Startseite Parabolic Lipschitz truncation for multi-phase problems: The degenerate case
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Parabolic Lipschitz truncation for multi-phase problems: The degenerate case

  • Bogi Kim , Jehan Oh ORCID logo EMAIL logo und Abhrojyoti Sen
Veröffentlicht/Copyright: 29. April 2025

Abstract

This article is devoted to exploring the Lipschitz truncation method for parabolic multi-phase problems. The method is based on Whitney decomposition and covering lemmas with a delicate comparison scheme of appropriate alternatives to distinguish phases, as introduced by the first and second authors in [B. Kim and J. Oh, Higher integrability for weak solutions to parabolic multi-phase equations, J. Differential Equations 409 2024, 223–298].

1 Introduction

In this paper, we prove energy estimates for weak solutions to parabolic multi-phase problems of type

u t - div ( | u | p - 2 u + a ( z ) | u | q - 2 u + b ( z ) | u | s - 2 u ) = - div ( | F | p - 2 F + a ( z ) | F | q - 2 F + b ( z ) | F | s - 2 F ) in  Ω T ,

where Ω T := Ω × ( 0 , T ) represents a space-time cylinder with a bounded open set Ω n for n 2 , 2 p < q < s < and the modulating coefficients a ( ) and b ( ) are nonnegative and Hölder continuous.

Energy estimates are very important for proving the existence and regularity results for partial differential equations. We prove the energy estimates using the Lipschitz truncation method for parabolic multi-phase problems. Here, the Lipschitz truncation is a method of redefining a given function so that it keeps its values on a specific “good set”, while redefining it in “bad sets” using a partition of unity related to a Whitney covering argument. Acerbi and Fusco [1, 2] have introduced the Lipschitz truncation for elliptic problems. Furthermore, Kinnunen and Lewis [30] and Kim, Kinnunen and Särkiö [28] have studied related methods for the parabolic p-Laplace system and for the parabolic double phase systems, respectively.

In this paper, we deal with a parabolic multi-phase system

(1.1) u t - div 𝒜 ( z , u , u ) = - div ( z , F ) in  Ω T ,

where Carathéodory vector fields 𝒜 : Ω T × N × N n N n and : Ω T × N n N n satisfy the following growth conditions: for any z Ω T , v N and ξ N n , there exist two constants 0 < ν L < such that

(1.2) ν H ( z , | ξ | ) 𝒜 ( z , v , ξ ) ξ , | 𝒜 ( z , v , ξ ) | | ξ | L H ( z , | ξ | )

and

(1.3) | ( z , ξ ) | | ξ | L H ( z , | ξ | ) ,

where the function H : Ω T × + + is defined by

H ( z , κ ) = κ p + a ( z ) κ q + b ( z ) κ s

for z Ω T and κ + . Furthermore, the source term F : Ω T N n satisfies

(1.4) Ω T H ( z , | F | ) 𝑑 z < + .

We assume that the modulating coefficients a : Ω T + and b : Ω T + satisfy

(1.5) q p + 2 α n + 2 , 0 a C α , α 2 ( Ω T ) for some  α ( 0 , 1 ] ,

and

(1.6) s p + 2 β n + 2 , 0 b C β , β 2 ( Ω T ) for some  β ( 0 , 1 ] .

Here, a C α , α 2 ( Ω T ) means that a L ( Ω T ) and there exists a Hölder constant [ a ] α := [ a ] α , α 2 ; Ω T > 0 such that

| a ( x 1 , t 1 ) - a ( x 2 , t 2 ) | [ a ] α ( | x 1 - x 2 | + | t 1 - t 2 | ) α

for all x 1 , x 2 Ω and t 1 , t 2 ( 0 , T ) .

Definition 1.1.

A function u : Ω × ( 0 , T ) N satisfying

u C ( 0 , T ; L 2 ( Ω , N ) ) L 1 ( 0 , T ; W 1 , 1 ( Ω , N ) )

and

Ω T [ H ( z , | u | ) + H ( z , | u | ) ] 𝑑 z <

is a weak solution to (1.1) if

Ω T [ - u φ t + 𝒜 ( z , u , u ) φ ] 𝑑 z = Ω T [ ( z , F ) φ ] 𝑑 z

for every φ C 0 ( Ω T , N ) .

The parabolic multi-phase problems derive from elliptic double phase problems. The elliptic double phase problems of type

- div ( | u | p - 2 u + a ( x ) | u | q - 2 u ) = - div ( | F | p - 2 F + a ( x ) | F | q - 2 F )

was first introduced in [35, 36, 37, 38]. These problems originate from the Lavrentiev phenomenon and the homogenization of strongly anisotropic materials. According to [13, 21], the conditions

(1.7) a ( ) C α ( Ω ) , α ( 0 , 1 ] and q p 1 + α n

and

(1.8) u L ( Ω ) , a ( ) C α ( Ω ) , α ( 0 , 1 ] and q p + α

are sharp for obtaining regularity results of weak solutions. In fact, when (1.7) or (1.8) holds, the gradient of a weak solution u is Hölder continuous, see [6, 14, 19, 13]. Moreover, the Calderón–Zygmund estimates have been discussed in [3, 17, 15]. Also, Baroni, Colombo and Mingione [5] have investigated the Harnack’s inequality. In addition, other regularity results for elliptic double phase problems have been discussed in [31, 32, 10, 9, 8, 11, 23, 22, 25]. The regularity results for elliptic multi-phase problems given by

- div ( | u | p - 2 u + i = 1 m a i ( x ) | u | p i - 2 u ) = - div ( | F | p - 2 F + i = 1 m a i ( x ) | F | p i - 2 F ) in  Ω ,

with 1 < p < p 1 p m and 0 a i ( ) C 0 , α i ( Ω ¯ ) , α i ( 0 , 1 ] , have also been discussed in [4, 16, 18, 20].

On the other hand, the regularity for parabolic double phase problems

u t - div ( | u | p - 2 u + a ( z ) | u | q - 2 u ) = - div ( | F | p - 2 F + a ( z ) | F | q - 2 F ) in  Ω T

with 2 p < q has been studied recently. If (1.5) holds, the existence of weak solutions has been discussed in [12], see also [28, 34]. The gradient higher integrability and the Calderón–Zygmund-type estimate have been studied in [26, 27]. Moreover, the gradient higher integrability results for degenerate and singular parabolic multi-phase problems have also been studied in [24] and [33], respectively.

In this paper, our goal is to prove the energy estimates of the weak solution to (1.1). For this, we denote parabolic cylinders U r , τ ( z 0 ) and Q r ( z 0 ) by

(1.9) U r , τ ( z 0 ) := B r ( x 0 ) × τ ( t 0 ) and Q r ( z 0 ) := B r ( x 0 ) × I r ( t 0 )

with

(1.10) τ ( t 0 ) := ( t 0 - τ , t 0 + τ ) and I r ( t 0 ) := ( t 0 - r 2 , t 0 + r 2 ) .

Our main theorem of this paper is as follows.

Theorem 1.2.

Let u be a weak solution to (1.1). Then, for U R 2 , S 2 ( z 0 ) Ω T , R 1 [ R 2 2 , R 2 ) and S 1 [ S 2 2 2 , S 2 ) , there exists a constant c depending on n , p , q , s , ν and L such that the following inequality holds:

sup t ( t 0 - S 1 , t 0 + S 1 ) B R 1 ( x 0 ) | u - ( u ) U R 1 , S 1 ( z 0 ) | 2 S 1 𝑑 x + ⨍⨍ U R 1 , S 1 ( z 0 ) H ( z , | u | ) 𝑑 z
c ⨍⨍ U R 2 , S 2 ( z 0 ) H ( z , | u - ( u ) U R 2 , S 2 ( z 0 ) | R 2 - R 1 ) 𝑑 z + c ⨍⨍ U R 2 , S 2 ( z 0 ) | u - ( u ) U R 2 , S 2 ( z 0 ) | 2 S 2 - S 1 𝑑 z + c ⨍⨍ U R 2 , S 2 ( z 0 ) H ( z , | F | ) 𝑑 z .

Remark 1.3.

In [24], the energy estimate presented in their Lemma 3.1 is derived under the assumption that | u | L s ( Ω T ) . Consequently, the gradient higher integrability result [24, Theorem 1.2] also relies on this assumption. However, in light of the energy estimate above, [24, Theorem 1.2] can be established under the weaker assumption specified in Definition 1.1.

Remark 1.4.

We would like to mention that the existence and uniqueness results for (1.1) with Dirichlet boundary condition can be obtained by closely following the proofs of [28, Theorem 2.6 and Theorem 2.7]. Therefore, we choose not to pursue such aspects in this paper.

To prove Theorem 1.2, we will construct the Whitney decomposition by dividing into p-, ( p , q ) -, ( p , s ) - and ( p , q , s ) -phases in Section 2. Furthermore, in this section, we prove the Vitali covering argument by dividing it into a total of sixteen cases as in [24] and establish the related properties, which we summarize in Lemma 2.18. In Section 3, we define the Lipschitz truncation and establish the related properties. We then prove the energy estimates in Section 4.

2 Whitney decomposition and covering lemmas

In this section, we construct a family of Whitney decomposition and show that the family covers the bad set E ( Λ ) c via a Vitali covering argument. The construction of such a decomposition is heavily used in the subsequent sections.

2.1 Auxiliary definitions

Let z 0 = ( x 0 , t 0 ) n + 1 and ϱ > 0 . Parabolic cylinders with quadratic scaling in time are denoted as

Q ϱ ( z 0 ) = B ϱ × I ϱ ( t 0 ) ,

where B ϱ = B ϱ ( x 0 ) = { y n : | x 0 - y | < ϱ } and I ϱ ( t 0 ) = ( t 0 - ϱ 2 , t 0 + ϱ 2 ) .

Let us define the strong maximal function for f L loc 1 ( n + 1 ) as

(2.1) M f ( z ) = sup z Q ⨍⨍ Q | f | 𝑑 w ,

where the supremum is taken over cubes Q n + 1 . As in [28], using the Hardy–Littlewood–Wiener maximal function theorem with respect to space and time, we obtain the following lemma.

Lemma 2.1.

For 1 < σ < and f L σ ( R n + 1 ) , there exists a constant c = c ( n , σ ) such that

n + 1 | M f | σ 𝑑 z c n + 1 | f | σ 𝑑 z .

Without loss of generality we can assume that the modulating coefficient functions a ( ) and b ( ) are defined in n + 1 satisfying a ( ) C α , α 2 ( n + 1 ) and b ( ) C β , β 2 ( n + 1 ) , see for instance [28] and [29, Theorem 2.7]. Let f L p ( n + 1 ) , f 0 , be such that

n + 1 ( f p + a f q + b f s ) 𝑑 z < ,

and let d = s - 1 + p 2 . By Lemma 2.1, there exists Λ 0 > 1 + a L ( n + 1 ) + b L ( n + 1 ) such that

n + 1 ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( z ) ) p d 𝑑 z c ( n , p , s ) n + 1 ( f p + a f q + b f s ) 𝑑 z Λ 0 .

Let Λ > Λ 0 and we define the set

(2.2) E ( Λ ) = { z n + 1 : M ( f d + ( a f q ) d p + ( b f s ) d p ) ( z ) Λ d p } .

Chebyshev’s inequality implies that

(2.3) lim Λ Λ | E ( Λ ) c | lim Λ E ( Λ ) c ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( z ) ) p d 𝑑 z = 0 .

Now, we write K 2 as

K := 2 + 800 [ a ] α ( 1 | B 1 | n + 1 ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ) p d ( z ) 𝑑 z ) α n + 2
(2.4) + 800 [ b ] β ( 1 | B 1 | n + 1 ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ) p d ( z ) 𝑑 z ) β n + 2 .

Note that for each z E ( Λ ) c , there exists a unique λ z > 1 such that Λ = λ z p + a ( z ) λ z q + b ( z ) λ z s . We then consider a family of metrics { d z ( , ) } z E ( Λ ) c given by

d z ( z 1 , z 2 ) = { max { | x 1 - x 2 | , λ z p - 2 | t 1 - t 2 | } if  K 2 λ z p a ( z ) λ z q  and  K 2 λ z p b ( z ) λ z s , max { | x 1 - x 2 | , g q ( z , λ z ) λ z - 2 | t 1 - t 2 | } if  K 2 λ z p < a ( z ) λ z q  and  K 2 λ z p b ( z ) λ z s , max { | x 1 - x 2 | , g s ( z , λ z ) λ z - 2 | t 1 - t 2 | } if  K 2 λ z p a ( z ) λ z q  and  K 2 λ z p < b ( z ) λ z s , max { | x 1 - x 2 | , g q , s ( z , λ z ) λ z - 2 | t 1 - t 2 | } if  K 2 λ z p < a ( z ) λ z q  and  K 2 λ z p < b ( z ) λ z s

for z 1 = ( x 1 , t 1 ) , z 2 = ( x 2 , t 2 ) n + 1 , where the functions g q , g s , and g q , s , which are q, s, and ( q , s ) -growth functions, respectively, defined as

(2.5) g q ( z , κ ) := κ p + a ( z ) κ q ,
(2.6) g s ( z , κ ) := κ p + b ( z ) κ s ,
(2.7) g q , s ( z , κ ) := κ p + a ( z ) κ q + b ( z ) κ s

for z Ω T and κ + . For every z E ( Λ ) c , we define a distance function z to E ( Λ ) as

(2.8) 4 r z = d z ( z , E ( Λ ) ) = inf w E ( Λ ) d z ( z , w ) .

Using this distance, we construct a family of open subsets { U z Λ } z E ( Λ ) c defined as

U z Λ = { Q r z , λ z if  K 2 λ z p a ( z ) λ z q  and  K 2 λ z p b ( z ) λ z s , Q r z , λ z q if  K 2 λ z p < a ( z ) λ z q  and  K 2 λ z p b ( z ) λ z s , Q r z , λ z s if  K 2 λ z p a ( z ) λ z q  and  K 2 λ z p < b ( z ) λ z s , Q r z , λ z q , s if  K 2 λ z p < a ( z ) λ z q  and  K 2 λ z p < b ( z ) λ z s .

