1. Introduction and main results

SHARP SELF-IMPROVING PROPERTIES OF GENERALIZED ORLICZ-POINCARÉ INEQUALITIES IN CONNECTED METRIC MEASURE SPACES TONI HEIKKINEN We study the self-improving properties of (generalized) Φ-Poincaré inequalities in connected metric spaces equipped with a doubling measure. As a consequence we obtain results concerning integrability, continuity and dierentiability of Orlicz-Sobolev functions on spaces supporting a Φ-Poincaré inequality. Our results are optimal and generalize some of the results of Cianchi [4, 5], Hajªasz and Koskela [9, 10], MacManus and Pérez [19], Balogh, Rogovin and Zürcher [2] and Stein [22].

Abstract.

1.

Introduction and main results

𝑋 = (𝑋, 𝑑, 𝜇) be a metric measure space with 𝜇 a Borel regular outer measure 0 < 𝜇(𝑈 ) < ∞, whenever 𝑈 is nonempty, open and bounded. Suppose further that 𝜇 is doubling, that is, there exists a constant 𝐶𝑑 such that Let

satisfying

𝜇(2𝐵) ≤ 𝐶𝑑 𝜇(𝐵),

(1.1) whenever

𝐵

is a ball. It is easy to see that the doubling property is equivalent to

the existence of constants

𝐶𝑠

such that

(𝑢, 𝑔)

if there exist

𝑥 ∈ 𝐵(𝑥0 , 𝑟0 )

(

𝑟 𝑟0

)𝑠

𝑟 ≤ 𝑟0 . 𝑔 ≥ 0, satises the 𝑝-Poincaré constants 𝐶𝑃 and 𝜏 ≥ 1 such that (∫ )1/𝑝 ∫ 𝑝 − ∣𝑢 − 𝑢𝐵 ∣ 𝑑𝜇 ≤ 𝐶𝑃 𝑟𝐵 − 𝑔 𝑑𝜇

holds, whenever

(1.3)

and

𝜇(𝐵(𝑥, 𝑟)) ≥ 𝐶𝑠−1 𝜇(𝐵(𝑥0 , 𝑟0 ))

(1.2)

A pair

𝑠

and

of measurable functions,

𝐵 for every ball

𝐵 = 𝐵(𝑥, 𝑟) ⊂ 𝑋 .

inequality,

𝜏𝐵 Hajªasz and Koskela [9, 10] proved the following

𝜇 satises (1.2), and that a pair 𝑝-Poincaré inequality (1.3). Let 𝛿 > 0 and constant 𝐶 = 𝐶(𝐶𝑠 , 𝑠, 𝐶𝑃 , 𝜏, 𝛿) such that the

self-improving properties of (1.3): Assume that

(𝑢, 𝑔), where 𝑔 ∈ 𝐿𝑝loc (𝑋) satises ˆ = (1 + 𝛿)𝜏 𝐵 . There exists a 𝐵

the

following holds.

Mathematics Subject Classication. 46E35, 26B05. Key words and phrases. Metric measure space, doubling measure, Poincaré inequality, Sobolev

2000

space, Orlicz space, Sobolev embedding, dierentiability. This paper is part of my PhD thesis written under the supervision of professor Pekka Koskela. I would like to thank him, Amiran Gogatishvili, Jana Björn and Heli Tuominen for interesting discussions and valuable comments on the manuscript. The research was supported by Vilho, Yrjö and Kalle Väisälä Foundation.

1

2

TONI HEIKKINEN 1) If

𝑝 < 𝑠,

then

( sup 𝑡

(1.4)

𝑡>0 where

𝜇({𝑥 : ∣𝑢(𝑥) − 𝑢𝐵 ∣ > 𝑡} 𝜇(𝐵)

𝑝𝑠 =

)1/𝑝𝑠

𝑠𝑝 𝑠−𝑝 . Consequently, for

(∫ )1/𝑝 𝑝 ≤ 𝐶𝑟𝐵 − 𝑔 𝑑𝜇 , ˆ 𝐵

𝑞 < 𝑝𝑠 ,

we have

(∫ )1/𝑞 (∫ − ∣𝑢 − 𝑢𝐵 ∣𝑞 𝑑𝜇 ≤ 𝐶 ′ 𝑟𝐵 −

(1.5)

𝑔 𝑝 𝑑𝜇

)1/𝑝 ,

𝜏 ′𝐵

𝐵

𝐶 ′ depends on 𝐶 and 𝑞 . In general, (1.3) does not yield (1.5) with 𝑞 = 𝑝𝑠 . However, if a pair (𝑢, 𝑔) has the truncation property, which means that for every 𝑏 ∈ ℝ, 0 < 𝑡1 < 𝑡2 < ∞ and 𝜀 ∈ {−1, 1}, the pair (𝑣𝑡𝑡12 , 𝑔𝜒{𝑡1 1 and 𝑋 is connected, then where

2)

∥𝑢 − 𝑢𝐵 ∥−𝐿Φ (𝐵) ≤ 𝐶𝑟𝐵 ∥𝑔∥−𝐿𝑠 (𝐵) ˆ ,

(1.6)

∥ ⋅ ∥−𝐿Φ (𝐵) is the normalized Luxemburg norm generated by the func′ 𝑠 . Φ(𝑡) = exp(𝑡𝑠 ) − 1 (see Section 2) and 𝑠′ = 𝑠−1 𝑝 > 𝑠, then 𝑢 has a locally Hölder continuous representative, for which (∫ )1/𝑝 𝑠/𝑝 ∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ 𝐶𝑟𝐵 𝑑(𝑥, 𝑦)1−𝑠/𝑝 − 𝑔 𝑝 𝑑𝜇

where tion 3) If

(1.7)

ˆ 𝐵

for

𝑥, 𝑦 ∈ 𝐵 .

In this paper we are interested in the self-improving properties of the following

Φ-Poincaré inequality, introduced recently in [23].

For the denition and properties

of Young functions and Orlicz spaces, see Section 2.

Denition 1.1.

Φ be a Young function. A pair (𝑢, 𝑔) of measurable functions, 𝑔 ≥ 0, satises the Φ-Poincaré inequality (in an open set 𝑈 ), if constants 𝐶𝑃 and 𝜏 such that (∫ ) ∫ −1 − ∣𝑢 − 𝑢𝐵 ∣ 𝑑𝜇 ≤ 𝐶𝑃 𝑟𝐵 Φ − Φ (𝑔) 𝑑𝜇

𝑢 ∈ 𝐿1loc (𝑋) there are (1.8)

Let

and

𝐵 for every ball

𝐵⊂𝑋

𝜏𝐵

(such that

𝜏 𝐵 ⊂ 𝑈 ).

Assuming that the underlying space is connected, we obtain results which are sharp in the sense that they reproduce a version of Cianchi's optimal embedding theorem for Orlicz-Sobolev spaces on

ℝ𝑛

[4, 5].

Notice that a pair

(𝑢, ∣∇𝑢∣)

of a

weakly dierentiable function and the length of its weak gradient satises the Poincaré inequality, and so, by Jensen's inequality, the every Young function Let

𝑠 > 1.

(1.9)

0

𝑡 Φ(𝑡)

Φ

satisfying

)𝑠′ −1

∫ 𝑑𝑡 < ∞

∞

and

dene (1.10)

Φ𝑠 = Φ ∘ Ψ−1 𝑠 ,

(

𝑡 Φ(𝑡)

1-

inequality for

Φ.

For a Young function

∫ (

Φ-Poincaré

)𝑠′ −1 𝑑𝑡 = ∞,

3 where

(∫

𝑟

(

Ψ𝑠 (𝑟) =

(1.11)

0

𝑡 Φ(𝑡)

)1/𝑠′

)𝑠′ −1

𝑑𝑡

.

If

∞

∫

(

(1.12)

𝑡 Φ(𝑡)

)𝑠′ −1 𝑑𝑡 < ∞,

dene ′

where

′

𝜔𝑠 (𝑡) = (𝑡Θ−1 (𝑡𝑠 ))𝑠 ,

(1.13)

Θ−1

is the left-continuous inverse of the function given by

′

∫

∞

Θ(𝑟) = 𝑠

(1.14)

𝑟

ˆ Φ

ˆ Φ(𝑡) 𝑑𝑡 𝑡1+𝑠′

Φ. ∫𝑐 𝑓 (𝑡) 𝑑𝑡 < ∞ means that 𝑓 (𝑡) 𝑑𝑡 < ∞ for some 𝑐 > 0. Similarly, we 0 ∫ ∞0 ∫∞ ∫∞ denote 𝑓 (𝑡) 𝑑𝑡 < ∞, if 𝑐 𝑓 (𝑡) 𝑑𝑡 < ∞ for some 𝑐 > 0, and 𝑓 (𝑡) 𝑑𝑡 = ∞, if ∫∞ 𝑓 (𝑡) 𝑑𝑡 = ∞ for all 𝑐 > 0. 𝑐 We wish to point out that, under (1.9), functions Φ, Ψ𝑠 and Φ𝑠 are bijections. Notice also that one can modify any Young function Φ near zero so that the conand

is the conjugate of

Here,

∫

dition

∫ ( 0

𝑡 ˜ Φ(𝑡)

is satised for the modied function

˜ Φ

)𝑠′ −1 𝑑𝑡 < ∞ and that

˜

Φ 𝐿Φ loc (𝑋) = 𝐿loc (𝑋).

We will state Cianchi's result only for balls, but it actually holds for much more

Let 𝑠 ≥ 2, let 𝐵 ⊂ ℝ𝑠 be a ball, and let 𝑢 be a weakly dierentiable function such that ∣∇𝑢∣ ∈ 𝐿Φ (𝐵). Then there is a constant 𝐶 depending only on 𝑠 such that 1) If (1.9) holds, then general domains (see [46]):

∥𝑢 − 𝑢𝐵 ∥𝐿Φ𝑠 (𝐵) ≤ 𝐶∥∣∇𝑢∣∥𝐿Φ (𝐵) .

