Home Sobolev extension in a simple case
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Sobolev extension in a simple case

  • Marjorie Drake , Charles Fefferman , Kevin Ren and Anna Skorobogatova EMAIL logo
Published/Copyright: May 24, 2024

Abstract

In this paper, we establish the existence of a bounded, linear extension operator T : L 2 , p ( E ) L 2 , p ( R 2 ) when 1 < p < 2 and E is a finite subset of R 2 contained in a line.

MSC code: 46E35

1 Introduction

Continuing from [1], [2], [3], [4], [5], [6], we study the Sobolev extension problem: given a finite set E R n and a function f : E R , our task is to find a function F in the homogeneous Sobolev space L m , p ( R n ) satisfying F = f on E such that F L m , p ( R n ) has the least possible order of magnitude. Moreover, we want to compute the order of magnitude of F L m , p ( R n ) . Here, L m , p ( R n ) denotes the homogeneous Sobolev spaces consisting of functions F on R n whose distributional derivatives of order m belong to L p , with seminorm

F L m , p ( R n ) | α | = m R n | α F | p 1 / p .

Notice that when p > n m , any function F L m , p ( R n ) is continuous, so the restriction F| E to any finite subset E R n is well-defined. Thus, our requirement that F = f on E makes sense for any choice of f and our problem is well-posed provided that p > n m .

In the narrower parameter range p > n, the problem is well-understood; see Israel [3], Fefferman-Israel-Luli [1], [2] and Drake [7]. We recall the relevant results as follows. Let L m,p (E) denote the space of functions f : E R equipped with the seminorm

f L m , p ( E ) inf { F L m , p ( R n ) : F L m , p ( R n ) , F = f  on  E } .

For p > n and E R n finite, there exist a linear map T : L m , p ( E ) L m , p ( R n ) , and linear functionals 1 , , ν max : L m , p ( E ) R , with the following properties:

  1. T f = f  on  E  and  T f L m , p ( R n ) C m , n , p f L m , p ( E )  for all  f L m , p ( E ) .

  2. Let f p ν = 1 ν max | ν ( f ) | p 1 / p . Then

    c m , n , p f p f L m , p ( E ) C m , n , p f p  for all  f L m , p ( E ) .

    Here, the constants C m,n,p and c m,n,p depend only on m, n, p, the parameters specifying the Sobolev space.

  3. Moreover, the number ν max of linear functionals ν grows only linearly with N, the number of points of E, and the ν have a sparse structure that lets us evaluate f p in O(N) computer operations. The functionals ν and the operator T can be computed efficiently. See [1] for details.

When n m < p n , the proofs of the above results break down, and we know almost nothing. In this paper, we analyze the simplest non-trivial case. We work in the homogeneous Sobolev space L 2 , p ( R 2 ) for 1 < p < 2. We suppose our finite set E is contained in a line. Let (x, y) denote the canonical orthogonal coordinates in R 2 ; by rotating our coordinates, we may assume that E lies on the x-axis. Our main results are as follows.

Fix p (1 < p < 2). Let E R × { 0 } R 2 be finite, with N = #E ≥ 2. We write c, C, C′, etc. to denote absolute constants, and we write c p , C p , C p , etc. to denote constants depending only on p.

Theorem 1.

There exists a linear map T : L 2 , p ( E ) L 2 , p ( R 2 ) such that Tf = f on E and T f L 2 , p ( R 2 ) C p f L 2 , p ( E ) for every fL 2,p (E). Moreover, we can take T independent of p, such that for each point z R 2 , the value of Tf at z is determined by the value of f at at most C points of E.

Further, under the same assumptions as in Theorem 1, we can efficiently compute the order of magnitude of f L 2 , p ( E ) :

Theorem 2.

There exist positive constants λ 1 , , λ ν max R and linear functionals 1 , , ν max : L 2 , p ( E ) R such that the quantity f p ν = 1 ν max λ ν | ν ( f ) | p 1 / p satisfies

c p f p f L 2 , p ( E ) C p f p

for all fL 2,p (E).

Moreover, we can take λ ν , ν independent of p such that ν maxCN for N = #E, and for each ν, the quantity ν (f) is determined by the value of f at at most C points of E.

The remainder of the paper is dedicated to the proof of Theorem 1 and Theorem 2.

Remark 1.1.

The conclusions of Theorem 1 and Theorem 2 remain open for an arbitrary choice of finite subset E R 2 . See Carruth-Israel [8] for a class of examples in which E is contained in a union of two closely-spaced parallel lines and substantial new issues arise. See also Feerman-Klartag [9].

Before proceeding, let us first outline an important issue arising in the proof of Theorem 1 and Theorem 2 and the corresponding key new ingredient introduced here to deal with this. Note that this discussion is merely an overview; see the relevant succeeding sections for a more precise description. Fix 1 < p < 2 and suppose we are given f : E R . We will make a Calderón–Zygmund (henceforth abbreviated by CZ) decomposition in the plane. Figure 1 depicts two possible scenarios for a CZ square Q, whose distance to the x-axis is comparable to its side length δ Q , together with nearby points in E. In Figure 1a, |z 1z 2| is comparable to δ Q , while in Figure 1b, |z 1z 2| is much smaller than δ Q , while |z* − z 1| is much larger than δ Q .

Figure 1: 
Two possible scenarios for a CZ square Q. (a) z
1, z
2 ∈ E near Q with |z
1 − z
2|≃ δ

Q
. (b) z
1, z
2, z* ∈ E near Q with |z
1 − z
2|≪ δ

Q
 and |z* − z
1|≫ δ

Q
.
Figure 1:

Two possible scenarios for a CZ square Q. (a) z 1, z 2E near Q with |z 1z 2|≃ δ Q . (b) z 1, z 2, z* ∈ E near Q with |z 1z 2|≪ δ Q and |z* − z 1|≫ δ Q .

More precisely, the points z 1, z 2, z* belong to E and all other points of E lie much farther away from Q than those depicted. Let F L 2 , p ( R 2 ) satisfy F = f on E with F L 2 , p ( R 2 ) being of least possible order of magnitude. What can we infer about the behavior of F on Q? In particular, what can we say about x F on Q?

If z 1, z 2 and Q are as in Figure 1a, one expects that x F| Q is closely approximated by the difference quotient

DQ ( z 1 , z 2 ) f ( z 2 ) f ( z 1 ) x 2 x 1 .

The comparability of |z 1z 2| to δ Q combined with a Sobolev inequality allows one to control x F DQ ( z 1 , z 2 ) L p ( Q ) by the product δ Q 2 F L p ( C Q ) , where CQ is the dilation of Q about its center by an appropriate constant factor C > 1 such that CQ contains z 1, z 2.

On the other hand, suppose that z 1, z 2, z* and Q are as in Figure 1b. This time, since |z 1z 2|≪ δ Q , DQ(z 1, z 2) is not an effective approximation for x F| Q . Instead, the difference quotient DQ z 1 , z * is a better choice of approximant for x F| Q . We may again control x F DQ z 1 , z * L p ( Q ) by the product δ Q 2 F L p ( Q * ) for a square Q* containing z 1, z*, but this time Q* is much larger than Q, since |z* − z 1|≫ δ Q . Therefore, in order to avoid having unbounded overlaps when summing over all squares Q, we cannot simply consider a single such square Q associated to Q*.

