t)|}\,dt \leq 8 \mathcal{D},$

thus

$\displaystyle (|A|-\gamma)^2 - 8 \mathcal{D} \leq \int_{\gamma}^{+\infty}{|S_{A,-A}(t)|}\,dt \leq \int_{0}^{+\infty}{|S_{A,-A}(t)|}\,dt - \gamma |S_{A,-A}(\gamma)|$

$\displaystyle = |A|^2 - \gamma |S_{A,-A}(\gamma)|.$

If ${|S_{A,-A}(\gamma)|> 2|A|}$ then it also is

$\displaystyle (|A|-\gamma)^2 - 8 \mathcal{D} < A|^2 - 2\gamma |A|,$

but then ${|A|-|C| = \gamma < 2 \sqrt{2} \mathcal{D}^{1/2} < 4 \mathcal{D}^{1/2}}$, contradicting the hypotheses of the proposition.

Therefore we can safely use (7) on ${\mathcal{D}'(A,-A,\gamma)}$ too, and hence by the same calculations as before

$\displaystyle \frac{1}{4}\left(2|C| - |S_{A,-A}(\gamma)|\right)^2 = (\nu - \gamma)^2 \leq \mathcal{D}'(A,-A,\gamma) \leq 8 \mathcal{D},$

or

$\displaystyle ||S_{A,-A}(\gamma)| - 2|C||\lesssim \mathcal{D}^{1/2}.$

$\Box$

By using the above proposition we then have that

$\displaystyle |S_{A,-A}(\gamma)| - 2|S_{A,B}(\tau)| \leq |S_{A,-A}(\gamma)| - 2|C| + 2 ||S_{A,B}(\tau)| - |C|| \leq 4\sqrt{2} \mathcal{D}^{1/2} + 8 \mathcal{D}^{1/2} < 16 \mathcal{D}^{1/2}.$

Now, we want ${|S_{A,-A}(\gamma)| < 3 |S_{A,B}(\tau)|}$, which we can enforce: as seen in the proof ${|C| - 4 \mathcal{D}^{1/2} < |S_{A,B}(\tau)|}$, and therefore it suffices to assume that ${|C|> 20 \mathcal{D}^{1/2}}$.

Proposition 9 Let ${A,B,C}$ be ${\eta}$-strictly admissible, s.t.

• ${|A|=|B|}$;
• ${|A|-|C| \geq 4 \mathcal{D}^{1/2}}$;
• ${20\mathcal{D}^{1/2} < \eta |A|}$.

Then there exists an interval ${I}$ s.t.

$\displaystyle |C\; \Delta \; I| \lesssim \mathcal{D}^{1/2}.$

Proof: By the remark before the last statement and the previous proposition, we have

$\displaystyle |S_{A,B} (\tau) - S_{A,B}(\tau)| < 3|S_{A,B}(\tau)|.$

Now apply Freiman’s theorem 1, which yields an interval ${I}$ s.t. ${S_{A,B} \subset I }$ and

$\displaystyle |S_{A,B}(\tau) \;\Delta \; I | \leq |S_{A,B} (\tau) - S_{A,B}(\tau)| - 2 |S_{A,B} (\tau)|\leq 16 \mathcal{D}^{1/2}.$

Finally, since ${|C \; \Delta \; S_{A,B}(\tau)| \leq 4 \mathcal{D}^{1/2}}$ by Proposition 5, we have

$\displaystyle |C\; \Delta \; I| \leq |C \; \Delta \; S_{A,B}(\tau)| + |S_{A,B}(\tau) \;\Delta \; I | \leq 20 \mathcal{D}^{1/2}.$

$\Box$

One small thing to be noticed is that, besides the assumption ${|A|=|B|}$ we had to introduce another assumption, namely that ${|A|-|C|> 4\mathcal{D}^{1/2}}$, or that ${|A|}$ is “measurably” bigger than ${|C|}$ (remember we consider quantities ${O(\mathcal{D}^{1/2})}$ to be small, so if the opposite inequality were to hold, we wouldn’t be able to distinguish between ${|A|}$ and ${|C|}$, informally). This means that the statement is not symmetric w.r.t. to permutations, and indeed the conclusion is reached only for set ${C}$. A symmetric statement would yields the result for all three sets, because the form ${\left\langle \chi_A \ast \chi_B , \chi_C \right\rangle}$ is essentially symmetric:

$\displaystyle \left\langle \chi_A \ast \chi_B , \chi_C \right\rangle = \left\langle \chi_A \ast \chi_{-C} , \chi_{-B} \right\rangle, \ \ \ \ \ (9)$

$\displaystyle \left\langle \chi_A \ast \chi_B , \chi_C \right\rangle= \left\langle \chi_B \ast \chi_A , \chi_C \right\rangle. \ \ \ \ \ (10)$

2.5. Truncations

In this section we deal with the final point of the strategy.

I’ll recall what truncations are:

Definition 10 A truncation of a set ${E \subset \mathbb{R}}$ of positive measure with parameters ${\xi, \eta > 0}$ that satisfy ${\xi + \eta < |E|}$, is the subset ${E_{\xi, \eta} \subset E}$ given by ${E_{\xi, \eta} = E \cap [a,b]}$, where ${a,b}$ are such that

$\displaystyle \left|(-\infty, a) \,\cap\, E\right| = \xi$

$\displaystyle \left|E \, \cap\, (b, +\infty)\right| = \eta.$

Of course there is ambiguity in the choice of ${a,b}$, but it’s irrelevant – say one takes the highest ${a}$ and the lowest ${b}$. Notice also that it is an intrinsically ${1}$-dimensional device.

These objects have nice properties, which I list in lemmas following [ChRS].

Lemma 11 Let ${A,B,C}$ be measurable sets and ${\xi, \eta\geq 0}$ s.t. ${\xi + \eta < \min(|A|,|B|)}$. Then

$\displaystyle \left\langle \chi_A \ast \chi_B, \chi_C \right\rangle \leq \left\langle \chi_{A_{\xi,\eta}} \ast \chi_{B_{\eta,\xi}}, \chi_C \right\rangle + (\xi + \eta) |C|.$

Notice that the positions of ${\xi, \eta}$ in the pedices are inverted.

Lemma 12 Let ${\xi, \eta\geq 0}$, and let ${I,J,K}$ be intervals centered at ${0}$ s.t.

• ${|I|> \xi + \eta}$;
• ${|J|> \xi + \eta}$;
• ${|K| < |I|+|J|}$.

Then

$\displaystyle \left\langle \chi_I \ast \chi_J, \chi_K \right\rangle = \left\langle \chi_{(I_{\xi, \eta})^{\ast}} \ast \chi_{(J_{\eta,\xi})^{\ast}}, \chi_K \right\rangle + (\xi + \eta) |K|.$

Notice the order of ${\xi, \eta}$ is inverted again. This lemma is the one that allows to pass to the truncations: indeed, it implies that

Corollary 13

$\displaystyle \mathcal{D}(A_{\xi,\eta}, B_{\eta,\xi},C) \leq \mathcal{D}(A,B,C).$

That is, ${(A_{\xi,\eta}, B_{\eta,\xi},C)}$ is a tighter near-extremizer of Riesz-Sobolev. This is because, by Lemma 11,

$\displaystyle \left\langle \chi_{A_{\xi,\eta}} \ast \chi_{B_{\eta,\xi}}, \chi_C \right\rangle \geq \left\langle \chi_A \ast \chi_B, \chi_C \right\rangle - (\xi + \eta) |C|$

$\displaystyle = \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle - \mathcal{D} - (\xi + \eta) |C|$

and by Lemma 12, since ${(A^\ast)_{\xi, \eta}^{\ast} = (A_{\xi, \eta})^{\ast}}$ this is

$\displaystyle =\left\langle \chi_{(A_{\xi, \eta})^{\ast}} \ast \chi_{(B_{\eta,\xi})^{\ast}}, \chi_C \right\rangle - \mathcal{D}.$

I won’t prove the lemmas because this post would get too long. You can find the easy proofs in [ChRS].

Remark 2 Lemma 11 and Lemma 12 can be used to give a proof of the Riesz-Sobolev inequality in one dimension.

Before addressing the removal of the hypothesis ${|A|=|B|}$, I need a further lemma with some combinatorial flavour. It essentially says that if all the truncations which chop off a fixed portion of a set ${A}$ are nearly intervals, then ${A}$ itself must nearly coincide with some interval.

Lemma 14 Let ${A}$ be a measurable set of positive measure, ${\varepsilon >0}$ and ${0< \lambda < 1}$. Suppose that for every truncation ${A_{\xi,\eta}}$, with ${\xi + \eta = (1-\lambda)|A|}$, there exists an interval (not necessarily the same for everyone) ${I}$ s.t.

$\displaystyle |A_{\xi,\eta} \;\Delta \; I| \leq \varepsilon |A|.$

Then there exists ${\mathcal{I}}$ interval s.t.