Here, the intrinsic cylinders denoted as follows:

Q r z , λ z ( z ) := B r z ( x ) × I r z , λ z ( t ) ,
Q r z , λ z q ( z ) := B r z ( x ) × I r z , λ z q ( t ) ,
Q r z , λ z s ( z ) := B r z ( x ) × I r z , λ z s ( t ) ,
Q r z , λ z q , s ( z ) := B r z ( x ) × I r z , λ z q , s ( t )

for any z = ( x , t ) n × ( 0 , T ) , r z > 0 , λ z 1 , where

I r z , λ z ( t ) := ( t - λ z 2 - p r z 2 , t + λ z 2 - p r z 2 ) ,
I r z , λ z q ( t ) := ( t - λ z 2 g q ( z , λ z ) r z 2 , t + λ z 2 g q ( z , λ z ) r z 2 ) ,
I r z , λ z s ( t ) := ( t - λ z 2 g s ( z , λ z ) r z 2 , t + λ z 2 g s ( z , λ z ) r z 2 ) ,
I r z , λ z q , s ( t ) := ( t - λ z 2 g q , s ( z , λ z ) r z 2 , t + λ z 2 g q , s ( z , λ z ) r z 2 ) .

Let f L 1 ( Ω T ) be a function, and let Q Ω T be a measurable set with finite positive measure. We define the integral average of f over Q by

( f ) Q := ⨍⨍ Q f 𝑑 z .

If Q is one of the four intrinsic cylinders defined above, we denote it as follows:

( f ) z 0 ; ρ , λ := ⨍⨍ Q ρ , λ ( z 0 ) f 𝑑 z ,
( f ) z 0 ; ρ , λ ( q ) := ⨍⨍ Q ρ , λ q ( z 0 ) f 𝑑 z ,
( f ) z 0 ; ρ , λ ( s ) := ⨍⨍ Q ρ , λ s ( z 0 ) f 𝑑 z ,
( f ) z 0 ; ρ , λ ( q , s ) := ⨍⨍ Q ρ , λ q , s ( z 0 ) f 𝑑 z .

The integral average on a ball B ρ ( x 0 ) Ω is denoted by

( f ) x 0 ; ρ ( t ) := B ρ ( x 0 ) f ( x , t ) 𝑑 x , t ( 0 , T ) .

We intend to use a Vitali-type argument to find a countable collection of points z i E ( Λ ) c such that the corresponding subfamily satisfies some properties. From now on, we denote

λ i = λ z i , d i ( , ) = d z i ( , ) Q i = Q r z i , λ z i , Q i q = Q r z i , λ z i q , Q i s = Q r z i , λ z i s , Q i q , s = Q r z i , λ z i q , s

and

(2.9) U i = B i × I i = { Q i if  K 2 λ i p a ( z i ) λ i q  and  K 2 λ i p b ( z i ) λ i s , Q i q if  K 2 λ i p < a ( z i ) λ i q  and  K 2 λ i p b ( z i ) λ i s , Q i s if  K 2 λ i p a ( z i ) λ i q  and  K 2 λ i p < b ( z i ) λ i s , Q i q , s if  K 2 λ i p < a ( z i ) λ i q  and  K 2 λ i p < b ( z i ) λ i s .

Moreover, we denote

(2.10) d i ( U i , E ( Λ ) ) = inf z U i , w E ( Λ ) d i ( z , w ) , = { j : 2 K U i 2 K U j } , K i = 200 K 3 for  U i = Q i , Q i q  or  Q i s    and    K i = 200 for  U i = Q i q , s .

2.2 Some preliminary estimates

We start with some basic estimates.

Lemma 2.2.

Let z E ( Λ ) c . Assume K 2 λ z p < a ( z ) λ z q and K 2 λ z p < b ( z ) λ z s . Then a ( z ) 2 a ( z ~ ) 2 a ( z ) and b ( z ) 2 b ( z ~ ) 2 b ( z ) for any z ~ 200 K Q r z ( z ) .

Proof.

We claim that 2 [ a ] α ( 200 K r z ) α < a ( z ) and 2 [ b ] β ( 200 K r z ) β < b ( z ) . We will only prove the second statement, and the proof of the first statement is similar. On the contrary, let us assume b ( z ) 2 [ b ] β ( 200 K r z ) β . Since Q r z , λ z q , s ( z ) E ( Λ ) c and a ( z ) λ z q + b ( z ) λ z s < Λ , we get

a ( z ) λ z q + b ( z ) λ z s < ⨍⨍ Q r z , λ z q , s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w .

Now using Λ = λ z p + a ( z ) λ z q + b ( z ) λ z s < 2 [ a ( z ) λ z q + b ( z ) λ z s ] , we obtain

a ( z ) λ z q + b ( z ) λ z s < ⨍⨍ Q r z , λ z q , s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w
= Λ λ z - 2 2 | B 1 | r z n + 2 Q r z , λ z q , s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w
< ( a ( z ) λ z q + b ( z ) λ z s ) λ z - 2 | B 1 | r z n + 2 Q r z , λ z q , s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w .

Raising the power to β n + 2 in both of sides to the above expression, we have

r z β λ z 2 β n + 2 < ( 1 | B 1 | Q r z , λ z q , s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w ) β n + 2 K 800 [ b ] β .

Now using K 2 λ z p < b ( z ) λ z s , s p + 2 β n + 2 and the counter-assumption, we get

K 2 λ z p < b ( z ) λ z s 2 [ b ] β ( 200 K r z ) β λ z p λ z 2 β n + 2 1 2 K 2 λ z p ,

which is a contradiction. This completes the proof of 2 [ b ] β ( 200 K r z ) β < b ( z ) . Now using the Hölder continuity of b ( z ) , we get

2 [ b ] β ( 200 K r z ) β < b ( z ) inf z ~ 200 K Q r z ( z ) b ( z ~ ) + [ b ] β ( 200 K r z ) β ,

and hence

[ b ] β ( 200 K r z ) β inf z ~ 200 K Q r z ( z ) b ( z ~ ) .

It follows that

sup z ~ 200 K Q r z ( z ) b ( z ~ ) inf z ~ 200 K Q r z ( z ) b ( z ~ ) + [ b ] β ( 200 K r z ) β < 2 inf z ~ 200 K Q r z ( z ) b ( z ~ )

and this proves b ( z ) 2 b ( z ~ ) 2 b ( z ) for any z ~ 200 K Q r z ( z ) . The statement for a ( ) can be proved similarly. ∎

Lemma 2.3.

Let z E ( Λ ) c . Assume K 2 λ z p a ( z ) λ z q and K 2 λ z p < b ( z ) λ z s . Then we have b ( z ) 2 b ( z ~ ) 2 b ( z ) for any z ~ 200 K Q r z ( z ) . Moreover, we have

(2.11) [ a ] α ( 800 K r z ) α λ z q ( K 2 - 1 ) λ z p .

Proof.

First, we note that Q r z , λ z s ( z ) E ( Λ ) c , and therefore

g s ( z , λ z ) < ⨍⨍ Q r z , λ z s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w
= g s ( z , λ z ) λ z - 2 2 | B 1 | r z n + 2 Q r z , λ z s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w .

Raising power α n + 2 in the above expression, we have

r z α λ z 2 α n + 2 < ( 1 | B 1 | Q r z , λ z s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w ) α n + 2 K - 1 800 [ a ] α .

Using q p + 2 α n + 2 , we get

[ a ] α ( 800 K r z ) α λ z q [ a ] α 800 K r z α λ z p λ z 2 α n + 2 [ a ] α 800 K λ z p K - 1 800 [ a ] α ( K 2 - 1 ) λ z p .

Moreover, the proof of the first statement follows from the previous lemma. This completes the proof. ∎

Lemma 2.4.

Let z E ( Λ ) c . Assume K 2 λ z p < a ( z ) λ z q and K 2 λ z p b ( z ) λ z s . Then we have a ( z ) 2 a ( z ~ ) 2 a ( z ) for any z ~ 200 K Q r z ( z ) . Furthermore, we have

(2.12) [ b ] β ( 800 K r z ) β λ z s ( K 2 - 1 ) λ z p .

Proof.

Since Q r z , λ z q E ( Λ ) c , we have

g q ( z , λ z ) < ⨍⨍ Q r z , λ z s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w
= g q ( z , λ z ) λ z - 2 2 | B 1 | r z n + 2 Q r z , λ z s ( z ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) p d 𝑑 w .

Moreover, following the previous lemma, we get

[ b ] β ( 800 K r z ) β λ z s ( K 2 - 1 ) λ z p .

Also, the first statement follows from Lemma 2.2. This completes the proof. ∎

Lemma 2.5.

Let z , z ~ E ( Λ ) c . Assume K 2 λ z p a ( z ) λ z q and K 2 λ z p < b ( z ) λ z s , and z ~ 200 K Q r z ( z ) . Then we have K - 2 p λ z ~ λ z K 2 p λ z ~ .

Proof.

From Lemma 2.2, we get

(2.13) b ( z ) 2 b ( z ~ ) 2 b ( z ) .

Now, note that it is enough to prove K - 2 p λ z ~ λ z . On contrary, let us assume λ z < K - 2 p λ z ~ . Then using (2.11), (2.13) and the counter assumption, we obtain

Λ = λ z p + a ( z ) λ z q + b ( z ) λ z s
λ z p + [ a ] α ( 200 K r z ) α λ z q + a ( z ~ ) λ z q + b ( z ) λ z s
< λ z p + ( K 2 - 1 ) λ z p + a ( z ~ ) K - 2 q p λ z ~ q + 2 b ( z ~ ) K - 2 s p λ z ~ s
K 2 λ z p + a ( z ~ ) K - 2 q p λ z ~ q + b ( z ~ ) K 1 - 2 s p λ z ~ s
< λ z ~ p + a ( z ~ ) λ z ~ q + b ( z ~ ) λ z ~ s = Λ ,

which is a contradiction. ∎

Lemma 2.6.

Let z , z ~ E ( Λ ) c . Assume K 2 λ z p < a ( z ) λ z q and K 2 λ z p b ( z ) λ z s , and z ~ 200 K Q r z ( z ) . Then we have K - 2 p λ z ~ λ z K 2 p λ z ~ .

Proof.

From Lemma 2.4, we have

(2.14) a ( z ) 2 a ( z ~ ) a ( z )

for any z ~ 200 K Q r z ( z ) . We will show K - 2 p λ z ~ . On contrary, let us assume K - 2 p > λ z ~ . Then using (2.12) and (2.14), we get

Λ = λ z p + a ( z ) λ z q + b ( z ) λ z s
λ z p + a ( z ) λ z q + [ b ] β ( 200 K r z ) β λ z s + b ( z ~ ) λ z s
< λ z p + ( K 2 - 1 ) λ z p + 2 a ( z ~ ) K - 2 q p λ z ~ q + b ( z ~ ) K - 2 s p λ z ~ s
K 2 λ z p + a ( z ~ ) K 1 - 2 q p λ z ~ q + b ( z ~ ) K - 2 s p λ z ~ s
λ z ~ p + a ( z ~ ) λ z ~ q + b ( z ~ ) λ z ~ s = Λ ,

which gives a contradiction. This completes the proof. ∎

Lemma 2.7.

Let z , z ~ E ( Λ ) c . Assume a ( z ) 2 a ( z ~ ) 2 a ( z ) and b ( z ) 2 b ( z ~ ) 2 b ( z ) . Then we have 2 - 1 p λ z ~ λ z 2 1 p λ z ~ . Moreover, the above inequality holds provided that K 2 λ z p < a ( z ) λ z q and K 2 λ z p < b ( z ) λ z s , and z ~ 200 K Q r z ( z ) .

Proof.

It is enough to prove the first statement. The second statement of the lemma follows from Lemma 2.2. We claim λ z 2 1 p λ z ~ . In contrast, assume λ z > 2 1 p λ z ~ . Using the hypothesis and the counter assumption, we get

Λ = λ z p + a ( z ) λ z q + b ( z ) λ z s > 2 λ z ~ p + 2 q p - 1 a ( z ~ ) λ z ~ q + 2 s p - 1 b ( z ~ ) λ z ~ s
λ z ~ p + a ( z ~ ) λ z ~ q + b ( z ~ ) λ z ~ s = Λ ,

which is a contradiction. ∎

Lemma 2.8.

Let z E ( Λ ) c . Assume that K 2 λ z p a ( z ) λ z q and K 2 λ z p b ( z ) λ z s . Then [ a ] α ( 50 K r z ) α λ z q ( K 2 - 1 ) λ z p and [ b ] β ( 50 K r z ) β λ z s ( K 2 - 1 ) λ z p .

Proof.

The proof can be completed analogously to the argument presented in the proof of Lemma 2.2. ∎

Lemma 2.9.

Let z E ( Λ ) c . Assume that K 2 λ z p a ( z ) λ z q and K 2 λ z p b ( z ) λ z s . If z ~ 50 K Q r z ( z ) , then λ z K 2 p λ z ~ .

Proof.

The proof can be completed from Lemma 2.5 and Lemma 2.6. ∎

2.3 Vitali covering and their properties

In this subsection, we want to choose a countable collection of intrinsic cylinders from { U z Λ } z E ( Λ ) c such that

E ( Λ ) c z E ( Λ ) c 1 K U z λ and 1 6 K 6 U z 1 Λ 1 6 K 6 U z 2 Λ = for any  z 1 , z 2 E ( Λ ) c .

First, we claim that { r z : z E ( Λ ) c } is uniformly bounded. Indeed, since, by (2.3), 0 < | E ( Λ ) c | < and Λ > 1 + a L ( n + 1 ) + b L ( n + 1 ) , there exist λ > 1 and R > 0 , which are in particular independent of z E ( Λ ) c , such that

(2.15) λ p + a L ( n + 1 ) λ q + b L ( n + 1 ) λ s = Λ

and | B R × ( - λ 2 Λ - 1 R 2 , λ 2 Λ - 1 R 2 ) | = | E ( Λ ) c | . It clear that λ λ z Λ for any z E ( Λ ) c . Hence, if r z > R for some z E ( Λ ) c , then

| E ( Λ ) c | = | B R × ( - λ 2 Λ - 1 R 2 , λ 2 Λ - 1 R 2 ) | < | U z Λ | | E ( Λ ) c | .