Instead, for a fixed z* ∈ E, we group together all CZ squares Q for which a scenario like that in Figure 1b holds for z*. We denote this subcollection of the CZ squares by GP(z*). Thanks to a lemma in Section 8, the structure of the squares in GP(z*) looks like that depicted in Figure 2.

Figure 2: 
CZ squares Q
1, Q
2, Q
3 ∈ GP(z*) for a point z* ∈ E.
Figure 2:

CZ squares Q 1, Q 2, Q 3GP(z*) for a point z* ∈ E.

This allows us to estimate

Q GP ( z * ) δ Q p x F DQ z 1 , z * L p ( Q ) p

by 2 F L p ( Q * ) p for a fixed z* ∈ E. This is a fundamental difference between the case 1 < p < 2 in consideration here, and the aforementioned case p > 2. Namely, when p > 2 we can make use of the difference quotient DQ(z 1, z 2) even in the setting of Figure 1b, without taking into account z*. This allows us to study each CZ square Q in isolation, then sum over all Q together.

So our main new ingredient is the introduction of GP(z*). By making a translation and dilation of E, we may assume without loss of generality that E = E ̲ × { 0 } R 2 for a finite subset E ̲ with max E ̲ = 2 11 and min E ̲ = 2 11 . We then work with the set E + = E ∪ {(−1, 0)}, so that we may guarantee the existence of a point z* ∈ E as in Figure 1b to the left of any CZ square Q. At the end of the proof in Section 11, we remove this additional point (−1, 0).

2 Main result

We will prove the following theorem, which is equivalent to Theorems 1 and 2.

Theorem 3.

Let E R × { 0 } R 2 be finite with N = #E ≥ 2. Let X(E) denote the vector space of real-valued functions on E. Then there exists a linear map T : X ( E ) p ( 1,2 ) L 2 , p ( R 2 ) , and there exist linear functionals ν : X ( E ) R and positive coefficients λ ν for ν = 1, , ν max, with the following properties:

  1. For every fX(E), T f = f on E;

  2. For every fX(E) and p ∈ (1, 2),

    T f L 2 , p ( R 2 ) p C p ν = 1 ν max λ ν | ν ( f ) | p ,

    where the constant C p depends only on p;

  3. For fX(E), p ∈ (1, 2) and F L 2 , p ( R 2 ) with F = f on E, we have

    F L 2 , p ( R 2 ) p c p ν = 1 ν max λ ν | ν ( f ) | p ,

    where the constant c p depends only on p;

  4. For every point z R 2 , there exists a subset S(z) ⊂ E with #S(z) ≤ C, such that T f(z) depends only on f| S(z);

  5. For each ν ∈ {1, , ν max}, there exists a subset S ν E with #S ν C, such that ν (f) depends only on f | S ν ;

  6. ν maxCN.

The constants C in (d), (e), and (f), are universal, so independent of the set E and the function f.

3 Constants

Let 1 < p < 2. Fix p ̃ = 1 + p 2 ( 1 , p ) . We write c, C, C′ to denote absolute constants and we write c p , C p , C p to denote constants depending only on p. We may use the same symbol to denote different constants in different occurrences.

Let K > 0 be larger than some absolute constant. We write

c ( K ) , C ( K ) , C ( K ) ,

to denote constants depending only on K. Once again, the same symbol could be used to denote different occurrences.

At the end of Section 8, we choose K to be an absolute constant, taken sufficiently large. Once we do so, K and any instances of C(K) become absolute constants C.

4 Dyadic squares and intervals

Here, dyadic squares are Q 0 = [ 1,1 ] 2 R 2 and all closed squares arising from Q 0 by successive bisection. More precisely, any dyadic square Q is either Q 0 or of the form [i2k , (i + 1)2k ] × [j2k , (j + 1)2k ], for some k N { 0 } and i, j ∈ {−2 k , , 2 k − 1}. Similarly, dyadic intervals consist of the interval I 0 = [ 1,1 ] R and all the closed intervals [i2k , (i + 1)2k ] for i ∈ {−2 k , , 2 k − 1} and k N , obtained by successive bisection of I 0. Note that dyadic squares and intervals are closed.

If Q is any dyadic square, we denote its side length by δ Q and if I is a dyadic interval, we denote its length by |I|. Moreover, we denote the area δ Q 2 of a dyadic square Q by |Q|. Each dyadic square Q of side length δ Q ≤ 1 arises by bisecting a dyadic square Q +Q of side length 2δ Q ; we call Q + the dyadic parent of Q. Similarly, each dyadic interval I of length |I| ≤ 1 arises by bisecting a dyadic interval I +I of length 2|I|; we call I + the dyadic parent of I.

If I is a dyadic interval and C > 1, then CI denotes the open interval of length C|I| whose center coincides with that of I. Similarly, if Q is a dyadic square and C > 1/4 (C ≠ 1), then CQ denotes the open square of side length Q whose center coincides with that of Q.

Throughout this article, we will make frequent use of the square Q inner [ 2 10 , 2 10 ] 2 Q 0 .

5 Some elementary inequalities

Suppose F L 2 , p ( R 2 ) . In this section, we let Q denote a closed square in R 2 . We will henceforth simply refer to such Q as a square. For any such Q R 2 , we have the Sobolev Inequality [9], Lemma 7.16]: F may be expressed in the form F 1 + L on Q, where L is a first degree polynomial and F 1 is continuous on Q; moreover,

δ Q 2 2 p max z Q { | F 1 ( z ) | } p + δ Q p z Q | F 1 ( z ) | p d z C p 1 | Q | z Q | 2 F | p ̃ d z p / p ̃ δ Q 2 .

For instance, given any point z 0Q, one may take L ( z ) = F ( z 0 ) + 1 | Q | w Q F ( w ) d w ( z z 0 ) . This inequality is not sharp, but it’s enough for our purposes.

We introduce the (non-centered) maximal function

M φ ( z ) = sup Q z 1 | Q | w Q | φ ( w ) | p ̃ d w 1 / p ̃

for functions φ : R 2 R . Here the supremum is taken over all squares Q containing z.

We recall the Hardy–Littlewood Maximal Theorem [10], Chapter I, §3, Theorem 1]:

R 2 ( M φ ) p d z C p R 2 | φ | p d z , 1 < p , φ L p ( R 2 )

For F L 2 , p ( R 2 ) and a square Q R 2 , we write Av Q F to denote the mean 1 | Q | Q F ( z ) d z . Similarly, we write Av Q x F and Av Q y F to denote the components of Av Q F.

Applying the Sobolev Inequality to the open square CQ (C > 1.1) defined as in Section 4, we obtain the following corollaries:

Corollary 5.1.

Let z ̲ C Q , and let L Q ( z ) = F ( z ̲ ) + ( Av Q F ) ( z z ̲ ) . Then for |α| ≤ 1, we have

δ Q | α | p 2 p 1.1 Q | α ( F L Q ) | p d z C p δ Q 2 1 | C Q | C Q | 2 F | p ̃ d z p / p ̃ C p Q M ( | 2 F | ) p d z .

Remark 5.1.

In fact, if Q, Q′ are squares such that δ Q = δ Q and 100Q ∩ 100Q′ ≠ ∅, then we have C p δ Q 2 1 | C Q | C Q | 2 F | p ̃ d z p / p ̃ C p Q M ( | 2 F | ) p d z .

Corollary 5.2.