2)

Moreover, 𝐿Φ𝑠 (𝐵) is the smallest Orlicz space into which 𝑊 1,Φ (𝐵) can be continuously embedded ([5, Theorem 2], [6, Lemma 2]). If (1.12) holds, then 𝑢 has a continuous representative for which ∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ 𝐶∥∣∇𝑢∣∥𝐿Φ (𝐵) 𝜔𝑠−1 (∣𝑥 − 𝑦∣−𝑠 ),

for 𝑥, 𝑦 ∈ 𝐵 ([4, Theorem 3], [7, Lemma 2.3]). Theorems 1.2 and 1.4 below generalize the result of Cianchi.

Theorem 1.2. Assume that

𝑋 is connected, 𝜇 satises (1.2) with 1 < 𝑠 < ∞, ˆ , and that a pair (ˆ ˆ = (1 + 𝛿)𝜏 𝐵 , 𝑔 ∈ 𝐿Φ (𝐵) 𝑢, 𝑔ˆ), where 𝐵 ⊂ 𝑋 is a ball, 𝛿 > 0, 𝐵 −1 −1 ˆ. 𝑢 ˆ = ∥𝑔∥𝐿Φ (𝐵) 𝑢 and 𝑔 ˆ = ∥𝑔∥ 𝑔 , satises the Φ -Poincaré inequality in 𝐵 ˆ ˆ 𝐿Φ (𝐵) 1) (1.15)

If

(1.9)

holds, then

∥𝑢 − 𝑢𝐵 ∥𝐿Φ𝑤𝑠 (𝐵) ≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 ∥𝑔∥𝐿Φ (𝐵) ˆ ,

where Φ𝑠 is dened by

-

.

(1.10) (1.11)

4

TONI HEIKKINEN 2)

If

holds, then, for Lebesgue points 𝑥, 𝑦 ∈ 𝐵 of 𝑢,

(1.12)

−1 𝑠 −1 ∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 ∥𝑔∥𝐿Φ (𝐵) 𝑑(𝑥, 𝑦)−𝑠 ), ˆ 𝜔𝑠 (𝑟𝐵 𝜇(𝐵)

(1.16)

where 𝜔𝑠 is dened by Here, 𝐶 = 𝐶(𝐶𝑠 , 𝑠, 𝐶𝑃 , 𝜏, 𝛿). If the

-

.

(1.13) (1.14)

Φ-Poincaré inequality is stable under truncations, the weak estimate (1.15)

turns into a strong one.

Denition 1.3.

A pair

0 < 𝑡1 < 𝑡2 < ∞

and

(𝑢, 𝑔) has the truncation property, if for every 𝑏 ∈ ℝ, 𝜀 ∈ {−1, 1}, the pair (𝑣𝑡𝑡12 , 𝑔𝜒{𝑡1 0 and 𝐶 ≥ 1 depending only on 𝐶𝑑 such that 𝜇(𝐵(𝑥, 𝑟)) ≤𝐶 𝜇(𝐵(𝑥0 , 𝑟0 ))

(2.1)

(

𝑟 𝑟0

)𝛼 ,

whenever 𝑥 ∈ 𝐵(𝑥0 , 𝑟0 ) and 𝑟 ≤ 𝑟0 . For a proof, see for example [19].

Young functions and Orlicz spaces.

In this subsection we give a brief

review of Young functions and Orlicz spaces.

A more detailed treatment of the

2.2.

subject can be found for example in [20]. A function

Φ : [0, ∞) → [0, ∞]

is called a Young function if it has the form

∫ Φ(𝑡) =

𝑡

𝜙(𝑠) 𝑑𝑠, 0

where

𝜙 : [0, ∞) → [0, ∞]

is increasing, left-continuous function, which is neither

identically zero nor identically innite on

(0, ∞).

A Young function is convex and,

in particular, satises

Φ(𝜀𝑡) ≤ 𝜀Φ(𝑡)

(2.2) for

0 𝑡}.

We have that

Φ(Φ−1 (𝑡)) ≤ 𝑡 ≤ Φ−1 (Φ(𝑡)) for

𝑡 ≥ 0.

Φ

is

7 The conjugate of a Young function

Φ

is the Young function dened by

ˆ Φ(𝑡) = sup{𝑡𝑠 − Φ(𝑠) : 𝑠 > 0} 𝑡 ≥ 0. Φ be a Young function. The Orlicz space 𝐿Φ (𝑋) functions 𝑢 for which there exists 𝜆 > 0 such that ( ) ∫ ∣𝑢(𝑥)∣ Φ 𝑑𝜇(𝑥) < ∞. 𝜆 𝑋 for

Let

The Luxemburg norm of

𝑢 ∈ 𝐿Φ (𝑋)

is the set of all measurable

is

{ ) } ( ∫ ∣𝑢(𝑥)∣ ∥𝑢∥𝐿Φ (𝑋) = ∥𝑢∥𝐿Φ (𝑋;𝜇) = inf 𝜆 > 0 : 𝑑𝜇(𝑥) ≤ 1 . Φ 𝜆 𝑋 If

∥𝑢∥𝐿Φ (𝑋) ∕= 0,

we have that

(

∫ Φ 𝑋

∣𝑢(𝑥)∣ ∥𝑢∥𝐿Φ (𝑋)

) 𝑑𝜇(𝑥) ≤ 1.

The following generalized Hölder inequality holds for Luxemburg norms:

∫ 𝑋

𝑢(𝑥)𝑣(𝑥) 𝑑𝜇(𝑥) ≤ 2∥𝑢∥𝐿Φ (𝑋) ∥𝑣∥𝐿Φˆ (𝑋) .

The weak Orlicz space

𝐿Φ 𝑤 (𝑋)

is dened to be the set of all those measurable

functions for which the weak Luxemburg norm

}) } { ({ ∣𝑢(𝑥)∣ >𝑡 ≤1 ∥𝑢∥𝐿Φ𝑤 (𝑋) = inf 𝜆 > 0 : sup Φ(𝑡)𝜇 𝑥∈𝑋: 𝜆 𝑡>0 is nite. If

∥𝑢∥𝐿Φ𝑤 (𝑋) ∕= 0,

it follows that

({ sup Φ(𝑡)𝜇 𝑥∈𝑋: 𝑡>0 If

𝜇(𝑋) < ∞,

}) ∣𝑢(𝑥)∣ >𝑡 ≤ 1. ∥𝑢∥𝐿Φ𝑤 (𝑋)

the normalized (weak) Luxemburg norm, that is, the (weak) Luxem-

burg norm taken with respect to measure (∥

𝜇(𝑋)−1 𝜇,

will be denoted by

∥ ⋅ ∥−𝐿Φ (𝑋)

⋅ ∥−𝐿Φ𝑤 (𝑋) ). A function

constant

for all

𝐶

Φ

dominates a function

Ψ

globally (resp. near innity), if there is a

such that

𝑡≥0

Functions

(resp. for

Φ

and

Ψ

𝑡

larger

Ψ(𝑡) ≤ Φ(𝐶𝑡) than some 𝑡0 ).

are equivalent globally (near innity), if each dominates the

other globally (near innity).

𝜇(𝑋) < ∞

If

and

Φ

Ψ

near innity, we have that

∥𝑢∥−𝐿Ψ (𝑋) ≤ 𝐶(Φ, Ψ)∥𝑢∥−𝐿Φ (𝑋) .

(2.3) 2.3.

dominates

Weak upper gradients.

curve family

Γ

Let

Φ

be a Young function. The

Φ-modulus

of a

is

ModΦ (Γ)

{ ∫ = inf ∥𝑔∥𝐿Φ (𝑋) : 𝑔 𝑑𝑠 ≥ 1

for all

} 𝛾∈Γ .

𝛾 If

𝑋

Φ-Poincaré inequality, then (1.8) holds for locally integrable Φ-weak upper gradients. This is an immediate consequence of

supports the

functions and their

the following lemma ([23], Lemma 4.3).

8

TONI HEIKKINEN

Lemma 2.2. Let

be a Young function and let 𝑔 ∈ 𝐿Φ (𝑋) be a Φ-weak upper gradient of a function 𝑢. Then there is a decreasing sequence (𝑔𝑖 ) of upper gradients of 𝑢 such that 𝑔𝑖 → 𝑔 in 𝐿Φ (𝑋). Φ

An important property of

Φ-weak

upper gradients is the following ([23], Lemma

4.11).

Lemma 2.3. Let Φ be a Young function. Assume that 𝑢 ∈ 𝐴𝐶𝐶Φ (𝑋) and that the functions 𝑣 and 𝑤 have Φ-weak upper gradients 𝑔𝑣 , 𝑔𝑤 ∈ 𝐿Φ (𝑋). If 𝐸 is a Borel set such that 𝑢∣𝐸 = 𝑣 and 𝑢∣𝑋∖𝐸 = 𝑤, then the function 𝑔 = 𝑔𝑣 𝜒𝐸 + 𝑔𝑤 𝜒𝑋∖𝐸

is a Φ-weak upper gradient of 𝑢. Here  𝑢

∈ 𝐴𝐶𝐶Φ (𝑋)

Γ of rectiable curves for which [0, 𝑙(𝛾)] has zero Φ-modulus. Φ It follows from the lemma above that if 𝑔 ∈ 𝐿 (𝑋) is a Φ-weak upper gradient of a measurable function 𝑣 , and 𝐸 ⊃ {𝑡1 < 𝑣 ≤ 𝑡2 } is a Borel set, then 𝑔𝜒𝐸 is a Φ-weak upper gradient of the function 𝑣𝑡𝑡12 = min{max{0, 𝑣 − 𝑡1 }, 𝑡2 − 𝑡1 }. Since 𝜇 is Borel regular, we can choose 𝐸 such that 𝜇(𝐸 ∖ {𝑡1 < 𝑣 ≤ 𝑡2 }) = 0. Thus, we 𝑢∘𝛾

means that the family

is not absolutely continuous on

have the following.