This is a contradiction, and we conclude r z R for any z E ( Λ ) c .

Let = { 1 6 K 6 U z Λ } z E ( Λ ) c and, for each j ,

j = { 1 6 K 6 U z Λ : R 2 j < r z R 2 j - 1 } .

Since r z R for any z E ( Λ ) c , we obtain

= j j .

Then, in the same way as [28, Subsection 3.3], we have a countable subcollection 𝒢 of pairwise disjoint cylinders in . From now on, we show that the 6 K 5 -times the cylinders in 𝒢 covers E ( Λ ) c . Fix 1 6 K 6 U z ~ Λ . Then there exists i such that 1 6 K 6 U z ~ Λ i and by the construction of 𝒢 in [28] there exists a cylinder 1 6 K 6 U z Λ j = 1 i 𝒢 j such that

(2.16) 1 6 K 6 U z ~ Λ 1 6 K 6 U z Λ .

Since 1 6 K 6 U z ~ Λ i and 1 6 K 6 U z Λ j for some j { 1 , , i } , we obtain

(2.17) r z ~ 2 r z .

We claim that

(2.18) 1 6 K 6 U z ~ Λ 1 K U z Λ .

Let z = ( x , t ) and z ~ = ( x ~ , t ~ ) . By (2.16), (2.17) and the proof of the standard Vitali covering lemma, we obtain

1 6 K 6 B r z ~ ( y ) 1 K B r z ( x ) .

Moreover, by (2.17) and the standard Vitali covering argument, we obtain that Q r z ~ ( z ~ ) 5 Q r z ( z ) and hence, we conclude

(2.19) z ~ 5 Q r z ( z ) 200 K Q r z ( z ) .

Now, we prove the inclusion in (2.18) in the time direction, by considering the sixteen cases depicted in Table 1.

Table 1

The combinations of U z ~ Λ and U z Λ .

U z ~ Λ U z Λ Q r z , λ z ( z ) Q r z , λ z q ( z ) Q r z , λ z s ( z ) Q r z , λ z q , s ( z )
Q r z ~ , λ z ~ ( z ~ ) (1-1) (1-2) (1-3) (1-4)
Q r z ~ , λ z ~ q ( z ~ ) (2-1) (2-2) (2-3) (2-4)
Q r z ~ , λ z ~ s ( z ~ ) (3-1) (3-2) (3-3) (3-4)
Q r z ~ , λ z ~ q , s ( z ~ ) (4-1) (4-2) (4-3) (4-4)

Case ( 1 - 1 ) . By (2.16) and (2.17), for τ 1 6 K 6 I r z ~ , λ z ~ ( t ~ ) , we obtain

| τ - t | | τ - t ~ | + | t ~ - t |
2 λ z ~ 2 - p ( r z ~ 6 K 6 ) 2 + λ z 2 - p ( r z 6 K 6 ) 2
( 8 λ z ~ 2 - p + λ z 2 - p ) ( r z 6 K 6 ) 2 .

Lemma 2.9 and (2.19) imply

| τ - t | ( 8 K 2 ( p - 2 ) p + 1 ) λ z 2 - p ( r z 6 K 6 ) 2
( 9 K 2 ) λ z 2 - p ( r z 6 K 6 ) 2
λ z 2 - p ( r z K ) 2 ,

and hence τ 1 K I r z , λ z ( t ) .

Cases ( 2 - 1 ) , ( 3 - 1 ) and ( 4 - 1 ) . Since Q r z ~ , λ z ~ q ( z ~ ) Q r z ~ , λ z ~ ( z ~ ) , Q r z ~ , λ z ~ s ( z ~ ) Q r z ~ , λ z ~ ( z ~ ) and Q r z ~ , λ z ~ q , s ( z ~ ) Q r z ~ , λ z ~ ( z ~ ) , by the previous argument, we obtain the conclusion.

Case ( 1 - 2 ) . By (2.16) and (2.17), for τ 1 6 K 6 I r z ~ , λ z ~ ( t ~ ) , we obtain

| τ - t | | τ - t ~ | + | t ~ - t |
2 λ z ~ 2 - p ( r z ~ 6 K 6 ) 2 + λ z 2 g q ( z , λ z ) ( r z 6 K 6 ) 2
( 8 λ z ~ 2 - p + λ z 2 g q ( z , λ z ) ) ( r z 6 K 6 ) 2 .

From a ( z ~ ) λ z ~ q K 2 λ z ~ p , we have

λ z ~ 2 - p = 2 λ z ~ 2 λ z ~ p + λ z ~ p 2 K 2 λ z ~ 2 g q ( z ~ , λ z ~ ) .

Since (1.5) implies q p 2 , we obtain from Lemma 2.4, Lemma 2.6, (1.5) and (2.19) that

(2.20) λ z ~ 2 g q ( z ~ , λ z ~ ) K 6 λ z 2 λ z p + a ( z ~ ) λ z q 2 K 6 λ z 2 g ( z , λ z ) .

Thus, we conclude

| τ - t | ( 32 K 8 + 1 ) λ z 2 g q ( z , λ z ) ( r z 6 K 6 ) 2 λ z 2 g q ( z , λ z ) ( r z K ) 2

and hence τ 1 K I r z , λ z q ( t ) .

Cases ( 1 - 3 ) , ( 1 - 4 ) , ( 2 - 3 ) , ( 3 - 2 ) , ( 2 - 4 ) and ( 3 - 4 ) . By Lemmas 2.4, 2.3, 2.5, 2.6 and 2.7, we obtain from the above argument the conclusion.

Case ( 2 - 2 ) . By (2.16) and (2.17), for τ 1 6 K 6 , we obtain

| τ - t | | τ - t ~ | + | t ~ - t |
2 λ z ~ 2 g q ( z ~ , λ z ~ ) ( r z ~ 6 K 6 ) 2 + λ z 2 g q ( z , λ z ) ( r z 6 K 6 ) 2
( 8 λ z ~ 2 g q ( z ~ , λ z ~ ) + λ z 2 g q ( z , λ z ) ) ( r z 6 K 6 ) 2 .

Then (2.20) implies

| τ - t | ( 16 K 8 + 1 ) λ z 2 g q ( z , λ z ) ( r z 6 K 6 ) 2 λ z 2 g q ( z , λ z ) ( r z K ) 2

and hence τ 1 K I r z , λ z q ( t ) .

Cases ( 3 - 3 ) and ( 4 - 4 ) . In the same way as the above case, we easily obtain the conclusion.

Cases ( 4 - 2 ) and ( 4 - 3 ) . We obtain the conclusion for Cases (2-2) and (2-3) in the same way for Cases (2-1), (3-1) and (4-1).

Thus, we have established a countable covering family { 1 K U i } i of intrinsic cylinders defined as in (2.9) and with pairwise disjoint 1 6 K 6 U i . Now, we prove some properties of the collection { U i } i that will be summarized in Lemma 2.18 at the end.

Lemma 2.10.

We have 3 r i d i ( U i , E ( Λ ) ) 4 r i for every i N .

Proof.

Since, by the definition of r i in (2.8),

d i ( U i , E ( Λ ) ) d i ( z i , E ( Λ ) ) = 4 r i ,

the second inequality is satisfied. Moreover, from the triangle inequality we obtain that, for any z U i ,

d i ( z , E ( Λ ) ) d i ( z i , E ( Λ ) ) - d i ( z , z i ) 4 r i - r i = 3 r i .

Since z U i is arbitrary, we have d i ( U i , E ( Λ ) ) 3 r i . ∎

Next, we prove the property (v) in Lemma 2.18.

Lemma 2.11.

We have ( 12 K 2 ) - 1 r j r i 12 K 2 r j for every i N and j I . Moreover, if U j = Q j q , s , then r i 12 r j

Proof.

It is enough to prove that

r i 12 K 2 r j

for j . If r i < r j , then clearly the conclusion is satisfied. Thus, we assume r j r i . Then

(2.21) z j 4 Q r i ( z i ) .

Let w 2 K U i 2 K U j . Since d i ( z i , w ) 2 K r i 2 r i and d j ( z j , w ) 2 K r j 2 r j , using Lemma 2.10 and the triangle inequality as in [28, Lemma 3.8], we get

(2.22) r i d i ( w , E ( Λ ) ) and d j ( w , E ( Λ ) ) 6 r j .

To complete the proof, we consider 16 cases in Table 1 with z = z i and z ~ = z j .

Case ( 1 - 1 ) , ( 2 - 1 ) , ( 3 - 1 ) , ( 4 - 1 ) , ( 1 - 2 ) , ( 1 - 3 ) and ( 1 - 4 ) . Using Lemma 2.5, Lemma 2.6 and Lemma 2.7 and following the proof of Case 1, Case 2 and Case 3 in [28, Lemma 3.8], we have the conclusion.

Case ( 2 - 2 ) . We obtain from Lemma 2.6 and K 2 λ j p b ( z j ) λ j s that

g q ( z i , λ i ) λ i - 2 g q , s ( z i , λ i ) λ i - 2
K 4 p g q , s ( z j , λ j ) λ j - 2
2 K 2 + 4 p g q ( z j , λ j ) λ j - 2
4 K 4 g q ( z j , λ j ) λ j - 2 ,

where the last inequality follows from 1 p 1 2 . Therefore, we obtain

d i ( z , w ) 2 K 2 d j ( z , w ) for any  z E ( Λ ) ,

and hence, by (2.22), we have the conclusion.

Case ( 2 - 3 ) , ( 2 - 4 ) , ( 3 - 3 ) , ( 3 - 2 ) and ( 3 - 4 ) . Proceed similarly to the proof above.

Case ( 4 - 4 ) . Using Lemma 2.7 and following the proof of [28, Case 3 in Lemma 3.8], we have r i 12 r j .

Case ( 4 - 2 ) and ( 4 - 3 ) . Note that, by Lemma 2.7,

g q ( z i , λ i ) λ i - 2 g q , s ( z i , λ i ) λ i - 2
= g q , s ( z j , λ j ) λ i - 2
2 2 p g q , s ( z j , λ j ) λ j - 2

and, similarly,

g s ( z i , λ i ) λ i - 2 2 2 p g q , s ( z i , λ i ) λ j - 2 .

Thus, we obtain d i ( z , w ) 2 d j ( z , w ) and, therefore, conclude r i 12 r j . ∎

By this lemma, we get

(2.23) 2 K Q r j ( z j ) 200 K Q r i ( z i )

for all j . Using this, we summarize Lemma 2.5, Lemma 2.6, Lemma 2.7 and Lemma 2.9.

Lemma 2.12.

For any i N and j I , we have K - 2 p λ j λ i K 2 p λ j . Moreover, if U i = Q i q , s , then 2 - 1 p λ j λ i 2 1 p λ j .

We show from the previous two lemmas that the measures of the neighboring cylinders are comparable.

Lemma 2.13.

There exists c depending on n and K such that

sup i j | U i | | U j | c .

Proof.

We divide it into sixteen cases as in Lemma 2.11.

Case ( 1 - 1 ) , ( 1 - 2 ) , ( 1 - 3 ) and ( 1 - 4 ) . We obtain from I i I r i , λ i , Lemma 2.11 and Lemma 2.12 that

| U i | | U j | | Q i | | Q j | = 2 | B 1 | r i n + 2 λ i 2 - p 2 | B 1 | r j n + 2 λ j 2 - p ( 12 K 2 ) n + 2 r j n + 2 K 2 ( p - 2 ) p λ j 2 - p r j n + 2 λ j 2 - p ( 12 K 2 ) n + 3

for all i and j .

Case ( 2 - 2 ) , ( 2 - 4 ) , ( 3 - 3 ) , ( 3 - 4 ) and ( 4 - 4 ) . We conclude in a similar way from Lemma 2.11 and Lemma 2.12.

Case ( 2 - 1 ) . We get from a ( z i ) λ i q K 2 λ i p , b ( z i ) λ i s K 2 λ i p , Lemma 2.11 and Lemma 2.12 that

| U i | | U j | = | Q i | | Q j q | = 2 | B 1 | r i n + 2 λ i 2 g q ( z j , λ j ) 2 | B 1 | r j n + 2 λ j 2 λ i p = r i n + 2 λ i 2 g q ( z j , λ j ) r j n + 2 λ j 2 λ i p
( 12 K 2 ) n + 2 K 4 p λ j 2 r j n + 2 g q ( z j , λ j ) λ j 2 r j n + 2 λ i p = 3 ( 12 K 2 ) n + 3 g q ( z j , λ j ) λ i p + λ i p + λ i p
2 ( 12 K 2 ) n + 3 K 2 Λ λ i p + a ( z i ) λ i q + b ( z i ) λ i s c ( n , K )

for all i and j .

Case ( 3 - 1 ) , ( 4 - 1 ) , ( 3 - 2 ) , ( 2 - 3 ) , ( 4 - 2 ) , ( 4 - 3 ) . Similarly, we obtain the conclusion. ∎

Now, we establish the inclusion property of U i and U j for any i and j .

Lemma 2.14.

Let i N be such that U i = Q i . We have 2 K U j 50 K 2 Q i for every j I .

Proof.

Since 2 K B i 2 K B j and r j 12 K 2 r i , we clearly obtain 2 K B j 50 K B i . It remains to prove the inclusion in the time direction. As I j I r j , λ j ( t j ) , for τ 2 K I j , we have from Lemma 2.11 and Lemma 2.12 that

| τ - t i | | τ - t j | + | t j - t i | 2 λ j 2 - p ( 2 K r j ) 2 + λ i 2 - p ( 2 K r i ) 2
2 ( K - 2 p λ i ) 2 - p ( 24 K r i ) 2 + λ i 2 - p ( 2 r i ) 2 λ i 2 - p ( 50 K 2 r i ) 2

and hence 2 K U j 50 K 2 Q i . ∎

Lemma 2.15.