Let z 1 = (x 1, 0) and z 2 = (x 2, 0) belong to CQ, with |x 1x 2|≥ Q . Then

δ Q 2 p Av Q x F F ( z 2 ) F ( z 1 ) x 2 x 1 p C p δ Q 2 1 | C Q | C Q | 2 F | p ̃ d z p / p ̃ C p Q M ( | 2 F | ) p d z .

Corollary 5.3.

Let z 1 = (x 1, y 1) and z 2 = (x 1, y 2) belong to CQ, with |y 1y 2|≥ Q . Then

δ Q 2 p Av Q y F F ( z 2 ) F ( z 1 ) y 2 y 1 p C p δ Q 2 1 | C Q | C Q | 2 F | p ̃ d z p / p ̃ C p Q M ( | 2 F | ) p d z .

Corollary 5.4.

Let Q 1, Q 2, Q be squares, and suppose Q 1, Q 2CQ and δ Q i c δ Q for i = 1, 2. Then

δ Q 2 p Av Q 1 F Av Q 2 F p C p δ Q 2 1 | C Q | C Q | 2 F | p ̃ d z p / p ̃ C p Q M ( | 2 F | ) p d z .

We use also the following estimates:

Lemma 5.5.

Let Q, Q′ be squares. Suppose QQ′ ≠ ∅, and suppose that their side lengths differ by at most a factor of two. Let z ̲ C Q . Let P be a first degree polynomial on R 2 . Then the quantities | α | 1 δ Q | α | p 2 p 1.1 Q 1.1 Q | α P | p d z and δ Q 2 | δ Q 2 P ( z ̲ ) | p + | δ Q 1 x P | p + | δ Q 1 y P | p differ by at most a factor C p , where C p is determined by C and p.

The elementary proof is left to the reader.

Lemma 5.6.

Let x 0 , x 1 , , x L 1 R , and let 1 < p < 2. Then

(5.1) = 0 L 1 k = L 1 x k p 2 ( 2 p ) C p = 0 L 1 | x | p 2 ( 2 p ) .

Proof

To see this, let β = 22/p−1−ɛ/p ∈ (1, 22/p−1) for ɛ ∈ (0, 2 − p); note that

k = L 1 x k p k = L 1 | x k | p β ( k ) p k = L 1 β ( k ) p p / p

by Hölder’s inequality, where p = p p 1 . Consequently,

= 0 L 1 k = L 1 x k p 2 ( 2 p ) C p = 0 L 1 k = L 1 | x k | p β ( k ) p 2 ( 2 p ) = C p k = 0 L 1 | x k | p = 0 k β ( k ) p 2 ( 2 p ) = C p k = 0 L 1 | x k | p = 0 k 2 ( k ) ( 2 p ε ) 2 ( 2 p ) ,

and thus

= 0 L 1 k = L 1 x k p 2 ( 2 p ) C p k = 0 L 1 | x k | p 2 ( 2 p ) k = 0 k 2 ( k ) ( ε ) C p k = 0 L 1 | x k | p 2 ( 2 p ) k .

Lemma 5.7.

Let L be a first degree polynomial on R 2 , and let L ̃ ( x , y ) = L ( x , 0 ) . Suppose Q R 2 is a square of the form I × J, with 0 ∈ CJ. Then

(5.2) Q | L ̃ ( x , y ) | p d x d y C p Q | L ( x , y ) | p d x d y .

We leave the proof of this inequality to the reader.

6 The set E

Let E = E ̲ × { 0 } R 2 , where E ̲ R is a finite set. Let N = #E ≥ 2. We suppose max E ̲ = 2 11 and min E ̲ = 2 11 .

We will find it convenient to add the point (−1, 0) to E; we set E ̲ + = E ̲ { 1 } , and let E + = E ̲ + × { 0 } .

We will work with functions f : E R and with functions f + : E + R .

7 The Calderón–Zygmund decomposition

Starting with the square Q 0, we perform repeated bisection à la Calderón–Zygmund, stopping at a square Q provided #(E ∩ 3Q) ≤ 1. Recall 3Q denotes an open square of side length 3δ Q . This process results in a decomposition of Q 0 consisting of finitely many closed Calderón–Zygmund squares. We write CZ to denote the set of Calderón–Zygmund squares. Equivalently, the CZ squares are precisely the maximal dyadic squares Q such that #(E ∩ 3Q) ≤ 1.

Remark 7.1.

All the CZ squares Q that we encounter have one of the following forms:

“Contact Squares” I × [ 0 , | I | ]  or  I × [ | I | , 0 ] “Contactless Squares” I × [ | I | , 2 | I | ]  or  I × [ 2 | I | , | I | ] ,

for a dyadic interval I. This follows from an obvious induction, together with the observation that E ∩ 3Q = ∅ for contactless squares Q (because E R × { 0 } , while the open square 3Q is disjoint from R × { 0 } ).

We write QQ′ for CZ squares Q, Q′ to indicate that QQ′ ≠ ∅. If the CZ squares Q, Q′ satisfy QQ′ ≠ ∅, then 1 2 δ Q δ Q 2 δ Q . We recall the standard proof: Suppose QQ′ ≠ ∅ with δ Q 1 4 δ Q . Then the dyadic parent (Q′)+ of Q′ satisfies 3(Q′)+ ⊂ 3Q. Since #(E ∩ 3Q) ≤ 1, it follows that #(E ∩ 3(Q′)+) ≤ 1, and consequently we should not have bisected (Q′)+ to arrive at the CZ square Q′. This contradiction shows that δ Q 1 2 δ Q . By the same reasoning, δ Q 1 2 δ Q . In light of this “good geometry” we recall a familiar bound:

Patching Estimate [1], Lemma 9.1], [8], Lemma 5]: Let ( θ Q ) Q C Z C ( Q 0 ) be a smooth partition of unity satisfying for QCZ, 0 ≤ θ Q ≤ 1; θ Q vanishes on Q 0 \ 1.1Q; | α θ Q | C δ Q | α | for |α| ≤ 2; θ Q | 1 2 Q 1 | 1 2 Q ; and QCZ θ Q ≡ 1 on Q inner. Suppose G Q L 2,p (1.1Q) for each QCZ satisfying 1.1QQ inner ≠ ∅. We define the function G : Q inner R as G(z) : = QCZ θ Q (z)G Q (z). We have

(7.1) G L 2 , p ( Q inner ) p C p Q C Z : 1.1 Q Q inner G Q L 2 , p ( 1.1 Q ) p + | α | 1 Q , Q C Z : Q Q 1.1 Q Q inner , 1.1 Q Q inner δ Q | α | p 2 p 1.1 Q 1.1 Q | α G Q G Q | p d z .

Let π : R 2 R be the projection (x, y) ↦ x. The shadow of a dyadic square Q is the dyadic interval π(Q).

We make the following observation regarding shadows:

Lemma 7.1.

Let I = π(Q) for a CZ square Q. If | I | 1 4 , then I + = π(Q′) for a CZ square Q′.

Proof.

We note that Q arises by bisecting its dyadic parent Q +, which in turn arises by bisecting Q ++. Both Q + and Q ++ are contact squares, else we would not have bisected them. Then one of the two contactless squares Q′ obtained by bisecting Q ++ belongs to CZ and satisfies π(Q′) = I +. See Figure 3 for a depiction of this.□

Figure 3: 
Two possible scenarios for a contactless child Q′ of Q
++.
Figure 3:

Two possible scenarios for a contactless child Q′ of Q ++.