Lemma 2.4. If

𝑋 supports the Φ-Poincaré inequality, then every pair (𝑢, 𝑔) of a locally integrable function and its Φ-weak upper gradient 𝑔 ∈ 𝐿Φ (𝑋) has the truncation property.

Lipschitz functions.

A function 𝑢 : 𝑋 → ℝ is Lipschitz, if there exists a 𝐿 such that ∣𝑓 (𝑥) − 𝑓 (𝑦)∣ ≤ 𝐿𝑑(𝑥, 𝑦) for all 𝑥, 𝑦 ∈ 𝑋 . The lower and upper pointwise Lipschitz constants of a function 𝑢 : 𝑋 → ℝ are

2.4.

constant

lip 𝑢(𝑥) = lim inf 𝑟→0

sup 𝑦∈𝐵(𝑥,𝑟)

∣𝑢(𝑥) − 𝑢(𝑦)∣ 𝑟

and

Lip 𝑢(𝑥) = lim sup 𝑟→0

∣𝑢(𝑥) − 𝑢(𝑦)∣ . 𝑟 𝑦∈𝐵(𝑥,𝑟) sup

Clearly,

Lip 𝑢(𝑥) = lim sup 𝑦→𝑥 The lower Lipschitz constant locally Lipschitz function 2.5.

Dierentiability.

𝑢

lip 𝑢,

∣𝑢(𝑥) − 𝑢(𝑦)∣ . 𝑑(𝑥, 𝑦)

and hence also

Lip 𝑢,

is an upper gradient of a

(cf. [3]).

Let

𝒞 ⊂

= {𝑓 : 𝑋 → ℝ : 𝑓 is Lipschitz} be a {(𝑋𝛼 , 𝜑𝛼 )} be a countable collection of pairs, that each 𝑋𝛼 ⊂ 𝑋 is measurable with positive LIP(𝑋)

vector space of functions, and let called

coordinate patches,

such

measure and that

𝑁 (𝛼) 𝜑𝛼 = (𝜑1𝛼 , . . . , 𝜑𝛼 ) : 𝑋𝛼 → ℝ𝑁 (𝛼) is a function, where

{(𝑋𝛼 , 𝜑𝛼 )}

is a

following holds.

𝑁 (𝛼) ∈ ℕ ∪ {0}

and

𝜑𝑖𝛼 ∈ 𝒞

for every

strong measurable dierentiable structure

1 ≤ 𝑖 ≤ 𝑁 (𝛼). 𝒞

with respect to

Then if the

9

∙

The sets

𝑋𝛼

are disjoint and

(

)

𝜇 𝑋∖

∪

𝑋𝛼

= 0.

𝛼

∙ ∙

There exists a number For every

𝑓 ∈𝒞

𝑁

𝑁 (𝛼) ≤ 𝑁 for every 𝛼. (𝑋𝛼 , 𝜑𝛼 ) there exists a unique (up 𝛼 𝑁 (𝛼) measurable function 𝑑 𝑓 : 𝑋𝛼 → ℝ such that such that

and coordinate patch

to a set of zero measure)

∣𝑓 (𝑦) − 𝑓 (𝑥) − 𝑑𝛼 𝑓 (𝑥) ⋅ (𝜑𝛼 (𝑦) − 𝜑𝛼 (𝑥))∣ =0 𝑦→𝑥 𝑑(𝑥, 𝑦) lim

(2.4)

𝑦∕=𝑥

for almost every

𝑥 ∈ 𝑋𝛼 .

The following  Liplip condition of Keith [16] guarantees the existence of a strong measurable dierentiable structure.

Theorem 2.5. Let (𝑋, 𝑑, 𝜇) be doubling, let 𝐾 ≥ 1, and let 𝒞 ⊂ LIP(𝑋) be a vector space of functions such that for every 𝑓 ∈ 𝒞

Lip 𝑓 (𝑥) ≤ 𝐾 lip 𝑓 (𝑥)

(2.5)

for almost every 𝑥 ∈ 𝑋 . Then (𝑋, 𝑑, 𝜇) admits a strong measurable dierentiable structure with respect to 𝒞 . We say that a function 𝑓 : 𝑋 → ℝ is dierentiable at 𝑥 with respect to the structure

{(𝑋𝛼 , 𝜑𝛼 )},

if (2.4) holds.

The following generalization of Stepanov's

theorem was recently proved by Balogh, Rogovin and Zürcher [2].

Theorem 2.6. Assume that (𝑋, 𝑑, 𝜇) is doubling and admits a strong measurable dierentiable structure {(𝑋𝛼 , 𝜑𝛼 )} with respect to LIP(𝑋). Then a function 𝑓 : 𝑋 → ℝ is almost everywhere dierentiable in the set 𝑆(𝑓 ) = {𝑥 : Lip 𝑓 (𝑥) < ∞} with respect to the structure {(𝑋𝛼 , 𝜑𝛼 )}. 3.

Generalized Poincaré inequalities

Franchi, Pérez and Wheeden [8] and MacManus and Pérez [18, 19] studied the self-improving properties of inequalities of type

∫ − ∣𝑢 − 𝑢𝐵 ∣ 𝑑𝜇 ≤ ∥𝑢∥𝑎 𝑎(𝜏 𝐵),

(3.1)

𝐵 where

∥𝑢∥𝑎 > 0, 𝜏 ≥ 1

and

𝑎 : {𝐵 ⊂ 𝑋 : 𝐵

is a ball}

→ [0, ∞)

is a functional that

satises certain summability conditions. In [18] MacManus and Pérez showed that if

𝛿>0

is xed, and the functional

∑

(3.2) whenever the balls

𝐵𝑖

𝑎

satises condition

𝑎(𝐵𝑖 )𝑟 𝜇(𝐵𝑖 ) ≤ 𝑐𝑟 𝑎(𝐵)𝑟 𝜇(𝐵),

are disjoint and contained in the ball

𝐵,

then the Poincaré-

type inequality (3.1) improves to

( (3.3)

sup 𝜆 𝜆>0

𝜇({𝑥 ∈ 𝐵 : ∣𝑢(𝑥) − 𝑢𝐵 ∣ > 𝜆}) 𝜇(𝐵)

)1/𝑟

ˆ = (1+𝛿)𝜏 𝐵 . ∥𝑎∥ is the minimum of the constants 𝑐 so that (3.2) holds and 𝐵 [19], they proved that if 𝑋 is connected, 𝑟 > 1, and 𝑎 satises the stronger

where In

ˆ ≤ 𝐶∥𝑎∥∥𝑢∥𝑎 𝑎(𝐵),

condition (3.4)

∑

𝑎(𝐵𝑖 )𝑟 ≤ 𝑐𝑟 𝑎(𝐵)𝑟 ,

10

TONI HEIKKINEN

whenever the balls

𝐵𝑖

are disjoint and contained in the ball

𝐵,

then

ˆ ∥𝑢 − 𝑢𝐵 ∥−𝐿Φ (𝐵) ≤ 𝐶𝑎(𝐵),

(3.5) ′

𝑟 Φ(𝑡) = exp(𝑡𝑟 )−1 and 𝑟′ = 𝑟−1 . MacManus and Pérez also gave an example of a disconnected space 𝑋 , where (3.1) holds, and 𝑎 satises condition (3.4), but 𝑢 ∕∈ 𝐿Φ for all Φ(𝑡) = exp(𝑡𝛼 ), 𝛼 > 1. where

To see that the results of MacManus and Pérez generalize those of Hajªasz and Koskela, simply note that if

𝜇

satises (1.2), then the functional

(∫ )1/𝑝 𝑝 𝑎(𝐵) = 𝑟𝐵 − 𝑔 𝑑𝜇 , 𝐵 𝑝

0 ≤ 𝑔 ∈ 𝐿loc (𝑋), satises condition (3.2) with 𝑟 = 𝑠𝑝/(𝑠 − 𝑝), if 𝑝 < 𝑠, and 𝑟 = 𝑠, if 𝑝 = 𝑠. Let us now state our result. Let 𝑈 be an open set, Φ a Young function, 𝜏 ≥ 1 and 0 < 𝑠 ≤ ∞. Denote

) ∫

∑ (

−1/𝑠 ∥𝑢∥𝐴Φ,𝑠 = sup 𝜇(𝐵) − ∣𝑢 − 𝑢 ∣ 𝑑𝜇 𝜒 ,

𝐵 𝐵 (𝑈 ) 𝜏

ℬ∈ℬ𝜏 (𝑈 ) 𝐵 where

condition (3.4) with

𝐵∈ℬ

𝐿Φ (𝑈 ;𝜇)

where

ℬ𝜏 (𝑈 ) = {{𝐵𝑖 } : balls 𝜏 𝐵𝑖

are disjoint and contained in

𝑈 }.

Theorem 3.1. Let

𝑋 be connected, 𝜇 doubling, Φ a Young function, 𝐵 ⊂ 𝑋 a ball, 1 < 𝑠 < ∞, 𝜏 ≥ 1 and 𝛿 > 0. Denote 𝐵ˆ = (1 + 𝛿)𝜏 𝐵 . 1) If (1.9) holds, then ∥𝑢 − 𝑢𝐵 ∥𝐿Φ𝑤𝑠 (𝐵) ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ , (𝐵) 𝜏

(3.6)

2)

where Φ𝑠 is dened by (1.10)-(1.11). If (1.12) holds, then, for Lebesgue points 𝑥, 𝑦 ∈ 𝐵 of 𝑢, −1 −1 ∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ), ˆ 𝜔𝑠 (𝜇(𝐵𝑥𝑦 ) (𝐵) 𝜏

(3.7)

where 𝐵𝑥𝑦 = 𝐵(𝑥, 2𝑑(𝑥, 𝑦)), and 𝜔𝑠 is dened by Here, 𝐶 = 𝐶(𝐶𝑑 , 𝜏, 𝛿).

-

.

(1.13) (1.14)

It is easy to see that the rst part of Theorem 1.2 is a consequence of inequality (3.6).

In Section 5 we will show that it also implies the generalized Trudinger

inequality (3.5). As the example mentioned above shows, it is essential that connected. In the general case Theorem 3.1 still holds for those too close to the function

𝑡 7→ 𝑡𝑠 .