Let i N be such that either U i = Q i q or U i = Q i s holds. Then 2 K U j 100 K 3 U i for every j I .

Proof.

We may assume that U i = Q i q . Since 2 K B i 2 K B j and r j 12 K 2 r i , clearly, we obtain 2 K B j 50 K B i . It remains to prove the inclusion in the time direction. Since I j I r j , λ j ( t j ) , it is enough to check only when U j = Q j . For τ 2 K I r j , λ j ( t j ) , we have

| τ - t i | | τ - t j | + | t j - t i | 2 λ j 2 - p ( 2 K r j ) 2 + λ i 2 g q ( z i , λ i ) ( 2 K r i ) 2 .

Since K 2 λ j p a ( z j ) λ j q , K 2 λ j p b ( z j ) λ j s and Lemma 2.11 and Lemma 2.12 give r j 12 K 2 r i and λ j K 2 p λ i , we obtain

λ j 2 - p ( 2 K r j ) 2 = 3 λ j 2 λ j p + λ j p + λ j p ( 2 K r j ) 2 12 λ j 2 λ j p + a ( z j ) λ j q + b ( z j ) λ j s r j 2
12 K 4 p λ i 2 λ i p + a ( z i ) λ i q + b ( z i ) λ i s ( 12 K 2 r i ) 2 λ i 2 g q ( z i , λ i ) ( 48 K 3 r i ) 2 .

Thus, we have

| τ - t i | 2 λ i 2 g q ( z i , λ i ) ( 48 K 3 r i ) 2 + λ i 2 g q ( z i , λ i ) ( 2 r i ) 2 λ i 2 g q ( z i , λ i ) ( 100 K 3 r i ) 2 ,

and hence 2 K U j 100 K 3 U i . ∎

Lemma 2.16.

Let i N be such that U i = Q i q , s . Then we have 2 K U j 200 U i for each j I .

Proof.

The proof can be obtained by using r j 12 r i and 2 - 1 p λ j λ i 2 1 p λ j instead of ( 12 K 2 ) - 1 r j r i 12 K 2 r j and K - 2 p λ j λ i K 2 p λ j in the proof of the above lemma. ∎

Through the above three lemmas, we obtain the condition

(2.24) 2 K U j K i U i for all  i  and  j ,

where K i is defined in (2.10). Finally, we prove that the cardinality of is uniformly bounded.

Lemma 2.17.

There exists a constant c depending only on n and K such that | I | c for every i N .

Proof.

Since the cylinders { 1 6 K 6 U j } j are disjoint, we get (2.24) and Lemma 2.13 that

| 200 K 4 U i | | j 1 6 K 6 U j | = j | 1 6 K 6 U j | j c ( n , K ) | U i | = c ( n , K ) | | | U i | ,

and hence | | c ( n , K ) . ∎

We summarize the above results below.

Lemma 2.18.

Let K be as in (2.1) and E ( Λ ) as in (2.2). There exists a collection { 1 K U i } i N of cylinders defined as in (2.9) satisfying the following properties:

  1. E ( Λ ) c i 1 K U i .

  2. 1 6 K 6 U i 1 6 K 6 U j = for every i , j with i j .

  3. 3 r i d i ( U i , E ( Λ ) ) 4 r i for every i .

  4. 4 U i E ( Λ ) c and 5 U i E ( Λ ) for every i .

  5. ( 12 K 2 ) - 1 r j r i 12 K 2 r j for every i , j .

  6. K - 2 p λ j λ i K 2 p λ j for every j .

  7. There exists a constant c = c ( n , K ) such that | U i | c | U j | for every i and j .

  8. 2 K U j K i U i for every i , j .

  9. If U i = Q i , then there exists a constant c = c ( [ a ] α , α , K ) such that r i α λ i q c λ i p and r i β λ i s c λ i p .

  10. If U i = Q i q , then a ( z i ) 2 a ( z ) 2 a ( z i ) for every z 200 K Q r i ( z i ) . If U i = Q i s , then b ( z i ) 2 b ( z ) 2 b ( z i ) for every z 200 K Q r i ( z i ) and if U i = Q i q , s , then a ( z i ) 2 a ( z ) 2 a ( z i ) and b ( z i ) 2 b ( z ) 2 b ( z i ) for every z 200 K Q r i ( z i ) .

  11. For any i , the cardinality of , denoted by | | , is finite. Moreover, there exists a constant c = c ( n , K ) such that | | c .

2.4 Partition of unity

The following lemma demonstrates the construction of a partition of unity subordinate to Whitney decomposition { 2 K U i } i .

Lemma 2.19.

There exists a partition of unity { ω i } i N subordinate to the Whitney decomposition { 2 K U i } i N with the following properties:

  1. 0 ω i 1 , ω i C 0 ( 2 K U i ) for every i and i ω i = 1 on E ( Λ ) c .

  2. There exists a constant c = c ( n , K ) such that ω j c r i - 1 for every i and j .

  3. There exists a constant c = c ( n , K ) such that

    t ω j { c r i - 2 λ i p - 2 if  U i = Q i , c r i - 2 g q ( z i , λ i ) λ i - 2 if  U i = Q i q , c r i - 2 g s ( z i , λ i ) λ i - 2 if  U i = Q i s , c r i - 2 Λ λ i - 2 if  U i = Q i q , s

    for any i and j .

Proof.

In the previous subsection, we obtain the Whitney decomposition { 2 K U i } i . Then, for each i , we choose ψ i C 0 ( 2 K U i ) satisfying

0 ψ i 1 , ψ i 1  in  U i , ψ i 2 K r i - 1

and

t ψ i { 2 K r i - 2 λ i p - 2 if  U i = Q i , 2 K r i - 2 g q ( z i , λ i ) λ i - 2 if  U i = Q i q , 2 K r i - 2 g s ( z i , λ i ) λ i - 2 if  U i = Q i s , 2 K r i - 2 Λ λ i - 2 if  U i = Q i q , s .

Since E ( Λ ) c i 2 K U i and | | is finite for each i , the function

ω i ( z ) = ψ i ( z ) j ψ j ( z ) = ψ i ( z ) j ψ j ( z )

is well-defined and satisfies

ω i C 0 ( 2 K U i ) , 0 ω i 1  in  2 K U i , i ω i ( z ) = 1 .

Thus, the collection { ω i } i is a partition of unity subordinate to { 2 K U i } i . By Lemmas 2.11, 2.12 and 2.17, we have

ω j c r i - 1 and t ω j { c r i - 2 λ i p - 2 if  U i = Q i , c r i - 2 g q ( z i , λ i ) λ i - 2 if  U i = Q i q , c r i - 2 g s ( z i , λ i ) λ i - 2 if  U i = Q i s , c r i - 2 Λ λ i - 2 if  U i = Q i q , s

for any i and j . ∎

3 Construction of test function via Lipschitz truncation

The main goal of this section is to construct a Lipschitz function v h Λ which can be used as a test function in the proof of energy estimate Theorem 1.2. We begin by defining v h and v h Λ , establishing a crucial Poincaré-type inequality for v h . By combining this result with the properties of the Whitney decomposition, we conclude the Lipschitz regularity of v h Λ .

3.1 Definition of test function

Take f = χ U R 2 , S 2 ( z 0 ) ( | u | + | u - u 0 | + | F | ) L p ( n + 1 ) and u 0 = ( u ) U R 2 , S 2 ( z 0 ) , where χ is the characteristic function and u , F are extended to zero outside U R 2 , S 2 ( z 0 ) .

Let 0 < h 0 < S 2 - S 1 4 be a sufficiently small number, and let η C 0 ( B R 2 ( x 0 ) ) and ζ C 0 ( S 2 - h 0 ( t 0 ) ) be standard cutoff functions satisfying 0 η 1 , 0 ζ 1 , η 1 in B R 1 ( x 0 ) , ζ 1 in S 1 ( t 0 ) and

(3.1) η 3 R 2 - R 1 , t ζ 3 S 2 - S 1 .

Now for 0 < h < h 0 , we define the truncated solution

(3.2) v h ( z ) = [ u ( z ) - u 0 ] h η ( x ) ζ ( t ) , z = ( x , t ) n + 1 ,

as a suitable candidate for test function to be used to prove energy estimates. For z n + 1 , we define the Lipschitz truncation of v h as

(3.3) v h Λ ( z ) = v h ( z ) - i ( v h ( z ) - v h i ) ω i ( z ) ,

where

v h i = { ⨍⨍ 2 K U i v h ( z ) 𝑑 z if  2 K U i U R 2 , S 2 ( z 0 ) , 0 elsewhere .

Similarly, we define

(3.4) v ( z ) = ( u ( z ) - u 0 ) η ( x ) ζ ( t ) and v Λ ( z ) = v ( z ) - i ( v ( z ) - v i ) ω i ( z ) ,

where

v i = { ⨍⨍ 2 K U i v ( z ) 𝑑 z if  2 K U i U R 2 , S 2 ( z 0 ) , 0 elsewhere .

3.2 Preliminary lemmas

In this subsection, we discuss some preparatory machinery to prove Lemma 3.5. We denote a family of parameters as

data data ( n , N , p , q , s , α , β , ν , L , a L , b L , [ a ] α , [ b ] β , R 1 , R 2 , S 1 , S 2 , K ) .

First, we recall the following lemma from [7, Lemma 8.1].

Lemma 3.1.

Let f L 1 ( Ω T ) and h > 0 . Then there exists a constant c = c ( n ) such that

⨍⨍ U r 1 , r 2 ( z 0 ) [ f ] h 𝑑 z c ⨍⨍ [ U r 1 , r 2 ( z 0 ) ] h f 𝑑 z ,

where [ U r 1 , r 2 ( z 0 ) ] h = U r 1 , r 2 + h ( z 0 ) .

Lemma 3.2.

Let 1 γ d . Then for any cylinder Q R n + 1 such that Q E ( Λ ) , we have

(3.5) ⨍⨍ Q f γ 𝑑 z Λ γ p .

Moreover, there exists a constant c = c ( data ) such that

(3.6) ⨍⨍ 4 K i U i f γ 𝑑 z c λ i γ ,

where K i U i is defined in (2.9)-(2.10).

Proof.

Let w Q E ( Λ ) . Then by (2.1) and (2.2), we have

⨍⨍ Q f γ 𝑑 z ( ⨍⨍ Q f d 𝑑 z ) γ d ( M ( f d ) ( w ) ) γ d ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) γ d Λ γ p

and that proves (3.5). By Lemma 2.18(iv), we have 4 K i U i E ( Λ ) and it follows that

⨍⨍ 4 K i U i f γ 𝑑 z Λ γ p .

Case I: U i = Q i . In this case, we have K 2 λ i p a ( z i ) λ i q and K 2 λ i p b ( z i ) λ i s . Using this, we have

Λ = λ i p + a ( z i ) λ i q + b ( z i ) λ i s ( 2 K 2 + 1 ) λ i p .

Hence, we get

⨍⨍ 4 K i U i f γ 𝑑 z Λ γ p ( 2 K 2 + 1 ) γ p λ i γ = c λ i γ

and this proves (3.6) for this case.

Case II: U i = Q i q . In this case, we have K 2 λ i p < a ( z i ) λ i q and K 2 λ i p b ( z i ) λ i s . By Lemma 2.18(x), we also have a ( z i ) 2 a ( z ) 2 a ( z i ) for every z 200 K Q r i ( z i ) . Then we have

(3.7) ( a ( z i ) ) d p ⨍⨍ 4 K i U i ( f q ) d p 𝑑 z = ⨍⨍ 4 K i U i ( a ( z i ) f q ) d p 𝑑 z 2 d p ⨍⨍ 4 K i U i ( a ( z ) f q ) d p 𝑑 z .

By Lemma 2.18(iv), there exists a w 4 K i U i E ( Λ ) and we obtain

⨍⨍ 4 K i U i ( a ( z ) f q ) d p 𝑑 z M ( ( a f q ) d p ) ( w ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) Λ d p .

Hence, from (3.7), we have

a ( z i ) d p ⨍⨍ 4 K i U i ( f q ) d p 𝑑 z c Λ d p = c ( λ i p + a ( z i ) λ i q + b ( z i ) λ i s ) d p
c ( 2 a ( z i ) λ i q + K 2 λ i p ) d p ( 3 c ) d p ( a ( z i ) λ i q ) d p .

Since a ( z i ) > 0 , we get

⨍⨍ 4 K i U i ( f q ) d p 𝑑 z c λ i q d p

and finally, we have

⨍⨍ 4 K i U i f γ 𝑑 z ( ⨍⨍ 4 K i U i f q d p 𝑑 z ) γ p q d c λ i γ ,

which proves (3.6) for this case.

Case III: U i = Q i s . In this case, we have K 2 λ i p a ( z i ) λ i q and K 2 λ i p < b ( z i ) λ i s . By Lemma 2.18(x), we also have b ( z i ) 2 b ( z ) 2 b ( z i ) for every z 200 K Q r i ( z i ) . Then we have

(3.8) ( b ( z i ) ) d p ⨍⨍ 4 K i U i ( f s ) d p 𝑑 z = ⨍⨍ 4 K i U i ( b ( z i ) f s ) d p 𝑑 z 2 d p ⨍⨍ 4 K i U i ( b ( z ) f s ) d p 𝑑 z .

By Lemma 2.18(iv), there exists a w 4 K i U i E ( Λ ) and we obtain

⨍⨍ 4 K i U i ( b ( z ) f s ) d p 𝑑 z M ( ( b f s ) d p ) ( w ) ( M ( f d + ( a f q ) d p + ( b f s ) d p ) ( w ) ) Λ d p .