The square Q 0 satisfies #(E ∩ 3Q 0) ≥ 2 since (−2−11, 0), (2−11, 0) ∈ E. Hence every CZ square Q arises by bisecting Q +, so #(E ∩ 3Q +) ≥ 2, else we wouldn’t have bisected Q +. This means that # ( E ̲ 3 I + ) 2 , where I + = π(Q +).

For each shadow I, we fix a point x ̂ ( I ) E ̲ 3 I + E ̲ 7 I .

Let Q be a CZ square. We say that Q is relevant if 1.1QQ inner ≠ ∅, see Section 4, and we denote the set of relevant CZ squares by CZ rel

We note that every relevant CZ square Q satisfies δ Q ≤ 2−8. Indeed if δ Q ≥ 2−7 and 1.1QQ inner ≠ ∅, then 3QQ inner. Hence #(E ∩ 3Q) ≥ 2, so Q cannot be a CZ square.

A relevant shadow is defined to be the shadow of a relevant CZ square. relevant shadows I thus satisfy 1.1I ∩ [−2−10, 2−10] ≠ ∅ and |I| ≤ 2−8.

8 Groups of shadows and squares

Let I = π(Q) be the shadow of QCZ.

Recall, from Section 3 we assume K > 0 larger than a universal constant. We say that I is easy if

(8.1) diam ( E ̲ + K I ) K 1 | I | .

Otherwise, we say that I is hard.

Fix x * E ̲ + . Then we define Gp(x*) (the “group” of x*) to be the set of all shadows I such that

  1. x* ≤ inf  41I, and

  2. E ̲ + ( x * , inf 9 I ] = .

The following elementary lemma will be useful when decomposing shadows into disjoint subfamilies.

Lemma 8.1.

Let I be a shadow. Then IGp(x*) for at most one x * E ̲ + . If in addition I is a hard relevant shadow, then provided K ≥ 41, there exists x * E ̲ + such that IGp(x*).

Proof.

Let I be a shadow; and suppose (a) and (b) hold for x * E ̲ + . Because x* is the maximal element of ( , inf 9 I ] E ̲ + , (a) and (b) cannot also hold for another element of E ̲ + .

Now let I be a hard relevant shadow. Then since 1.1I ∩ [−2−10, 2−10] ≠ ∅ and |I| ≤ 2−8, we have −1 < inf  9I, hence x ̃ < inf 9 I for some x ̃ E ̲ + (we added −1 to E ̲ to form E ̲ + for this reason). Let

x * max { x ̃ E ̲ + : x ̃ inf 9 I } .

We have two possibilities; either x* ∈ 41I \ 9I or x* ≤ inf  41I. Suppose x* ∈ 41I\9I and recall that there exists x ̂ ( I ) E ̲ 7 I . Thus, x * , x ̂ ( I ) E ̲ + 41 I and | x * x ̂ ( I ) | | I | , implying I is easy. This contradicts the assumption I is hard, so we must have x* ≤ inf  41I. By our choice of x*, property (b) also holds.

Thus, IGp(x*).□

Until further notice, we fix a point x * E ̲ + and study its group Gp(x*). We assume that Gp(x*) ≠ ∅.

Lemma 8.2.

Let J, J′ ∈ Gp(x*), then 9J ∩ 9J′ ≠ ∅.

Proof.

Suppose not. Without loss of generality, we may suppose that 9J lives to the left of 9J′, or rigorously:

x * inf 41 J < inf 9 J < sup 9 J inf 9 J .

Since J is a shadow, x ̂ ( J ) E ̲ 7 J . In particular, x ̂ ( J ) E ̲ + ( x * , inf 9 J ] , contradicting condition (b) for J′.□

Fix a shadow I 0Gp(x*) with

| I 0 | = min { | I | : I Gp ( x * ) } .

Lemma 8.3.

Let c > 0 and suppose that JGp(x*) of length |J| > cdist(x*, I 0). Then J is easy for K > 2 c + 19 > 0 .

Proof.

Fix c > 0 and J as in the statement of the lemma. We have 9J ∩ 9I 0 ≠ ∅ by Lemma 8.2, and |J|≥|I 0| by the definition of I 0. Since also |J| > c dist(x*, I 0), it follows that x* ∈ KJ for any K > 2 c + 19 . Thus, x ̂ ( J ) and x* both belong to E ̲ + K J . Moreover property (a) guarantees that x* ≤ inf  41J, whereas x ̂ ( J ) 7 J . Therefore | x ̂ ( J ) x * | | J | . So diam ( K J E ̲ + ) K 1 | J | , proving that J is easy.□

Now consider the family I 0I 1 ⊂ … ⊂ I L , where each I ( ≥ 1) is the dyadic parent of I −1, and L is the largest integer for which all the I (L) belong to Gp(x*). We first demonstrate that I L satisfies the hypothesis of Lemma 8.3 with c = 1 43 .

Lemma 8.4.

The interval I L satisfies | I L | 1 42 dist ( x * , I 0 ) .

Proof.

If |I L | > 1/4, then we must have x* ∈ 7I L because [−1, 1] ⊂ 7I L , a contradiction of property (a) above. Thus |I L | ≤ 1/4. Then by Lemma 7.1, I L + is a shadow. Since I L Gp(x*), we have x* ≤ inf  41I L . Suppose that x * inf 41 I L + . Since inf 9 I L + < inf 9 I 0 and E ̲ + ( x * , inf 9 I 0 ] = , it follows that E ̲ + ( x * , inf 9 I L + ] = . Thus, I L + satisfies conditions (a) and (b) above, i.e., I L + Gp ( x * ) , contradicting the maximality of the index L. Thus we must have x * 41 I L + , and

dist ( x * , I 0 ) 20 | I L + | + | I L + | 42 | I L | .

We will henceforth fix c = 1 43 in Lemma 8.3. Then, by Lemma 8.3, if JGp(x*) is hard, we must have | J | 1 42 dist ( x * , I 0 ) < | I L | . Consequently, for some (0 ≤ L − 1), we have |J| = |I |. Since J and I belong to Gp(x*), Lemma 8.2 tells us that 9J ∩ 9I ≠ ∅.

We now pass from groups of shadows to groups of CZ squares.

Definition 8.1.

Let z* = (x*, 0) ∈ E +. We define

GP ( z * ) { Q CZ : π ( Q ) Gp ( x * ) } ,

namely GP(z*) consists of all CZ squares whose shadows belong to Gp(x*). Corresponding to the sequence of shadows I 0I 1 ⊂ ⋯ ⊂ I L defined as above, we introduce a sequence of CZ squares Q 0, Q 1, …, Q L by choosing Q such that π(Q ) = I for ∈ 0, …, L. For each we have at least one and at most four possible choices for Q ; see Remark 7.1 above.

Collecting the above results regarding the properties of Gp(x*), and noting Lemma 7.1, we obtain the following:

Lemma 8.5.

Let z* ∈ E + with GP(z*) ≠ ∅. Then there exist CZ squares Q 0, Q 1, …, Q L GP(z*) and a constant C > 0 with the following properties:

  1. δ Q = 2 δ Q 0  for  0 L ,

  2. Q CQ  for  < ′,

  3. Q L is easy,

  4. given any hard QGP(z*), there exists  (0 ≤ L − 1), with δ Q = δ Q  and  100 Q 100 Q .

