Introduction
Let p ( ⋅ ) p(\cdot) p ( ⋅ ) R \mathbb{R} R M M M [6 ] proved that M M M L p ( ⋅ ) ( Ω ) L^{p(\cdot)}(\Omega) L p ( ⋅ ) ( Ω ) Ω \Omega Ω 1 < p − ≤ p + < ∞ 1 < p_- \le p_+ < \infty 1 < p − ≤ p + < ∞ p ( ⋅ ) p(\cdot) p ( ⋅ ) L H 0 LH_0 L H 0
∣ p ( x ) − p ( y ) ∣ ≤ C 0 log ( e + 1 / ∣ x − y ∣ ) , x , y ∈ Ω . |p(x)-p(y)| \le\frac{C_0}{\log(e+1/|x-y|)}, \qquad x,y\in\Omega. ∣ p ( x ) − p ( y ) ∣ ≤ log ( e + 1/∣ x − y ∣ ) C 0 , x , y ∈ Ω. On the whole space, Cruz-Uribe, Fiorenza and Neugebauer [4 ] showed that it suffices to add the log-Hölder decay condition at infinity (L H ∞ LH_\infty L H ∞ p ∞ p_\infty p ∞ ∣ p ( x ) − p ∞ ∣ ≤ C ∞ / log ( e + ∣ x ∣ ) |p(x)-p_\infty| \le C_\infty/\log(e+|x|) ∣ p ( x ) − p ∞ ∣ ≤ C ∞ / log ( e + ∣ x ∣ ) [20 ] showed that L H ∞ LH_\infty L H ∞ N ∞ N_\infty N ∞ Definition 2.7 below). These conditions are sharp among pointwise moduli of continuity [21 , 4 ] . We refer to the monographs [3 , 7 ] for a detailed account.
These pointwise regularity assumptions, however, are not necessary. In the two-sided setting, the local log-Hölder condition can be replaced by the weaker K 0 K_0 K 0 N ∞ N_\infty N ∞ [3 ] . The conditions K 0 K_0 K 0 N ∞ N_\infty N ∞ [25 ] .
The purpose of this paper is to develop an analogous approach for the one-sided Hardy–Littlewood maximal operator
M + f ( x ) = sup h > 0 1 h ∫ x x + h ∣ f ( y ) ∣ d y . M^+f(x)=\sup_{h>0}\frac{1}{h}\int_x^{x+h}|f(y)|\,dy. M + f ( x ) = h > 0 sup h 1 ∫ x x + h ∣ f ( y ) ∣ d y . 2020 Mathematics Subject Classification. Primary 42B25, Secondary 42E30.
The left-sided maximal operator is defined by
M − f ( x ) = sup h > 0 1 h ∫ x − h x ∣ f ( y ) ∣ d y . M^{-}f(x)=\sup_{h>0}\frac{1}{h}\int_{x-h}^{x}|f(y)|\,\mathrm{d}y. M − f ( x ) = h > 0 sup h 1 ∫ x − h x ∣ f ( y ) ∣ d y . The theory of one-sided maximal operators has a long history in weighted harmonic analysis. Sawyer [24 ] introduced the one-sided Muckenhoupt classes A p + A_p^{+} A p + A p − A_p^{-} A p − [16 ] , and by Martín-Reyes [17 ] . In particular, the classes A ∞ + A_{\infty}^{+} A ∞ + [18 ] . One-sided reverse Hölder inequalities and their relationship with one-sided Muckenhoupt classes were further studied by Cruz-Uribe [2 ] , Cruz-Uribe, Neugebauer and Olesen [5 ] , and Riveros and de la Torre [22 ] .
One-sided operators have also been studied in the variable exponent setting. Edmunds, Kokilashvili and Meskhi [9 , 10 ] obtained boundedness results for one-sided maximal functions, potentials and singular integrals on variable Lebesgue spaces under directional conditions on the exponent which are weaker than the usual symmetric log-Hölder assumptions. Weighted one- and two-weight inequalities in variable Lebesgue spaces were subsequently studied by Kokilashvili, Meskhi and Sarwar [12 ] . A general self-improvement criterion for maximal operators associated with bases was obtained by Lerner and Ombrosi [15 ] . Their framework includes the one-dimensional right-sided maximal operator. In fact, the one-sided basis generating M + M^{+} M +
The geometry of M + M^{+} M + a < b < c a<b<c a < b < c ∣ f ∣ |f| ∣ f ∣ ( b , c ) (b,c) ( b , c ) M + f M^{+}f M + f ( a , b ) (a,b) ( a , b ) K 0 K_{0} K 0 K 0 + K_{0}^{+} K 0 +
Our main result is the following.
Theorem 1.1. Let p ( ⋅ ) ∈ P ( R ) p(\cdot)\in\mathcal{P}(\mathbb{R}) p ( ⋅ ) ∈ P ( R ) 1 < p − ≤ p + < ∞ 1<p_{-}\leq p_{+}<\infty 1 < p − ≤ p + < ∞ p ( ⋅ ) ∈ K 0 + ( R ) ∩ N ∞ ( R ) p(\cdot)\in K_{0}^{+}(\mathbb{R})\cap N_{\infty}(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R ) ∩ N ∞ ( R ) L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R )
∥ M + f ∥ p ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) . \|M^{+}f\|_{p(\cdot)}\leq C\|f\|_{p(\cdot)}. ∥ M + f ∥ p ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) . for every f ∈ L p ( ⋅ ) ( R ) f\in L^{p(\cdot)}(\mathbb{R}) f ∈ L p ( ⋅ ) ( R )
Since L H 0 ∩ L H ∞ ⊂ K 0 ∩ N ∞ LH_{0}\cap LH_{\infty}\subset K_{0}\cap N_{\infty} L H 0 ∩ L H ∞ ⊂ K 0 ∩ N ∞ [3 Propositions 4.9 and 4.57] ) and K 0 ⊂ K 0 + K_{0}\subset K_{0}^{+} K 0 ⊂ K 0 + (43) , Theorem 1.1 contains the log-Hölder case. We note that in this case the boundedness of M + M^{+} M + M M M M + f ≤ M f M^{+}f\leq Mf M + f ≤ M f N ∞ N_{\infty} N ∞
The main new ingredients in the proof are one-sided counterparts of the modular and reverse Hölder estimates associated with the classical K 0 K_{0} K 0 Lemmas 5.1 and 5.3.