Hence, from (3.8), we have

b ( z i ) d p ⨍⨍ 4 K i U i ( f q ) d p 𝑑 z c Λ d p = c ( λ i p + a ( z i ) λ i q + b ( z i ) λ i s ) d p
c ( 2 b ( z i ) λ i q + K 2 λ i p ) d p ( 3 c ) d p ( b ( z i ) λ i q ) d p .

Since b ( z i ) > 0 , we get

⨍⨍ 4 K i U i ( f s ) d p 𝑑 z c λ i s d p

and finally, we have

⨍⨍ 4 K i U i f γ 𝑑 z ( ⨍⨍ 4 K i U i f s d p 𝑑 z ) γ p s d c λ i γ ,

which proves (3.6) for this case.

Case IV: U i = Q i q , s . In this case, we have K 2 λ i p < a ( z i ) λ i q and K 2 λ i p < b ( z i ) λ i s . By Lemma 2.18(x), we also have a ( z i ) 2 a ( z ) 2 a ( z i ) and b ( z i ) 2 b ( z ) 2 b ( z i ) for every z 200 K Q r i ( z i ) . Now we consider two cases: either a ( z i ) λ i q b ( z i ) λ i s or a ( z i ) λ i q b ( z i ) λ i s . The first case corresponds to case III and the second case corresponds to case II and we arrive at the same conclusion. ∎

Now we prove a parabolic Poincaré-type result.

Lemma 3.3.

Let U = B r 1 × r 2 be any cylinder defined in (1.9) and (1.10) satisfying B r 1 B R 2 ( x 0 ) . The the following estimates hold:

  1. There exists a constant c = c ( data ) such that

    (3.9) ⨍⨍ U | v h - ( v h ) U | 𝑑 z c r 2 r 1 ⨍⨍ [ U ] h ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z + c ( r 1 + r 2 ) ⨍⨍ [ U ] h f 𝑑 z .

  2. If in addition r 2 S 2 ( t 0 ) c , then there exists a constant c = c ( data ) such that

    (3.10) ⨍⨍ U | v h | 𝑑 z c r 2 r 1 ⨍⨍ [ U ] h ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z + c ( r 1 + r 2 ) ⨍⨍ [ U ] h f 𝑑 z .

In addition, the above estimates hold with v h and [ U ] h replaced by v and U.

Proof.

For t 1 , t 2 r 2 , t 1 t 2 , let ζ δ W 0 1 , ( r 2 ) be a piecewise linear cut-off function defined by

ζ δ ( t ) = { 0 , t ( - , t 1 - δ ) , 1 + t - t 1 δ , t [ t 1 - δ , t 1 ] , 1 , t ( t 1 , t 2 ) , 1 - t - t 2 δ , t [ t 2 , t 2 + δ ] , 0 , t ( t 2 + δ , ) .

Furthermore, let φ C 0 ( B r 1 ) be a nonnegative function satisfying

B r 1 φ 𝑑 x = 1 , φ c ( n ) r 1 , φ c ( n ) .

By taking a test function ψ = φ η ζ ζ δ W 0 1 , ( U U R 2 , S 2 - h ( z 0 ) ) in the Steklov averaged weak formulation of (1.1), where standard cutoff functions η and ζ are defined in Section 3.1, we obtain

U - [ u - u 0 ] h φ η t ( ζ ζ δ ) d z = U [ - 𝒜 ( , u ) + ( , F ) ] h ψ d z .

By the definition of f, (1.2) and (1.3), we have

I = | U - [ u - u 0 ] h φ η ζ t ζ δ d z |
| U [ f ] h η φ ζ δ t ζ d z | + c ( L ) | U [ f p - 1 + a ( ) f q - 1 + b ( ) f s - 1 ] h ψ d z |
= II + III .

First, we get from Lemma 3.1 that

II U [ f ] h | t ζ | φ 𝑑 z c ( n , S 1 , S 2 ) r 1 n r 2 ⨍⨍ [ U ] h f 𝑑 z

and

III c ( L ) U [ f p - 1 + a ( ) f q - 1 + b ( ) f s - 1 ] h | ψ | 𝑑 z
c ( L ) U [ f p - 1 + a ( ) f q - 1 + b ( ) f s - 1 ] h | φ η + η φ | 𝑑 z
c ( L ) U [ f p - 1 + a ( ) f q - 1 + b ( ) f s - 1 ] h ( φ + η ) | η + φ | 𝑑 z
c ( n , L ) r 1 n r 2 ( 1 R 2 - R 1 + 1 r 1 ) ⨍⨍ [ U ] h ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z .

Note that r 1 R 2 , and hence

1 R 2 - R 1 + 1 r 1 2 max { R 2 R 2 - R 1 , 1 } 1 r 1 .

Therefore, we have

III c ( n , L , R 1 , R 2 ) r 1 n - 1 r 2 ⨍⨍ [ U ] h ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z .

Next, we obtain from the one-dimensional Lebesgue differentiation theorem that

lim δ 0 + I = | lim δ 0 + U [ u - u 0 ] h φ η ζ t ζ δ d z |
= | B r 1 × { t 1 } v h φ 𝑑 z - B r 1 × { t 2 } v h φ 𝑑 z |
= | B 1 | r 1 n | ( v h φ ) B r 1 ( t 1 ) - ( v h φ ) B r 1 ( t 2 ) | .

Combining the above inequalities, we conclude that

(3.11) ess sup t 1 , t 2 r 2 | ( v h φ ) B r 1 ( t 1 ) - ( v h φ ) B r 1 ( t 2 ) | c r 2 ⨍⨍ [ U ] h f 𝑑 z + c r 1 - 1 r 2 ⨍⨍ [ U ] h ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z ,

where c depends only on n , L , S 1 , S 2 , R 1 and R 2 .

Now, we estimate the left-hand sides of (3.9) and (3.10). First, we prove (3.9). By the standard Poincaré inequality, we have

(3.12)

⨍⨍ U | v h - ( v h ) U | 𝑑 z ⨍⨍ U | v h - ( v h ) B r 1 | 𝑑 z + ⨍⨍ U | ( v h ) B r 1 - ( v h ) U | 𝑑 z
c ( n ) r 1 ⨍⨍ U | v h | d z + ⨍⨍ U | ( v h ) B r 1 - ( v h ) U | 𝑑 z .

It follows from the definition of f, Lemma 3.1 and (3.1) that

(3.13) ⨍⨍ U | v h | 𝑑 z = ⨍⨍ U | [ u ] h η ζ + [ u - u 0 ] h η ζ | 𝑑 z c ( n , R 1 , R 2 ) ⨍⨍ [ U ] h f 𝑑 z .

Moreover, we obtain

(3.14)

⨍⨍ U | ( v h ) B r 1 ( τ ) - ( v h ) U | 𝑑 z = r 2 | ( v h ) B r 1 ( τ ) - ( v h ) U | 𝑑 τ
= r 2 | r 2 ( v h ) B r 1 ( τ ) - ( v h ) B r 2 ( σ ) d σ | 𝑑 τ
2 r 2 | ( v h ) B r 1 ( τ ) - ( v h φ ) B r 1 ( τ ) | 𝑑 τ + ess sup t 1 , t 2 r 2 | ( v h φ ) B r 1 ( t 1 ) - ( v h φ ) B r 1 ( t 2 ) | .

To estimate the first term, we use the fact that ( φ ) B r 1 = 1 , the standard Poincaré inequality and (3.13) to obtain

(3.15)

r 2 | ( v h ) B r 1 ( τ ) - ( v h φ ) B r 1 ( τ ) | 𝑑 τ = r 2 | ( v h ) B r 1 ( τ ) B r 1 φ 𝑑 x - B r 1 × { τ } v h φ 𝑑 x | 𝑑 τ
= r 2 | B r 1 × { τ } φ ( v h - ( v h ) B r 1 ) 𝑑 x | 𝑑 τ
c ( n ) φ r 1 ⨍⨍ U | v h | 𝑑 z
c ( n , R 1 , R 2 ) r 1 ⨍⨍ [ U ] h f 𝑑 z .

Combining (3.11), (3.12), (3.13), (3.14) and (3.15), we have the conclusion.

Finally, we prove (3.10). As in (3.12), we obtain

(3.16) ⨍⨍ U | v h | 𝑑 z ⨍⨍ U | v h - ( v h ) B r 1 | 𝑑 z + r 2 | ( v h ) B r 1 | 𝑑 τ c r 1 ⨍⨍ U | v h | d z + r 2 | ( v h ) B r 1 | 𝑑 τ .

Then we have

(3.17) r 2 | ( v h ) B r 1 | 𝑑 τ r 2 | ( v h ) B r 1 - ( v h φ ) B r 1 | 𝑑 t + r 2 | ( v h φ ) B r 1 | 𝑑 t .

Since r 2 S 2 ( t 0 ) c , there exists t 2 r 2 such that t 2 S 2 ( t 0 ) c . Since ζ 0 in S 2 ( t 0 ) , we have

(3.18) r 2 | ( v h φ ) B r 1 | 𝑑 t ess sup t r 2 | ( v h φ ) B r 1 ( t ) | ess sup t 1 , t 2 r 2 | ( v h φ ) B r 1 ( t 1 ) - ( v h φ ) B r 1 ( t 2 ) | .

Thus, combining (3.11), (3.13), (3.15), (3.16), (3.17) and (3.18), we have the conclusion. ∎

Next, we recall the boundary version of the Poincaré inequality from [29, Theorem 6.22].

Lemma 3.4.

Let B ρ ( x 0 ) R n and B r R n be balls that satisfy B r B ρ ( x 0 ) c . Assume that v W 0 1 , η ( B ρ ( x 0 ) ) with 1 < η < . Moreover, let 1 σ n η n - η for 1 < η < n and 1 σ < for n η < . Then there exists a constant c = c ( n , η , σ ) such that

( B 4 r | v | σ 𝑑 x ) 1 σ c r ( B 4 r | v | η 𝑑 x ) 1 η .

3.3 Poincaré-type inequality for the test function

In this subsection, we prove the Poincaré-type inequality for v h .

Lemma 3.5.

Let K i U i be defined in (2.10). Then the following estimates hold:

  1. If K i U i U R 2 , S 2 ( z 0 ) , then there exists a constant c such that

    (3.19) ⨍⨍ K i U i | v h - ( v h ) K i U i | 𝑑 z c ( data , Λ ) r i ,
    (3.20) ⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c ( data ) λ i r i .

  2. If K i U i U R 2 , S 2 ( z 0 ) , then there exists a constant c such that

    (3.21) ⨍⨍ K i U i | v h | 𝑑 z c ( data , Λ ) r i ,
    (3.22) ⨍⨍ K i U i | v | 𝑑 z c ( data ) λ i r i .

Proof.

We start by proving (3.20). Note that we assume K i U i U R 2 , S 2 ( z 0 ) , and hence we can apply Lemma 3.3 with Q = K i U i . We consider the following cases.

Case I: U i = Q i . Using Lemma 3.3, we get

⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c λ i 2 - p r i ⨍⨍ K i U i ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z + c ( r i + λ i 2 - p r i 2 ) ⨍⨍ K i U i f 𝑑 z .

Since r i R 2 and p 2 , we have

(3.23) c ( r i + λ i 2 - p r i 2 ) ⨍⨍ K i U i f 𝑑 z c ( R 2 ) r i ⨍⨍ K i U i f 𝑑 z .

Plugging (3.23) in the above estimate, we get

(3.24) ⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c λ i 2 - p r i ⨍⨍ K i U i ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z + c r i ⨍⨍ K i U i f 𝑑 z .

Since U i = Q i , using K 2 λ i p a ( z i ) λ i q and K 2 λ i p b ( z i ) λ i s , we estimate

⨍⨍ K i U i a ( z ) f q - 1 + b ( z ) f s - 1 d z ⨍⨍ K i U i | a ( z ) - a ( z i ) | f q - 1 𝑑 z + a ( z i ) ⨍⨍ K i U i f q - 1 𝑑 z
+ ⨍⨍ K i U i | b ( z ) - b ( z i ) | f s - 1 𝑑 z + b ( z i ) ⨍⨍ K i U i f s - 1 𝑑 z
[ a ] α ( K i r i ) α ⨍⨍ K i U i f q - 1 𝑑 z + K 2 λ i p - q ⨍⨍ K i U i f q - 1 𝑑 z
+ [ b ] β ( K i r i ) β ⨍⨍ K i U i f s - 1 𝑑 z + K 2 λ i p - s ⨍⨍ K i U i f s - 1 𝑑 z .

Using Hölder’s inequality and applying (3.6), we further estimate

⨍⨍ K i U i a ( z ) f q - 1 + b ( z ) f s - 1 d z [ a ] α ( K i r i ) α ( ⨍⨍ K i U i f s - 1 𝑑 z ) q - 1 s - 1 + K 2 λ i p - q ( ⨍⨍ K i U i f s - 1 𝑑 z ) q - 1 s - 1
+ [ b ] β ( K i r i ) β λ i s - 1 + K 2 λ i p - s λ i s - 1
[ a ] α ( K i r i ) α λ i q - 1 + [ b ] β ( K i r i ) β λ i s - 1 + 2 K 2 λ i p - 1
c λ i p - 1 ,

where the last inequality follows from Lemma 2.18(ix). Plugging the above estimate in (3.24), we obtain

⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c ( data ) λ i r i .

Case II: U i = Q i q . Note that, in this case we have K 2 λ i p < a ( z i ) λ i q and K 2 λ i p b ( z i ) λ i s . Using Lemma 3.3, we get

⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c λ i 2 r i g q ( z i , λ i ) ⨍⨍ K i U i ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z J 1 + c ( r i + λ i 2 r i 2 g q ( z i , λ i ) ) ⨍⨍ K i U i f 𝑑 z J 2 .