Φ

𝑋

is

which are not

This as well as generalizations of the rst result

of MacManus and Pérez will be investigated in the forthcoming paper [12]. 4.

Proofs of main theorems

The proof of Theorem 3.1 requires several lemmas.

In the rst three lemmas

equivalent representations of conditions (1.9) and (1.12) and of functions

𝜔𝑠

Φ𝑠

and

are given.

Lemma 4.1 ([6, Lemma 2], [7, Lemma 2.3]). Let Φ be a Young function. We have (4.1)

∫ ˆ Φ(𝑡) 𝑑𝑡 < ∞ 1+𝑠′ 𝑡 0

if and only if

∫ ( 0

𝑡 Φ(𝑡)

)𝑠′ −1 𝑑𝑡 < ∞

11

and ∫ (4.2)

∞

ˆ Φ(𝑡) 𝑑𝑡 < ∞ 𝑡1+𝑠′

if and only if

∫

∞

(

𝑡 Φ(𝑡)

)𝑠′ −1 𝑑𝑡 < ∞.

Moreover, the function Φ𝑠 is globally equivalent to the function 𝐷𝑠 given by ′

′

𝐷𝑠 (𝑡) = (𝑡𝐽 −1 (𝑡𝑠 ))𝑠

(4.3)

for 𝑡 ≥ 0, where 𝐽 −1 is the left-continuous inverse of the function given by ′

∫

𝐽(𝑟) = 𝑠

(4.4)

0

𝑟

ˆ Φ(𝑡) 𝑑𝑡. 𝑡1+𝑠′

Lemma 4.2 ([6, Lemma 1]). Let Φ be a Young function. Then ∥𝑟−1/𝑠 ∥𝐿 ′

∞

ˆ Φ (𝑡,∞)

for every 𝑡 > 0, if and only if


0, where 𝐷𝑠−1 is the right-continuous inverse of 𝐷𝑠 . (4.7)

Lemma 4.3 ([4, Lemma 2]). Let Φ be a Young function. Then ∥𝑟−1/𝑠 ∥𝐿 ′

for every 𝑡 > 0, if and only if

∫

∞

ˆ Φ (0,𝑡)

0, where 𝜔𝑠−1 is the right-continuous inverse of 𝜔𝑠 . It is easy to see that, for

𝐶 ≥ 1, 𝐷𝑠−1 (𝐶𝑡) ≤ 𝐶𝐷𝑠−1 (𝑡)

(4.10) and

𝜔𝑠−1 (𝐶 −1 𝑡) ≤ 𝐶𝜔𝑠−1 (𝑡).

(4.11)

Lemma 4.4. Let Φ be a Young function. Then Φ(𝑟)−1/𝑠 𝑟 ≤ Φ−1 𝑠 (Φ(𝑟))

for 𝑟 ≥ 0. Proof. Since Φ is convex, the function 𝑡 7→ 𝑡/Φ(𝑡) is decreasing. Φ−1 𝑠 (Φ(𝑟)) =

(∫

𝑟

0

(

′

𝑡 Φ(𝑡)

)𝑠 −1

)1/𝑠′ 𝑑𝑡

Hence

( ( )𝑠′ −1 )1/𝑠′ 𝑟 ≥ 𝑟 = Φ(𝑟)−1/𝑠 𝑟. Φ(𝑟) □

The next lemma is a variant of Lemma 6.2 in [10]. It is the part of the proofs of theorems 3.1 and

1.2,

where the connectedness of the space comes into play.

12

TONI HEIKKINEN

Lemma 4.5. Assume that 𝑋 is connected, 𝜇 doubling, 𝜏 ≥ 1 and 𝛿 > 0. Let 𝐵 be a ball, 𝑥 ∈ 𝐵 and 0 < 𝑟 < 𝛿𝑟𝐵 . Then there is a sequence {𝐵0 , . . . , 𝐵𝑘 } of balls contained in (1 + 𝛿)𝐵 such that 𝜇(𝐵0 ) is comparable to 𝜇(𝐵), 𝜇(𝐵𝑘 ) is comparable ˆ , to 𝜇(𝐵(𝑥, 𝑟)), {𝐵1 , . . . , 𝐵𝑘 } ∈ ℬ𝜏 (𝐵) 2𝜇(𝐵𝑖+1 ) ≤ 𝜇(𝐵𝑖 ) ≤ 𝐶𝜇(𝐵𝑖+1 ),

(4.12)

for 1 ≤ 𝑖 < 𝑘, and 𝑘 ∫ ∑ ∣𝑢𝐵(𝑥,𝑟) − 𝑢𝐵0 ∣ ≤ 𝐶 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇,

(4.13)

𝑖=1 𝐵𝑖

where 𝐶 = 𝐶(𝐶𝑑 , 𝜏, 𝛿). Proof. Fix 𝑥 ∈ 𝐵 and 0 < 𝑟 < 𝛿𝑟𝐵 .

𝐴𝑗 = 𝐵(𝑥, 2−𝑗 𝛿𝑟𝐵 ) ∖ 𝐵(𝑥, 2 𝛿𝑟𝐵 ) 𝐴𝑗 such that the balls 1 𝐷 , 𝐷 ∈ 𝒞𝑗 , are disjoint. It follows easily from the doubling property of 𝜇 that 2 ′ #𝒞𝑗 ≤ 𝐶 . Since 𝑋 is connected, there must be a sequence {𝐵0′ , . . . , 𝐵𝑘−1 } ⊂ ∪𝑚 𝑗=1 𝒞𝑗 ′ ′ ′ ′ ′ so that 𝐵0 ∈ 𝒞1 , 𝐵𝑖 ∩ 𝐵𝑖+1 ∕= ∅ for all 𝑖, 𝐵𝑘−1 ⊂ 𝐵(𝑥, 𝑟) and 𝜇(𝐵𝑘−1 ) is comparable ′ ′ ′ to 𝜇(𝐵(𝑥, 𝑟)). Denote 𝐵0 = 𝐵0 , 𝐵𝑘 = 𝐵𝑘 = 𝐵(𝑥, 𝑟) and 𝐵𝑖 := 5𝐵𝑖 for 1 ≤ 𝑖 < 𝑘 . ′ Then 𝐵𝑖 ⊂ 𝐵𝑖+1 , and so ∫ ′ ′ ∣𝑢𝐵𝑖′ − 𝑢𝐵𝑖+1 ∣ ≤ ∣𝑢𝐵𝑖′ − 𝑢𝐵𝑖+1 ∣ + ∣𝑢𝐵𝑖+1 − 𝑢𝐵𝑖+1 ∣ ≤ 𝐶− ∣𝑢 − 𝑢𝐵𝑖+1 ∣ 𝑑𝜇. Let 𝒞𝑗 be a cover of −1 −𝑗 by balls of radius (20𝜏 ) 2 𝛿𝑟𝐵 centered at

−𝑗−1

𝐵𝑖+1

Thus

∣𝑢𝐵(𝑥,𝑟) − 𝑢𝐵0 ∣ ≤

𝑘−1 ∑

∣𝑢𝐵𝑖′

𝑘 ∫ ∑ ′ − 𝑢𝐵𝑖+1 ∣≤𝐶 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇. 𝑖=1 𝐵𝑖

𝑖=0

ˆ and satises (4.12) {𝐵𝑖 } has a subsequence that belongs to ℬ𝜏 (𝐵) 1 ≤ 𝑗 ≤ 𝑚, choose 𝐷𝑗 ∈ {𝐵𝑖 } centered at 𝐴𝑗 such that {∫ } ∫ − ∣𝑢 − 𝑢𝐷𝑗 ∣ 𝑑𝜇 = max − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 : 𝑥𝐵𝑖 ∈ 𝐴𝑗 ,

We will show that and (4.13). For

𝐵𝑖

𝐷𝑗 where

𝑥 𝐵𝑖

denotes the center of

𝐵𝑖 .

Then

𝑚 ∫ ∑ ∣𝑢𝐵(𝑥,𝑟) − 𝑢𝐵0 ∣ ≤ 𝐶 − ∣𝑢 − 𝑢𝐷𝑗 ∣ 𝑑𝜇. 𝑗=1 𝐷𝑗 If

∣𝑖 − 𝑗∣ ≥ 2,

then

𝜏 𝐷𝑖 ∩ 𝜏 𝐷𝑗 = ∅.

By (1.2) and (2.1) there are constants

𝛼>0

and

𝛽>0

depending on

𝐶𝑑

such

that

𝐶 −1 2−𝛽𝑛 ≤

(4.14)

𝜇(𝐷𝑗+𝑛 ) ≤ 𝐶2−𝛼𝑛 𝜇(𝐷𝑗 )

𝑗, 𝑛 ≥ 1. Let 𝑛 ≥ 2 be such that 𝐶2−𝛼𝑛 ≤ 2−1 . For 𝑝 + (𝑖 − 1)𝑛 ≤ 𝑚, denote = 𝐷𝑝+(𝑖−1)𝑛 . Then the sequence {𝐵1𝑝 , 𝐵2𝑝 . . . } satises (4.12) and belongs to ˆ . By choosing 1 ≤ 𝑝 < 𝑛 such that ℬ𝜏 (𝐵) ∑∫ ∑∫ − ∣𝑢 − 𝑢𝐵𝑖𝑝 ∣ 𝑑𝜇 = max − ∣𝑢 − 𝑢𝐵𝑖𝑞 ∣ 𝑑𝜇,

for

𝐵𝑖𝑝

𝑖 we obtain

𝐵𝑖𝑝

1≤𝑞 𝜆.

such that

∫ 𝜇(𝐵𝑥 )−1/𝑠 − ∣𝑢 − 𝑢𝐵𝑥 ∣ 𝑑𝜇 > 𝜆. 𝐵𝑥 So,

−1

𝜇(𝐵𝑥 ) ≤ Φ(𝜆)

(4.16)