The squares Q 0, Q 1, …, Q L and the number L in Lemma 8.5 depend on z*; we will thus sometimes write Q (z*) to denote the square Q and emphasize the dependence on z*. We write Q small(z*) to denote Q 0 and Q big(z*) to denote Q L . We let

Gang ( , z * ) Q GP ( z * ) : δ Q = δ Q , 100 Q 100 Q .

The CZ squares Q (z*) may or may not be hard. We write L(z*) to denote the number L associated to z* in Lemma 8.5.

Lemma 8.6.

Let QGP(z*) and Q ̃ GP ( z ̃ * ) . If Q Q ̃ , then z * = z ̃ * .

Proof.

Suppose that the shadows J , J ̃ satisfy JGp(x*) and J ̃ Gp ( x ̃ * ) for two points x * , x ̃ * E ̲ + . Suppose that J J ̃ , and that 1 2 | J | | J ̃ | 2 | J | . Then we claim x * = x ̃ * , and the Lemma follows immediately as the shadows π(Q) and π ( Q ̃ ) satisfy these conditions.

To see the claim is true, suppose without loss of generality that inf 9 J inf 9 J ̃ . We know that x ̃ * inf 41 J ̃ , and our assumptions on J , J ̃ imply that inf 41 J ̃ < inf 9 J . Hence, x ̃ * < inf 9 J . Moreover, by the defining property (b) for J ̃ we have E ̲ + ( x ̃ * , inf 9 J ̃ ] = , and therefore E ̲ + ( x ̃ * , inf 9 J ] = also. Thus x ̃ * is the maximal point of E ̲ + ( , inf 9 J ] . However, since JGp(x*), we know that x* is the maximal point of E ̲ + ( , inf 9 J ] . We conclude that x * = x ̃ * .□

We conclude this section by estimating the number of easy CZ squares. Recall that N = #E.

Lemma 8.7.

The number of easy CZ squares is at most C(K) ⋅ N. Moreover, the number of CZ squares Q such that 1.1QE ≠ ∅ is at most CN.

Proof.

For each zE, there are at most C distinct CZ squares Q for which 1.1Qz. Summing over zE, we see at once that the number of CZ squares Q such that 1.1QE ≠ ∅ is at most CN.

To estimate the number of easy CZ squares, we use the Well-Separated Pairs Decomposition, due to Callahan and Kosaraju [11]. As a consequence of the main result of [11], there exist points x ν , x ν ν = 1 ν max E ̲ + × E ̲ + \ Diagonal with the following properties:

  1. Given ( x , x ) E ̲ + × E ̲ + \ Diagonal , there exists ν with | x x ν | + | x x ν | < 1 100 | x x |

  2. ν maxCN.

Now let Q be an easy CZ square with shadow I = π(Q). By definition, there exist x , x E ̲ + K I such that |x′ − x″|≥ K −1|I|. For such x′, x″, we pick ν as in (i). Thus

(8.2) | x ν x ν | 1 2 K 1 | I |  and  x ν , x ν 2 K I .

For fixed x ν , x ν , there are most C(K) dyadic intervals I satisfying (8.2). Summing over ν and applying (ii), we find that there are at most C(K)N distinct shadows of easy CZ squares. Since each shadow arises from at most four CZ squares, we conclude there are at most C(K)N easy CZ squares.□

We now take K to be an absolute constant, large enough that the preceding arguments work. The constant C(K) in Lemma 8.7 is therefore also an absolute constant, henceforth denoted by C. Moreover, we now write that an easy shadow I satisfies diam ( E ̲ + C I ) c | I | , and an easy CZ square Q satisfies diam(E +CQ) ≥ Q .

9 Assigning special points to CZ squares

Let Q be a relevant CZ square. We will associate to Q three points of E +, denoted z 1 ( Q ) , z 2 ( Q ) , z ̲ ( Q ) , as follows.

Recall from Lemma 8.1 that a relevant CZ square is either easy, or else it belongs to GP(z*) for a unique z* ∈ E +. Using this dichotomy, we proceed to define z 1(Q), z 2(Q) ∈ E + for any relevant CZ square Q.

  1. If Q is easy, then by the defining property (8.1), there exist two points z 1, z 2E +CQ such that |z 1z 2|≥ Q . Define z 1(Q) ≔ z 1 and z 2(Q) ≔ z 2.

  2. If Q is hard and belongs to GP(z*) for some z* ∈ E +, then recall from Lemma 8.5 that the CZ square Q big(z*) is easy. Hence, z 1(Q big(z*)) and z 2(Q big(z*)) have already been defined in case (a). Define z 1(Q) ≔ z 1(Q big(z*)) and z 2(Q) ≔ z 2(Q big(z*)). Note that since Q big(z*) is easy, | z 1 ( Q ) z 2 ( Q ) | c δ Q big ( z * ) .

Now, we define the points z ̲ ( Q ) E . We proceed by cases. Let Q be a relevant CZ square.

  1. If 1.1QE ≠ ∅, then because Q is a CZ square, there is a unique point in 1.1QE. Let z ̲ ( Q ) be this unique point.

  2. If 1.1QE = ∅ and Q is easy, then we take z ̲ ( Q ) to be any point of E ∩ 7Q (this set contains E ∩ 3Q + and #(E ∩ 3Q +) ≥ 2)

  3. If 1.1QE = ∅ and Q is hard with QGP(z*) for some z* ∈ E +, let z ̲ ( Q ) = z ̲ ( z * ) , where z ̲ ( z * ) E 7 Q small ( z * ) is fixed for each z* ∈ E +. Thus z ̲ ( Q ) only depends on z* and not the particular choice of Q.

We have thus defined z 1 ( Q ) , z 2 ( Q ) , z ̲ ( Q ) for all relevant CZ squares. We make the following simple observations.

Lemma 9.1.

Let Q be a relevant CZ square. Then:

  1. z ̲ ( Q ) E and if 1.1QE ≠ ∅, then z ̲ ( Q ) is the unique point in 1.1QE.

  2. z ̲ ( Q ) C Q .

Proof.

Part (a) is clear, so we tackle part (b). This is clear in cases (1) and (2). Suppose we are in case (3), so 1.1QE = ∅ and Q is hard, so QGP(z*) for some z* ∈ E +. Let Q 0, Q 1, …, Q L be as in Lemma 8.5 for this choice of z*. Then Q small(z*) = Q 0, and for some 0 ≤ L we have δ Q = δ Q and 100Q ∩ 100Q ≠ ∅. By that same lemma, we have Q 0CQ , hence 7Q 0 ⊂ 7CQ . Since δ Q = δ Q and 100Q ∩ 100Q ≠ ∅, we have 7CQ CQ. Thus, z ̲ ( Q ) 7 Q small ( z * ) = 7 Q 0 C Q , completing the proof of the lemma.□

10 The norm of an interpolant

Let f + : E + R , and let F L 2 , p ( R 2 ) satisfy F = f + on E +. We will derive a lower bound on F L 2 , p ( R 2 ) .

Let QCZ be relevant and easy.

Recall that z i (Q) = (x i (Q), 0) for i = 1, 2 and z ̲ ( Q ) = ( x ̲ ( Q ) , 0 ) belong to E +CQ and satisfy |z 1(Q) − z 2(Q)|≥ Q .