Note that J 2 can be estimated as previous since g q ( z i , λ i ) > λ i p . Now we estimate J 1 . First we note that, from Lemma 2.4 we have a ( z i ) 2 a ( z ) 2 a ( z i ) and [ b ] β ( 50 K r i ) β λ i s ( K 2 - 1 ) λ i p . Hence we have

J 1 λ i 2 r i λ i p ⨍⨍ K i U i f p - 1 𝑑 z + λ i 2 r i a ( z i ) λ i q ⨍⨍ K i U i a ( z ) f q - 1 𝑑 z + λ i 2 r i λ i p ⨍⨍ K i U i b ( z ) f s - 1 𝑑 z
c λ i r i + 2 λ i 2 r i λ i q ⨍⨍ K i U i f q - 1 𝑑 z + λ i 2 r i λ i p ( [ b ] β ( K i r i ) β ⨍⨍ K i U i f s - 1 𝑑 z + K 2 λ i p - s ⨍⨍ K i U i f s - 1 𝑑 z ) c λ i r i .

The last part of the above calculation follows from Case I.

Case III: U i = Q i s . Note that, in this case we have K 2 λ i p a ( z i ) λ i q and K 2 λ i p < b ( z i ) λ i s . Using Lemma 3.3, we get

⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c λ i 2 r i g s ( z i , λ i ) ⨍⨍ K i U i ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z J 3 + c ( r i + λ i 2 r i 2 g s ( z i , λ i ) ) ⨍⨍ K i U i f 𝑑 z J 4 .

The estimates of J 3 and J 4 are similar to Case II. We use the Hölder regularity of a ( z ) and bounds on b ( z ) , i.e., b ( z i ) 2 b ( z ) 2 b ( z i ) , and we can arrive at the same conclusion.

Case IV: U i = Q i q , s . In this case, we have K 2 λ i p < a ( z i ) λ i q and K 2 λ i p < b ( z i ) λ i s . and again using Lemma 3.3, we get

⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c λ i 2 r i g q , s ( z i , λ i ) ⨍⨍ K i U i ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z J 5 + c ( r i + λ i 2 r i 2 g q , s ( z i , λ i ) ) ⨍⨍ K i U i f 𝑑 z J 6 .

From Lemma 2.18(x), we have a ( z i ) 2 a ( z ) 2 a ( z i ) and b ( z i ) 2 b ( z ) 2 b ( z i ) for every z 200 K Q r i ( z i ) . The estimate of J 6 is same as previous as λ i p + a ( z i ) λ i q + b ( z i ) λ i s > λ i p . Next, we estimate J 5 . Then we have

J 5 λ i 2 r i λ i p ⨍⨍ K i U i f p - 1 𝑑 z + λ i 2 r i a ( z i ) λ i q ⨍⨍ K i U i a ( z ) f q - 1 𝑑 z + λ i 2 r i b ( z i ) λ i s ⨍⨍ K i U i b ( z ) f s - 1 𝑑 z
λ i 2 r i λ i p ⨍⨍ K i U i f p - 1 𝑑 z + 2 λ i 2 r i λ i q ⨍⨍ K i U i f q - 1 𝑑 z + 2 λ i 2 r i λ i s ⨍⨍ K i U i f s - 1 𝑑 z
c λ i r i ,

where the last inequality follows from (3.6). This completes the proof of (3.20).

Now we prove (3.22). If K i U i U R 2 , S 2 ( z 0 ) , then either K i B i B R 2 ( x 0 ) and K i I i l S 2 c ( t 0 ) , or K i B i B R 2 c . In the first case, we may apply (3.10) of Lemma 3.3. Note that, since the right-hand side of (3.9) and (3.10) are the same, the proof follows from the previous case. In the second case, we apply Lemma 3.4 with σ = 1 and η = d since v ( , t ) W 0 1 , d ( B R 2 ( x 0 ) , N ) . Indeed, we have

⨍⨍ K i U i | v | 𝑑 z c r i ( ⨍⨍ 4 K i U i | v | d 𝑑 z ) 1 d

and using | v | c f , and (3.6) we obtain

( ⨍⨍ 4 K i U i | v | d 𝑑 z ) 1 d c ( ⨍⨍ 4 K i U i f d 𝑑 z ) 1 d c λ i ,

which completes the proof of (3.22).

Next, we focus on the estimates involving Steklov averages, i.e., (3.19) and (3.21). If K i U i U R 2 , S 2 ( z 0 ) , then from Lemma 3.3, we get

⨍⨍ K i U i | v h - ( v h ) K i U i | 𝑑 z c | I i | r i ⨍⨍ [ K i U i ] h ( f p - 1 + a ( z ) f q - 1 + b ( z ) f s - 1 ) 𝑑 z + c ( | I i | + r i ) ⨍⨍ [ K i U i ] h f 𝑑 z .

We note that | I i | r i 2 r i R 2 and a ( z ) a , b ( z ) b . Moreover, we observe that for small h > 0 , [ K i U i ] h may intersect E ( Λ ) , i.e., [ K i U i ] h E ( Λ ) . Using these observations and applying (3.5), we obtain

⨍⨍ K i U i | v h - ( v h ) K i U i | 𝑑 z c r i ⨍⨍ [ K i U i ] h f p - 1 𝑑 z + c r i a ⨍⨍ [ K i U i ] h f q - 1 𝑑 z + c r i b ⨍⨍ [ K i U i ] h f s - 1 𝑑 z
c r i Λ p - 1 p + c r i a ( ⨍⨍ [ K i U i ] h f s - 1 𝑑 z ) q - 1 s - 1 + c r i b Λ s - 1 p
c r i ( Λ p - 1 p + a Λ q - 1 p + b Λ s - 1 p ) = c ( data , Λ ) r i .

This completes the proof of (3.19). To prove (3.21) in the case K i B i B R 2 ( x 0 ) and K i I i l S 2 c , we need to estimate (3.10) which is the same as estimating (3.9). When K i B i B R 2 , we may apply Lemma 3.4 to get

⨍⨍ K i U i | v h | 𝑑 z c r i ( ⨍⨍ 4 K i U i | v h | d 𝑑 z ) 1 d .

Now using Lemma 3.1 and (3.5), we estimate the right-hand side of the above expression as

( ⨍⨍ 4 K i U i | v h | d 𝑑 z ) 1 d c ( ⨍⨍ [ 4 K i U i ] h | v | d 𝑑 z ) 1 d c ( ⨍⨍ [ 4 K i U i ] h f d 𝑑 z ) 1 d c Λ 1 p ,

which gives the proof of (3.21). ∎

Corollary 3.6.

We have the following estimates on v h i , v h j and v i , v j for every i N and j I :

  1. (3.25) ⨍⨍ 2 K U i | v - v i | 𝑑 z c ( data ) λ i r i ,

  2. (3.26) ⨍⨍ 2 K U i | v h - v h i | 𝑑 z c ( data , Λ ) r i ,

  3. (3.27) | v h i - v h j | c ( data , Λ ) r i ,

  4. (3.28) | v i - v j | c ( data ) λ i r i .

Proof.

We start by proving (i). Indeed, we obtain

⨍⨍ 2 K U i | v - v i | 𝑑 z ⨍⨍ 2 K U i | v - ( v ) K i U i | 𝑑 z + ⨍⨍ 2 K U i | ( v ) K i U i - ( v ) 2 K U i | 𝑑 z
c ⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z .

If K i U i U R 2 , S 2 ( z 0 ) , then from (3.20) we get

⨍⨍ 2 K U i | v - v i | 𝑑 z c ⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c ( data ) λ i r i .

If K i U i U R 2 , S 2 ( z 0 ) , from (3.22) we have

⨍⨍ 2 K U i | v - v i | 𝑑 z 2 ⨍⨍ 2 K U i | v | 𝑑 z c ⨍⨍ K i U i | v | 𝑑 z c λ i r i .

The proof of (ii) follows from (3.19) and (3.21) as above. Now, let us prove (iv) and the proof of (iii) is similar. First, we assume K i U i , K j U j U R 2 , S 2 ( z 0 ) . Using Lemma 2.18(vii) and (viii), we estimate

| v i - v j | | v i - ( v ) K i U i | + | v j - ( v ) K i U i |
= | ⨍⨍ 2 K U i v - ( v ) K i U i d z | + | ⨍⨍ 2 K U j v - ( v ) K i U i d z |
c ⨍⨍ K i U i | v - ( v ) K i U i | 𝑑 z c λ i r i ,

where the last inequality follows from (3.20). On the other hand, let K i U i U R 2 , S 2 ( z 0 ) . In this case, again using Lemma 2.18(vii) and (viii), we obtain

| v i - v j | ⨍⨍ 2 K U i | v | 𝑑 z + ⨍⨍ 2 K U j | v | 𝑑 z c ⨍⨍ K i U i | v | 𝑑 z c λ i r i .

The other case K j U j U R 2 , S 2 ( z 0 ) follows from the fact that r i , r j and λ i , λ j are comparable. ∎

3.4 Bounds on Lipschitz truncation and its derivatives

In this subsection, we show that v h Λ , v Λ and their gradients are bounded.

Lemma 3.7.

We have | v h Λ ( z ) | c ( data , Λ ) and | v Λ ( z ) | c ( data ) λ i for every z U i .

Proof.

Fix z U i . Then, for each j , we obtain from Lemma 2.18(viii), Lemma 3.2 and the definition of v j that

| v j | ⨍⨍ 2 K U j | v | 𝑑 z c ⨍⨍ K i U i f 𝑑 z c λ i .

Moreover, we have from Lemma 2.19 that

(3.29) v Λ ( z ) = j v j ω j ( z ) = j v j ω j ( z ) .

Therefore, we conclude from Lemma 2.18(xi) and Lemma 2.19 that

| v Λ ( z ) | j | v j | | ω j ( z ) | j | v j | c λ i .

On the other hand, for each j , we get from Lemma 3.1 and Lemma 3.2 that

| v h i | ⨍⨍ 2 K U i | v h ( z ) | 𝑑 z c ⨍⨍ 2 K [ U i ] h f 𝑑 z c ( data , Λ ) .

As in (3.29), we have

(3.30) v h Λ ( z ) = j v h j ω j ( z ) ,

and hence we have the conclusion. ∎

Lemma 3.8.

For any z U i , we have

(3.31) | v h Λ ( z ) | c ( data , Λ ) 𝑎𝑛𝑑 | v Λ ( z ) | c ( data ) λ i .

Furthermore, we have

(3.32) | t v h Λ ( z ) | c ( data , Λ ) r i - 1

and

(3.33) | t v Λ ( z ) | { c ( data ) r i - 1 λ i p - 1 if  U i = Q i , c ( data ) r i - 1 g q ( z i , λ i ) λ i - 1 if  U i = Q i q , c ( data ) r i - 1 g s ( z i , λ i ) λ i - 1 if  U i = Q i s , c ( data ) r i - 1 Λ λ i - 1 if  U i = Q i q , s .

Proof.

Fix z U i . Then, by (3.29) and (3.30), we get

v h Λ ( z ) = ( j v h j ω j ( z ) ) = j v h j ω j ( z )

and

v Λ ( z ) = j v j ω j ( z ) .

Note that Lemma 2.19 implies

0 = ( j ω j ( z ) ) = ( j ω j ( z ) ) = j ω j ( z ) .

Thus, we get from Lemma 2.18(xi), Lemma 2.19 and Corollary 3.6 that

| v h Λ ( z ) | = | j ( v h j - v h i ) ω j ( z ) | j | v h j - v h i | | ω j | c ( data , Λ )

and

| v Λ ( z ) | j | v j - v i | | ω j | c ( data ) λ i .

Next, by the above arguments, we get

t v h Λ ( z ) = j v h j t ω j ( z ) = j ( v h j - v h i ) t ω j ( z )

and

t v Λ ( z ) = j v j t ω j ( z ) = j ( v j - v i ) t ω j ( z ) .

Therefore, by Lemma 2.18(xi), Lemma 2.19 and Corollary 3.6, we have the conclusion. ∎

Lemma 3.9.

Let E ( Λ ) be defined as in (2.2). Then v h Λ , v Λ , the Lipschitz truncation defined in (3.3)–(3.4), satisfies the following estimates:

  1. E ( Λ ) c | v h - v h Λ | | t v h Λ | 𝑑 z c ( data , Λ ) | E ( Λ ) c | ,

  2. E ( Λ ) c | v - v Λ | | t v Λ | 𝑑 z c ( data ) Λ | E ( Λ ) c | ,

  3. H ( z , | v Λ | ) + H ( z , | v Λ | ) c ( data ) Λ

    for almost every z n + 1 .

Proof.

We start with proving (i):

(3.34) E ( Λ ) c | v h - v h Λ | | t v h Λ | 𝑑 z E ( Λ ) c i | v h - v h i | | ω i | | t v h Λ | d z
i 2 K U i | v h - v h i | | ω i | | t v h Λ | 𝑑 z
(3.35) i t v h Λ L ( 2 K U i ) 2 K U i | v h - v h i | 𝑑 z .

Now using Corollary 3.6(ii) and (3.32), we can estimate the last term of the above inequality to obtain

E ( Λ ) c | v h - v h Λ | | t v h Λ | 𝑑 z c i | U i | = c i | 1 6 K 6 U i | = c ( data , Λ ) | E ( Λ ) c | .

The proof of (ii) can be obtained similarly. To prove (iii), first let z E ( Λ ) . In this case,

| v Λ ( z ) | p + a ( z ) | v Λ ( z ) | q + b ( z ) | v Λ ( z ) | s + | v Λ ( z ) | p + a ( z ) | v Λ ( z ) | q + b ( z ) | v Λ ( z ) | s
= | v ( z ) | p + a ( z ) | v ( z ) | q + b ( z ) | v ( z ) | s + | v ( z ) | p + a ( z ) | v ( z ) | q + b ( z ) | v ( z ) | s
( f p + a ( z ) f q + b ( z ) f s ) c Λ .

Now let us consider z E ( Λ ) c . Then z U i for some i . Using Lemma 3.7 and Lemma 3.8, we get

| v Λ ( z ) | p + a ( z ) | v Λ ( z ) | q + b ( z ) | v Λ ( z ) | s + | v Λ ( z ) | p + a ( z ) | v Λ ( z ) | q + b ( z ) | v Λ ( z ) | s
c ( λ i p + a ( z ) λ i q + b ( z ) λ i s ) .