$\displaystyle |A \;\Delta \; \mathcal{I}| \lesssim \lambda^{-1} \varepsilon |A|.$

Proof: Suppose without loss of generality that ${\lambda = N^{-1}}$, where ${N}$ is integer. Then denote ${\xi_j = \frac{j}{2N}|A|}$ for ${j\leq 2N-2}$, ${\eta_j}$ accordingly as a consequence of ${\xi_j + \eta_j = (1-\lambda)|A|}$, and ${A_j := A_{\xi_j, \eta_j}}$. That is, ${A_j}$ has been chopped off a fraction of ${j/{2N}}$ on the left and ${(2N -2 -j)/{2N}}$ on the right, which leaves a portion ${1/N}$: ${|A_j| = \frac{1}{N}|A|}$. For every ${A_j}$ there is an interval ${I_j}$ s.t. ${|A_j \;\Delta \;I_j | \leq \varepsilon |A|}$, which implies

$\displaystyle |I_j|\geq \left(\frac{1}{N} - \varepsilon\right) |A|.$

On the other hand, we have control over the symmetric differences:

$\displaystyle |I_j \;\Delta \;I_{j+1}| \leq |I_j \;\Delta \;A_j | + |A_j \;\Delta \;A_{j+1} | + |A_{j+1} \;\Delta \;I_{j+1} | \leq 2\varepsilon |A| + \frac{1}{N}|A|.$

If ${\varepsilon}$ is sufficiently small, say ${4 \varepsilon < 1/N}$, then

$\displaystyle |I_j \;\Delta \;I_{j+1}| \leq |A| \left(\frac{1}{N} + 2\varepsilon \right)< 2\left(\frac{1}{N} + \varepsilon \right)|A| \leq |I_j| +|I_{j+1}|,$

and for the inequality to be strict then ${I_j}$ and ${I_{j+1}}$ must intersect. But this holds for all ${j}$, and therefore all the intervals form a chain, and ${\mathcal{I}:= \bigcup_{j = 1}^{2N-2} {I_j}}$ is an interval itself. Moreover

$\displaystyle |A \backslash \mathcal{I}| \leq |\bigcup_{j}{A_j \backslash I_j}| \leq \sum_{j}{|A_j \backslash I_j|}\leq 2N \varepsilon |A|,$

and same for ${|\mathcal{I} \backslash A|}$. This concludes the proof. $\Box$

With this last lemma, we’re ready to complete the proof. We’ll have to do a bit of jiggling, and it’s best to divide into cases.

Case 1: Let’s define ${M:= \max(|A|,|B|,|C|)}$ for shortness. Suppose

• ${\mathcal{D}^{1/2} \lesssim \eta^2 M}$ (with constant small enough; the square is necessary);
• ${|A|>\max(|B|,|C|) - \frac{1}{4}\eta M}$;

we want to prove that there exists an interval ${I}$ s.t. ${|A\; \Delta\; I|\lesssim \eta^{-1} \mathcal{D}^{1/2}}$.

All we have to do is show that we can apply Proposition 9 to any truncation of a fixed portion, and then invoke 14 to conclude that ${A}$ itself must be approximated by an interval. Remember one condition of Proposition 9 is that the third set is smaller than the other two by ${\Omega(\mathcal{D}^{1/2})}$. Thus we want to remove from two sets enough mass: if we remove ${|A|-|C|}$ from ${B}$ and ${|A|-|B|}$ from ${C}$ they will have the same size. Notice we are removing portions of different sizes, and this can only be achieved if we take two truncations, one for ${(A,B)}$ and one for ${(A,C)}$. But this is not enough, since we want the truncated ${B}$ and ${C}$ to be sensibly larger than what’s left of ${A}$, and thus we remove slightly more from ${B}$ and ${C}$, namely an additional ${\delta \gtrsim \mathcal{D}^{1/2}}$. More precisely, choose ${\delta}$ s.t.

$\displaystyle 4 \mathcal{D}^{1/2} \leq \delta \leq \frac{1}{8}\eta M,$

which we can do (we have the freedom to choose a small enough constant in the hypotheses above, and ${\eta^2 < \eta}$). Then we’ll remove

$\displaystyle \rho^\ast = |A|-|C| + \delta$

from ${B}$ and

$\displaystyle \sigma^\ast = |A| -|B| + \delta$

from ${C}$. Take ${\rho,\rho'>0}$ s.t. ${\rho+\rho' = \rho^\ast}$ and ${\sigma,\sigma' >0}$ s.t. ${\sigma + \sigma' = \sigma^\ast}$, and define the truncations

$\displaystyle \mathcal{A}:= A_{\rho + \sigma, \rho' + \sigma'} \qquad \mathcal{B}:= B_{\rho', \rho} \qquad \mathcal{C}:= C_{\sigma',\sigma}.$

We have to verify we can apply Proposition 9 to the triplet ${(\mathcal{C}, \mathcal{B}, \mathcal{A}) }$. As said above, ${|\mathcal{B}| = |\mathcal{C}|}$, and then

$\displaystyle |\mathcal{A}| = |A| - \rho^\ast - \sigma^\ast = |B|+|C| - |A| - 2\delta = |\mathcal{C}| - \delta,$

from which it follows that ${|\mathcal{C}| - |\mathcal{A}| > 4 \mathcal{D}^{1/2}}$. Remember ${\mathcal{D} = \mathcal{D}(A,B,C)}$ and not ${\mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})}$, which is the triplet we’re working on, but by Corollary 13

$\displaystyle \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A}) \leq \mathcal{D}(A_{\sigma,\sigma'}, B,C_{\sigma',\sigma}) \leq \mathcal{D}(A,B,C).$

Another thing to verify is that the triplet ${(\mathcal{C}, \mathcal{B}, \mathcal{A})}$ is ${\frac{1}{2}\eta}$-strictly admissible, and this is true because

$\displaystyle |\mathcal{A}|+|\mathcal{B}|-|\mathcal{C}| = |\mathcal{A}| = |B|+|C|-|A|-2\delta \geq \eta M - \frac{2}{8}\eta M > \frac{1}{2}\eta M.$

Then we have to verify ${20 \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})^{1/2} < \frac{1}{2}\eta |\mathcal{C}|}$, and this is true because

$\displaystyle |\mathcal{C}| = |B|+|C|-|A|-\delta>\eta M - \frac{1}{8}\eta M > \frac{1}{2}\eta M,$

and therefore

$\displaystyle \frac{1}{2}\eta |\mathcal{C}|>\frac{1}{4}\eta^2 M > 20 \mathcal{D}^{1/2} \geq 20 \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})^{1/2},$

as long as we fix the constant in the hypothesis to be ${80 \mathcal{D}^{1/2} < \eta^2 M}$.

Thus we can apply Proposition 9 to ${(\mathcal{C}, \mathcal{B}, \mathcal{A})}$, which yields an interval ${\mathcal{I}}$ s.t.

$\displaystyle |\mathcal{C}\;\Delta \; \mathcal{I}| \leq 20 \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})^{1/2} \leq 20 \mathcal{D}^{1/2}.$

This holds for all truncation parameters ${\rho,\rho', \sigma,\sigma'}$ as above, and then there exists an interval ${I}$ s.t.

$\displaystyle |A \;\Delta \; I| \lesssim \lambda^{-1} \mathcal{D}^{1/2},$

where ${\lambda}$ is given by ${\rho^\ast + \sigma^\ast = (1-\lambda) |A|}$, thus

$\displaystyle \lambda = 1 - \frac{\rho^\ast + \sigma^\ast}{|A|} = \frac{|B|+|C|-|A|-2\delta}{|A|}$

$\displaystyle \geq \eta - \frac{2\delta}{|A|} \geq \eta - \frac{2 \eta M }{8|A|} \geq \frac{1}{2}\eta,$

because ${|A|\geq \frac{1}{2}M}$: indeed, this is true if ${|A|=M}$, otherwise ${\max(|B|,|C|) = M}$ and by the hypothesis ${|A|>M - \frac{1}{4}\eta M\geq M - \frac{1}{4} M> \frac{1}{2}M}$. Then the bound on ${|A \;\Delta \; I|}$ is

$\displaystyle |A \;\Delta \; I|\lesssim \eta^{-1} \mathcal{D}^{1/2},$

like we wanted.

By the symmetry relations (9), (10), we can permute ${A,B,C}$ and assume ${M=|A|\geq |B|\geq |C|}$. Then the result above holds for ${A}$ clearly, but for ${B}$ and ${C}$ too, provided ${|B|> |A| - \frac{1}{4}\eta |A|}$ and likewise for ${C}$. If this condition fails, we have the next case.

Case 2: Assume again ${M=|A|\geq |B|\geq |C|}$, thus ${A}$ is nearly an interval as above, but suppose

$\displaystyle |B|\leq |A| - \frac{1}{4}\eta |A|.$

In this case we can reduce ${A}$ and ${B}$ to having the same size, and therefore we must trim ${A}$ by ${|A|-|B|}$: let ${\rho^\ast = |A|-|B|}$, ${\rho, \rho'>0}$ s.t. ${\rho+ \rho' = \rho^\ast}$, define

$\displaystyle \mathcal{A}:= B \qquad \mathcal{B}:= A_{\rho, \rho'} \qquad \mathcal{C}:= C_{\rho', \rho}.$

As said, ${|\mathcal{A}|=|\mathcal{B}|\geq |\mathcal{C}|}$. We have to verify the hypotheses of previous case hold again, and then by the same reasoning we will have that ${\mathcal{A}}$ is nearly an interval, but ${\mathcal{A}=B}$.