Let

(10.1) L ̂ Q ( x , y ) = f + ( z ̲ ( Q ) ) + f + z 2 ( Q ) f + z 1 ( Q ) x 2 ( Q ) x 1 ( Q ) ( x x ̲ ( Q ) ) + Av Q y F y .

Applying Corollaries 5.1 and 5.2 from Section 5, we see that for all QCZ which are both relevant and easy,

(10.2) | α | 1 δ Q | α | p 2 p 1.1 Q | α ( F L ̂ Q ) | p d z C p Q [ M ( | 2 F | ) ] p d z .

We want an analogue of estimate (10.2) when QCZ is relevant and hard. We won’t be able to derive such an estimate for an individual Q. Instead, we consider together all the hard Q in GP(z*) for a given z* ∈ E +.

Fix z* ∈ E + with GP(z*) ≠ ∅, and let Q 0, Q 1, …, Q L be as in Lemma 8.5. By definition, Q 0 = Q small(z ) and Q L = Q big(z*), and recall that Q L is easy by Lemma 8.5(iii). Since Q −1CQ for = 1, …, L, we see from Corollary 5.4 in Section 5 that

| Av Q 1 x F Av Q x F | p δ Q 2 p C p Q [ M ( | 2 F | ) ] p d z .

Thus, by Lemma 5.6 with x = Av Q x F Av Q + 1 x F , we see that

(10.3) = 0 L 1 | Av Q x F Av Q L x F | p δ Q 2 p C p = 1 L Q [ M ( | 2 F | ) ] p d z .

Recall that Gang(, z*) consists of dyadic squares Q such that δ Q = δ Q and 100Q ∩ 100Q ≠ ∅. Since #Gang(, z*) ≤ C, it follows from (10.3) that

(10.4) = 0 L 1 Q Gang ( , z * ) | Av Q x F Av Q L x F | p δ Q 2 p C p = 1 L Q [ M ( | 2 F | ) ] p d z .

Also, by Corollary 5.4, we have for all QGang(, z*),

| Av Q x F Av Q x F | p δ Q 2 p C p Q [ M ( | 2 F | ) ] p d z .

Summing over Q and again recalling that #Gang(, z*) ≤ C, for each = 0, , L − 1 we obtain

Q Gang ( , z * ) | Av Q x F Av Q x F | p δ Q 2 p C p Q [ M ( | 2 F | ) ] p d z .

Summing this over and combining with (10.4) via the triangle inequality, we get

(10.5) = 0 L 1 Q Gang ( , z * ) | Av Q x F Av Q L x F | p δ Q 2 p C p = 0 L Q [ M ( | 2 F | ) ] p d z .

Again recalling that #Gang(, z*) ≤ C, and that δ Q = δ Q = 2 δ Q 0 for QGang(, z*), we see that

(10.6) = 0 L 1 Q Gang ( , z * ) δ Q 2 p C p δ Q L 2 p .

Note that the condition p < 2 is used to sum the geometric series. Now since Q L is easy, we know that | z 1 ( Q L ) z 2 ( Q L ) | c δ Q L and z 1(Q L ), z 2(Q L ) ∈ E +CQ L . From Corollary 5.2, we have

δ Q L 2 p Av Q L x F f + ( z 2 ( Q L ) ) f + ( z 1 ( Q L ) ) x 2 ( Q L ) x 1 ( Q L ) p C p Q L M ( | 2 F | ) p d z .

Together with (10.6), this yields

= 0 L 1 Q Gang ( , z * ) Av Q L x F f + ( z 2 ( Q L ) ) f + ( z 1 ( Q L ) ) x 2 ( Q L ) x 1 ( Q L ) p δ Q 2 p C p Q L [ M ( | 2 F | ) ] p d z .

This equation combined with (10.5) and the triangle inequality gives

(10.7) = 0 L 1 Q Gang ( , z * ) Av Q x F f + ( z 2 ( Q L ) ) f + ( z 1 ( Q L ) ) x 2 ( Q L ) x 1 ( Q L ) p δ Q 2 p C p = 0 L Q [ M ( | 2 F | ) ] p d z .

For each QGP(z*) we have fixed a point z ̲ ( Q ) = ( x ̲ ( Q ) , 0 ) E C Q . Let L Q ( 1 ) = f + ( z ̲ ( Q ) ) + ( Av Q F ) ( z z ̲ ( Q ) ) . From Remark 5.1 and Corollary 5.1, we have for QGang(, z*),

(10.8) | α | 1 δ Q | α | p 2 p 1.1 Q | α ( F L Q ( 1 ) ) | p d z C p Q M ( | 2 F | ) p d z .

Now let

(10.9) L Q ( 2 ) = f + ( z ̲ ( Q ) ) + f + ( z 2 ( Q L ) ) f + ( z 1 ( Q L ) ) x 2 ( Q L ) x 1 ( Q L ) ( x x ̲ ( Q ) ) + ( Av Q y F ) y .

Note that L Q ( 1 ) L Q ( 2 ) = Av Q x F f + ( z 2 ( Q L ) ) f + ( z 1 ( Q L ) ) x 2 ( Q L ) x 1 ( Q L ) ( x x ̲ ( Q ) ) , so an elementary integration shows that

(10.10) | α | 1 δ Q | α | p 2 p 1.1 Q | α ( L Q ( 1 ) L Q ( 2 ) ) | p d z C p δ Q 2 p Av Q x F f + ( z 2 ( Q L ) ) f + ( z 1 ( Q L ) ) x 2 ( Q L ) x 1 ( Q L ) p .

From (10.7), (10.8), (10.10), and #Gang(, z*) ≤ C, we have

(10.11) = 0 L 1 Q Gang ( , z * ) | α | 1 δ Q | α | p 2 p 1.1 Q | α ( F L Q ( 2 ) ) | p d z C p = 0 L Q [ M ( | 2 F | ) ] p d z .

Recall from Lemma 8.5 that Q for 0 ≤ L − 1 belongs to GP(z*) and every hard QGP(z*) belongs to Gang(, z*) for some 0 ≤ L − 1. Consequently, (10.11) implies the estimate

(10.12) Q GP ( z * ) Q hard | α | 1 δ Q | α | p 2 p 1.1 Q | α ( F L Q ( 2 ) ) | p d z C p Q GP ( z * ) Q [ M ( | 2 F | ) ] p d z .

Let Q be a relevant CZ square. If Q is easy, we set L Q # = L ̂ Q with L ̂ Q as in (10.1). If Q is hard, then set L Q # = L ̂ Q ( 2 ) as in (10.9), arising from the unique z* such that QGP(z*). In this case, we recall that z i (Q) = z i (Q big(z*)) = z i (Q L ) for i = 1, 2, with Q L as in (10.9). Hence, comparing (10.1) with (10.9), we find that

(10.13) L Q # = f + ( z ̲ ( Q ) ) + f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) x 2 ( Q ) x 1 ( Q ) ( x x ̲ ( Q ) ) + ( Av Q y F ) y .

for all relevant CZ squares Q.

Now every relevant CZ square is either easy, or is hard and belongs to GP(z*) for exactly one z* ∈ E +. Summing (10.2) over all easy relevant Q, and summing (10.12) over all z* ∈ E +, we obtain that

(10.14) Q C Z rel | α | 1 δ Q | α | p 2 p 1.1 Q | α F L Q # | p d z = Q C Z rel , easy , Q G P ( z * ) z * E + | α | 1 δ Q | α | p 2 p 1.1 Q | α F L Q # | p d z + z * E + Q C Z rel , Q GP ( z * ) | α | 1 δ Q | α | p 2 p 1.1 Q | α F L Q # | p d z C p Q CZ Q [ M ( | 2 F | ) ] p d z = C p Q 0 [ M ( | 2 F | ) ] p d z C p R 2 | 2 F | p d z .