The paper is organized as follows. In § 2 , we collect the known preliminary results used throughout the paper. In § 3 , we introduce the condition K 0 + K_{0}^{+} K 0 + § 4 , we prove that K 0 + K_{0}^{+} K 0 + M + M^{+} M + L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R ) § 5 is devoted to the proof of our main result, Theorem 1.1, preceded by four auxiliary lemmas that provide the key ingredients of the argument. Finally, in Section 6, we present examples of exponents satisfying the condition K 0 + K_{0}^{+} K 0 +
Preliminaries
Let P ( R ) \mathcal{P}(\mathbb{R}) P ( R ) p ( ⋅ ) : R → [ 1 , ∞ ) p(\cdot):\mathbb{R}\to[1,\infty) p ( ⋅ ) : R → [ 1 , ∞ ) Ω ⊂ R \Omega\subset\mathbb{R} Ω ⊂ R
p − ( Ω ) = ess inf x ∈ Ω p ( x ) , p + ( Ω ) = ess sup x ∈ Ω p ( x ) . p_{-}(\Omega)=\operatorname*{ess\,inf}_{x\in\Omega}p(x),\qquad p_{+}(\Omega)=\operatorname*{ess\,sup}_{x\in\Omega}p(x). p − ( Ω ) = x ∈ Ω ess inf p ( x ) , p + ( Ω ) = x ∈ Ω ess sup p ( x ) . When Ω = R \Omega=\mathbb{R} Ω = R p − = p − ( R ) p_{-}=p_{-}(\mathbb{R}) p − = p − ( R ) p + = p + ( R ) p_{+}=p_{+}(\mathbb{R}) p + = p + ( R )
Throughout the paper we assume 1 < p − ≤ p + < ∞ 1<p_{-}\leq p_{+}<\infty 1 < p − ≤ p + < ∞ p ′ ( ⋅ ) p'(\cdot) p ′ ( ⋅ )
1 p ( x ) + 1 p ′ ( x ) = 1. \frac{1}{p(x)}+\frac{1}{p'(x)}=1. p ( x ) 1 + p ′ ( x ) 1 = 1. The modular associated with p ( ⋅ ) p(\cdot) p ( ⋅ )
ρ p ( ⋅ ) ( f ) = ∫ R ∣ f ( x ) ∣ p ( x ) d x . \rho_{p(\cdot)}(f)=\int_{\mathbb{R}}|f(x)|^{p(x)}\,dx. ρ p ( ⋅ ) ( f ) = ∫ R ∣ f ( x ) ∣ p ( x ) d x . Given p ( ⋅ ) ∈ P ( R ) p(\cdot)\in\mathcal{P}(\mathbb{R}) p ( ⋅ ) ∈ P ( R ) L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R ) f f f R \mathbb{R} R λ > 0 \lambda>0 λ > 0 ρ p ( ⋅ ) ( f / λ ) < ∞ \rho_{p(\cdot)}(f/\lambda)<\infty ρ p ( ⋅ ) ( f / λ ) < ∞
∥ f ∥ p ( ⋅ ) = inf { λ > 0 : ρ p ( ⋅ ) ( f / λ ) ≤ 1 } . \|f\|_{p(\cdot)}=\inf\{\lambda>0:\rho_{p(\cdot)}(f/\lambda)\leq1\}. ∥ f ∥ p ( ⋅ ) = inf { λ > 0 : ρ p ( ⋅ ) ( f / λ ) ≤ 1 } . These spaces are known as variable Lebesgue spaces and generalize the classical Lebesgue spaces L p ( R ) L^{p}(\mathbb{R}) L p ( R ) [3 ] .
Lemma 2.1. Let p ( ⋅ ) : R → [ 1 , ∞ ) p(\cdot):\mathbb{R}\to[1,\infty) p ( ⋅ ) : R → [ 1 , ∞ ) p + < ∞ p_{+}<\infty p + < ∞ ∥ f ∥ p ( ⋅ ) \|f\|_{p(\cdot)} ∥ f ∥ p ( ⋅ ) ρ p ( ⋅ ) ( f ) \rho_{p(\cdot)}(f) ρ p ( ⋅ ) ( f ) C 1 , C 2 > 0 C_{1},C_{2}>0 C 1 , C 2 > 0
∥ f ∥ p ( ⋅ ) ≤ C 1 ⟹ ρ p ( ⋅ ) ( f ) ≤ max { C 1 p − , C 1 p + } , \|f\|_{p(\cdot)}\leq C_{1}\quad\Longrightarrow\quad\rho_{p(\cdot)}(f)\leq\max\{C_{1}^{p_{-}},C_{1}^{p_{+}}\}, ∥ f ∥ p ( ⋅ ) ≤ C 1 ⟹ ρ p ( ⋅ ) ( f ) ≤ max { C 1 p − , C 1 p + } , and
ρ p ( ⋅ ) ( f ) ≤ C 2 ⟹ ∥ f ∥ p ( ⋅ ) ≤ max { C 2 1 / p − , C 2 1 / p + } . \rho_{p(\cdot)}(f)\leq C_{2}\quad\Longrightarrow\quad\|f\|_{p(\cdot)}\leq\max\{C_{2}^{1/p_{-}},C_{2}^{1/p_{+}}\}. ρ p ( ⋅ ) ( f ) ≤ C 2 ⟹ ∥ f ∥ p ( ⋅ ) ≤ max { C 2 1/ p − , C 2 1/ p + } . In particular,
∥ f ∥ p ( ⋅ ) ≤ 1 ⟺ ρ p ( ⋅ ) ( f ) ≤ 1. \|f\|_{p(\cdot)}\leq1\quad\Longleftrightarrow\quad\rho_{p(\cdot)}(f)\leq1. ∥ f ∥ p ( ⋅ ) ≤ 1 ⟺ ρ p ( ⋅ ) ( f ) ≤ 1. Lemma 2.2. Let p ( ⋅ ) : R → [ 1 , ∞ ) p(\cdot):\mathbb{R}\to[1,\infty) p ( ⋅ ) : R → [ 1 , ∞ ) p + < ∞ p_{+}<\infty p + < ∞ Ω ⊂ R \Omega\subset\mathbb{R} Ω ⊂ R ∣ Ω ∣ > 0 |\Omega|>0 ∣Ω∣ > 0
(a) If ∥ f χ Ω ∥ p ( ⋅ ) ≤ 1 \|f\chi_{\Omega}\|_{p(\cdot)}\leq1 ∥ f χ Ω ∥ p ( ⋅ ) ≤ 1
∥ f χ Ω ∥ p ( ⋅ ) p + ( Ω ) ≤ ∫ Ω ∣ f ( x ) ∣ p ( x ) d x ≤ ∥ f χ Ω ∥ p ( ⋅ ) p − ( Ω ) . \|f\chi_{\Omega}\|_{p(\cdot)}^{p_{+}(\Omega)}
\leq\int_{\Omega}|f(x)|^{p(x)}\,dx
\leq\|f\chi_{\Omega}\|_{p(\cdot)}^{p_{-}(\Omega)}. ∥ f χ Ω ∥ p ( ⋅ ) p + ( Ω ) ≤ ∫ Ω ∣ f ( x ) ∣ p ( x ) d x ≤ ∥ f χ Ω ∥ p ( ⋅ ) p − ( Ω ) . (b) If ∥ f χ Ω ∥ p ( ⋅ ) ≥ 1 \|f\chi_{\Omega}\|_{p(\cdot)}\geq1 ∥ f χ Ω ∥ p ( ⋅ ) ≥ 1
∥ f χ Ω ∥ p ( ⋅ ) p − ( Ω ) ≤ ∫ Ω ∣ f ( x ) ∣ p ( x ) d x ≤ ∥ f χ Ω ∥ p ( ⋅ ) p + ( Ω ) . \|f\chi_{\Omega}\|_{p(\cdot)}^{p_{-}(\Omega)}
\leq\int_{\Omega}|f(x)|^{p(x)}\,dx