) ( ∫ −1/𝑠 Φ 𝜇(𝐵𝑥 ) − ∣𝑢 − 𝑢𝐵𝑥 ∣ 𝑑𝜇 𝜇(𝐵𝑥 ). 𝐵𝑥

By the standard

5𝑟-covering

lemma ([13, Theorem 1.16]), we can cover the set

# {𝑥 ∈ 𝐵 : 𝑀𝑠,𝐵 (𝑥) > 𝜆} by balls

𝐵

5𝜏 𝐵𝑖

𝜏 𝐵𝑖

such that the balls

(2.2), and the fact that

𝐵𝑖 is contained in 𝜇, estimate (4.16), inequality

are disjoint and that each

and satises (4.16). Using the doubling property of

{𝐵𝑖 } ∈ ℬ𝜏 (𝜏 𝐵),

we obtain

# 𝜇({𝑥 ∈ 𝐵 : 𝑀𝑠,𝐵 𝑢(𝑥) > 𝜆}) ≤

∑

𝜇(5𝜏 𝐵𝑖 ) ≤ 𝐶(𝐶𝑑 , 𝜏 )

∑

𝑖 −1

≤ 𝐶(𝐶𝑑 , 𝜏 )Φ (𝜆)

∑

(

) ∫ − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 𝜇(𝐵𝑖 )

−1/𝑠

Φ 𝜇(𝐵𝑖 )

𝐵𝑖

𝑖

( ≤Φ ( ≤Φ

𝜆 𝐶(𝐶𝑑 , 𝜏 )

)−1 ∑

𝜆 𝐶(𝐶𝑑 , 𝜏 )

)−1

(

) ∫ −1/𝑠 Φ 𝜇(𝐵𝑖 ) − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 𝜇(𝐵𝑖 ) 𝐵𝑖

𝑖

.

The claim follows by the denition of

Proof of Theorem 3.1.

𝜇(𝐵𝑖 )

𝑖

1) Denote

∥ ⋅ ∥𝐿Φ𝑤 .

□

𝐵 ′ = (1 + 𝛿)𝐵 .

It suces to show that the point-

wise inequality

( ( (4.17)

−1 ∣𝑢(𝑥) − 𝑢𝐵 ∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ Φ𝑠 (𝐵) 𝜏

Φ

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥) # ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

))

14

TONI HEIKKINEN

holds for Lebesgue points

(

𝑥 ∈ 𝐵. )

∣𝑢(𝑥) − 𝑢𝐵 ∣ 𝜇 𝑥∈𝐵: >𝑡 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ (𝐵) 𝜏

Indeed, if (4.17) holds, then

(

( Φ−1 𝑠

≤𝜇 𝑥∈𝐵:

≤𝜇 𝑥∈𝐵:

𝑥∈𝐵

Fix a Lebesgue point

of

−1

)

# ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

(

≤ Φ𝑠 (𝑡)

∘Φ

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

) >𝑡 )

>Φ

# ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

−1

∘ Φ𝑠 (𝑡)

.

𝑢 and 0 < 𝑟 ≤ 𝛿𝑟𝐵 . Let {𝐵0 , . . . 𝐵𝑘 } be the chain 𝑥 and 𝑟. Since the balls 𝐵𝑖 , 𝑖 ≥ 1, are disjoint,

from Lemma 4.5 corresponding to we have that

𝑘 ∫

𝑘 ∫

∑ ∑ 𝜒𝐵𝑖

− ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 =

𝜇(𝐵𝑖 ) 𝐵𝑖 𝐵𝑖 𝑖=1

𝑖=1

𝐿1 (𝑋)

and

∫ 𝑘 ∫ 𝑘 𝑘 ∑ ∑ ∑ ′ 𝜒𝐵𝑖 −1/𝑠 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 = 𝜇(𝐵𝑖 ) − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇𝜒𝐵𝑖 ⋅ 𝜇(𝐵𝑖 )−1/𝑠 𝜒𝐵𝑖 . 𝜇(𝐵 ) 𝑖 𝐵𝑖 𝑖=1 𝐵𝑖 𝑖=1 𝑖=1 Hence, by the Hölder inequality,

𝑘 ∫ ∑ − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 𝑖=1 𝐵𝑖



𝑘

∫ 𝑘

∑

∑

′



𝜇(𝐵𝑖 )−1/𝑠 𝜒𝐵𝑖 ≤ 2 𝜇(𝐵𝑖 )−1/𝑠 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇𝜒𝐵𝑖



Φˆ 𝐵𝑖 𝑖=1 𝐿Φ (𝑋) 𝑖=1 𝐿 (𝑋)

𝑘

∑

′

𝜇(𝐵𝑖 )−1/𝑠 𝜒𝐵𝑖 ≤ 2∥𝑢∥𝐴Φ,𝑠 . ˆ (𝐵) 𝜏

Φˆ 𝑖=1

𝐿 (𝑋)

By the denition of Luxemburg norm

𝑘

∑

−1/𝑠′ 𝜇(𝐵𝑖 ) 𝜒 𝐵𝑖



= inf

𝜆>0:

ˆ 𝐿Φ (𝑋)

𝑖=1

For each

{

𝑖,

𝑘 ∑

( ˆ Φ

𝑖=1

′

𝜇(𝐵𝑖 )−1/𝑠 𝜆

)

} 𝜇(𝐵𝑖 ) ≤ 1 .

we have that

( ˆ Φ

𝜇(𝐵𝑖 )−1/𝑠 𝜆

′

)

∫ 𝜇(𝐵𝑖 ) ≤ 2 ∫ ≤

𝜇(𝐵𝑖 ) 𝜇(𝐵𝑖 ) 2

𝜇(𝐵𝑖 ) 𝜇(𝐵𝑖 ) 2

( ˆ Φ (

ˆ Φ

𝑡−1/𝑠 𝜆

′

′

2𝑡−1/𝑠 𝜆

) 𝑑𝑡 ) 𝑑𝑡,

where the rst inequality follows from the fact that the function

ˆ −1/𝑠′ /𝜆) 𝑡 7→ Φ(𝑡 is decreasing, and the second from (2.2). Since

𝜇(𝐵𝑖+1 ) ≤

𝜇(𝐵𝑖 ) , 2

15 we obtain

𝑘 ∑

( ˆ Φ

𝑖=1

′

𝜇(𝐵𝑖 )−1/𝑠 𝜆

)

𝜇(𝐵1 )

∫ 𝜇(𝐵𝑖 ) ≤

𝜇(𝐵𝑘 ) 2

( ˆ Φ

′

2𝑡−1/𝑠 𝜆

) 𝑑𝑡,

which implies that

𝑘

∑

′

𝜇(𝐵𝑖 )−1/𝑠 𝜒𝐵𝑖



{ ≤ inf

∫ 𝜆>0:

ˆ 𝐿Φ (𝑋)

𝑖=1

𝜇(𝐵1 ) 𝜇(𝐵𝑘 ) 2

( ˆ Φ

′

= 2∥𝑡−1/𝑠 ∥𝐿Φˆ ( 𝜇(𝐵𝑘 ) ,𝜇(𝐵 2

1 ))

′

2𝑡−1/𝑠 𝜆

)

} 𝑑𝑡 ≤ 1

.

Thus

(4.18)

𝑘 ∫ ∑ −1/𝑠′ − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ∥𝐿Φˆ ( 𝜇(𝐵𝑘 ) ,𝜇(𝐵 ˆ ∥𝑡 (𝐵) 𝜏

1 ))

2

𝑖=1 𝐵𝑖

.

By similar reasoning,

∣𝑢𝐵0 − 𝑢𝐵 ∣ ≤ ∣𝑢𝐵0 − 𝑢 ∣ + ∣𝑢 𝐵′

(4.19)

𝐵′

∫ − 𝑢𝐵 ∣ ≤ 𝐶− ∣𝑢 − 𝑢𝐵 ′ ∣ 𝑑𝜇 𝐵′

−1/𝑠′

≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ ∥𝑡 (𝐵) 𝜏

∥𝐿Φˆ ( 𝜇(𝐵′ ) ,𝜇(𝐵 ′ )) . 2

It follows from estimates (4.18) and (4.19) that (4.20)

′

−1/𝑠 ∣𝑢𝐵(𝑥,𝑟) − 𝑢𝐵 ∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ∥𝐿Φˆ (𝐶 −1 𝜇(𝐵(𝑥,𝑟)),𝐶𝜇(𝐵)) . ˆ ∥𝑡 (𝐵) 𝜏

Hence, by lemmas 4.1 and 4.2, and by (4.10), (4.21)

−1 −1 ∣𝑢𝐵(𝑥,𝑟) − 𝑢𝐵 ∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ). ˆ Φ𝑠 (𝜇(𝐵(𝑥, 𝑟)) (𝐵) 𝜏

∣𝑢(𝑥) − 𝑢𝐵(𝑥,𝑟) ∣ in terms of maximal function (4.15). For 𝑖 ≥ 𝐵𝑖 = 𝐵(𝑥, 2−𝑖 𝑟). By the Lebesgue dierentiation theorem ([13, Theorem → 𝑢(𝑥), as 𝑖 → ∞. Thus, by (1.1) and (2.1), ∑ ∣𝑢(𝑥) − 𝑢𝐵(𝑥,𝑟) ∣ ≤ ∣𝑢𝐵𝑖 − 𝑢𝐵𝑖+1 ∣

Next, we will estimate

0,

denote

1.8]),

𝑢𝐵𝑖

𝑖≥0

≤𝐶

∑∫ − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇

≤𝐶

∑

𝑖≥0 𝐵𝑖 # 𝜇(𝐵𝑖 )1/𝑠 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

𝑖≥0 # ≤ 𝐶𝜇(𝐵(𝑥, 𝑟))1/𝑠 𝑀𝑠,𝐵 ′ 𝑢(𝑥). So, by Lemma 4.6,

(4.22)

1/𝑠 ∣𝑢(𝑥) − 𝑢𝐵(𝑥,𝑟) ∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ 𝜇(𝐵(𝑥, 𝑟)) (𝐵) 𝜏

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥) # ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

.