We have that the triplet ${(\mathcal{A},\mathcal{B},\mathcal{C})}$ is ${\eta}$-strictly admissible:

$\displaystyle |\mathcal{C}| = |A|+|C|-|B|> \eta |A| \geq \eta \max(|\mathcal{A}|,|\mathcal{B}|,|\mathcal{C}|).$

Moreover, condition ${|\mathcal{A}| > |\mathcal{B}| - \frac{1}{4}\eta \max(|\mathcal{A}|,|\mathcal{B}|,|\mathcal{C}|)}$ is trivially satisfied. As for the last condition, by Corollary 13 it is ${\mathcal{D}(\mathcal{A},\mathcal{B},\mathcal{C})\leq \mathcal{D}}$. Notice that by ${\eta}$-strict admissibility of ${(A,B,C)}$ one has

$\displaystyle |B|=|\mathcal{A}|=\max(\mathcal{A},\mathcal{B},\mathcal{C})\geq \eta \max(A,B,C) = \eta |A|.$

Thus to have the condition satisfied it suffices to assume ${\mathcal{D}^{1/2} \lesssim \eta^3 |A|}$ with a sufficiently small constant, and then

$\displaystyle \mathcal{D}(\mathcal{A},\mathcal{B},\mathcal{C})^{1/2} \leq \mathcal{D}^{1/2} \lesssim \eta^3 |A| \leq \eta^2 |\mathcal{A}|.$

This way, we’ve reduced to the previous case, and ${\mathcal{A}=B}$ is nearly an interval, with error ${O(\mathcal{D}(\mathcal{A},\mathcal{B},\mathcal{C})^{1/2}) = O(\mathcal{D}^{1/2})}$.

Case 3: Suppose now that ${|C|\leq |A| - \frac{1}{4}\eta |A|}$. This case is the same as Case 2, but now we work on triplet

$\displaystyle \mathcal{A}= A_{\rho,\rho'} \qquad \mathcal{B}= B \qquad \mathcal{C}= C_{\rho',\rho}.$

The reasoning is identical, and this exhausts all the possibilities. The theorem is proved.

Footnotes:

[1] To be precise, ${A}$ is essentially contained in ${I}$, in the sense that ${|A\backslash I|=0}$.

[2] by the names of the people who studied it in a variety of contexts: Kemperman, Pollard, Rusza, Green, Tao.

[3] note we have to prove ${|S_{A,-A}(\gamma)| \leq 2|A|}$ for this to work.

References:

[ChRS] M. Christ, Near equality in the Riesz-Sobolev inequality, arXiv:1309.5856 [math.CA], 2013.

It’s finally time to address one of Christ’s papers I talked about in the previous two blogposts. As mentioned there, I’ve chosen to read the one about the near-equality in the Riesz-Sobolev inequality because it seems the more approachable, while still containing one very interesting idea: exploiting the additive structure lurking behind the inequality via Freiman’s theorem.

1. Elaborate an attack strategy

Everything is in dimension ${d=1}$ and some details of the proof are specific to this dimension and don’t extend to higher dimensions. I’ll stick to Christ’s notation.

Recall that the Riesz-Sobolev inequality is

$\displaystyle \boxed{\left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle \leq \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle} \ \ \ \ \ (1)$

and its extremizers – which exist under the hypothesis that the sizes are all comparable – are intervals, i.e. the intervals are the only sets that realize equality in (1). See previous post for further details. The aim of paper [ChRS] is to prove that whenever ${\left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle}$ is suitably close to ${\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle}$ (i.e. we nearly have equality) then the sets ${A,B,C}$ are nearly intervals themselves.

As explained in the previous post, there is one powerful tool to prove that a set must be close to an interval: the continuum version of Freiman’s theorem, which asserts that if ${A}$ is a measurable set in ${\mathbb{R}}$ s.t. ${|A+A| < 3|A|}$ then ${A}$ nearly coincides with an interval (and this is bounded in size). I’ll restate it for convenience:

Theorem 1 (continuum Freiman’s theorem) Let ${A\subset \mathbb{R}}$ be a measurable set with finite measure ${>0}$. If

$\displaystyle |A+A|< 3|A|,$

then there exists an interval ${I}$ s.t. ${A\subset I}$ [1] and

$\displaystyle |I| \leq |A+A|-|A|.$

The problem is that we have no other information about ${A,B,C}$ other than the fact that (1) is nearly an equality when applied to these three sets – and we definitely need some strong information about the additive behaviour of the sets to prove something like ${|A+A|< 3|A|}$ out of the blue. Thus we are forced to reduce to well behaved sets. The most natural direction to look for is the superlevel sets of ${\chi_A \ast \chi_B}$, defined by

$\displaystyle S_{A,B}(t):=\{x \in \mathbb{R}\,:\, \chi_A \ast \chi_B (x)>t\},$

and at this point the (naive) strategy to prove ${C}$ is close to an interval is the following:

Strategy draft \# 1

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(t)}$ for some ${t}$ that will likely depend on all ${A,B,C}$, by proving that ${C}$ nearly coincides with ${S_{A,B}(t)}$;
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(t)+S_{A,B}(t)|< 3|S_{A,B}(t)|}$.

This would suffice, because then ${S_{A,B}(t)}$ would nearly coincide with an interval by Freiman’s theorem 1 and so ${C}$ would nearly coincide with an interval itself. This is obviously too naive to be expected to work, but there is something already there: indeed, one can prove that if ${A,B,C}$ nearly realizes inequality then ${C}$ nearly coincides with ${S_{A,B}(\tau)}$, where ${\tau}$ is s.t.

$\displaystyle |C| = |A| + |B| - 2 \tau,$

i.e. ${\tau = (|A|+|B|-|C|)/2}$. Thus point (1.) isn’t a problem. I should also mention that ${A,B,S_{A,B}(\tau)}$ realizes near-equality better than ${A,B,C}$ (the difference between the RHS and LHS of (1) is smaller).

The strategy will be refined and updated in these notes. Turns out there is no direct relationship between ${S_{A,B}(t)+S_{A,B}(t)}$ and ${S_{A,B}(t)}$, but there is a relationship between ${S_{A,B}-S_{A,B}}$ and ${S_{A,-A}}$ of the form

$\displaystyle S_{A,B} (t) - S_{A,B} (t') \subset S_{A,-A}(t+t' - |B|). \ \ \ \ \ (2)$

Remark 1 I haven’t discussed this point in the related post, but discrete Freiman’s theorem conclusion (and thus the continuum one) applies as well to the case ${|A-A|< 3|A|-3}$, and this follows from two facts: first, that the quantitative lemma where one estimates ${|A+A|}$ from below by partitioning in accordance to the remainders modulus some ${N}$ holds as well for ${|A-A|}$; second, by the fact that Kneser’s theorem is stated for two general sets ${A,B}$. If you don’t know what I’m talking about, check the blogpost.

Thus we have a new refined strategy:

Strategy draft \# 2

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(\tau)}$ for ${\tau}$ as above, by proving that ${C}$ nearly coincides with ${S_{A,B}(\tau)}$; ✓
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(\tau)-S_{A,B}(\tau)|< |S_{A,-A}(2\tau-|B|)|}$; ✓
3. prove that ${|S_{A,-A}(2\tau-|B|)| < 3|S_{A,B}(\tau)|}$ under the hypotheses made;

then the result would follow from Freiman’s theorem as above. There isn’t a direct obvious way to prove something like (3.): point (2.) is tackled with a set theoretic inclusion, which is geometric information after all, but no such inclusion holds for those sets. Moreover, by Brunn-Minkowski, we seem to be going in the opposite direction: since ${|S_{A,B}(\tau)-S_{A,B}(\tau)| \geq 2|S_{A,B}(\tau)|}$, we have

$\displaystyle 2|S_{A,B}(\tau)| \leq |S_{A,-A}(2\tau-|B|)|.$

Do we have to discard the strategy? no, maybe we’re rushing. Since ${S_{A,B}(\tau)}$ is about the same size as ${C}$, we can contempt ourselves with proving that ${|S_{A,-A}(2\tau-|B|)|}$ isn’t much larger than ${2|C|}$.

Let’s shorten ${2\tau-|B|}$ to ${\gamma}$ and notice that ${\gamma = |A|-|C|}$. A moment’s thought reminds us that the inclusion relation (2) holds for all values of ${t}$, not just for our ${\tau}$. It might be worth it to integrate instead, as the integral over the superlevel sets has a clear interpretation, and see if we can squeeze out some information this way: since ${2|S_{A,B}(t)| \leq |S_{A,-A}(2t-|B|)|}$, integrating from ${\tau}$ to ${\infty}$ one gets

$\displaystyle \int_{\gamma}^{\infty}{|S_{A,-A}(t)|}\,dt \geq 4 \int_{\tau}^{\infty}{|S_{A,B}(t)|}\,dt. \ \ \ \ \ (3)$

Now, for a generic function ${F}$ and ${t>0}$ it is

$\displaystyle \int{|F(x)|}\,dx =\int_{t}^{\infty}{\left|\left\{|F|>u\right\}\right|}\,du + t\left|\left\{|F|>t\right\}\right| + \int_{\left\{|F|\leq t\right\}}{|F(x)|}\,dx$

$\displaystyle = \int_{t}^{\infty}{\left|\left\{|F|>u\right\}\right|}\,du + \int{\min(|F(x)|, t)}\,dx,$

and since ${\int_{\mathbb{R}}{\chi_A \ast \chi_B}\,dx = |A||B|}$ we can write

$\displaystyle \int_{\tau}^{\infty}{|S_{A,B}(t)|}\,dt = |A||B| - \int_{\mathbb{R}}{\min(\chi_A \ast \chi_B (x), \tau)}\,dx \ \ \ \ \ (4)$