Case I: U i = Q i . Using the Hölder regularity of a ( z ) and b ( z ) , and Lemma 2.18, (ix) we obtain

λ i p + a ( z ) λ i q + b ( z ) λ i s λ i p + a ( z i ) λ i q + [ a ] α r i α λ i q + b ( z i ) λ i s + [ b ] β r i β λ i s
c ( λ i p + a ( z i ) λ i q + b ( z i ) λ i s ) = c Λ .

Case II: U i = Q i q . In this case, we use Lemma 2.4 to get

λ i p + a ( z ) λ i q + b ( z ) λ i s λ i p + 2 a ( z i ) λ i q + b ( z i ) λ i s + [ b ] β r i β λ i s
c ( λ i p + a ( z i ) λ i q + b ( z i ) λ i s ) = c Λ .

Case III: U i = Q i s . In this case, we use Lemma 2.3 to get

λ i p + a ( z ) λ i q + b ( z ) λ i s λ i p + a ( z i ) λ i q + 2 b ( z i ) λ i s + [ a ] α r i α λ i q
c ( λ i p + a ( z i ) λ i q + b ( z i ) λ i s ) = c Λ .

Case IV: U i = Q i q , s . In this case we use Lemma 2.2 and obtain

λ i p + a ( z ) λ i q + b ( z ) λ i s λ i p + 2 a ( z i ) λ i q + 2 b ( z i ) λ i s 2 Λ .

This completes the proof. ∎

3.5 Lipschitz regularity of v h Λ

In this subsection, we show that v h Λ is Lipschitz continuous with respect to the metric

d λ p ( z , w ) = max { | x - y | , λ p - 2 | t - s | } ,

where z , w U R 2 , S 2 ( z 0 ) with z = ( x , t ) and w = ( y , s ) and λ is chosen such that Λ = λ p + a λ q + b λ s . Let us recall the definition of Q l , λ ( w ) , that is,

Q l , λ ( w ) := B l ( y ) × ( s - λ 2 - p l 2 , s + λ 2 - p l 2 ) .

Lemma 3.10 (Campanato characterization).

Assume that f L loc 1 ( R n + 1 ) . Then there exist a constant c = c ( n ) and a set E R n + 1 with | E | = 0 such that

| f ( z ) - f ( w ) | c ( n ) d λ p ( z , w ) sup l > 0 ⨍⨍ Q l , λ ( w ) | f ( w ~ ) - ( f ) Q l , λ ( w ) | l 𝑑 w ~ + c ( n ) d λ p ( z , w ) sup l > 0 ⨍⨍ Q l , λ ( z ) | f ( z ~ ) - ( f ) Q l , λ ( z ) | l 𝑑 z ~

for every z , w R n + 1 E .

Proof.

Since Q d λ p ( z , w ) , λ ( z ) Q 2 d λ p ( z , w ) , λ ( w ) , by replacing B ( x , r ) with Q l , λ ( z ) and | x - y | with d λ p ( z , w ) , and taking β = 1 in the proof of [29, Lemma 4.13], we obtain that

| f ( z ) - f ( w ) | c ( n ) d λ p ( z , w ) sup 0 < l < 4 d λ p ( z , w ) ⨍⨍ Q l , λ ( w ) | f ( w ~ ) - ( f ) Q l , λ ( w ) | l 𝑑 w ~
+ c ( n ) d λ p ( z , w ) sup 0 < l < 4 d λ p ( z , w ) ⨍⨍ Q l , λ ( z ) | f ( z ~ ) - ( f ) Q l , λ ( z ) | l 𝑑 z ~
c ( n ) d λ p ( z , w ) sup l > 0 ⨍⨍ Q l , λ ( w ) | f ( w ~ ) - ( f ) Q l , λ ( w ) | l 𝑑 w ~
+ c ( n ) d λ p ( z , w ) sup l > 0 ⨍⨍ Q l , λ ( z ) | f ( z ~ ) - ( f ) Q l , λ ( z ) | l 𝑑 z ~ .

This completes the proof. ∎

We remark that the above conclusion holds for every z , w n + 1 by [29, Remark 4.14]. Now we are ready to prove the Lipschitz regularity of v h Λ .

Lemma 3.11.

There exists a constant c Λ = c ( data , Λ ) such that

| v h Λ ( z ) - v h Λ ( w ) | c Λ d λ p ( z , w )

for every z , w R n + 1 .

Proof.

We first note that, since λ i λ from the definition of d, we have

(3.36) d λ p ( z , w ) d i ( z , w ) for all  i .

Applying Lemma 3.10, we get

(3.37)

| v h Λ ( z ) - v h Λ ( w ) | c ( n ) d λ p ( z , w ) sup l > 0 ⨍⨍ Q l , λ ( w ) | v h Λ ( w ~ ) - ( v h Λ ) Q l , λ ( w ) | l 𝑑 w ~
+ c ( n ) d λ p ( z , w ) sup l > 0 ⨍⨍ Q l , λ ( z ) | v h Λ ( z ~ ) - ( v h Λ ) Q l , λ ( z ) | l 𝑑 z ~ .

Note that v h Λ L loc 1 ( n + 1 ) . Therefore from (3.37), it is enough to show that there exists a constant c Λ = c Λ ( data , Λ ) such that

(3.38) ⨍⨍ Q l , λ ( w ) | v h Λ ( w ~ ) - ( v h Λ ) Q l , λ ( w ) | l 𝑑 w ~ c Λ for all  Q l , λ ( w ) n + 1 .

We fix the cube Q l , λ ( w ) and prove the above estimate (3.38) when 2 Q l , λ ( w ) completely lies on the bad set E ( Λ ) c , or Q l , λ ( w ) lies inside the bad set E ( Λ ) c , but 2 Q l , λ ( w ) meets the good set E ( Λ ) or both Q l , λ ( w ) and 2 Q l , λ ( w ) meets the good set E ( Λ ) .

Case 1: 2 Q l , λ ( ω ) E ( Λ ) c . Let z Q l , λ ( ω ) . Then, by Lemma 2.18(i) there exists i such that z U i . Using (3.36), we have

l d λ p ( z , E ( Λ ) ) d λ p ( z , z i ) + d λ p ( z i , E ( Λ ) ) d i ( z , z i ) + d i ( z i , E ( Λ ) ) .

Here, z i is the center of U i . Since U i is a ball of radius r i with respect to the metric d i ( , ) , we get from Lemma 2.18(iv) that

l d i ( z , z i ) + d i ( z i , E ( Λ ) ) r i + 5 r i = 6 r i .

Lemma 3.8 implies the uniform estimate

(3.39) | t v h Λ ( z ) | c ( data , Λ ) r i - 1 c ( data , Λ ) l - 1

for all z Q l , λ ( w ) .

To prove (3.38), let z 1 , z 2 Q l , λ ( w ) with z 1 = ( x 1 , t 1 ) and z 2 = ( x 2 , t 2 ) . Since v h Λ C ( E ( Λ ) c , N ) and Q l , λ ( w ) E ( Λ ) c , by the intermediate value theorem, (3.31) and (3.39), we have

| v h Λ ( z 1 ) - v h Λ ( z 2 ) | | v h Λ ( x 1 , t 1 ) - v h Λ ( x 2 , t 1 ) | + | v h Λ ( x 2 , t 1 ) - v h Λ ( x 2 , t 2 ) |
c l sup z Q l , λ ( ω ) | v h Λ ( z ) | + c λ 2 - p l 2 sup z Q l , λ ( ω ) | t v h Λ ( z ) | c ( data , Λ ) l .

Thus, we conclude

⨍⨍ Q l , λ ( w ) | v h Λ - ( v h Λ ) Q l , λ ( w ) | l 𝑑 w ~ ⨍⨍ Q l , λ ( w ) ⨍⨍ Q l , λ ( w ) | v h Λ ( z 1 ) - v h Λ ( z 2 ) | l 𝑑 z 2 𝑑 z 1 c Λ ( data , Λ ) .

Case 2: 2 Q l , λ ( w ) E ( Λ ) and Q l , λ ( w ) E ( Λ ) c = . In this case v h Λ = v h and hence

⨍⨍ Q l , λ ( ω ) | v h Λ - ( v h Λ ) Q l , λ ( w ) | l 𝑑 w ~ = ⨍⨍ Q l , λ ( ω ) | v h - ( v h ) Q l , λ ( w ) | l 𝑑 w ~ .

We denote w = ( y , s ) . If B l ( y ) B R 2 ( x 0 ) , then l R 2 is satisfied and Lemma 3.2 and Lemma 3.3 imply that

⨍⨍ Q l , λ ( w ) | v h - ( v h ) Q l , λ ( w ) | l 𝑑 w ~ c ( data , Λ ) ( λ 2 - p + λ 2 - p l + 1 ) c ( data , Λ ) .

On the other hand, if B l ( y ) B R 2 ( x 0 ) , we apply Lemma 3.4 with σ = 1 and η = d to get

⨍⨍ Q l , λ ( w ) | v h - ( v h ) Q l , λ ( w ) | l 𝑑 w ~ 2 ⨍⨍ Q l , λ ( w ) | v h | l 𝑑 w ~ c ⨍⨍ 4 Q l , λ ( w ) | v h | 𝑑 w ~ .

Recalling that in this case 2 Q l , λ ( w ) E ( Λ ) , we conclude with Lemma 3.2 that

⨍⨍ 4 Q l , λ ( w ) | v h | 𝑑 w ~ c ⨍⨍ [ 4 Q l , λ ( w ) ] h f 𝑑 w ~ c ( data , Λ ) .

Case 3: 2 Q l , λ ( w ) E ( Λ ) and Q l , λ E ( Λ ) . We define the index set

P = { i : Q l , λ ( w ) 2 K U i } .

We want to show that the radii r i are bounded uniformly by l with respect to i P . Let i P , w 1 Q l , λ ( w ) 2 K U i and w 2 2 Q l , λ ( w ) E ( Λ ) with w 1 = ( y 1 , s 1 ) and w 2 = ( y 2 , s 2 ) . By Lemma 2.18(iii) and w 1 2 K U i , we obtain

3 r 1 d i ( U i , E ( Λ ) ) d i ( z i , w 2 ) d i ( z i , w 1 ) + d i ( w 1 , w 2 ) 2 r i + d i ( w 1 , w 2 ) ,

and hence r i d i ( w 1 , w 2 ) . Moreover, since λ λ i and w 1 , w 2 2 Q l , λ ( w ) , we get

d i ( w 1 , w 2 ) max { | y 1 - y 2 | , Λ λ i - 2 | s 1 - s 2 | }
Λ 1 2 max { | y 1 - y 2 | , λ p - 2 | s 1 - s 2 | }
= Λ 1 2 d λ p ( w 1 , w 2 ) 4 Λ 1 2 l .

Thus, we have r i c ( Λ ) l .

Note that

⨍⨍ Q l , λ ( w ) | v h Λ - ( v h Λ ) Q l , λ ( w ) | l 𝑑 w ~ 2 ⨍⨍ Q l , λ ( w ) | v h Λ - ( v h ) Q l , λ ( w ) | l 𝑑 w ~
2 ⨍⨍ Q l , λ ( w ) | v h Λ - v h | l 𝑑 w ~ + 2 ⨍⨍ Q l , λ ( w ) | v h - ( v h ) Q l , λ ( w ) | l 𝑑 w ~ .

As in Case 2, we can estimate the second term on the right-hand side. To estimate the first term, we obtain from the definition of v h Λ and the fact that w i is supported in 2 K U i that

⨍⨍ Q l , λ ( w ) | v h Λ - v h | l 𝑑 w ~ = ⨍⨍ Q l , λ ( w ) | i P ( v h - v h i ) w i | l 𝑑 w ~
⨍⨍ Q l , λ ( w ) i P | v h - v h i | w i l d w ~
= 1 | Q l , λ ( w ) | i P Q i , λ ( w ) 2 K U i | v h - v h i | w i l 𝑑 w ~ .

Since w i 1 and r i c ( Λ ) l , we have

1 | Q l , λ ( w ) | i P Q i , λ ( w ) 2 K U i | v h - v h i | w i l 𝑑 w ~ c ( Λ ) | Q l , λ ( w ) | i P 2 K U i | v h - v h i | r i 𝑑 w ~ .

Combining the previous inequalities with (3.25), we obtain

⨍⨍ Q l , λ ( w ) | v h Λ - v h | l 𝑑 w ~ c ( data , Λ ) | Q l , λ ( w ) | i P | U i | .

Since r i c ( Λ ) l , we get U i c ( data , Λ ) Q i , Λ ( w ) for every i P . By Lemma 2.18(ii), we have that

⨍⨍ Q l , λ ( w ) | v h Λ - v h | l 𝑑 w ~ c ( data , Λ ) | Q l , λ ( w ) | i P | 1 6 K 6 U i | c ( data , Λ ) .

Thus, the proof is completed. ∎

Corollary 3.12.

Let E ( Λ ) be defined in (2.2). Then v h satisfies the estimate

H ( z , | v h Λ ( z ) | ) + H ( z , | v h Λ ( z ) | ) c ( data , Λ ) for a.e.  z n + 1 .

Proof.

Since v h Λ ( z ) = 0 in U R 2 , S 2 ( z 0 ) c and v h Λ is Lipschitz continuous for almost every z n + 1 , we get that | v h Λ ( z ) | c ( data , Λ ) and | v h Λ ( z ) | c ( data , Λ ) for almost every z n + 1 . Then we have

| v h Λ ( z ) | p + a ( z ) | v h Λ ( z ) | q + b ( z ) | v h Λ ( z ) | s + | v h Λ ( z ) | p + a ( z ) | v h Λ ( z ) | q + b ( z ) | v h Λ ( z ) | s
c ( data , Λ ) ( 1 + a + b ) .

This completes the proof. ∎

3.6 Some more properties of Lipschitz truncation

In the following proposition, we collect some more properties of Lipschitz truncation.

Proposition 3.13.