\leq\|f\chi_{\Omega}\|_{p(\cdot)}^{p_{+}(\Omega)}. ∥ f χ Ω ∥ p ( ⋅ ) p − ( Ω ) ≤ ∫ Ω ∣ f ( x ) ∣ p ( x ) d x ≤ ∥ f χ Ω ∥ p ( ⋅ ) p + ( Ω ) . In particular, if ∥ f ∥ p ( ⋅ ) ≤ 1 \|f\|_{p(\cdot)}\leq1 ∥ f ∥ p ( ⋅ ) ≤ 1
∫ R ∣ f ( x ) ∣ p ( x ) d x ≤ ∥ f ∥ p ( ⋅ ) . \int_{\mathbb{R}}|f(x)|^{p(x)}\,dx\leq\|f\|_{p(\cdot)}. ∫ R ∣ f ( x ) ∣ p ( x ) d x ≤ ∥ f ∥ p ( ⋅ ) . Lemma 2.3. Let p ( ⋅ ) : R → [ 1 , ∞ ) p(\cdot):\mathbb{R}\to[1,\infty) p ( ⋅ ) : R → [ 1 , ∞ ) p + < ∞ p_{+}<\infty p + < ∞ ∥ f ∥ L p ( ⋅ ) = 1 \|f\|_{L^{p(\cdot)}}=1 ∥ f ∥ L p ( ⋅ ) = 1
∫ R ∣ f ( x ) ∣ p ( x ) d x = 1. \int_{\mathbb{R}} |f(x)|^{p(x)}\,dx=1. ∫ R ∣ f ( x ) ∣ p ( x ) d x = 1. As a consequence of Lemma 2.3, for every nonzero f ∈ L p ( ⋅ ) ( R ) f\in L^{p(\cdot)}(\mathbb{R}) f ∈ L p ( ⋅ ) ( R )
ρ p ( ⋅ ) ( f ∥ f ∥ p ( ⋅ ) ) = 1. \rho_{p(\cdot)}\left(\frac{f}{\|f\|_{p(\cdot)}}\right)=1. ρ p ( ⋅ ) ( ∥ f ∥ p ( ⋅ ) f ) = 1. In particular, for every measurable set E ⊂ R E\subset\mathbb{R} E ⊂ R 0 < ∣ E ∣ < ∞ 0<|E|<\infty 0 < ∣ E ∣ < ∞
∫ E ( 1 ∥ χ E ∥ p ( ⋅ ) ) p ( x ) d x = 1. (1) \int_E\left(\frac{1}{\|\chi_E\|_{p(\cdot)}}\right)^{p(x)}\,dx=1. \tag*{(1)} ∫ E ( ∥ χ E ∥ p ( ⋅ ) 1 ) p ( x ) d x = 1. (1) Lemma 2.4. Let p ( ⋅ ) ∈ P ( R ) p(\cdot)\in\mathcal{P}(\mathbb{R}) p ( ⋅ ) ∈ P ( R ) s s s 1 / p − ≤ s < ∞ 1/p_{-}\le s<\infty 1/ p − ≤ s < ∞
∥ ∣ f ∣ s ∥ p ( ⋅ ) = ∥ f ∥ s p ( ⋅ ) s . \| |f|^s\|_{p(\cdot)}=\|f\|_{sp(\cdot)}^s. ∥∣ f ∣ s ∥ p ( ⋅ ) = ∥ f ∥ s p ( ⋅ ) s . Lemma 2.5. Let p ( ⋅ ) p(\cdot) p ( ⋅ ) p ′ ( ⋅ ) p'(\cdot) p ′ ( ⋅ ) C > 0 C>0 C > 0
∫ R ∣ f ( x ) g ( x ) ∣ d x ≤ C ∥ f ∥ p ( ⋅ ) ∥ g ∥ p ′ ( ⋅ ) \int_{\mathbb{R}} |f(x)g(x)|\,dx\le C\|f\|_{p(\cdot)}\|g\|_{p'(\cdot)} ∫ R ∣ f ( x ) g ( x ) ∣ d x ≤ C ∥ f ∥ p ( ⋅ ) ∥ g ∥ p ′ ( ⋅ ) for every f ∈ L p ( ⋅ ) ( R ) f\in L^{p(\cdot)}(\mathbb{R}) f ∈ L p ( ⋅ ) ( R )
Lemma 2.6. Let p ( ⋅ ) p(\cdot) p ( ⋅ ) c , C > 0 c,C>0 c , C > 0 f f f f ∈ L p ( ⋅ ) ( R ) f\in L^{p(\cdot)}(\mathbb{R}) f ∈ L p ( ⋅ ) ( R ) g ∈ L p ′ ( ⋅ ) ( R ) g\in L^{p'(\cdot)}(\mathbb{R}) g ∈ L p ′ ( ⋅ ) ( R ) ∥ g ∥ p ′ ( ⋅ ) ≤ 1 \|g\|_{p'(\cdot)}\le1 ∥ g ∥ p ′ ( ⋅ ) ≤ 1
c ∥ f ∥ p ( ⋅ ) ≤ ∣ ∫ R f ( x ) g ( x ) d x ∣ ≤ C ∥ f ∥ p ( ⋅ ) . c\|f\|_{p(\cdot)}\le\left|\int_{\mathbb{R}}f(x)g(x)\,dx\right|\le C\|f\|_{p(\cdot)}. c ∥ f ∥ p ( ⋅ ) ≤ ∫ R f ( x ) g ( x ) d x ≤ C ∥ f ∥ p ( ⋅ ) . The condition at infinity
We retain the usual N ∞ N_{\infty} N ∞
Definition 2.7. We say that
p ( ⋅ ) ∈ N ∞ ( R ) p(\cdot)\in N_{\infty}(\mathbb{R}) p ( ⋅ ) ∈ N ∞ ( R ) if there exist constants Λ ∞ > 0 \Lambda_{\infty}>0 Λ ∞ > 0 p ∞ ∈ [ 1 , ∞ ) p_{\infty}\in[1,\infty) p ∞ ∈ [ 1 , ∞ )
∫ Ω ∞ exp ( − Λ ∞ ∣ 1 p ( x ) − 1 p ∞ ∣ − 1 ) d x < ∞ , \int_{\Omega_{\infty}}\exp\left(-\Lambda_{\infty}\left|\frac{1}{p(x)}-\frac{1}{p_{\infty}}\right|^{-1}\right)\,dx<\infty, ∫ Ω ∞ exp ( − Λ ∞ p ( x ) 1 − p ∞ 1 − 1 ) d x < ∞ , where
Ω ∞ = { x ∈ R : ∣ 1 p ( x ) − 1 p ∞ ∣ > 0 } . \Omega_{\infty}=\left\{x\in\mathbb{R}:\left|\frac{1}{p(x)}-\frac{1}{p_{\infty}}\right|>0\right\}. Ω ∞ = { x ∈ R : p ( x ) 1 − p ∞ 1 > 0 } . Since p + < ∞ p_{+}<\infty p + < ∞
∫ Ω ∞ exp ( − Λ ∞ ∣ p ( x ) − p ∞ ∣ − 1 ) d x < ∞ . \int_{\Omega_{\infty}}\exp\left(-\Lambda_{\infty}|p(x)-p_{\infty}|^{-1}\right)\,dx<\infty. ∫ Ω ∞ exp ( − Λ ∞ ∣ p ( x ) − p ∞ ∣ − 1 ) d x < ∞. One-sided weights
A weight is a non-negative locally integrable function w : R → [ 0 , ∞ ) w : \mathbb{R} \to[0,\infty) w : R → [ 0 , ∞ ) E E E
w ( E ) = ∫ E w ( x ) d x . w(E)=\int_E w(x)\,dx. w ( E ) = ∫ E w ( x ) d x . We shall use a classical one-sided A ∞ A_\infty A ∞
Definition 2.8. We say that a weight w w w A ∞ + A_\infty^+ A ∞ + η ∈ ( 0 , 1 ) \eta\in(0,1) η ∈ ( 0 , 1 ) β = β ( η ) ∈ ( 0 , 1 ) \beta=\beta(\eta)\in(0,1) β = β ( η ) ∈ ( 0 , 1 ) a < b < c a<b<c a < b < c E ⊂ ( b , c ) E\subset(b,c) E ⊂ ( b , c ) w ( E ) ≤ β w ( a , b ) w(E)\leq\beta w(a,b) w ( E ) ≤ β w ( a , b ) ∣ E ∣ c − a < η \frac{|E|}{c-a}<\eta c − a ∣ E ∣ < η
This is one of the standard equivalent formulations of the one-sided A ∞ A_\infty A ∞ [18 , 2 ] .