(and an analogous expression for ${S_{A,-A}}$). We’ve just been rewriting our expression, but doing so we’ve introduced a quantity that has actually been studied before. I’m talking here about the last integral I’ve written: one has the so-called KPRGT inequality [2]

$\displaystyle \int_{\mathbb{R}}{\min(\chi_A \ast \chi_B (x), t)}\,dx \geq t \left(|A|+ |B| - t\right).$

But, because of (4), this can be recast as

$\displaystyle \int_{t}^{\infty}{|S_{A,B}(u)|}\,du \leq (|A|-t)(|B|-t).$

If one denotes by ${\mathcal{D}' (A,B,t)}$ the difference

$\displaystyle \mathcal{D}' (A,B,t)= (|A|-t)(|B|-t)-\int_{t}^{\infty}{|S_{A,B}(u)|}\,du,$

then (3) can be rewritten as

$\displaystyle (|A|-\gamma)^2 - \mathcal{D}'(A,-A,\gamma) \geq 4(|A|-\tau)(|B|-\tau) - 4\mathcal{D}'(A,B,\tau),$

and since by the definitions it is ${|A|-\gamma = |C|}$ and ${4(|A|-\tau)(|B|-\tau) = |C|^2 - (|A|-|B|)^2}$, the above is equivalent to

$\displaystyle (|A|-|B|)^2 + 4\mathcal{D}'(A,B,\tau) \geq \mathcal{D}'(A,-A,\gamma). \ \ \ \ \ (5)$

It’s important to notice that this inequality is tighter if ${|A|=|B|}$. Now, our hope is that we can extract information from this quantity ${\mathcal{D}'}$, and it is indeed the case. It’s easily seen to be related to the Riesz-Sobolev inequality in the following way: define analogously

$\displaystyle \mathcal{D}(A,B,C) :=\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle - \left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle,$

which is a non-negative quantity. Now, for ${C = S_{A,B}(t)}$, we have ${\left\langle \chi_{A} \ast \chi_{B}, \chi_{S_{A,B}(t)}\right\rangle = \int_{S_{A,B}(t)}{\chi_A \ast \chi_B}}$, a quantity we know how to relate to ${\int{\min(\chi_A \ast \chi_B, t)}}$; but what about ${\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{{S_{A,B}(t)}^\ast}\right\rangle}$? we need to work out what this is. A little algebra shows that, if we define ${\sigma}$ by ${|S_{A,B}(t)| = |A|+|B|-2\sigma}$ (and thus suppose ${\left||A|-|B|\right|\leq |S_{A,B}(t)| \leq |A|+|B|}$), we have

$\displaystyle \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{{S_{A,B}(t)}^\ast}\right\rangle = |A||B|-\sigma^2. \ \ \ \ \ (6)$

Then

$\displaystyle \mathcal{D}'(A,B,t) = (|A|-t)(|B|-t) - \int_{t}^{\infty}{|S_{A,B}(u)|}\,du$

$\displaystyle = |A||B| - t (|A|+|B|-t) - \int_{S_{A,B}(t)}{\chi_A \ast \chi_B}\,dx + t |S_{A,B}(t)|$

$\displaystyle = \mathcal{D}(A,B,S_{A,B}(t)) + \sigma^2 - t (|A|+|B|-t) + t |S_{A,B}(t)|$

$\displaystyle = \mathcal{D}(A,B,S_{A,B}(t)) + \sigma^2 - t (|S_{A,B}(t)|+2\sigma -t) + t |S_{A,B}(t)|,$

and therefore

$\displaystyle \mathcal{D}'(A,B,t) = \mathcal{D}(A,B,S_{A,B}(t)) + (\sigma - t)^2$

(notice ${\sigma=\sigma(t)}$). This tells us plenty of things: first of all,

$\displaystyle \mathcal{D}'(A,B,t) \geq \mathcal{D}(A,B,S_{A,B}(t)),$

but also that ${\mathcal{D}'(A,B,t) \sim \mathcal{D}(A,B,S_{A,B}(t))}$ if we have some control of the kind ${|\sigma - t| \lesssim \mathcal{D}(A,B,S_{A,B}(t))^{1/2}}$. Turns out that this is the case in our situation above. Indeed,

$\displaystyle (\sigma - \tau)^2 = \left(\frac{|A|+|B|-|S_{A,B}(\tau)|}{2} - \frac{|A|+|B|-|C|}{2} \right)^2 = \frac{1}{4}(|S_{A,B}(\tau)|- |C|)^2,$

and as said we’ve chosen ${\tau}$ so that the difference ${|S_{A,B}(\tau)|- |C|}$ is small, in particular (but I haven’t mentioned it before) one has ${||S_{A,B}(\tau)|- |C||\lesssim \mathcal{D}(A,B,C)}$ (which is assumed small because we’re assuming near-equality); therefore ${\mathcal{D}'(A,B,\tau) \lesssim \mathcal{D}(A,B,S_{A,B}(\tau)) + \mathcal{D}(A,B,C)}$. Luckily though, it’s easy to prove that ${\mathcal{D}(A,B,S_{A,B}(\tau)) \leq \mathcal{D}(A,B,C)}$ (as one would naively expect), so using all of the above we can write for some constant ${c_0>0}$

$\displaystyle (|A|-|B|)^2 + c_0 \mathcal{D}(A,B,C) \geq \mathcal{D}'(A,-A,\gamma) \geq (\nu -\gamma )^2,$

where ${\nu}$ is defined by [3]

$\displaystyle |S_{A,-A}(\gamma)| = |A| + |-A| - 2\nu.$

Then

$\displaystyle \nu - \gamma = \frac{2|A| - |S_{A,-A}(\gamma)| }{2} - (|A|-|C|) = \frac{1}{2}(2|C|-|S_{A,-A}(\gamma)| ),$

and therefore

$\displaystyle (|A|-|B|)^2 + c_0 \mathcal{D}(A,B,C) \gtrsim \left|2|C| - |S_{A,-A}(\gamma)|\right|.$

If ${|A|=|B|}$ this is what we wanted: ${\mathcal{D}(A,B,C)}$ is assumed to be small, and in particular is smaller than ${c_0^{-1}|C|}$, and therefore adjusting the constants we can have ${|S_{A,-A}(\gamma)| < 3|S_{A,B}(\tau)|}$, which was the original goal! On the other hand, if ${|A|\neq |B|}$ this miserably fails, and ${|A|=|B|}$ is an incredibly strong condition to ask for.

We revise the strategy as follows:

Strategy draft \# 3

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(\tau)}$ for ${\tau}$ as above, by proving that ${C}$ nearly coincides with ${S_{A,B}(\tau)}$; ✓
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(\tau)-S_{A,B}(\tau)|< |S_{A,-A}(2\tau-|B|)|}$; ✓
3. prove that ${|S_{A,-A}(2\tau-|B|)| < 3|S_{A,B}(\tau)|}$ under the hypotheses made and the further assumption that ${|A|=|B|}$; ✓
4. find a way to remove the assumption that ${|A|=|B|}$.

To deal with point (4.), Christ resorted to the machinery of truncations. Truncations are defined as follows: a truncation of a set ${E \subset \mathbb{R}}$ of positive measure with parameters ${\xi, \eta > 0}$ that satisfy ${\xi + \eta < |E|}$, is the subset ${E_{\xi, \eta} \subset E}$ given by ${E_{\xi, \eta} = E \cap [a,b]}$, where ${a,b}$ are such that

$\displaystyle \left|(-\infty, a) \,\cap\, E\right| = \xi$

$\displaystyle \left|E \, \cap\, (b, +\infty)\right| = \eta.$

In other words, ${E_{\xi, \eta}}$ is the middle portion of a trisection of ${E}$ s.t. the left section has measure ${\xi}$ and the right section has measure ${\eta}$ (and therefore the ${E_{\xi,\eta}}$ has measure ${|E|-\xi - \eta}$). This device allows to chop off portions of the sets ${A,B,C}$ in such a way that two of them are reduced to having the same size, but the triplet is still a near-extremizer of Riesz-Sobolev inequality. This is natural but not entirely obvious, and I don’t think I can sketch this part as I did for the previous three. Perhaps is time to move on to the rigourous proof. But first let me write the final

Strategy

1. reduce from the case ${A,B,C}$ to the case of ${A,B, S_{A,B}(\tau)}$ for ${\tau}$ as above, by proving that ${C}$ nearly coincides with ${S_{A,B}(\tau)}$;
2. exploit the definition of superlevel set and the hypothesis of the theorem to prove that ${|S_{A,B}(\tau)-S_{A,B}(\tau)|< |S_{A,-A}(2\tau-|B|)|}$;
3. prove that ${|S_{A,-A}(2\tau-|B|)| < 3|S_{A,B}(\tau)|}$ under the hypotheses made and the further assumption that ${|A|=|B|}$;
4. Remove the assumption that ${|A|=|B|}$ by using truncations.

2. The actual proof

I will first of all state the theorem of Christ, but that requires a definition. We introduce a parameter to control how the sizes of the sets are comparable:

Definition 2 A triplet of sets ${A_1, A_2, A_3}$ is said to be ${\eta}$-strictly admissible for some ${0<\eta<1}$ if

$\displaystyle |A_i|+|A_j|> |A_k| + \eta \max(|A_1|,|A_2|,|A_3|)$

holds for all ${(i,j,k)}$ that are permutations of ${(1,2,3)}$.