Recall that for each QCZ, there are at most C distinct Q′ ∈ CZ such that QQ′; those Q′ satisfy 1 2 δ Q δ Q 2 δ Q . Consequently, (10.14) implies that

| α | 1 Q , Q CZ rel Q Q δ Q | α | p 2 p 1.1 Q | α F L Q # | p d z + 1.1 Q | α F L Q # | p d z C p R 2 | 2 F | p d z ,

which in turn implies (by the triangle inequality) that

(10.15) | α | 1 Q , Q CZ rel Q Q δ Q | α | p 2 p 1.1 Q 1.1 Q | α L Q # L Q # | p d z C p R 2 | 2 F | p d z .

Next, we define

(10.16) L Q ( x , y ) f + ( z ̲ ( Q ) ) + f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) x 2 ( Q ) x 1 ( Q ) ( x x ̲ ( Q ) )

for QCZ relevant. In particular, L Q (x, y) is independent of y, and L Q ( x , y ) = L Q # ( x , 0 ) .

Recall that each QCZ has the form I × [0, |I|], I × [|I|, 2|I|], I × [−|I|, 0], or I × [−2|I|, − |I|] for an interval I. Hence, Lemma 5.7 applies to Q, and combined with (10.15), (10.16), this yields

(10.17) | α | 1 Q , Q CZ rel Q Q δ Q | α | p 2 p 1.1 Q 1.1 Q | α L Q L Q | p d z C p R 2 | 2 F | p d z .

By inspecting (10.16), we see that the left-hand side of (10.17) is determined by f +, independently of F.

Conversely, it is easy to exhibit a function F ̃ L 2 , p ( Q inner ) such that F ̃ = f + on E, and F ̃ L 2 , p ( Q inner ) p is dominated by the left-hand side of (10.17). To produce F ̃ , we take a Whitney partition of unity { θ Q } Q CZ adapted to our Calderón–Zygmund decomposition. Specifically, the C 2 functions θ Q satisfy

Q CZ rel θ Q = 1  on  Q inner ,

and each θ Q is supported on 1.1Q with | α θ Q | C δ Q | α | for |α| ≤ 2. Given f + : E + R , we then define

F ̃ Q CZ rel θ Q L Q  on  Q inner ,

with L Q given by (10.16).

Note that F ̃ depends linearly on f +, and that for each zQ inner, F ̃ ( z ) is determined by f +| S(z) for a subset S(z) ⊂ E + with #(S(z)) ≤ C.

Let us check that F ̃ = f + on E. Let QCZ rel. If 1.1QE ≠ ∅, then we have defined z ̲ ( Q ) to be the one and only point of 1.1QE. Moreover, L Q ( z ̲ ( Q ) ) = f + ( z ̲ ( Q ) ) , by its definition (10.16). Therefore, L Q = f + on E ∩ 1.1Q for each relevant Q. In particular, L Q = f + on Esuppθ Q . Since Q θ Q = 1 on Q inner, it follows that

(10.18) F ̃ ( z ) = Q θ Q ( z ) L Q ( z ) = Q θ Q ( z ) f + ( z ) = f + ( z ) ( z E Q inner ) ,

as claimed.

Now, by the Patching Estimate (7.1), we see that

F ̃ L 2 , p ( Q inner ) p C p Q CZ rel L Q L 2 , p ( 1.1 Q ) p + C p | α | 1 Q , Q CZ rel Q Q δ Q | α | p 2 p 1.1 Q 1.1 Q | α L Q L Q | p d z .

Since each L Q is a first-degree polynomial, we have L Q L 2 , p ( 1.1 Q ) p = 0 for all QCZ rel. Hence,

(10.19) F ̃ L 2 , p ( Q inner ) p C p | α | 1 Q , Q CZ rel Q Q δ Q | α | p 2 p 1.1 Q 1.1 Q | α L Q L Q | p d z .

Thus, T + : f + F ̃ is a linear map from functions f + on E + to functions F ̃ L 2 , p ( Q inner ) , with F ̃ ( z ) determined by the values of f + from at most C points of E +. Moreover, T + satisfies F ̃ = f + on E, as well as (10.19).

Next, we investigate circumstances in which a summand in (10.17), (10.19) is identically zero. Those summands arise from (Q, Q′) with QQ′ and Q, Q′ ∈ CZ rel; suppose such (Q, Q′) are also both hard and 1.1QE = 1.1Q′ ∩ E = ∅. By Lemma 8.6, we have Q, Q′ ∈ GP(z*) for some z*. According to our discussion in Section 9, we have z i (Q) = z i (Q big(z*)) = z i (Q′) and z ̲ ( Q ) = z ̲ ( Q small ( z * ) ) = z ̲ ( Q ) . So x i (Q) = x i (Q′) and x ̲ ( Q ) = x ̲ ( Q ) , which means by (10.16), we have L Q = L Q. So the summand in (10.17), (10.19) arising from such a pair (Q, Q′) is identically zero. In summary, combining the estimates (10.17) and (10.19) with the discussion herein, we have proved the following.

Lemma 10.1.

Let

(10.20) f + + p | α | 1 Q , Q CZ rel , Q Q Q  or  Q easy ,  or  1.1 Q E  or  1.1 Q E δ Q | α | p 2 p 1.1 Q 1.1 Q | α L Q L Q | p d z .

Suppose F ̃ L 2 , p ( R 2 ) with F ̃ = f + on E +. Then

f + + p C p F ̃ L 2 , p ( R 2 ) p .

Conversely, the linear map T + : f + F ̃ taking functions f + on E + to functions F ̃ L 2 , p ( Q inner ) defined by (10.18) satisfies

T + f + ( z ) = f + ( z ) ( z E ) ,  and  T + f + L 2 , p ( Q inner ) p C p f + + p .

Moreover, for any zQ inner, T + f +(z) is determined by f + S(z), for a subset S(z) ⊂ E + with #(S(z)) ≤ C.

In the definition of f + in the above lemma, there are at most C#E summands thanks to Lemma 8.7. Moreover, from Lemma 9.1, we have z ̲ ( Q ) C Q , so we can apply Lemma 5.5 to control each summand as follows.

Since y L Q L Q = 0 ,

x L Q L Q = f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) x 2 ( Q ) x 1 ( Q ) f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) x 2 ( Q ) x 1 ( Q ) ,  and L Q L Q ( z ̲ ( Q ) ) = f + ( z ̲ ( Q ) ) f + ( z ̲ ( Q ) ) x ̲ ( Q ) x ̲ ( Q ) x 2 ( Q ) x 1 ( Q ) f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) ,

we find that

| α | 1 δ Q | α | p 2 p 1.1 Q 1.1 Q | α L Q L Q | p d z

and

δ Q 2 δ Q 2 f + ( z ̲ ( Q ) ) f + ( z ̲ ( Q ) ) x ̲ ( Q ) x ̲ ( Q ) x 2 ( Q ) x 1 ( Q ) f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) p + δ Q 2 δ Q 1 f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) x 2 ( Q ) x 1 ( Q ) f + ( z 2 ( Q ) ) f + ( z 1 ( Q ) ) x 2 ( Q ) x 1 ( Q ) p

differ by at most a multiplicative factor C p . Combining this with the conclusions of Lemma 10.1, we have the following result. We let X(E +) denote the vector space of real-valued functions on E +.