Combining the above estimates, we obtain

( ∣𝑢(𝑥) − 𝑢𝐵 ∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ (𝐵) 𝜏

−1 Φ−1 ) + 𝜇(𝐵𝑟 )1/𝑠 𝑠 (𝜇(𝐵𝑟 )

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥) # ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

) ,

16

TONI HEIKKINEN

( where

𝐵𝑟 = 𝐵(𝑥, 𝑟).

If

Φ

)

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥) # ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ )

≥ 𝜇(𝐵𝛿𝑟𝐵 )−1 ,

𝑟 ≤ 𝛿𝑟𝐵

we can choose

𝑤

such that

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

( 𝜇(𝐵𝑟 )

−1

≤Φ

) ≤ 𝐶𝜇(𝐵𝑟 )−1 .

# ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

Then

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

1/𝑠

𝜇(𝐵𝑟 ) (4.23)

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

(

# ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

≤ 𝐶Φ

)−1/𝑠

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

# # ∥𝑀𝑠,𝐵 ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤 𝑤 ( ( )) # 𝑀𝑠,𝐵 ′ 𝑢(𝑥) ≤ 𝐶Φ−1 Φ , 𝑠 # ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

where the last inequality comes from Lemma 4.4. Thus, we obtain (4.17). If

( Φ

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

)

# ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

< 𝜇(𝐵𝛿𝑟𝐵 )−1 ,

𝑟 = 𝛿𝑟𝐵 , ∫ − 𝑢𝐵 ∣ ≤ 𝐶− ∣𝑢 − 𝑢𝐵 ′ ∣ 𝑑𝜇

it suces to combine estimate (4.22), where

∣𝑢𝐵𝛿𝑟𝐵

with the estimate

𝐵′ # ≤ 𝐶𝜇(𝐵 ′ )1/𝑠 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ 𝜇(𝐵𝛿𝑟𝐵 ) (𝐵) 𝜏

# 𝑀𝑠,𝐵 ′ 𝑢(𝑥)

1/𝑠

# ∥𝑀𝑠,𝐵 ′ 𝑢∥𝐿Φ (𝐵 ′ ) 𝑤

and argue as in (4.23). 2) Letting

𝑟

tend to zero in (4.20) and using Lemma 4.3 and (4.11), we obtain

(4.24)

−1 −1 ∣𝑢(𝑥) − 𝑢𝐵 ∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ). ˆ 𝜔𝑠 (𝜇(𝐵) (𝐵) 𝜏

Let 𝑥, 𝑦 ∈ 𝐵 be Lebesgue points of 1 𝛿𝑟 𝐵 , then 𝜇(𝐵) ≤ 𝐶𝜇(𝐵𝑥𝑦 ). So 3

𝑢.

Denote

𝐵𝑥𝑦 = 𝐵(𝑥, 2𝑑(𝑥, 𝑦)).

If

𝑑(𝑥, 𝑦) >

∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ ∣𝑢(𝑥) − 𝑢𝐵 ∣ + ∣𝑢(𝑦) − 𝑢𝐵 ∣ −1 −1 ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ) ˆ 𝜔𝑠 (𝜇(𝐵) (𝐵) 𝜏 −1 −1 ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ). ˆ 𝜔𝑠 (𝜇(𝐵𝑥𝑦 ) (𝐵) 𝜏 If

𝑑(𝑥, 𝑦) ≤ 31 𝛿𝑟𝐵 ,

then (4.24), applied to the ball

𝐵𝑥𝑦 ,

yields

∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ ∣𝑢(𝑥) − 𝑢𝐵𝑥𝑦 ∣ + ∣𝑢(𝑦) − 𝑢𝐵𝑥𝑦 ∣ −1 −1 ). ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ 𝑥𝑦 ) 𝜔𝑠 (𝜇(𝐵𝑥𝑦 ) (𝐵 𝜏 Since we may assume that

𝛿 < 1/2,

it follows that

ˆ𝑥𝑦 ⊂ 𝐵 ˆ. 𝐵

Hence

−1 −1 ∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ). ˆ 𝜔𝑠 (𝜇(𝐵𝑥𝑦 ) (𝐵) 𝜏

□

Proof of Theorem 1.2.

1) By Theorem 3.1, it suces to show that

(4.25)

−1/𝑠 ∥𝑢∥𝐴Φ,𝑠 ∥𝑔∥𝐿Φ (𝐵) ˆ . ˆ ≤ 𝐶𝑟𝐵 𝜇(𝐵) (𝐵) 𝜏

17 We may assume that

∥𝑔∥𝐿Φ (𝐵) ˆ = 1.

Let

𝐷

be a ball such that

ˆ. 𝜏𝐷 ⊂ 𝐵

Then, by

(1.8) and (1.2),

(∫ ∫ −1 − ∣𝑢 − 𝑢𝐷 ∣ 𝑑𝜇 ≤ 𝐶𝑝 𝑟𝐷 Φ − 𝐷

) Φ(𝑔) 𝑑𝜇

𝜏𝐷 −1/𝑠

≤ 𝐶𝑟𝐵 𝜇(𝐵)

1/𝑠

𝜇(𝐷)

Φ

−1

(∫ −

) Φ(𝑔) 𝑑𝜇 .

𝜏𝐷

ˆ , Hence, for 𝒟 ∈ ℬ𝜏 (𝐵) ) ( ∫ ∫ ∑∫ ∑ 𝜇(𝐷)−1/𝑠 −𝐷 ∣𝑢 − 𝑢𝐷 ∣ 𝑑𝜇 𝜇(𝐷) ≤ Φ(𝑔) 𝑑𝜇 ≤ Φ(𝑔) 𝑑𝜇 ≤ 1, Φ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 ˆ 𝐵 𝐷∈𝒟 𝜏 𝐷 𝐷∈𝒟 which implies that

) ∫

∑ (

𝜇(𝐷)−1/𝑠 − ∣𝑢 − 𝑢𝐷 ∣ 𝑑𝜇 𝜒𝐷

𝐷 𝐷∈𝒟

≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 .

ˆ 𝐿Φ (𝐵)

ˆ , we ℬ𝜏 (𝐵) that 𝛿 < 1/2.

By taking supremum over

obtain (4.25).

ˆ = 𝐷 be a ball centered at 𝐵 so that 𝐷 Fix a Lebesgue point 𝑥 ∈ 𝐷 , 0 < 𝑟 < 𝛿𝑟𝐷 and let {𝐵𝑖 } be the chain from Lemma 4.5 corresponding to 𝐷 , 𝑥 and 𝑟 . Clearly, the chain can be chosen so 𝑟𝐵 𝑖 that 𝑟𝐵𝑖+1 ≤ 2 . Since the balls 𝐵𝑖 , 𝑖 ≥ 1, are disjoint, we have that

𝑘 ∫ ∫ 𝑘

∑ ∑

𝜇(𝐵𝑖 )−1 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇𝜒𝐵𝑖 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 =

1

𝐵𝑖 𝐵𝑖 2) We may assume

Let

ˆ. (1 + 𝛿)𝜏 𝐷 ⊂ 𝐵

𝑖=1

𝑖=1

𝐿 (𝑋)

and

𝑘 ∑

∫ ∫ 𝑘 𝑘 ∑ ∑ 𝑟𝑖 𝜇(𝐵𝑖 )−1 𝜒𝐵𝑖 . 𝑟𝑖−1 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇𝜒𝐵𝑖 ⋅ 𝜇(𝐵𝑖 )−1 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇𝜒𝐵𝑖 = 𝐵𝑖

𝑖=1

𝐵𝑖

𝑖=1

𝑖=1

So, by the Hölder inequality,

𝑘 ∫ ∑ − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇 𝑖=1 𝐵𝑖

𝑘

∫

∑

≤ 2 𝑟𝑖−1 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇𝜒𝐵𝑖

𝐵𝑖 𝑖=1

Since the pair

𝐿Φ (𝑋)

∥𝑔∥−1 ˆ (𝑢, 𝑔) 𝐿Φ (𝐵)

ˆ ⊂ ℬ𝜏 (𝐵) ˆ , ℬ𝜏 (𝐷)

satises the

𝑘

∑

𝑟𝑖 𝜇(𝐵𝑖 )−1 𝜒𝐵𝑖



.

ˆ 𝐿Φ (𝑋)

𝑖=1

Φ-Poincaré

inequality in

ˆ 𝐵

and

{𝐵𝑖 } ∈

we have that

∫ 𝑘

∑

𝑟𝑖−1 − ∣𝑢 − 𝑢𝐵𝑖 ∣ 𝑑𝜇𝜒𝐵𝑖

𝐵𝑖 𝑖=1

≤ 𝐶∥𝑔∥𝐿Φ (𝐵) ˆ .

𝐿Φ (𝑋)

By the denition of Luxemburg norm

𝑘

∑

−1 𝑟𝑖 𝜇(𝐵𝑖 ) 𝜒𝐵𝑖

𝑖=1

ˆ 𝐿Φ (𝑋)

{ = inf

𝜆>0:

𝑘 ∑

ˆ Φ

(

𝑖=1

By (1.2),

𝜇(𝐵𝑖 )−1 ≤ (𝐶𝐵 𝑟𝑖 )−𝑠 ,

𝑟𝑖 𝜇(𝐵𝑖 )−1 𝜆

)

} 𝜇(𝐵𝑖 ) ≤ 1 .

18

TONI HEIKKINEN

where

−1 𝐶𝐵 = 𝐶𝑟𝐵 𝜇(𝐵)1/𝑠 .