This has as a consequence that the sets are all comparable in sizes, and in particular

$\displaystyle \min(|A_1|,|A_2|,|A_3|) \geq \eta \max(|A_1|,|A_2|,|A_3|),$

because suppose ${|A_1|\geq|A_2|\geq |A_3|}$, then

$\displaystyle |A_3| + |A_2| > |A_1| + \eta |A_1|>|A_2| + \eta |A_1|.$

Theorem 3 (Christ, [ChRS]) Let ${A,B,C}$ be a ${\eta}$-strictly admissible triplet of measurable sets for some ${0<\eta <1}$; if

$\displaystyle \mathcal{D}(A,B,C)^{1/2} \lesssim \eta^3 \max(|A|,|B|,|C|),$

then there exist intervals ${I,J,K}$ s.t.

$\displaystyle |A\;\Delta \;I|,|B\;\Delta \;J|,|C\;\Delta \;K| \lesssim \eta^{-1} \mathcal{D}^{1/2}.$

Before we start, we prove (6):

Lemma 4 For ${||A|-|B||<|C|<|A|+|B|}$,

$\displaystyle \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle = |A||B|-\tau^2,$

where ${\tau}$ is defined by ${|C| = |A|+|B|-2\tau}$.

Proof: We write ${a}$ for ${|A|}$ and ${b,c}$ similarly. Assume without loss of generality ${b\leq a}$. The rearranged sets are intervals centered in 0. Write ${\chi_a}$ for ${\chi_{A^\ast} = \chi_{[-a/2,a/2]}}$, and similarly for ${\chi_b}$, ${\chi_c}$. Then

$\displaystyle \chi_a \ast \chi_b (x) = \int_{[-a/2,a/2]}{\chi_b (x+y)}\,dy = |A^\ast \cap (x- B^\ast)|=\begin{cases} b \qquad \text{ if } 2|x| < a-b, \\ \frac{a+b}{2}-|x| \qquad \text{ if } a-b \leq 2|x| \leq a+b,\\ 0 \qquad \text{ if } 2|x| > a+b. \end{cases}$

Integrating from ${-c/2}$ to ${c/2}$,

$\displaystyle \left\langle \chi_a \ast \chi_b , \chi_c\right\rangle = b (a-b) + 2 \int_{\frac{a-b}{2}}^{c/2}{\left(\frac{a+b}{2}-|x|\right)}\,dx$

$\displaystyle = b (a-b) + \frac{1}{2} \int_{a-b}^{c}{\left(a+b-x\right)}\,dx$

$\displaystyle = b (a-b) + (a+b)\frac{c-(a-b)}{2} - \frac{c^2}{4} + \frac{(a-b)^2}{4},$

and doing the algebra this is

$\displaystyle = a b - \left(\frac{a+b-c}{2}\right)^2 = |A||B| - \tau^2.$

$\Box$

Very nice formula indeed. Let me expand it for future reference:

$\displaystyle \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle = |A||B|-\left(\frac{|A|+|B|-|C|}{2}\right)^2.$

2.1. Reduction to superlevel sets

Remember that we defined

$\displaystyle \mathcal{D}(A,B,C) :=\left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle - \left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle$

$\displaystyle = |A||B| - \tau^2 - \left\langle \chi_{A} \ast \chi_{B}, \chi_{C}\right\rangle.$

In the following, let ${\mathcal{D} = \mathcal{D}(A,B,C)}$ for shortness. Our parameter for “small” as used in the previous section will typically be ${\mathcal{D}^{1/2}}$. The proposition addresses point (1.) of the strategy directly.

Proposition 5 If ${A,B,C}$ are such that

$\displaystyle ||A| -|B|| + 2\mathcal{D}^{1/2} < |C|< |A|+|B| - 2\mathcal{D}^{1/2},$

then for ${\tau}$ as before (i.e. ${|C| =|A|+|B|-2\tau}$) it is

$\displaystyle |S_{A,B}(\tau) \; \Delta \; C| \lesssim \mathcal{D}^{1/2}.$

It obviously follows that ${||S_{A,B}(\tau)|- |C|| \lesssim \mathcal{D}^{1/2}}$ as well.

Proof: We prove it separately for ${|S_{A,B}(\tau) \backslash C|}$ and ${|C \backslash S_{A,B}(\tau)|}$. Write ${S}$ for ${S_{A,B}(\tau)}$ for shortness. Suppose ${|C \backslash S| \geq 2\mathcal{D}^{1/2}}$; since ${|C| > ||A|-|B||+2\mathcal{D}^{1/2}}$, we can find a measurable set ${T}$ s.t. ${C \cap S \subseteq T \subset C}$, with size ${|T|\geq ||A|-|B||}$ but ${|C\backslash T|>2\mathcal{D}^{1/2}}$ (e.g. ${T=C\cap S}$ if ${|C\cap S|> ||A|-|B||}$ too). Then, in ${C\backslash T}$ it is ${\chi_A \ast \chi_B \leq \tau}$ by definition of ${S}$; moreover, by the assumption on the size of ${T}$ we can apply the formula in Lemma 4, therefore

$\displaystyle \left\langle\chi_A \ast \chi_B, \chi_C \right\rangle = \left\langle\chi_A \ast \chi_B, \chi_T \right\rangle + \left\langle\chi_A \ast \chi_B, \chi_{C\backslash T} \right\rangle \leq \left\langle\chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{T^\ast} \right\rangle + \tau |C\backslash T|$

$\displaystyle = |A||B| - \left(\frac{|A|+|B|-|T|}{2}\right)^2 + \tau |C\backslash T|= |A||B| - \left(\frac{|C| + 2\tau - |T|}{2}\right)^2 + \tau |C\backslash T|$

$\displaystyle = |A||B| - \left(\tau + \frac{|C\backslash T|}{2}\right)^2 + \tau |C\backslash T|=|A||B| - \tau^2 - \frac{1}{4}\left(|C\backslash T|\right)^2 - \tau |C\backslash T| + \tau |C\backslash T|$

$\displaystyle = |A||B| - \tau^2 - \frac{1}{4}\left(|C\backslash T|\right)^2,$

and thus

$\displaystyle |C\backslash T|^2 \leq 4\mathcal{D},$

The proof for ${|S\backslash C|}$ is similar: one takes any measurable ${T}$ s.t. ${C \subset T \subset C \cup S}$ and s.t. ${|T| \leq |A|+|B|}$, then for all of them proves ${|T \backslash C| \leq 2\mathcal{D}^{1/2}}$ as above, and the result follows for ${S}$ as well. $\Box$

Thus we’ve proved that ${C}$ is nearly a superlevel set, with error of size ${O(\mathcal{D}^{1/2})}$.

Since we’ll need it in the following (it was mentioned above), we also prove

Proposition 6 Under the assumptions above,

$\displaystyle \mathcal{D}(A,B,S_{A,B}(\tau)) \leq \mathcal{D}(A,B,C).$

In other words, ${(A,B,S_{A,B}(\tau))}$ is a tighter near-extremizer than ${(A,B,C)}$.

Proof: One has ${\chi_A \ast \chi_B > \tau}$ on ${S\backslash C}$ and viceversa ${\chi_A \ast \chi_B \leq \tau}$ on ${C \backslash S}$. Therefore

$\displaystyle \left\langle\chi_A \ast \chi_B, \chi_S \right\rangle = \left\langle\chi_A \ast \chi_B, \chi_C + \chi_{S \backslash C} - \chi_{C\backslash S} \right\rangle \geq \left\langle\chi_A \ast \chi_B, \chi_C \right\rangle + \tau |S \backslash C| - \tau |C\backslash S|$

$\displaystyle = |A||B| - \tau^2 -\mathcal{D}(A,B,C) + \tau (|S| - |C|).$

Define ${\sigma}$ by ${|S_{A,B}(\tau)|=|A|+|B|-2\sigma}$ (the analogous of ${\tau}$ for the new triplet), and the last line can be rewritten as

$\displaystyle |A||B| - \tau^2 - \mathcal{D}(A,B,C) + \tau (-2\sigma + 2 \tau)$

$\displaystyle = |A||B| + (\sigma - \tau)^2 - \mathcal{D}(A,B,C) - \sigma^2$

$\displaystyle \geq |A||B| - \sigma^2 - \mathcal{D}(A,B,C),$

and rearranging we get exactly the conclusion. $\Box$

2.2. Additive structure in superlevel sets

This section addresses point (2.) of the strategy.