We shall also use the one-sided reverse Hölder classes.
Definition 2.9. Let r > 1 r>1 r > 1 w w w R H r + RH_r^+ R H r + C > 0 C>0 C > 0
I − = ( a − h , a ) , I + = ( a , a + h ) , I^-=(a-h,a),\qquad I^+=(a,a+h), I − = ( a − h , a ) , I + = ( a , a + h ) , one has
( 1 ∣ I − ∣ ∫ I − w ( x ) r d x ) 1 / r ≤ C ∣ I + ∣ ∫ I + w ( x ) d x . (2) \left(\frac{1}{|I^-|}\int_{I^-}w(x)^r\,dx\right)^{1/r}\leq\frac{C}{|I^+|}\int_{I^+}w(x)\,dx. \tag*{(2)} ( ∣ I − ∣ 1 ∫ I − w ( x ) r d x ) 1/ r ≤ ∣ I + ∣ C ∫ I + w ( x ) d x . (2) The formulation (2) is equivalent to the usual definition of R H r + RH_r^+ R H r + [22 ] .
A level-set estimate for M + M^+ M +
We shall use the following estimate due to Martín-Reyes, Rivera-Ríos and Rodríguez-Padilla.
Lemma 2.10 ([19 Lemma 4] ). Let v ∈ L l o c 1 ( R ) v\in L^1_{\mathrm{loc}}(\mathbb{R}) v ∈ L loc 1 ( R ) 0 < λ 1 < λ 2 0<\lambda_1<\lambda_2 0 < λ 1 < λ 2
F = { z ∈ R : M + v ( z ) ≤ λ 2 } . F=\{z\in\mathbb{R}:M^+v(z)\leq\lambda_2\}. F = { z ∈ R : M + v ( z ) ≤ λ 2 } . If x ∈ R x\in\mathbb{R} x ∈ R M + v ( x ) ≤ λ 1 M^+v(x)\leq\lambda_1 M + v ( x ) ≤ λ 1 y > x y>x y > x
∣ F ∩ ( x , y ) ∣ ≥ ( 1 − λ 1 λ 2 ) ( y − x ) . |F\cap(x,y)|\geq\left(1-\frac{\lambda_1}{\lambda_2}\right)(y-x). ∣ F ∩ ( x , y ) ∣ ≥ ( 1 − λ 2 λ 1 ) ( y − x ) . The condition K 0 + K_0^+ K 0 +
Our definition is also motivated by the classical K 0 K_0 K 0 K 0 K_0 K 0 [1 ] , and its importance in variable Lebesgue spaces was emphasized by Kopaliani [13 ] . The connection between K 0 K_0 K 0 [3 ] ; see also Diening [8 ] and Lerner [14 ] .
The directional nature of K 0 + K_0^+ K 0 + [11 ] ; see also the survey by Riveros and Vidal [23 ] .
Definition 3.1. Let p ( ⋅ ) ∈ P ( R ) p(\cdot) \in\mathcal{P}(\mathbb{R}) p ( ⋅ ) ∈ P ( R ) p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot) \in K_0^+(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R )
[ p ] K 0 + : = sup a < b < c ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) c − a < ∞ . [p]_{K_0^+}:=\sup_{a<b<c}\frac{\|\chi_{(a,b)}\|_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}}{c-a}<\infty. [ p ] K 0 + := a < b < c sup c − a ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) < ∞. The orientation in Definition 3.1 is essential. The L p ( ⋅ ) L^{p(\cdot)} L p ( ⋅ ) L p ′ ( ⋅ ) L^{p'(\cdot)} L p ′ ( ⋅ )
Proposition 3.2. Let p ( ⋅ ) ∈ P ( R ) p(\cdot) \in\mathcal{P}(\mathbb{R}) p ( ⋅ ) ∈ P ( R ) p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot) \in K_0^+(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R )
sup a ∈ R h > 0 ∥ χ ( a − h , a ) ∥ p ( ⋅ ) ∥ χ ( a , a + h ) ∥ p ′ ( ⋅ ) h < ∞ . (3) \sup_{\substack{a\in\mathbb{R}\\h>0}}\frac{\|\chi_{(a-h,a)}\|_{p(\cdot)}\|\chi_{(a,a+h)}\|_{p'(\cdot)}}{h}<\infty. \tag*{(3)} a ∈ R h > 0 sup h ∥ χ ( a − h , a ) ∥ p ( ⋅ ) ∥ χ ( a , a + h ) ∥ p ′ ( ⋅ ) < ∞. (3) Proof. It is immediate that p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot) \in K_0^+(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R ) (3) , up to an absolute factor, by taking a − h < a < a + h a-h<a<a+h a − h < a < a + h
Conversely, suppose that (3) holds and let a < b < c a<b<c a < b < c ℓ − = b − a \ell_- = b-a ℓ − = b − a ℓ + = c − b \ell_+ = c-b ℓ + = c − b ℓ − ≤ ℓ + \ell_- \leq\ell_+ ℓ − ≤ ℓ + ( a , b ) ⊂ ( b − ℓ + , b ) (a,b) \subset(b-\ell_+,b) ( a , b ) ⊂ ( b − ℓ + , b )
∥ χ ( a , b ) ∥ p ( ⋅ ) ≤ ∥ χ ( b − ℓ + , b ) ∥ p ( ⋅ ) . \|\chi_{(a,b)}\|_{p(\cdot)}\leq\|\chi_{(b-\ell_+,b)}\|_{p(\cdot)}. ∥ χ ( a , b ) ∥ p ( ⋅ ) ≤ ∥ χ ( b − ℓ + , b ) ∥ p ( ⋅ ) . Therefore,
∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C ℓ + ≤ C ( c − a ) . \|\chi_{(a,b)}\|_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}\leq C\ell_+\leq C(c-a). ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C ℓ + ≤ C ( c − a ) . If ℓ + ≤ ℓ − \ell_+ \leq\ell_- ℓ + ≤ ℓ − ( b , c ) ⊂ ( b , b + ℓ − ) (b,c) \subset(b,b+\ell_-) ( b , c ) ⊂ ( b , b + ℓ − )
∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C ℓ − ≤ C ( c − a ) . \|\chi_{(a,b)}\|_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}\leq C\ell_-\leq C(c-a). ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C ℓ − ≤ C ( c − a ) . This proves that p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot) \in K_0^+(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R )
One-sided averaging operators
Given a < b < c a<b<c a < b < c
A a , b , c + f ( x ) = χ ( a , b ) ( x ) c − a ∫ b c f ( y ) d y . (4) A^+_{a,b,c}f(x)=\frac{\chi_{(a,b)}(x)}{c-a}\int_b^c f(y)\,dy. \tag*{(4)} A a , b , c + f ( x ) = c − a χ ( a , b ) ( x ) ∫ b c f ( y ) d y . (4) These operators reflect exactly the geometry of M + M^+ M +
Proposition 3.3. The following statements are equivalent.