Lemma 10.2.

The map T + : X(E +) → L 2,p (Q inner) given by T + f + F ̃ as in (10.18) satisfies the following:

  1. F ̃ = f + on E;

  2. F ̃ L 2 , p ( Q inner ) p C p ν = 1 ν max λ ν | ν + ( f + ) | p for some λ ν > 0 and linear functionals ν + : X ( E + ) R ;

  3. Any F L 2 , p ( R 2 ) with F = f + on E + satisfies

    F L 2 , p ( R 2 ) p c p ν = 1 ν max λ ν | ν + ( f + ) | p ;

  4. λ ν and ν + depend only on E +; neither depends on p or f +.

  5. For every zQ inner, there exists a subset S(z) ⊂ E + with #S(z) ≤ C, such that F ̃ ( z ) depends only on f +| S(z);

  6. For each ν ∈ {1, , ν max}, there exists a subset S ν E + with #S ν C, such that ν + ( f + ) depends only on f + | S ν ;

  7. ν maxCN, where N = #E.

11 Removing the extra point from E +

Observe that Lemma 10.2 concludes Theorem 3, up to getting rid of the extra point (−1, 0) ∈ E + \ E. To do so, we introduce a trivial extension operator as follows.

Recall that E [ 2 11 , 2 11 ] × { 0 } R 2 and that (2−11, 0), (−2−11, 0) ∈ E. Given a function FL 2,p (Q inner), we let L F be the unique first degree polynomial that agrees with F at the points (−2−11, 0), (2−11, 0) and (2−11, 2−11). Then, using Corollaries 5.1–5.3, the Hardy–Littlewood Maximal Theorem, and that δ Q inner = 2 9 , we have

(11.1) α ( F L F ) L p ( Q inner ) C p F L 2 , p ( Q inner ) | α | 1

and L F | R × { 0 } is determined by F(2−11, 0) and F(−2−11, 0), with linear dependency. In particular,

(11.2) L F ( 1,0 ) = a + F ( 2 11 , 0 ) + a F ( 2 11 , 0 ) ,

for universal constants a +, a .

Now fix a cutoff χ C 2 ( R 2 ) supported in Q inner with χ = 1 on [−2−11, 2−11] × {0} and χ C 2 ( R 2 ) C . Define E : L 2 , p ( Q inner ) L 2 , p ( R 2 ) by

E F = χ F + ( 1 χ ) L F .

By (11.1), (11.2) and the definition of L F and χ, for every FL 2,p (Q inner) we have

  1. E F = F on [−2−11, 2−11] × {0};

  2. E F L 2 , p ( R 2 ) C p F L 2 , p ( Q inner ) ;

  3. E F ( 1,0 ) = a + F ( 2 11 , 0 ) + a F ( 2 11 , 0 ) .

Now suppose that f : E R is given. Define f + : E + R by f + = f on E and

f + ( 1,0 ) = a + f ( 2 11 , 0 ) + a f ( 2 11 , 0 ) .

Then define

F ̃ = T + f + L 2 , p ( Q inner ) , ν ( f ) = ν + ( f + ) ,

with T + and ν + as in Lemma 10.2. Then, the conclusion of Lemma 10.2 implies that F ̃ = f on E and

F ̃ L 2 , p ( Q inner ) p C p ν = 1 ν max λ ν | ν ( f ) | p .

Moreover, for any FL 2,p (Q inner) with F = f on E, letting F + E F and using properties (1), (2) above together with Lemma 10.2(iii) yields

ν = 1 ν max λ ν | ν ( f ) | p = ν = 1 ν max λ ν | ν + ( f + ) | p C p F + L 2 , p ( R 2 ) p C p F L 2 , p ( Q inner ) p .

Setting TfT + f +, we summarize what we have just shown in the following lemma.

Lemma 11.1.

Let E [ 2 11 , 2 11 ] × { 0 } R 2 be finite with N = #E and (2−11, 0), (−2−11, 0) ∈ E. Then there exist a linear map T : X(E) → L 2,p (Q inner), a positive integer ν max, positive coefficients λ 1 , , λ ν max and linear functionals 1 , , ν max : X ( E ) R such that the following holds. Given any fX(E),

  1. Tf = f on E;

  2. T f L 2 , p ( Q inner ) p C p ν = 1 ν max λ ν | ν ( f ) | p ;

  3. Any FL 2,p (Q inner) with F = f on E satisfies

    F L 2 , p ( Q inner ) p c p ν = 1 ν max λ ν | ν ( f ) | p ;

  4. λ ν and ν depend only on E, not on p or f;

  5. For every zQ inner, there exists a subset S(z) ⊂ E with #S(z) ≤ C, such that Tf(z) is determined only by f| S(z);

  6. For each ν, there exists a subset S ν E with #S ν C, such that ν (f) is determined only by f | S ν ;

  7. ν maxCN.

Proof of Theorem 3.

By translation and dilation, we can assume that the finite set E satisfies the hypotheses of Lemma 11.1. Given fX(E), let T f ( E T ) ( f ) , for T given by Lemma 11.1. Then T f L 2 , p ( R 2 ) and T f = Tf = f on E, with

(11.3) T f L 2 , p ( R 2 ) p C p T f L 2 , p ( Q inner ) p C p ν = 1 ν max λ ν | ν ( f ) | p .

In addition, for any F L 2 , p ( R 2 ) satisfying F = f on E we clearly have

F L 2 , p ( R 2 ) p F L 2 , p ( Q inner ) p c p ν = 1 ν max λ ν | ν ( f ) | p

This verifies conclusions (a), (b) and (c). The validity of property (d) follows from the fact that given any point z R 2 , T f(z) depends only on Tf evaluated at the points (±2−11, 0), (2−11, 2−11) and z (where the latter dependency is only if zQ inner). Thanks to conclusion (v) of Lemma 11.1, (d) indeed follows.

In light of (vi) and (vii) of Lemma 11.1, conclusions (e) and (f) follow. This completes the proof.□


Corresponding author: Anna Skorobogatova, Department of Mathematics, Fine Hall, Princeton University, Washington Road, Princeton, NJ 08540, USA, E-mail:

Dedicated to Bob Fefferman.


Acknowledgments

We are grateful to Alper Gunes, Jonathan Marty, Arie Israel, Jacob Carruth and Ignacio Uriarte-Tuero for valuable conversations. We are grateful to the NSF and the AFOSR for their generous support.

  1. Research ethics: Not applicable.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Competing interests: The authors state no conflict of interest.

  4. Research funding: Marjorie Drake was supported by NSF Award No. 2103209. Charles Fefferman was supported by the AFOSR through the grant FA9550-23-1-0273. Kevin Ren was supported by an NSF GRFP fellowship. Anna Skorobogatova was supported by the NSF through the grant FRG-1854147.

  5. Data availability: Not applicable.

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Received: 2024-02-18
Accepted: 2024-03-24
Published Online: 2024-05-24

© 2024 the author(s), published by De Gruyter, Berlin/Boston

This work is licensed under the Creative Commons Attribution 4.0 International License.

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