One has the following additive relation amongst superlevel sets of convolutions of characteristic functions

Proposition 7 Let ${U,V}$ be measurable sets, then

$\displaystyle S_{U,V}(\alpha) - S_{U,V}(\beta) \subset S_{U,-U}(\alpha+\beta - |V|).$

Proof: It will follow from the fact that

$\displaystyle S_{U,V}(\alpha) =\{ x \,:\, \|\chi_{U-x} - \chi_{-V}\|_{L^1} < |U|+|V| - 2\alpha\}.$

To see this, notice that ${\chi_U \ast \chi_V (x) = |(U-x) \cap (-V)|}$, and therefore ${\chi_U \ast \chi_V (x) > \alpha}$ is equivalent to

$\displaystyle 2\alpha < 2|(U-x) \cap (-V)| = |U| + |V| - |(U-x)\;\Delta\; (-V)| = |U|+|V| - \|\chi_{U-x} - \chi_{-V}\|_{L^1}.$

Now, take ${x \in S_{U,V}(\alpha)}$, ${y \in S_{U,V} (\beta)}$ and ${z= x-y}$. Notice the superlevel sets are open, so the points lie in the inside. We want to prove ${z \in S_{U,-U}(\alpha + \beta -|V|)}$. By the above

$\displaystyle \|\chi_{U-x} - \chi_{-V}\|_{L^1} < |U|+|V| - 2\alpha$

and analogously for ${y}$, thus

$\displaystyle \|\chi_{U-z} - \chi_{U}\|_{L^1} \leq \|\chi_{U-z} - \chi_{-V+y}\|_{L^1} +\|\chi_{U} - \chi_{-V+y}\|_{L^1}$

$\displaystyle =\|\chi_{U-x} - \chi_{-V}\|_{L^1} +\|\chi_{U-y} - \chi_{-V}\|_{L^1}< 2|U|+2|V| - 2\alpha - 2\beta = |U|+|-U| - 2(\alpha + \beta -|V|),$

which is equivalent to ${z \in S_{U,-U}(\alpha + \beta -|V|)}$. $\Box$

As we’ve seen before, then, we have

$\displaystyle S_{A,B} (\tau) - S_{A,B} (\tau) \subset S_{A,-A}(2\tau - |B|) = S_{A,-A}(|A|-|C|).$

The next step is then to estimate ${|S_{A,-A}(|A|-|C|)|}$.

2.3. The KPRGT inequality

As mentioned in the first section, we’re gonna need the KPRGT inequality, which I recall:

$\displaystyle \int_{\mathbb{R}}{\min(\chi_A\ast \chi_B (x) , t)}\,dx \geq t (|A|+|B| - t).$

As seen before, it can be recast as

$\displaystyle \int_{t}^{\infty}{|S_{A,B}(u)|}\,du \leq (|A|-t)(|B|-t);$

moreover, if we define the difference

$\displaystyle \mathcal{D}' (A,B,t)= (|A|-t)(|B|-t)-\int_{t}^{\infty}{|S_{A,B}(u)|}\,du,$

then we’ve seen that we have the relationship

$\displaystyle \mathcal{D}'(A,B,t) = \mathcal{D}(A,B,S_{A,B}(t)) + (\sigma - t)^2, \ \ \ \ \ (7)$

where ${|S_{A,B}(t)|=|A|+|B|-2\sigma}$, under the assumption that ${||A|-|B||< |S_{A,B}(t)|< |A|+|B|}$. For ${t = \tau}$ this requirement is satisfied. Notice that (7) is essentially a sharpened KPRGT inequality, since it means ${\mathcal{D}'(A,B,t) - (\sigma - t)^2 \geq 0}$, or

$\displaystyle \int_{t}^{\infty}{|S_{A,B}(u)|}\,du + (\sigma - t)^2 \leq (|A|-t)(|B|-t).$

2.4. Case ${|A|=|B|}$

This section addresses point (3.) of the strategy, under the assumption that ${|A|=|B|}$, assumption that will be removed in the next subsection. Remember that ${\mathcal{D}}$ is short for ${\mathcal{D}(A,B,C)}$.

We want to prove that ${|S_{A,-A} (\gamma)| < 3|S_{A,B}(\tau)|}$, where remember ${\gamma = |A|-|C|}$. We need to be rigorous, so we state:

Proposition 8 Let ${A,B,C}$ be ${\eta}$-strictly admissible, s.t.

• ${|A|=|B|}$;
• ${|A|-|C| \geq 4 \mathcal{D}^{1/2}}$;
• ${4\mathcal{D}^{1/2} < \eta |A|}$.

Then

$\displaystyle ||S_{A,-A} (\gamma)| - 2|C|| \lesssim \mathcal{D}^{1/2}.$

The hypotheses made here are so that the hypothesis in Proposition 5 applies. Indeed ${|C|\leq |A|\leq |A|+|A| - 2\mathcal{D}^{1/2}}$ (since ${\eta\leq 1}$), and ${|C|\geq \eta |A| > 4\mathcal{D}^{1/2}> 2 \mathcal{D}^{1/2}}$ by strict-admissibility. Thus ${||S_{A,B}(\tau)|-|C||\leq 4 \mathcal{D}^{1/2}}$ (check the proof for the factor ${4}$).

Proof: As seen above: the additive relation in Proposition 7 gives us (by Brunn-Minkowski)

$\displaystyle 2|S_{A,B}(t)|\leq |S_{A,B}(t) -S_{A,B}(t)| \leq |S_{A,-A}(2t -|B|)|,$

which when integrated from ${\tau}$ to ${+ \infty}$ becomes

$\displaystyle 4 \int_{\tau}^{\infty}{|S_{A,B}(t)|}\,dt \leq \int_{\gamma}^{\infty}{|S_{A,-A}(t)|}\,dt,$

which by the same algebraic manipulations done before can be equivalently rewritten as (5), that is

$\displaystyle \mathcal{D}'(A,-A,\gamma)\leq 4 \mathcal{D}'(A,B,\tau).$

Now, we want to prove ${\mathcal{D}'(A,B,\tau)\lesssim \mathcal{D}}$ by using (7), but to use that we need to show that ${0=||A|-|B||<|S_{A,B}(\tau)|<|A|+|B| = 2|A|}$. Indeed, as seen just before this proof, ${|S_{A,B}(\tau)|\geq |C| - 4 \mathcal{D}^{1/2}>0}$, and on the other hand ${|S_{A,B}(\tau)| \leq |C| + 4\mathcal{D}^{1/2} \leq |A|+|A|}$. So we can use (7) safely, and doing the calculations (see above)

$\displaystyle \mathcal{D}'(A,B,\tau) \leq \mathcal{D}(A,B,S_{A,B}(\tau)) + \frac{1}{4}\left(|S_{A,B}(\tau)|-|C|\right)^2 \leq \mathcal{D} + \mathcal{D} = 2\mathcal{D}.$

Therefore we have

$\displaystyle \mathcal{D}'(A,-A,\gamma)\leq 8\mathcal{D} \ \ \ \ \ (8)$

so far. We want to invoke (7) again, so that we can say ${\mathcal{D}'(A,-A,\gamma) \geq (\nu - \gamma)^2}$ with ${|S_{A,-A}(\gamma)|=2|A|-2\nu}$, and therefore we have to verify the hypothesis again: ${|S_{A,-A}(\gamma)|>0}$ follows from Brunn-Minkowski since ${|S_{A,B}(\tau)|>0}$, as for the other one suppose the contrary, i.e. that ${|S_{A,-A}(\gamma)|> 2|A|}$. By (8) expanded,

$\displaystyle (|A|-\gamma)^2 - \int_{\gamma}^{+\infty}{|S_{A,-A}(t)|}\,dt \leq 8 \mathcal{D},$

thus

$\displaystyle (|A|-\gamma)^2 - 8 \mathcal{D} \leq \int_{\gamma}^{+\infty}{|S_{A,-A}(t)|}\,dt \leq \int_{0}^{+\infty}{|S_{A,-A}(t)|}\,dt - \gamma |S_{A,-A}(\gamma)|$

$\displaystyle = |A|^2 - \gamma |S_{A,-A}(\gamma)|.$

If ${|S_{A,-A}(\gamma)|> 2|A|}$ then it also is

$\displaystyle (|A|-\gamma)^2 - 8 \mathcal{D} < A|^2 - 2\gamma |A|,$

but then ${|A|-|C| = \gamma < 2 \sqrt{2} \mathcal{D}^{1/2} < 4 \mathcal{D}^{1/2}}$, contradicting the hypotheses of the proposition.

Therefore we can safely use (7) on ${\mathcal{D}'(A,-A,\gamma)}$ too, and hence by the same calculations as before

$\displaystyle \frac{1}{4}\left(2|C| - |S_{A,-A}(\gamma)|\right)^2 = (\nu - \gamma)^2 \leq \mathcal{D}'(A,-A,\gamma) \leq 8 \mathcal{D},$

or

$\displaystyle ||S_{A,-A}(\gamma)| - 2|C||\lesssim \mathcal{D}^{1/2}.$

$\Box$

By using the above proposition we then have that

$\displaystyle |S_{A,-A}(\gamma)| - 2|S_{A,B}(\tau)| \leq |S_{A,-A}(\gamma)| - 2|C| + 2 ||S_{A,B}(\tau)| - |C|| \leq 4\sqrt{2} \mathcal{D}^{1/2} + 8 \mathcal{D}^{1/2} < 16 \mathcal{D}^{1/2}.$

Now, we want ${|S_{A,-A}(\gamma)| < 3 |S_{A,B}(\tau)|}$, which we can enforce: as seen in the proof ${|C| - 4 \mathcal{D}^{1/2} < |S_{A,B}(\tau)|}$, and therefore it suffices to assume that ${|C|> 20 \mathcal{D}^{1/2}}$.

Proposition 9 Let ${A,B,C}$ be ${\eta}$-strictly admissible, s.t.

• ${|A|=|B|}$;
• ${|A|-|C| \geq 4 \mathcal{D}^{1/2}}$;
• ${20\mathcal{D}^{1/2} < \eta |A|}$.

Then there exists an interval ${I}$ s.t.