p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot) \in K_0^+(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R )
the family { A a , b , c + : a < b < c } \{A^+_{a,b,c}:a<b<c\} { A a , b , c + : a < b < c } L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R )
Proof. Suppose first that p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot) \in K_0^+(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R ) Lemma 2.5 ,
∥ A a , b , c + f ∥ p ( ⋅ ) = 1 c − a ∣ ∫ b c f ( y ) d y ∣ ∥ χ ( a , b ) ∥ p ( ⋅ ) ≤ C p ( ⋅ ) ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) c − a ∥ f χ ( b , c ) ∥ p ( ⋅ ) ≤ C [ p ] K 0 + ∥ f ∥ p ( ⋅ ) . \begin{aligned}
\|A^+_{a,b,c}f\|_{p(\cdot)}
=\frac{1}{c-a}\left|\int_b^c f(y)\,dy\right|\|\chi_{(a,b)}\|_{p(\cdot)} \\
&\leq C_{p(\cdot)}\frac{\|\chi_{(a,b)}\|_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}}{c-a}\|f\chi_{(b,c)}\|_{p(\cdot)} \\
&\leq C[p]_{K_0^+}\|f\|_{p(\cdot)}.
\end{aligned} ∥ A a , b , c + f ∥ p ( ⋅ ) = c − a 1 ∫ b c f ( y ) d y ∥ χ ( a , b ) ∥ p ( ⋅ ) ≤ C p ( ⋅ ) c − a ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ∥ f χ ( b , c ) ∥ p ( ⋅ ) ≤ C [ p ] K 0 + ∥ f ∥ p ( ⋅ ) . Thus the family is uniformly bounded.
Conversely, suppose that
∥ A a , b , c + f ∥ p ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) \|A^+_{a,b,c}f\|_{p(\cdot)}\leq C\|f\|_{p(\cdot)} ∥ A a , b , c + f ∥ p ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) with C C C a a a b b b c c c Lemma 2.6 , of L p ′ ( ⋅ ) L^{p'(\cdot)} L p ′ ( ⋅ )
∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C sup ∥ f ∥ p ( ⋅ ) ≤ 1 ∫ b c ∣ f ( y ) ∣ d y . \|\chi_{(b,c)}\|_{p'(\cdot)}\leq C\sup_{\|f\|_{p(\cdot)}\leq1}\int_b^c|f(y)|\,dy. ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) ≤ 1 sup ∫ b c ∣ f ( y ) ∣ d y . Hence
∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C sup ∥ f ∥ p ( ⋅ ) ≤ 1 ∥ χ ( a , b ) ∥ p ( ⋅ ) ∫ b c ∣ f ( y ) ∣ d y = C ( c − a ) sup ∥ f ∥ p ( ⋅ ) ≤ 1 ∥ A a , b , c + ∣ f ∣ ∥ p ( ⋅ ) ≤ C ( c − a ) . \begin{aligned}
\|\chi_{(a,b)}\|_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}
&\le C\sup_{\|f\|_{p(\cdot)}\le1}\|\chi_{(a,b)}\|_{p(\cdot)}\int_b^c|f(y)|\,dy\\
&=C(c-a)\sup_{\|f\|_{p(\cdot)}\le1}\|A^+_{a,b,c}|f|\|_{p(\cdot)}\\
&\le C(c-a).
\end{aligned} ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) ≤ 1 sup ∥ χ ( a , b ) ∥ p ( ⋅ ) ∫ b c ∣ f ( y ) ∣ d y = C ( c − a ) ∥ f ∥ p ( ⋅ ) ≤ 1 sup ∥ A a , b , c + ∣ f ∣ ∥ p ( ⋅ ) ≤ C ( c − a ) . Therefore p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot)\in K^+_0(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R )
Necessity of the K 0 + K^+_0 K 0 +
We now show that K 0 + K^+_0 K 0 + M + M^+ M +
Theorem 4.1. Suppose that M + : L p ( ⋅ ) ( R ) ⟶ L p ( ⋅ ) ( R ) M^+:L^{p(\cdot)}(\mathbb{R})\longrightarrow L^{p(\cdot)}(\mathbb{R}) M + : L p ( ⋅ ) ( R ) ⟶ L p ( ⋅ ) ( R ) p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot)\in K^+_0(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R )
Proof. Fix a < b < c a<b<c a < b < c x ∈ ( a , b ) x\in(a,b) x ∈ ( a , b ) [ x , c ] [x,c] [ x , c ] ( b , c ) (b,c) ( b , c ) c − x ≤ c − a c-x\le c-a c − x ≤ c − a
M + f ( x ) ≥ 1 c − a ∫ b c ∣ f ( y ) ∣ d y . M^+f(x)\ge\frac{1}{c-a}\int_b^c|f(y)|\,dy. M + f ( x ) ≥ c − a 1 ∫ b c ∣ f ( y ) ∣ d y . Thus
A a , b , c + ∣ f ∣ ( x ) ≤ M + f ( x ) , x ∈ ( a , b ) . A^+_{a,b,c}|f|(x)\le M^+f(x),\qquad x\in(a,b). A a , b , c + ∣ f ∣ ( x ) ≤ M + f ( x ) , x ∈ ( a , b ) . Therefore,
∥ A a , b , c + ∣ f ∣ ∥ p ( ⋅ ) ≤ ∥ M + f ∥ p ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) . \|A^+_{a,b,c}|f|\|_{p(\cdot)}\le\|M^+f\|_{p(\cdot)}\le C\|f\|_{p(\cdot)}. ∥ A a , b , c + ∣ f ∣ ∥ p ( ⋅ ) ≤ ∥ M + f ∥ p ( ⋅ ) ≤ C ∥ f ∥ p ( ⋅ ) . The constant is independent of a , b , c a,b,c a , b , c p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot)\in K^+_0(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R )