𝑎 > 0,

we have

ˆ Φ where

(

𝑟𝑖 𝜇(𝐵𝑖 )−1 𝜆

)

Since the function

ˆ 𝜇(𝐵𝑖 ) ≤ Φ

𝑡𝑖 = (𝐶𝐵 𝑟𝑖 )𝑠 . It follows

𝑘

∑

−1 𝑟𝑖 𝜇(𝐵𝑖 ) 𝜒𝐵𝑖

Φˆ 𝑖=1

(

ˆ 𝑡 7→ Φ(𝑎𝑡)/𝑡

𝑟𝑖 (𝐶𝐵 𝑟𝑖 )−𝑠 𝜆

)

is increasing, for every

( ˆ (𝐶𝐵 𝑟𝑖 ) = Φ 𝑠

−1/𝑠′

𝑡𝑖 𝐶𝐵 𝜆

) 𝑡𝑖 ,

that

{ −1 𝐶𝐵

≤

inf

𝜆>0:

𝑘 ∑

( ˆ Φ

−1/𝑠′

𝑡𝑖

{ −1 2𝐶𝐵

≤

inf

∫

𝑡1

𝜆>0:

𝑡𝑖 ≤ 1

𝜆

𝑖=1

𝐿 (𝑋)

( ˆ Φ

0

}

)

′

𝑡−1/𝑠 𝜆

)

} ≤1

′

−1 −1/𝑠 = 2𝐶𝐵 ∥𝑡 ∥𝐿Φˆ (0,𝑡1 ) 𝑠 −𝑠 ≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 𝜔𝑠−1 (𝜇(𝐵)−1 𝑟𝐵 𝑟𝐷 ), where the last inequality comes from Lemma 4.3 and from (4.11). Thus

∣𝑢(𝑥) − 𝑢𝐵0 ∣ = lim ∣𝑢𝐵(𝑥,𝑟) − 𝑢𝐵0 ∣ 𝑟→0

−1 −1 𝑠 −𝑠 ≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 ∥𝑔∥𝐿Φ (𝐵) 𝑟𝐵 𝑟𝐷 ). ˆ 𝜔𝑠 (𝜇(𝐵) By similar reasoning,

∫ ∣𝑢𝐵0 − 𝑢𝐷 ∣ ≤ ∣𝑢𝐷′ − 𝑢𝐵0 ∣ + ∣𝑢𝐵0 − 𝑢𝐷′ ∣ ≤ 𝐶− ∣𝑢 − 𝑢𝐷′ ∣ 𝑑𝜇 −1/𝑠

≤ 𝐶𝑟𝐵 𝜇(𝐵)

𝐷′ −1 −1 𝑠 −𝑠 ∥𝑔∥𝐿Φ (𝐵) 𝑟𝐵 𝑟𝐷 ). ˆ 𝜔𝑠 (𝜇(𝐵)

So,

−1 −1 𝑠 −𝑠 ∣𝑢(𝑥) − 𝑢𝐷 ∣ ≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 ∥𝑔∥𝐿Φ (𝐵) 𝑟𝐵 𝑟𝐷 ). ˆ 𝜔𝑠 (𝜇(𝐵)

(4.26) Let

𝑥, 𝑦 ∈ 𝐵

be Lebesgue points of

𝑢.

If

𝑑(𝑥, 𝑦) > 31 𝛿𝑟𝐵 ,

then (4.26) with

𝐷=𝐵

yields

∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ ∣𝑢(𝑥) − 𝑢𝐵 ∣ + ∣𝑢(𝑦) − 𝑢𝐵 ∣ −1 −1 𝑠 ≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 ∥𝑔∥𝐿Φ (𝐵) 𝑟𝐵 𝑑(𝑥, 𝑦)−𝑠 ). ˆ 𝜔𝑠 (𝜇(𝐵) If

𝑑(𝑥, 𝑦) ≤ 13 𝛿𝑟𝐵 ,

then

ˆ ⊂𝐵 ˆ, 𝐷

for the ball

𝐷 = 𝐵(𝑥, 2𝑑(𝑥, 𝑦)),

and so by (4.26)

and (4.11),

∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ ∣𝑢(𝑥) − 𝑢𝐷 ∣ + ∣𝑢(𝑦) − 𝑢𝐷 ∣ −1 −1 𝑠 ≤ 𝐶𝑟𝐵 𝜇(𝐵)−1/𝑠 ∥𝑔∥𝐿Φ (𝐵) 𝑟𝐵 𝑑(𝑥, 𝑦)−𝑠 ). ˆ 𝜔𝑠 (𝜇(𝐵)

□

Remark

.

4.7

Theorem 3.1.

As shown above, the rst part of Theorem 1.2 is a consequence of More generally, suppose that (1.2) holds, and that a function

satises an inequality of type (4.27)

) ( ∫ 𝜈(𝜏 𝐷) 𝛼 −1 − ∣𝑢 − 𝑢𝐷 ∣ 𝑑𝜇 ≤ ∥𝑢∥𝜈 𝑟𝐷 Φ , 𝜇(𝜏 𝐷) 𝐷

𝑢

19

∑ 𝛼 > 0, and 𝜈 : {𝐵 : 𝐵 is a ball } → [0, ∞) satises 𝜈(𝐵𝑖 ) ≤ 1, whenever ˆ . Then, an argument similar to the balls 𝐵𝑖 are disjoint and contained in 𝐵

where the

proof of (4.25), shows that

𝛼 −𝛼/𝑠 ∥𝑢∥𝐴Φ,𝑠/𝛼 (𝐵) ∥𝑢∥𝜈 . ˆ ≤ 𝐶𝑟𝐵 𝜇(𝐵)

(4.28)

𝜏

Thus, if (1.9) holds, with

𝑠/𝛼

in place of

∥𝑢 − 𝑢𝐵 ∥

Φ𝑠/𝛼

𝐿𝑤

(𝐵)

𝑠,

Theorem 3.1 yields

𝛼 ≤ 𝐶𝑟𝐵 𝜇(𝐵)−𝛼/𝑠 ∥𝑢∥𝜈 .

The properties of functions satisfying inequalities of type (4.27) with

Φ(𝑡) = 𝑡𝑝

were

studied in [11].

Remark ∫

4.8. Suppose that (1.2) and (1.9) hold, and that a pair (𝑢, 𝑔), where 0 < ˆ . Then, for the measure Φ(𝑔) 𝑑𝜇 < ∞, satises the Φ-Poincaré inequality in 𝐵 ˆ 𝐵 (∫ )−1 𝜇 ˆ = 𝐵ˆ Φ(𝑔) 𝑑𝜇 𝜇, we have that ∥𝑔∥𝐿Φ (𝐵;ˆ ˆ 𝜇) = 1. Since (1.2) and (1.8) trivially hold for 𝜇 ˆ with the same constants as they hold for 𝜇, Theorem 1.2, for the measure 𝜇 ˆ, yields

ˆ(𝐵)−1/𝑠 , ∥𝑢 − 𝑢𝐵 ∥𝐿Φ𝑤𝑠 (𝐵;ˆ𝜇) ≤ 𝐶𝑟𝐵 𝜇 which is equivalent to

( sup Φ𝑠

(4.29)

𝑡>0 where

𝑡 𝐶𝑟𝜇(𝐵)−1/𝑠

)−1/𝑠 )

(∫

∫ 𝜇({∣𝑢 − 𝑢𝐵 ∣ > 𝑡}) ≤

Φ(𝑔) 𝑑𝜇 ˆ 𝐵

{∣𝑢 − 𝑢𝐵 ∣ > 𝑡} = {𝑥 ∈ 𝐵 : ∣𝑢(𝑥) − 𝑢𝐵 ∣ > 𝑡}.

Φ(𝑔) 𝑑𝜇, ˆ 𝐵

If (1.9) holds, the second part

of Theorem 1.2 yields

( 𝑠 ) ∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ 𝐶𝑟𝐵 𝜇 ˆ(𝐵)−1/𝑠 𝜔𝑠−1 𝑟𝐵 𝑑(𝑥, 𝑦)−𝑠 𝜇 ˆ(𝐵)−1 , which implies that

(∫ )1/𝑠 ( ) ∫ −1 𝑠 −𝑠 ∣𝑢(𝑥) − 𝑢(𝑦)∣ ≤ 𝐶𝑟𝐵 − Φ(𝑔) 𝑑𝜇 𝜔𝑠 𝑟𝐵 𝑑(𝑥, 𝑦) − Φ(𝑔) 𝑑𝜇 .

(4.30)

ˆ 𝐵

Proof of Theorem 1.4. ∫ where

0
0,

𝜈({𝑤 > 𝑡}) ≤ 2 inf 𝜈({∣𝑤 − 𝑐∣ > 𝑡/2}). 𝑐∈ℝ

Proof.

If ∣𝑐∣ ≤ 𝑡/2, then {𝑤 > 𝑡} ⊂ {∣𝑤 − 𝑐∣ > 𝑡/2}. {∣𝑤 − 𝑐∣ > 𝑡/2}, and so

Otherwise,

{𝑤 = 0} ⊂

𝜈({𝑤 > 𝑡}) ≤ 𝜈(𝑌 ) ≤ 2𝜈({𝑤 = 0}) ≤ 2𝜈({∣𝑤 − 𝑐∣ > 𝑡/2}). □

20

TONI HEIKKINEN

𝑣

Let

denote either

𝑔𝜒{2𝑘−1 2𝑘 }) ≤ 𝜇({𝑣𝑘 > 2𝑘−2 }) ≤ 2𝜇({∣𝑣𝑘 − (𝑣𝑘 )𝐵 ∣ > 2𝑘−3 })

(4.31) for

𝑣+

𝑘 ∈ ℤ.

Let

𝐶 = 25 𝐶0 ,

(

∫ Φ𝑠 𝐵

≤

∑∫

𝑣 𝐶𝑟𝜇(𝐵)−1/𝑠 ( Φ𝑠

{2𝑘 2𝑘 }) ˆ 𝐵

2𝑘−3 𝐶0 𝑟𝜇(𝐵)−1/𝑠

(∫ ˆ 𝐵

)−1/𝑠 ) Φ(𝑔𝑘 ) 𝑑𝜇 𝜇({∣𝑣𝑘 − (𝑣𝑘 )𝐵 ∣ > 2𝑘−3 })

Φ(𝑔𝑘 ) 𝑑𝜇

∫ ≤

Φ(𝑔) 𝑑𝜇. ˆ 𝐵

Thus

(

∫ Φ𝑠

inf

(4.32)

𝑏∈ℝ

𝐵

This, for the pair

∣𝑢 − 𝑏∣ 𝐶𝑟𝜇(𝐵)−1/𝑠

∥𝑔∥−1 ˆ (𝑢, 𝑔) 𝐿Φ (𝐵)

)−1/𝑠 ) ∫ Φ(𝑔) 𝑑𝜇 𝑑𝜇 ≤ Φ(𝑔) 𝑑𝜇.