$\displaystyle |C\; \Delta \; I| \lesssim \mathcal{D}^{1/2}.$

Proof: By the remark before the last statement and the previous proposition, we have

$\displaystyle |S_{A,B} (\tau) - S_{A,B}(\tau)| < 3|S_{A,B}(\tau)|.$

Now apply Freiman’s theorem 1, which yields an interval ${I}$ s.t. ${S_{A,B} \subset I }$ and

$\displaystyle |S_{A,B}(\tau) \;\Delta \; I | \leq |S_{A,B} (\tau) - S_{A,B}(\tau)| - 2 |S_{A,B} (\tau)|\leq 16 \mathcal{D}^{1/2}.$

Finally, since ${|C \; \Delta \; S_{A,B}(\tau)| \leq 4 \mathcal{D}^{1/2}}$ by Proposition 5, we have

$\displaystyle |C\; \Delta \; I| \leq |C \; \Delta \; S_{A,B}(\tau)| + |S_{A,B}(\tau) \;\Delta \; I | \leq 20 \mathcal{D}^{1/2}.$

$\Box$

One small thing to be noticed is that, besides the assumption ${|A|=|B|}$ we had to introduce another assumption, namely that ${|A|-|C|> 4\mathcal{D}^{1/2}}$, or that ${|A|}$ is “measurably” bigger than ${|C|}$ (remember we consider quantities ${O(\mathcal{D}^{1/2})}$ to be small, so if the opposite inequality were to hold, we wouldn’t be able to distinguish between ${|A|}$ and ${|C|}$, informally). This means that the statement is not symmetric w.r.t. to permutations, and indeed the conclusion is reached only for set ${C}$. A symmetric statement would yields the result for all three sets, because the form ${\left\langle \chi_A \ast \chi_B , \chi_C \right\rangle}$ is essentially symmetric:

$\displaystyle \left\langle \chi_A \ast \chi_B , \chi_C \right\rangle = \left\langle \chi_A \ast \chi_{-C} , \chi_{-B} \right\rangle, \ \ \ \ \ (9)$

$\displaystyle \left\langle \chi_A \ast \chi_B , \chi_C \right\rangle= \left\langle \chi_B \ast \chi_A , \chi_C \right\rangle. \ \ \ \ \ (10)$

2.5. Truncations

In this section we deal with the final point of the strategy.

I’ll recall what truncations are:

Definition 10 A truncation of a set ${E \subset \mathbb{R}}$ of positive measure with parameters ${\xi, \eta > 0}$ that satisfy ${\xi + \eta < |E|}$, is the subset ${E_{\xi, \eta} \subset E}$ given by ${E_{\xi, \eta} = E \cap [a,b]}$, where ${a,b}$ are such that

$\displaystyle \left|(-\infty, a) \,\cap\, E\right| = \xi$

$\displaystyle \left|E \, \cap\, (b, +\infty)\right| = \eta.$

Of course there is ambiguity in the choice of ${a,b}$, but it’s irrelevant – say one takes the highest ${a}$ and the lowest ${b}$. Notice also that it is an intrinsically ${1}$-dimensional device.

These objects have nice properties, which I list in lemmas following [ChRS].

Lemma 11 Let ${A,B,C}$ be measurable sets and ${\xi, \eta\geq 0}$ s.t. ${\xi + \eta < \min(|A|,|B|)}$. Then

$\displaystyle \left\langle \chi_A \ast \chi_B, \chi_C \right\rangle \leq \left\langle \chi_{A_{\xi,\eta}} \ast \chi_{B_{\eta,\xi}}, \chi_C \right\rangle + (\xi + \eta) |C|.$

Notice that the positions of ${\xi, \eta}$ in the pedices are inverted.

Lemma 12 Let ${\xi, \eta\geq 0}$, and let ${I,J,K}$ be intervals centered at ${0}$ s.t.

• ${|I|> \xi + \eta}$;
• ${|J|> \xi + \eta}$;
• ${|K| < |I|+|J|}$.

Then

$\displaystyle \left\langle \chi_I \ast \chi_J, \chi_K \right\rangle = \left\langle \chi_{(I_{\xi, \eta})^{\ast}} \ast \chi_{(J_{\eta,\xi})^{\ast}}, \chi_K \right\rangle + (\xi + \eta) |K|.$

Notice the order of ${\xi, \eta}$ is inverted again. This lemma is the one that allows to pass to the truncations: indeed, it implies that

Corollary 13

$\displaystyle \mathcal{D}(A_{\xi,\eta}, B_{\eta,\xi},C) \leq \mathcal{D}(A,B,C).$

That is, ${(A_{\xi,\eta}, B_{\eta,\xi},C)}$ is a tighter near-extremizer of Riesz-Sobolev. This is because, by Lemma 11,

$\displaystyle \left\langle \chi_{A_{\xi,\eta}} \ast \chi_{B_{\eta,\xi}}, \chi_C \right\rangle \geq \left\langle \chi_A \ast \chi_B, \chi_C \right\rangle - (\xi + \eta) |C|$

$\displaystyle = \left\langle \chi_{A^\ast} \ast \chi_{B^\ast}, \chi_{C^\ast}\right\rangle - \mathcal{D} - (\xi + \eta) |C|$

and by Lemma 12, since ${(A^\ast)_{\xi, \eta}^{\ast} = (A_{\xi, \eta})^{\ast}}$ this is

$\displaystyle =\left\langle \chi_{(A_{\xi, \eta})^{\ast}} \ast \chi_{(B_{\eta,\xi})^{\ast}}, \chi_C \right\rangle - \mathcal{D}.$

I won’t prove the lemmas because this post would get too long. You can find the easy proofs in [ChRS].

Remark 2 Lemma 11 and Lemma 12 can be used to give a proof of the Riesz-Sobolev inequality in one dimension.

Before addressing the removal of the hypothesis ${|A|=|B|}$, I need a further lemma with some combinatorial flavour. It essentially says that if all the truncations which chop off a fixed portion of a set ${A}$ are nearly intervals, then ${A}$ itself must nearly coincide with some interval.

Lemma 14 Let ${A}$ be a measurable set of positive measure, ${\varepsilon >0}$ and ${0< \lambda < 1}$. Suppose that for every truncation ${A_{\xi,\eta}}$, with ${\xi + \eta = (1-\lambda)|A|}$, there exists an interval (not necessarily the same for everyone) ${I}$ s.t.

$\displaystyle |A_{\xi,\eta} \;\Delta \; I| \leq \varepsilon |A|.$

Then there exists ${\mathcal{I}}$ interval s.t.

$\displaystyle |A \;\Delta \; \mathcal{I}| \lesssim \lambda^{-1} \varepsilon |A|.$

Proof: Suppose without loss of generality that ${\lambda = N^{-1}}$, where ${N}$ is integer. Then denote ${\xi_j = \frac{j}{2N}|A|}$ for ${j\leq 2N-2}$, ${\eta_j}$ accordingly as a consequence of ${\xi_j + \eta_j = (1-\lambda)|A|}$, and ${A_j := A_{\xi_j, \eta_j}}$. That is, ${A_j}$ has been chopped off a fraction of ${j/{2N}}$ on the left and ${(2N -2 -j)/{2N}}$ on the right, which leaves a portion ${1/N}$: ${|A_j| = \frac{1}{N}|A|}$. For every ${A_j}$ there is an interval ${I_j}$ s.t. ${|A_j \;\Delta \;I_j | \leq \varepsilon |A|}$, which implies

$\displaystyle |I_j|\geq \left(\frac{1}{N} - \varepsilon\right) |A|.$

On the other hand, we have control over the symmetric differences:

$\displaystyle |I_j \;\Delta \;I_{j+1}| \leq |I_j \;\Delta \;A_j | + |A_j \;\Delta \;A_{j+1} | + |A_{j+1} \;\Delta \;I_{j+1} | \leq 2\varepsilon |A| + \frac{1}{N}|A|.$

If ${\varepsilon}$ is sufficiently small, say ${4 \varepsilon < 1/N}$, then

$\displaystyle |I_j \;\Delta \;I_{j+1}| \leq |A| \left(\frac{1}{N} + 2\varepsilon \right)< 2\left(\frac{1}{N} + \varepsilon \right)|A| \leq |I_j| +|I_{j+1}|,$

and for the inequality to be strict then ${I_j}$ and ${I_{j+1}}$ must intersect. But this holds for all ${j}$, and therefore all the intervals form a chain, and ${\mathcal{I}:= \bigcup_{j = 1}^{2N-2} {I_j}}$ is an interval itself. Moreover

$\displaystyle |A \backslash \mathcal{I}| \leq |\bigcup_{j}{A_j \backslash I_j}| \leq \sum_{j}{|A_j \backslash I_j|}\leq 2N \varepsilon |A|,$

and same for ${|\mathcal{I} \backslash A|}$. This concludes the proof. $\Box$

With this last lemma, we’re ready to complete the proof. We’ll have to do a bit of jiggling, and it’s best to divide into cases.

Case 1: Let’s define ${M:= \max(|A|,|B|,|C|)}$ for shortness. Suppose

• ${\mathcal{D}^{1/2} \lesssim \eta^2 M}$ (with constant small enough; the square is necessary);
• ${|A|>\max(|B|,|C|) - \frac{1}{4}\eta M}$;

we want to prove that there exists an interval ${I}$ s.t. ${|A\; \Delta\; I|\lesssim \eta^{-1} \mathcal{D}^{1/2}}$.