Remark 4.2. Theorem 4.1 provides a functional justification for Definition 3.1. The condition K 0 + K^+_0 K 0 + M + M^+ M +
Proof of the main result
We now develop the two local estimates that replace the corresponding K 0 K_0 K 0
The following result is the one-sided counterpart of [3 Lemma 4.55] . In the classical setting, that lemma shows that the K 0 K_0 K 0 M + M^+ M + [3 Lemma 4.55] , with the essential modification that the usual K 0 K_0 K 0 K 0 + K^+_0 K 0 +
Lemma 5.1. Let p ( ⋅ ) ∈ P ( R ) p(\cdot)\in\mathcal{P}(\mathbb{R}) p ( ⋅ ) ∈ P ( R ) 1 < p − ≤ p + < ∞ 1<p_{-}\leq p_{+}<\infty 1 < p − ≤ p + < ∞ p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot)\in K_{0}^{+}(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R ) a < b < c a<b<c a < b < c f ∈ L p ( ⋅ ) ( R ) f\in L^{p(\cdot)}(\mathbb{R}) f ∈ L p ( ⋅ ) ( R ) c 1 , c 2 > 0 c_{1},c_{2}>0 c 1 , c 2 > 0 1 c − a ∫ b c ∣ f ( y ) ∣ d y ≥ c 1 \frac{1}{c-a}\int_{b}^{c}|f(y)|\,dy\geq c_{1} c − a 1 ∫ b c ∣ f ( y ) ∣ d y ≥ c 1 ∥ f ∥ p ( ⋅ ) ≤ c 2 \|f\|_{p(\cdot)}\leq c_{2} ∥ f ∥ p ( ⋅ ) ≤ c 2
∫ a b ( 1 c − a ∫ b c ∣ f ( y ) ∣ d y ) p ( x ) d x ≤ C ∫ b c ∣ f ( y ) ∣ p ( y ) d y . (5) \int_{a}^{b}\left(\frac{1}{c-a}\int_{b}^{c}|f(y)|\,dy\right)^{p(x)}dx\leq C\int_{b}^{c}|f(y)|^{p(y)}\,dy. \tag*{(5)} ∫ a b ( c − a 1 ∫ b c ∣ f ( y ) ∣ d y ) p ( x ) d x ≤ C ∫ b c ∣ f ( y ) ∣ p ( y ) d y . (5) where C C C p ( ⋅ ) p(\cdot) p ( ⋅ ) c 1 c_{1} c 1 c 2 c_{2} c 2 [ p ] K 0 + [p]_{K_{0}^{+}} [ p ] K 0 +
Proof. We adapt the argument of [3 Lemma 4.55] to the one-sided geometry. The main difference is that the variable x x x ( a , b ) (a,b) ( a , b ) y y y ( b , c ) (b,c) ( b , c ) D = c − a D=c-a D = c − a p − ′ = ( p + ) ′ > 1 p_{-}^{\prime}=(p_{+})^{\prime}>1 p − ′ = ( p + ) ′ > 1
Φ ( α ) = 1 D ∫ b c α p ′ ( y ) − 1 d y , α > 0 , \Phi(\alpha)=\frac{1}{D}\int_{b}^{c}\alpha^{p^{\prime}(y)-1}\,dy,\qquad\alpha>0, Φ ( α ) = D 1 ∫ b c α p ′ ( y ) − 1 d y , α > 0 , is continuous and strictly increasing. Moreover,
Φ ( α ) ⟶ 0 as α → 0 + , Φ ( α ) ⟶ ∞ as α → ∞ . \Phi(\alpha)\longrightarrow0\quad\text{as }\alpha\to0^{+},\qquad\Phi(\alpha)\longrightarrow\infty\quad\text{as }\alpha\to\infty. Φ ( α ) ⟶ 0 as α → 0 + , Φ ( α ) ⟶ ∞ as α → ∞. Hence there exists a unique α > 0 \alpha>0 α > 0
1 D ∫ b c α p ′ ( y ) − 1 d y = 1 D ∫ b c ∣ f ( y ) ∣ d y . (6) \frac{1}{D}\int_{b}^{c}\alpha^{p^{\prime}(y)-1}\,dy=\frac{1}{D}\int_{b}^{c}|f(y)|\,dy. \tag*{(6)} D 1 ∫ b c α p ′ ( y ) − 1 d y = D 1 ∫ b c ∣ f ( y ) ∣ d y . (6) Set
B = 1 D ∫ b c α p ′ ( y ) d y . B=\frac{1}{D}\int_{b}^{c}\alpha^{p^{\prime}(y)}\,dy. B = D 1 ∫ b c α p ′ ( y ) d y . Since p ′ ( x ) ( p ( x ) − 1 ) = p ( x ) p^{\prime}(x)(p(x)-1)=p(x) p ′ ( x ) ( p ( x ) − 1 ) = p ( x )
∫ a b ( 1 D ∫ b c ∣ f ( y ) ∣ d y ) p ( x ) d x = B ∫ a b ( 1 D ∫ b c α p ′ ( y ) − p ′ ( x ) d y ) p ( x ) − 1 d x . (7) \int_{a}^{b}\left(\frac{1}{D}\int_{b}^{c}|f(y)|\,dy\right)^{p(x)}dx
=B\int_{a}^{b}\left(\frac{1}{D}\int_{b}^{c}\alpha^{p^{\prime}(y)-p^{\prime}(x)}\,dy\right)^{p(x)-1}dx. \tag*{(7)} ∫ a b ( D 1 ∫ b c ∣ f ( y ) ∣ d y ) p ( x ) d x = B ∫ a b ( D 1 ∫ b c α p ′ ( y ) − p ′ ( x ) d y ) p ( x ) − 1 d x . (7) We estimate both factors. By (6),
∫ b c α p ′ ( y ) d y = α ∫ b c ∣ f ( y ) ∣ d y . \int_{b}^{c}\alpha^{p^{\prime}(y)}\,dy=\alpha\int_{b}^{c}|f(y)|\,dy. ∫ b c α p ′ ( y ) d y = α ∫ b c ∣ f ( y ) ∣ d y . Therefore,
∫ b c α p ′ ( y ) d y = 2 α ∫ b c ∣ f ( y ) ∣ d y − ∫ b c α p ′ ( y ) d y . \int_{b}^{c}\alpha^{p^{\prime}(y)}\,dy
=2\alpha\int_{b}^{c}|f(y)|\,dy-\int_{b}^{c}\alpha^{p^{\prime}(y)}\,dy. ∫ b c α p ′ ( y ) d y = 2 α ∫ b c ∣ f ( y ) ∣ d y − ∫ b c α p ′ ( y ) d y . It follows that
∫ b c α p ′ ( y ) d y ≤ 2 α ∫ { y ∈ ( b , c ) : 2 α ∣ f ( y ) ∣ > α p ′ ( y ) } ∣ f ( y ) ∣ d y . \int_{b}^{c}\alpha^{p^{\prime}(y)}\,dy
