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

Pdf: link.

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|.

Now we’re ready for

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},

which is a contradiction.

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.

Pdf: link.

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|.

Now we’re ready for

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},

which is a contradiction.

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.

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