All we have to do is show that we can apply Proposition 9 to any truncation of a fixed portion, and then invoke 14 to conclude that ${A}$ itself must be approximated by an interval. Remember one condition of Proposition 9 is that the third set is smaller than the other two by ${\Omega(\mathcal{D}^{1/2})}$. Thus we want to remove from two sets enough mass: if we remove ${|A|-|C|}$ from ${B}$ and ${|A|-|B|}$ from ${C}$ they will have the same size. Notice we are removing portions of different sizes, and this can only be achieved if we take two truncations, one for ${(A,B)}$ and one for ${(A,C)}$. But this is not enough, since we want the truncated ${B}$ and ${C}$ to be sensibly larger than what’s left of ${A}$, and thus we remove slightly more from ${B}$ and ${C}$, namely an additional ${\delta \gtrsim \mathcal{D}^{1/2}}$. More precisely, choose ${\delta}$ s.t.

$\displaystyle 4 \mathcal{D}^{1/2} \leq \delta \leq \frac{1}{8}\eta M,$

which we can do (we have the freedom to choose a small enough constant in the hypotheses above, and ${\eta^2 < \eta}$). Then we’ll remove

$\displaystyle \rho^\ast = |A|-|C| + \delta$

from ${B}$ and

$\displaystyle \sigma^\ast = |A| -|B| + \delta$

from ${C}$. Take ${\rho,\rho'>0}$ s.t. ${\rho+\rho' = \rho^\ast}$ and ${\sigma,\sigma' >0}$ s.t. ${\sigma + \sigma' = \sigma^\ast}$, and define the truncations

$\displaystyle \mathcal{A}:= A_{\rho + \sigma, \rho' + \sigma'} \qquad \mathcal{B}:= B_{\rho', \rho} \qquad \mathcal{C}:= C_{\sigma',\sigma}.$

We have to verify we can apply Proposition 9 to the triplet ${(\mathcal{C}, \mathcal{B}, \mathcal{A}) }$. As said above, ${|\mathcal{B}| = |\mathcal{C}|}$, and then

$\displaystyle |\mathcal{A}| = |A| - \rho^\ast - \sigma^\ast = |B|+|C| - |A| - 2\delta = |\mathcal{C}| - \delta,$

from which it follows that ${|\mathcal{C}| - |\mathcal{A}| > 4 \mathcal{D}^{1/2}}$. Remember ${\mathcal{D} = \mathcal{D}(A,B,C)}$ and not ${\mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})}$, which is the triplet we’re working on, but by Corollary 13

$\displaystyle \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A}) \leq \mathcal{D}(A_{\sigma,\sigma'}, B,C_{\sigma',\sigma}) \leq \mathcal{D}(A,B,C).$

Another thing to verify is that the triplet ${(\mathcal{C}, \mathcal{B}, \mathcal{A})}$ is ${\frac{1}{2}\eta}$-strictly admissible, and this is true because

$\displaystyle |\mathcal{A}|+|\mathcal{B}|-|\mathcal{C}| = |\mathcal{A}| = |B|+|C|-|A|-2\delta \geq \eta M - \frac{2}{8}\eta M > \frac{1}{2}\eta M.$

Then we have to verify ${20 \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})^{1/2} < \frac{1}{2}\eta |\mathcal{C}|}$, and this is true because

$\displaystyle |\mathcal{C}| = |B|+|C|-|A|-\delta>\eta M - \frac{1}{8}\eta M > \frac{1}{2}\eta M,$

and therefore

$\displaystyle \frac{1}{2}\eta |\mathcal{C}|>\frac{1}{4}\eta^2 M > 20 \mathcal{D}^{1/2} \geq 20 \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})^{1/2},$

as long as we fix the constant in the hypothesis to be ${80 \mathcal{D}^{1/2} < \eta^2 M}$.

Thus we can apply Proposition 9 to ${(\mathcal{C}, \mathcal{B}, \mathcal{A})}$, which yields an interval ${\mathcal{I}}$ s.t.

$\displaystyle |\mathcal{C}\;\Delta \; \mathcal{I}| \leq 20 \mathcal{D}(\mathcal{C}, \mathcal{B}, \mathcal{A})^{1/2} \leq 20 \mathcal{D}^{1/2}.$

This holds for all truncation parameters ${\rho,\rho', \sigma,\sigma'}$ as above, and then there exists an interval ${I}$ s.t.

$\displaystyle |A \;\Delta \; I| \lesssim \lambda^{-1} \mathcal{D}^{1/2},$

where ${\lambda}$ is given by ${\rho^\ast + \sigma^\ast = (1-\lambda) |A|}$, thus

$\displaystyle \lambda = 1 - \frac{\rho^\ast + \sigma^\ast}{|A|} = \frac{|B|+|C|-|A|-2\delta}{|A|}$

$\displaystyle \geq \eta - \frac{2\delta}{|A|} \geq \eta - \frac{2 \eta M }{8|A|} \geq \frac{1}{2}\eta,$

because ${|A|\geq \frac{1}{2}M}$: indeed, this is true if ${|A|=M}$, otherwise ${\max(|B|,|C|) = M}$ and by the hypothesis ${|A|>M - \frac{1}{4}\eta M\geq M - \frac{1}{4} M> \frac{1}{2}M}$. Then the bound on ${|A \;\Delta \; I|}$ is

$\displaystyle |A \;\Delta \; I|\lesssim \eta^{-1} \mathcal{D}^{1/2},$

like we wanted.

By the symmetry relations (9), (10), we can permute ${A,B,C}$ and assume ${M=|A|\geq |B|\geq |C|}$. Then the result above holds for ${A}$ clearly, but for ${B}$ and ${C}$ too, provided ${|B|> |A| - \frac{1}{4}\eta |A|}$ and likewise for ${C}$. If this condition fails, we have the next case.

Case 2: Assume again ${M=|A|\geq |B|\geq |C|}$, thus ${A}$ is nearly an interval as above, but suppose

$\displaystyle |B|\leq |A| - \frac{1}{4}\eta |A|.$

In this case we can reduce ${A}$ and ${B}$ to having the same size, and therefore we must trim ${A}$ by ${|A|-|B|}$: let ${\rho^\ast = |A|-|B|}$, ${\rho, \rho'>0}$ s.t. ${\rho+ \rho' = \rho^\ast}$, define

$\displaystyle \mathcal{A}:= B \qquad \mathcal{B}:= A_{\rho, \rho'} \qquad \mathcal{C}:= C_{\rho', \rho}.$

As said, ${|\mathcal{A}|=|\mathcal{B}|\geq |\mathcal{C}|}$. We have to verify the hypotheses of previous case hold again, and then by the same reasoning we will have that ${\mathcal{A}}$ is nearly an interval, but ${\mathcal{A}=B}$.

We have that the triplet ${(\mathcal{A},\mathcal{B},\mathcal{C})}$ is ${\eta}$-strictly admissible:

$\displaystyle |\mathcal{C}| = |A|+|C|-|B|> \eta |A| \geq \eta \max(|\mathcal{A}|,|\mathcal{B}|,|\mathcal{C}|).$

Moreover, condition ${|\mathcal{A}| > |\mathcal{B}| - \frac{1}{4}\eta \max(|\mathcal{A}|,|\mathcal{B}|,|\mathcal{C}|)}$ is trivially satisfied. As for the last condition, by Corollary 13 it is ${\mathcal{D}(\mathcal{A},\mathcal{B},\mathcal{C})\leq \mathcal{D}}$. Notice that by ${\eta}$-strict admissibility of ${(A,B,C)}$ one has

$\displaystyle |B|=|\mathcal{A}|=\max(\mathcal{A},\mathcal{B},\mathcal{C})\geq \eta \max(A,B,C) = \eta |A|.$

Thus to have the condition satisfied it suffices to assume ${\mathcal{D}^{1/2} \lesssim \eta^3 |A|}$ with a sufficiently small constant, and then

$\displaystyle \mathcal{D}(\mathcal{A},\mathcal{B},\mathcal{C})^{1/2} \leq \mathcal{D}^{1/2} \lesssim \eta^3 |A| \leq \eta^2 |\mathcal{A}|.$

This way, we’ve reduced to the previous case, and ${\mathcal{A}=B}$ is nearly an interval, with error ${O(\mathcal{D}(\mathcal{A},\mathcal{B},\mathcal{C})^{1/2}) = O(\mathcal{D}^{1/2})}$.

Case 3: Suppose now that ${|C|\leq |A| - \frac{1}{4}\eta |A|}$. This case is the same as Case 2, but now we work on triplet

$\displaystyle \mathcal{A}= A_{\rho,\rho'} \qquad \mathcal{B}= B \qquad \mathcal{C}= C_{\rho',\rho}.$

The reasoning is identical, and this exhausts all the possibilities. The theorem is proved.

Footnotes:

[1] To be precise, ${A}$ is essentially contained in ${I}$, in the sense that ${|A\backslash I|=0}$.

[2] by the names of the people who studied it in a variety of contexts: Kemperman, Pollard, Rusza, Green, Tao.

[3] note we have to prove ${|S_{A,-A}(\gamma)| \leq 2|A|}$ for this to work.

References:

[ChRS] M. Christ, Near equality in the Riesz-Sobolev inequality, arXiv:1309.5856 [math.CA], 2013.