(∫ ˆ 𝐵

in place of

ˆ 𝐵

(𝑢, 𝑔),

yields

inf ∥𝑢 − 𝑏∥𝐿Φ𝑠 (𝐵) ≤ 𝐶𝑟𝜇(𝐵)−1/𝑠 ∥𝑔∥𝐿Φ (𝐵) ˆ .

𝑏∈ℝ Since

∥𝑢 − 𝑢𝐵 ∥𝐿Φ (𝐴) ≤ 2 inf 𝑏∈ℝ ∥𝑢 − 𝑏∥𝐿Φ (𝐴)

for any set

𝐴

of nite measure, the

□

proof is complete.

Proof of Theorem 1.6.

We may assume that

∥𝑔∥𝐿Φ (𝑋) = 1.

Then

∫ Φ(𝑔) 𝑑𝜇 < ∞ 𝐵 for every ball

𝐵 ⊂ 𝑋.

Fix

𝑥, 𝑦 ∈ 𝑋

and denote

𝐵 = 𝐵(𝑥, 2𝑑(𝑥, 𝑦)).

Estimate (4.30)

implies that

(∫ )1/𝑠 ( ∫ ) ∣𝑢(𝑥) − 𝑢(𝑦)∣ −1 ≤ 𝐶 − Φ(𝑔) 𝑑𝜇 𝜔𝑠 𝐶− Φ(𝑔) 𝑑𝜇 . 𝑑(𝑥, 𝑦) ˆ ˆ 𝐵 𝐵 Thus, for a Lebesgue point

𝑥

of

Φ(𝑔),

we have that

Lip 𝑢(𝑥) = lim sup 𝑦→𝑥 The claim follows from Theorem 2.6.

∣𝑢(𝑥) − 𝑢(𝑦)∣ < ∞. 𝑑(𝑥, 𝑦) □

21

Proof of Theorem 1.8.

To show that

𝑋

structure, it suces to prove that for all

admits a strong measurable dierentiable

𝑢 ∈ LIP(𝑋)

Lip 𝑢(𝑥) ≤ 𝐶 lip 𝑢(𝑥) 𝑥 ∈ 𝑋 . Fix 𝑢 ∈ LIP(𝑋) and a Lebesgue point 𝑥 of lip 𝑢. By the Φ-Poincaré inequality, (∫ ) ∫ −1 −1 lim 𝑟 − ∣𝑢 − 𝑢𝐵(𝑥,𝑟) ∣ 𝑑𝜇 ≤ 𝐶 lim Φ − Φ(lip 𝑢) 𝑑𝜇 = 𝐶 lip 𝑢(𝑥).

for almost every

𝑟→0

𝑟→0

𝐵(𝑥,𝑟)

𝐵(𝑥,𝑟)

□

The rest of the proof follows from [16, Proposition 4.3.3]. 5.

Strong inequalities without truncation

In this section we will show how the weak estimate (3.6) implies strong ones. We begin with an easy lemma.

Lemma 5.1. Let 𝜇(𝑋) < ∞, and let Φ and Ψ be Young functions such that ∫

∞

(5.1)

1

Ψ′ (𝑡) 𝑑𝑡 < ∞. Φ(𝑡)

Then 𝐿Φ𝑤 (𝑋) ⊂ 𝐿Ψ (𝑋) and there is a constant 𝐶 = 𝐶(Ψ, Φ) such that ∥𝑢∥−𝐿Ψ (𝑋) ≤ 𝐶∥𝑢∥−𝐿Φ𝑤 (𝑋) .

(5.2)

Proof.

∥𝑢∥𝐿Φ𝑤 (𝑋) = 1. Denoting 𝜇 ˜ = 𝜇(𝑋)−1 𝜇, we obtain ∫ ∫ ∞ Ψ(∣𝑢∣) 𝑑˜ 𝜇= Ψ′ (𝑡)˜ 𝜇({𝑥 ∈ 𝑋 : ∣𝑢∣ > 𝑡}) 𝑑𝑡 𝑋 0 ∫ ∞ ≤ Ψ(1) + Ψ′ (𝑡)˜ 𝜇({𝑥 ∈ 𝑋 : ∣𝑢∣ > 𝑡}) 𝑑𝑡 1 ∫ ∞ ′ Ψ (𝑡) ≤ Ψ(1) + 𝑑𝑡 =: 𝐶 ′ , Φ(𝑡) 1

Assume

which implies (5.2) with For a measure

𝜈

on

∥𝑢∥𝐴Φ,𝑠 (𝑈 ;𝜈) 𝜏

𝐶 = max{𝐶 ′ , 1}.

𝑋,

□

denote

) ∫

∑ (

−1/𝑠 = sup 𝜈(𝐵) − ∣𝑢 − 𝑢𝐵 ∣ 𝑑𝜈 𝜒𝐵

ℬ∈ℬ𝜏 (𝑈 ) 𝐵 𝐵∈ℬ

For a ball

𝐵 ⊂ 𝑋,

denote

.

𝐿Φ (𝑈 ;𝜈)

𝜇𝐵 = 𝜇(𝐵)−1 𝜇.

Theorem 5.2. Suppose that the assumptions of Theorem 3.1 are in force, holds, and that Ψ is a Young function satisfying ∫ (5.3)

Then (5.4)

1

∞

(1.9)

Ψ′ (𝑡) 𝑑𝑡 < ∞. Φ𝑠 (𝑡)

∥𝑢 − 𝑢𝐵 ∥−𝐿Ψ (𝐵) ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ ˆ ), (𝐵;𝜇 𝜏 𝐵

ˆ , and a where 𝐶 = 𝐶(𝐶𝑑 , 𝑠, 𝜏, 𝛿, Φ, Ψ). Moreover, if 𝜇 satises (1.2), 𝑔 ∈ 𝐿Φ (𝐵) −1 ˆ pair ∥𝑔∥𝐿Φ (𝐵) ˆ (𝑢, 𝑔) satises the Φ-Poincaré inequality in 𝐵 , then (5.5)

∥𝑢 − 𝑢𝐵 ∥−𝐿Ψ (𝐵) ≤ 𝐶𝑟𝐵 ∥𝑔∥−𝐿Φ (𝐵) ˆ ,

22

TONI HEIKKINEN

where 𝐶 = 𝐶(𝐶𝑠 , 𝑠, 𝐶𝑃 , 𝜏, 𝛿, Φ, Ψ). Proof. Theorem 3.1, applied to the measure 𝜇𝐵 = 𝜇(𝐵)−1 𝜇, yields ∥𝑢 − 𝑢𝐵 ∥−𝐿Φ𝑤𝑠 (𝐵) ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ 𝐵 ). (𝐵;𝜇 𝜏 So, by Lemma 5.1,

∥𝑢 − 𝑢𝐵 ∥−𝐿Ψ (𝐵) ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ 𝐵 ). (𝐵;𝜇 𝜏 Since

∥ ⋅ ∥𝐿Φ (𝐵;𝜇 ˆ 𝐵 ) ≤ 𝐶∥ ⋅ ∥𝐿Φ (𝐵;𝜇 ˆ ˆ ), 𝐵

it follows that

∥𝑢∥𝐴Φ,𝑠 ˆ 𝐵 ) ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ ˆ ). (𝐵;𝜇 (𝐵;𝜇 𝜏 𝜏 𝐵

□

Inequality (5.5) follows from inequalities (5.4) and (4.25). Notice that if

Φ𝑠 (𝑡/2),

Φ𝑠 increases quickly enough, condition (5.3) is satised with Ψ(𝑡) =

and we have

∥𝑢 − 𝑢𝐵 ∥−𝐿Φ𝑠 (𝐵) ≤ 𝐶∥𝑢∥𝐴Φ,𝑠 ˆ ˆ ). (𝐵;𝜇 𝜏

(5.6)

𝐵

In particular, this is the case when

Φ

is equivalent to

Suppose now that (3.1) holds with a functional equivalent to

𝑡 7→ 𝑡𝑠

∥𝑢∥𝐴Φ,𝑠 ˆ ˆ) (𝐵;𝜇 𝜏 𝐵

𝑎

𝑡 7→ 𝑡𝑠

near innity.

satisfying (3.4), and that

Φ

is

near innity. Then

) ∫

∑ (

−1/𝑠 = sup − ∣𝑢 − 𝑢𝐵 ∣ 𝑑𝜇 𝜒𝐵 𝜇𝐵ˆ (𝐵)

ˆ 𝐵 ℬ∈ℬ𝜏 (𝐵) 𝐵∈ℬ ˆ − 𝐿Φ (𝐵)

( ) ∫

∑

≤ 𝐶 sup 𝜇𝐵ˆ (𝐵)−1/𝑠 − ∣𝑢 − 𝑢𝐵 ∣ 𝑑𝜇 𝜒𝐵

ˆ 𝐵 ℬ∈ℬ𝜏 (𝐵) 𝐵∈ℬ

( ≤𝐶

sup ˆ ℬ∈ℬ𝜏 (𝐵)

ˆ − 𝐿𝑠 (𝐵)

)𝑠 )1/𝑠 ∑ (∫ − ∣𝑢 − 𝑢𝐵 ∣ 𝑑𝜇 𝐵

𝐵∈ℬ

)1/𝑠

( ≤ 𝐶∥𝑢∥𝑎

sup

∑

ˆ ℬ∈ℬ𝜏 (𝐵)

𝐵∈ℬ

𝑠

𝑎(𝜏 𝐵)

ˆ ≤ 𝐶 ∥𝑢∥𝑎 𝑎(𝐵), where the rst inequality comes from (2.3).

Thus (5.6) implies the generalized

Trudinger inequality (3.5).

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