Let E ( Λ ) be as in (2.2), and let η , ζ be the cut-off functions. Then { v h Λ } h > 0 and a function v Λ satisfy the following properties:

  1. v h Λ W 0 1 , 2 ( supp ( ζ ) ; L 2 ( supp ( η ) , N ) ) L ( supp ( ζ ) ; W 0 1 , ( supp ( η ) , N ) ) .

  2. v Λ L ( supp ( ζ ) + h 0 ; W 0 1 , ( supp ( η ) , N ) ) .

  3. v h Λ = v h , v Λ = v , v h Λ = v h , v Λ = v a.e. in E ( Λ ) .

  4. v h Λ v Λ in L ( Ω , N ) as h 0 + , taking a subsequence if necessary.

  5. v h Λ v Λ and t v h Λ t v Λ a.e. in E ( Λ ) c as h 0 + .

Proof.

To show (i), we first note from Definition 1.1 and the definition of Steklov averages that

v h W 0 1 , 2 ( supp ( ζ ) ; L 2 ( supp ( η ) , N ) ) .

Moreover, the definition of v h Λ in (3.3) only matters for finite sum since | | c . Therefore, together with Lemma 3.11, we complete the proof of (i).

The proof of (ii) is obvious from Lemma 3.9(iii). Indeed, we have | v Λ ( z ) | c Λ 1 p and | v Λ ( z ) | c Λ 1 p for all z n + 1 .

The proof of (iii) follows from the definitions (3.3) and (3.4). In fact, since ω i ( z ) = 0 for z E ( Λ ) , we have the proof.

Since v h Λ ( z ) = 0 for z U R 2 , S 2 c ( z 0 ) , using Lemma 3.11 we see that { v h Λ } h > 0 is equicontinuous and uniformly bounded. By the properties of Steklov averages, we already have v h Λ v Λ as h 0 + . Hence, by Arzela–Ascoli theorem, we get v h Λ v Λ in L ( Ω , N ) . This proves (iv).

The proof of (v) follows from v h i v i as h 0 and the expressions computed in Lemma 3.8. ∎

4 Proof of energy estimate

In this section, we prove the energy estimate in Theorem 1.2 using the Lipschitz truncation v h Λ .

4.1 Proof of Theorem 1.2

We closely follow the proof of [28, Subsection 5.1]. For τ S 2 - h ( t 0 ) and sufficiently small δ > 0 , let

ζ δ ( t ) = { 1 , t ( - , τ - δ ) , 1 - t - τ + δ δ , t [ τ - δ , τ ] , 0 , t ( τ , t 0 + S 2 - h ) .

Note that, by Proposition 3.13(i), v h Λ η s ζ ζ δ ( , t ) W 0 1 , ( B R 2 ( x 0 ) , N ) for every t S 2 - h ( t 0 ) . Using this function as the test function of the Steklov averaged weak formulation in (1.1), we obtain

I + II = U R 2 , S 2 ( z 0 ) t [ u - u 0 ] h v h Λ η s ζ ζ δ d z + U R 2 , S 2 ( z 0 ) [ 𝒜 ( z , u ) ] h ( v h Λ η s ζ ζ δ ) d z
= U R 2 , S 2 ( z 0 ) [ ( z , F ) ] h ( v h Λ η s ζ ζ δ ) d z = III .

Now we estimate the each term above by dividing the integral domain into E ( Λ ) and E ( Λ ) c .

Estimate of I .

By integration by parts and the product rule, we obtain

I = U R 2 , S 2 ( z 0 ) ( - v h v h Λ η s - 1 t ζ δ - v h t v h Λ η s - 1 ζ δ ) 𝑑 z + U R 2 , S 2 ( z 0 ) - [ u - u 0 ] h v h Λ η s ζ δ t ζ d z = I 1 + I 2 .

First, we consider I 1 . Then we have

I 1 = U R 2 , S 2 ( z 0 ) - | v h | 2 η s - 1 t ζ δ d z + U R 2 , S 2 ( z 0 ) v h ( v h - v h Λ ) η s - 1 t ζ δ d z
- U R 2 , S 2 ( z 0 ) ( v h - v h Λ ) t v h Λ η s - 1 ζ δ d z - U R 2 , S 2 ( z 0 ) v h Λ t v h Λ η s - 1 ζ δ d z .

Note that integration by parts implies

U R 2 , S 2 ( z 0 ) v h Λ t v h Λ η s - 1 ζ δ d z = - 1 2 U R 2 , S 2 ( z 0 ) | v h Λ | 2 η s - 1 t ζ δ d z .

Since v h Λ = v h a.e. in E ( Λ ) and v h = v h Λ = 0 in U R 2 , S 2 ( z 0 ) c , we obtain

I 1 = U R 2 , S 2 ( z 0 ) - ( | v h | 2 - 1 2 | v h Λ | 2 ) η s - 1 t ζ δ d z + E ( Λ ) c v h ( v h - v h Λ ) η s - 1 t ζ δ d z - E ( Λ ) c ( v h - v h Λ ) t v h Λ η s - 1 ζ δ d z .

Letting h 0 + , we obtain from the properties of Steklov averages, Proposition 3.13(iv) and Lemma 3.9(i) that

lim h 0 + I 1 = U R 2 , S 2 ( z 0 ) - ( | v | 2 - 1 2 | v Λ | 2 ) η s - 1 t ζ δ d z + E ( Λ ) c v ( v - v Λ ) η s - 1 t ζ δ d z - E ( Λ ) c ( v - v Λ ) t v Λ η s - 1 ζ δ d z
= I 11 + I 12 + I 13 .

Since the Lipschitz truncation is done only in the bad set E ( Λ ) c , we get

I 11 = E ( Λ ) - 1 2 | v | 2 η s - 1 t ζ δ d z - E ( Λ ) c ( | v | 2 - 1 2 | v Λ | c ) η s - 1 t ζ δ d z

Since v L 2 ( U R 2 , S 2 ( z 0 ) , N ) , we obtain from the absolute continuity, Lemma 3.9(iii) and (2.3) that

lim Λ I 11 = U R 2 , S 2 ( z 0 ) - 1 2 | v | 2 η s - 1 t ζ δ d z .

For the same reason, we also have lim Λ I 12 = 0 . Finally, Lemma 3.9(ii) and (2.3) imply that lim Λ I 13 = 0 . Thus, we have

lim Λ lim h 0 + I 1 = U R 2 , S 2 ( z 0 ) - 1 2 | v | 2 η s - 1 t ζ δ d z .

Now, we estimate I 2 . By the properties of Steklov averages and Proposition 3.13(iv) and by dividing into the good and bad sets, we get

lim h 0 I 2 = - U R 2 , S 2 ( z 0 ) ( u - u 0 ) v Λ η s ζ δ t ζ d z
- U R 2 , S 2 ( z 0 ) E ( Λ ) | u - u 0 | 2 | t ζ | d z - E ( Λ ) c | u - u 0 | | v Λ | | t ζ | 𝑑 z .

Then Hölder’s inequality implies

E ( Λ ) c | u - u 0 | | v Λ | | t ζ | 𝑑 z ( U R 2 , S 2 ( z 0 ) E ( Λ ) c | u - u 0 | 2 | t ζ | 2 𝑑 z ) 1 2 ( E ( Λ ) c | v Λ | 2 𝑑 z ) 1 2 .

Since | u | L 2 ( Ω T ) , the first integral vanishes as Λ . By Lemma 3.9(iii) and (2.3), we get

lim Λ E ( Λ ) c | v Λ | 2 𝑑 z lim Λ c Λ 2 p | E ( Λ ) c | lim Λ c Λ | E ( Λ ) c | = 0 .

Thus, we get

lim Λ lim h 0 + I 2 - U R 2 , S 2 ( z 0 ) | u - u 0 | 2 | t ζ | 𝑑 z ,

and hence we conclude

lim Λ lim h 0 + I U R 2 , S 2 ( z 0 ) - 1 2 | v | 2 η s - 1 t ζ δ d z - U R 2 , S 2 ( z 0 ) | u - u 0 | 2 | t ζ | 𝑑 z .

Estimate of II .

As in [28], we obtain

lim h 0 + II = U R 2 , S 2 ( z 0 ) E ( Λ ) 𝒜 ( z , u ) ( ( u - u 0 ) η s + 1 ζ 2 ζ δ ) d z + U R 2 , S 2 ( z 0 ) E ( Λ ) c 𝒜 ( z , u ) ( v Λ η s ζ ζ δ ) d z = II 1 + II 2 .

Since the integral within the good set does not contain Λ, letting Λ , we see from (1.2) and (3.1) that

lim Λ lim h 0 + II 1 = U R 2 , S 2 ( z 0 ) ( 𝒜 ( z , u ) u ) η s + 1 ζ 2 ζ δ 𝑑 z + U R 2 , S 2 ( z 0 ) 𝒜 ( z , u ) ( u - u 0 ) ( η s + 1 ) ζ 2 ζ δ d z
c U R 2 , S 2 ( z 0 ) ( | u | p + a ( z ) | u | q + b ( z ) | u | s ) η s + 1 ζ 2 ζ δ 𝑑 z
- c U R 2 , S 2 ( z 0 ) ( | u | p - 1 + a ( z ) | u | q - 1 + b ( z ) | u | s - 1 ) | u - u 0 | R 2 - R 1 η s ζ 2 ζ δ 𝑑 z ,

where c = c ( s , ν , L ) . Then, Young’s inequality with conjugate ( p p - 1 , p ) , ( q q - 1 , q ) and ( s s - 1 , s ) , respectively, gives

c U R 2 , S 2 ( z 0 ) ( | u | p - 1 + a ( z ) | u | q - 1 + b ( z ) | u | s - 1 ) | u - u 0 | R 2 - R 1 η s ζ 2 ζ δ 𝑑 z
ν 2 U R 2 , S 2 ( z 0 ) ( | u | p + a ( z ) | u | q + b ( z ) | u | s ) η s + 1 ζ 2 ζ δ 𝑑 z
    + c U R 2 , S 2 ( z 0 ) ( | u - u 0 | p ( R 2 - R 1 ) p + a ( z ) | u - u 0 | q ( R 2 - R 1 ) q + b ( z ) | u - u 0 | s ( R 2 - R 1 ) s ) 𝑑 z ,

where c depends only on p , q , s , ν and L. Hence we have

lim Λ lim h 0 + II 1 ν 2 U R 2 , S 2 ( z 0 ) ( | u | p + a ( z ) | u | q + b ( z ) | u | s ) η s + 1 ζ 2 ζ δ 𝑑 z
- c U R 2 , S 2 ( z 0 ) ( | u - u 0 | p ( R 2 - R 1 ) p + a ( z ) | u - u 0 | q ( R 2 - R 1 ) q + b ( z ) | u - u 0 | s ( R 2 - R 1 ) s ) 𝑑 z .

Moreover, it is easy to show lim Λ lim h 0 + II 2 = 0 by using (1.2), Young’s inequality, Lemma 3.9(iii) and (2.3). For detailed calculations, refer to [28]. Thus, we conclude

lim Λ lim h 0 + II ν 2 U R 2 , S 2 ( z 0 ) H ( z , | u | ) η s + 1 ζ 2 ζ δ 𝑑 z - c U R 2 , S 2 ( z 0 ) H ( | u - u 0 | R 2 - R 1 ) 𝑑 z .

Estimate of III .

The estimate for III follows a process similar to that for II . Again, we divide into the good and bad parts to obtain

lim h 0 + III = U R 2 , S 2 ( z 0 ) E ( Λ ) ( z , F ) ( ( u - u 0 ) η s + 1 ζ 2 ζ δ ) d z + U R 2 , S 2 ( z 0 ) E ( Λ ) c ( z , F ) ( v h Λ η s ζ ζ δ ) d z = III 1 + III 2 .

Applying (1.3) and Young’s inequality, we get

lim Λ lim h 0 + III 1 c U R 2 , S 2 ( z 0 ) ( H ( z , | u - u 0 | R 2 - R 1 ) + H ( z , | F | ) ) d z + ν 4 U R 2 , S 2 ( z 0 ) H ( z , | u | ) η s + 1 ζ 2 ζ δ ) d z

for some c = c ( p , q , s , ν , L ) . In the same reason as II 2 , we get lim Λ lim h 0 + III 2 = 0 . Thus, we conclude that

lim Λ lim h 0 + III c U R 2 , S 2 ( z 0 ) ( H ( z , | u - u 0 | R 2 - R 1 ) + H ( z , | F | ) ) d z + ν 4 U R 2 , S 2 ( z 0 ) H ( z , | u | ) η s + 1 ζ 2 ζ δ ) d z .

Combining all the estimates for I , II and III , we obtain

U R 2 , S 2 ( z 0 ) - 1 2 | v | 2 η s - 1 t ζ δ d z + ν 4 U R 2 , S 2 ( z 0 ) H ( z , | u | ) η s + 1 ζ 2 ζ δ ) d z
c U R 2 , S 2 ( z 0 ) ( H ( z , | u - u 0 | R 2 - R 1 ) + | u - u 0 | 2 | t ζ | + H ( z , | F | ) ) 𝑑 z .

Finally, we complete the proof by letting δ 0 , recalling that τ S 1 ( t 0 ) is arbitrary, and replacing u 0 with ( u ) U R 1 , S 1 ( z 0 ) . ∎


Communicated by Juha Kinnunen


Award Identifier / Grant number: RS-2023-00217116

Award Identifier / Grant number: RS-2025-00555316

Funding statement: Bogi Kim is supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (Grant No. RS-2023-00217116). Jehan Oh is supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (Grant No. RS-2025-00555316). Abhrojyoti Sen is supported by research grants from the Alexander von Humboldt Foundation for postdocs.

Acknowledgements

The authors would like to express their sincere gratitude to the anonymous referee who provided valuable comments and suggestions on the earlier version, which have greatly improved the quality and clarity of the manuscript.

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Received: 2025-01-06
Accepted: 2025-03-28
Published Online: 2025-04-29
Published in Print: 2025-07-01

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