\leq2\alpha\int_{\{y\in(b,c):2\alpha|f(y)|>\alpha^{p^{\prime}(y)}\}}|f(y)|\,dy. ∫ b c α p ′ ( y ) d y ≤ 2 α ∫ { y ∈ ( b , c ) : 2 α ∣ f ( y ) ∣ > α p ′ ( y ) } ∣ f ( y ) ∣ d y . On the set appearing on the right-hand side, 2 ∣ f ( y ) ∣ > α p ′ ( y ) − 1 2|f(y)|>\alpha^{p^{\prime}(y)-1} 2∣ f ( y ) ∣ > α p ′ ( y ) − 1 α < ( 2 ∣ f ( y ) ∣ ) p ( y ) − 1 \alpha<(2|f(y)|)^{p(y)-1} α < ( 2∣ f ( y ) ∣ ) p ( y ) − 1
2 α ∣ f ( y ) ∣ ≤ 2 p ( y ) ∣ f ( y ) ∣ p ( y ) ≤ 2 p + ∣ f ( y ) ∣ p ( y ) . 2\alpha|f(y)|\leq2^{p(y)}|f(y)|^{p(y)}\leq2^{p_{+}}|f(y)|^{p(y)}. 2 α ∣ f ( y ) ∣ ≤ 2 p ( y ) ∣ f ( y ) ∣ p ( y ) ≤ 2 p + ∣ f ( y ) ∣ p ( y ) . Thus
B ≤ 2 p + 1 D ∫ b c ∣ f ( y ) ∣ p ( y ) d y . (8) B\leq2^{p_{+}}\frac{1}{D}\int_{b}^{c}|f(y)|^{p(y)}\,dy. \tag*{(8)} B ≤ 2 p + D 1 ∫ b c ∣ f ( y ) ∣ p ( y ) d y . (8) We now estimate the second factor in (7). By our hypotheses and (6),
c 1 ≤ 1 D ∫ b c α p ′ ( y ) − 1 d y . c_{1}\leq\frac{1}{D}\int_{b}^{c}\alpha^{p^{\prime}(y)-1}\,dy. c 1 ≤ D 1 ∫ b c α p ′ ( y ) − 1 d y . Since c − b ≤ D c-b\le D c − b ≤ D 0 < α ≤ 1 0<\alpha\le1 0 < α ≤ 1
c 1 ≤ 1 D ∫ b c α p ′ ( y ) − 1 d y ≤ α p ′ − 1 . c_1\le\frac{1}{D}\int_b^c \alpha^{p'(y)-1}\,dy\le\alpha^{p'-1}. c 1 ≤ D 1 ∫ b c α p ′ ( y ) − 1 d y ≤ α p ′ − 1 . Thus, in all cases,
α ≥ d 1 : = min { 1 , c 1 1 / ( p ′ − 1 ) } > 0. (9) \alpha\ge d_1:=\min\left\{1,c_1^{1/(p'-1)}\right\}>0. \tag*{(9)} α ≥ d 1 := min { 1 , c 1 1/ ( p ′ − 1 ) } > 0. (9) On the other hand, by the inequality Hölder and our hypothesis for f f f
∫ b c ∣ f ( y ) ∣ d y ≤ C p ( ⋅ ) ∥ f ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ c 2 C p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) . \int_b^c |f(y)|\,dy\le C_{p(\cdot)}\|f\|_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}
\le c_2C_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}. ∫ b c ∣ f ( y ) ∣ d y ≤ C p ( ⋅ ) ∥ f ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ c 2 C p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) . Let N = ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) N=\|\chi_{(b,c)}\|_{p'(\cdot)} N = ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) p ′ ( ⋅ ) p'(\cdot) p ′ ( ⋅ )
∫ b c N − p ′ ( y ) d y = 1. \int_b^c N^{-p'(y)}\,dy=1. ∫ b c N − p ′ ( y ) d y = 1. Choose d 2 ≥ 1 d_2\ge1 d 2 ≥ 1
d 2 p ′ − 1 ≥ c 2 C p ( ⋅ ) . d_2^{p'-1}\ge c_2C_{p(\cdot)}. d 2 p ′ − 1 ≥ c 2 C p ( ⋅ ) . Then
∫ b c ( d 2 N ) p ′ ( y ) − 1 d y = N ∫ b c d 2 p ′ ( y ) − 1 N − p ′ ( y ) d y ≥ c 2 C p ( ⋅ ) N ≥ ∫ b c ∣ f ( y ) ∣ d y . \begin{aligned}
\int_b^c\left(\frac{d_2}{N}\right)^{p'(y)-1}\,dy
&=N\int_b^c d_2^{p'(y)-1}N^{-p'(y)}\,dy\\
&\ge c_2C_{p(\cdot)}N\\
&\ge\int_b^c|f(y)|\,dy.
\end{aligned} ∫ b c ( N d 2 ) p ′ ( y ) − 1 d y = N ∫ b c d 2 p ′ ( y ) − 1 N − p ′ ( y ) d y ≥ c 2 C p ( ⋅ ) N ≥ ∫ b c ∣ f ( y ) ∣ d y . By (6) and monotonicity,
α ≤ d 2 N . (10) \alpha\le\frac{d_2}{N}. \tag*{(10)} α ≤ N d 2 . (10) For each x ∈ ( a , b ) x\in(a,b) x ∈ ( a , b )
E + ( x ) = { y ∈ ( b , c ) : p ′ ( y ) > p ′ ( x ) } E_+(x)=\{y\in(b,c):p'(y)>p'(x)\} E + ( x ) = { y ∈ ( b , c ) : p ′ ( y ) > p ′ ( x )} and
E − ( x ) = { y ∈ ( b , c ) : p ′ ( y ) ≤ p ′ ( x ) } . E_-(x)=\{y\in(b,c):p'(y)\le p'(x)\}. E − ( x ) = { y ∈ ( b , c ) : p ′ ( y ) ≤ p ′ ( x )} . Using (9) on E − ( x ) E_-(x) E − ( x ) E + ( x ) E_+(x) E + ( x )
∫ b c α p ′ ( y ) − p ′ ( x ) d y ≤ D 1 N p ′ ( x ) ∫ b c N − p ′ ( y ) d y + D 2 ( c − b ) = D 1 N p ′ ( x ) + D 2 ( c − b ) , \begin{aligned}
\int_b^c\alpha^{p'(y)-p'(x)}\,dy
&\le D_1N^{p'(x)}\int_b^cN^{-p'(y)}\,dy+D_2(c-b)\\
&=D_1N^{p'(x)}+D_2(c-b),
\end{aligned} ∫ b c α p ′ ( y ) − p ′ ( x ) d y ≤ D 1 N p ′ ( x ) ∫ b c N − p ′ ( y ) d y + D 2 ( c − b ) = D 1 N p ′ ( x ) + D 2 ( c − b ) , where D 1 , D 2 D_1,D_2 D 1 , D 2 p ( ⋅ ) p(\cdot) p ( ⋅ ) d 1 d_1 d 1 d 2 d_2 d 2
Hence
1 D ∫ b c α p ′ ( y ) − p ′ ( x ) d y ≤ D 1 N p ′ ( x ) D + D 2 . \frac{1}{D}\int_b^c\alpha^{p'(y)-p'(x)}\,dy\le D_1\frac{N^{p'(x)}}{D}+D_2. D 1 ∫ b c α p ′ ( y ) − p ′ ( x ) d y ≤ D 1 D N p ′ ( x ) + D 2 . Therefore,
∫ a b ( 1 D ∫ b c α p ′ ( y ) − p ′ ( x ) d y ) p ( x ) − 1 d x ≤ C D ∫ a b ( N D ) p ( x ) d x + C ( b − a ) . (11) \int_a^b \left(\frac{1}{D}\int_b^c \alpha^{p'(y)-p'(x)}\,dy\right)^{p(x)-1}\,dx
\le CD\int_a^b \left(\frac{N}{D}\right)^{p(x)}\,dx+C(b-a). \tag*{(11)} ∫ a b ( D 1 ∫ b c α p ′ ( y ) − p ′ ( x ) d y ) p ( x ) − 1 d x ≤ C D ∫ a b ( D N ) p ( x ) d x + C ( b − a ) . (11) Since p ( ⋅ ) ∈ K + 0 ( R ) p(\cdot)\in K_+^0(\mathbb{R}) p ( ⋅ ) ∈ K + 0 ( R )
∥ χ ( a , b ) ∥ p ( ⋅ ) N D ≤ [ p ] K + 0 . \frac{\|\chi_{(a,b)}\|_{p(\cdot)}N}{D}\le[p]_{K_+^0}. D ∥ χ ( a , b ) ∥ p ( ⋅ ) N ≤ [ p ] K + 0 . Thus
N D ≤ [ p ] K + 0 ∥ χ ( a , b ) ∥ p ( ⋅ ) . \frac{N}{D}\le\frac{[p]_{K_+^0}}{\|\chi_{(a,b)}\|_{p(\cdot)}}. D N ≤ ∥ χ ( a , b ) ∥ p ( ⋅ ) [ p ] K + 0 . Using (1),
D ∫ a b ( N D ) p ( x ) d x ≤ C D ∫ a b ( 1 ∥ χ ( a , b ) ∥ p ( ⋅ ) ) p ( x ) d x = C D . D\int_a^b\left(\frac{N}{D}\right)^{p(x)}\,dx
\le CD\int_a^b\left(\frac{1}{\|\chi_{(a,b)}\|_{p(\cdot)}}\right)^{p(x)}\,dx
=CD. D ∫ a b ( D N ) p ( x ) d x ≤ C D ∫ a b ( ∥ χ ( a , b ) ∥ p ( ⋅ ) 1 ) p ( x ) d x = C D . Since b − a ≤ D b-a\le D b − a ≤ D
∫ a b ( 1 D ∫ b c α p ′ ( y ) − p ′ ( x ) d y ) p ( x ) − 1 d x ≤ C D . (12) \int_a^b\left(\frac{1}{D}\int_b^c\alpha^{p'(y)-p'(x)}\,dy\right)^{p(x)-1}\,dx\le CD. \tag*{(12)} ∫ a b ( D 1 ∫ b c α p ′ ( y ) − p ′ ( x ) d y ) p ( x ) − 1 d x ≤ C D . (12) Combining (7), (8), and (12), we obtain
∫ a b ( 1 c − a ∫ b c ∣ f ( y ) ∣ d y ) p ( x ) d x ≤ C ∫ b c ∣ f ( y ) ∣ p ( y ) d y . \int_a^b\left(\frac{1}{c-a}\int_b^c|f(y)|\,dy\right)^{p(x)}\,dx
\le C\int_b^c|f(y)|^{p(y)}\,dy. ∫ a b ( c − a 1 ∫ b c ∣ f ( y ) ∣ d y ) p ( x ) d x ≤ C ∫ b c ∣ f ( y ) ∣ p ( y ) d y . This proves the lemma. □
The following lemma follows the argument in [18 Theorem 1, (c) ⇒ (d) ⇒ (e) ⇒ (f)] , with g ≡ 1 g\equiv1 g ≡ 1
Lemma 5.2. Let U = ( u , v ) U=(u,v) U = ( u , v ) w ∈ L ∞ ( U ) w\in L^\infty(U) w ∈ L ∞ ( U ) w > 0 w>0 w > 0 U U U η ∈ ( 0 , 1 ) \eta\in(0,1) η ∈ ( 0 , 1 ) β η ∈ ( 0 , 1 ) \beta_\eta\in(0,1) β η ∈ ( 0 , 1 )
w ( E ) ≤ β η w ( a , b ) ⟹ ∣ E ∣ < η ( c − a ) (13) w(E)\le\beta_\eta w(a,b)\quad\Longrightarrow\quad|E|<\eta(c-a) \tag*{(13)} w ( E ) ≤ β η w ( a , b ) ⟹ ∣ E ∣ < η ( c − a ) (13) for every u ≤ a < b < c ≤ v u\le a<b<c\le v u ≤ a < b < c ≤ v E ⊂ ( b , c ) E\subset(b,c) E ⊂ ( b , c ) r > 1 r>1 r > 1 C > 0 C>0 C > 0
( ∫ a − h a w ( x ) r d x ) 1 / r ≤ C ∫ a a + h w ( x ) d x (14) \left(\int_{a-h}^{a}w(x)^r\,dx\right)^{1/r}\le C\int_a^{a+h}w(x)\,dx \tag*{(14)} ( ∫ a − h a w ( x ) r d x ) 1/ r ≤ C ∫ a a + h w ( x ) d x (14) whenever h > 0 h>0 h > 0
u ≤ a − h < a < a + h ≤ v . u\le a-h<a<a+h\le v. u ≤ a − h < a < a + h ≤ v . The constants r r r C C C β 1 / 4 \beta_{1/4} β 1/4 β 1 / 2 \beta_{1/2} β 1/2 U U U ∥ w ∥ L ∞ ( U ) \|w\|_{L^\infty(U)} ∥ w ∥ L ∞ ( U )
Proof. We follow the proof of [18 Theorem 1, (c) $\Rightarrow$ (d) $\Rightarrow$ (e) $\Rightarrow$ (f)] , with g ≡ 1 g \equiv1 g ≡ 1 U U U
Set
κ = β 1 / 4 4 . \kappa= \frac{\beta^{1/4}}{4}. κ = 4 β 1/4 . We first show that
∣ { x ∈ J : w ( x ) > κ λ } ∣ ≥ 1 2 ∣ J ∣ (15) \lvert\{x \in J : w(x) > \kappa\lambda\}\rvert\ge\frac{1}{2}\lvert J\rvert \tag*{(15)} ∣{ x ∈ J : w ( x ) > κλ }∣ ≥ 2 1 ∣ J ∣ (15) whenever J = ( a , b ) ⊂ U J = (a,b) \subset U J = ( a , b ) ⊂ U λ > 0 \lambda> 0 λ > 0
w ( a , x ) x − a ≥ λ ( a < x < b ) . \frac{w(a,x)}{x-a} \ge\lambda\qquad(a < x < b). x − a w ( a , x ) ≥ λ ( a < x < b ) . Let x 0 = b x_0=b x 0 = b k k k x k ∈ ( a , x k + 1 ) x_k \in(a,x_{k+1}) x k ∈ ( a , x k + 1 )
w ( a , x k ) = 1 2 w ( a , x k + 1 ) . w(a,x_k) = \frac{1}{2}w(a,x_{k+1}). w ( a , x k ) = 2 1 w ( a , x k + 1 ) . These points exist because w > 0 w>0 w > 0
w ( a , x k ) = 2 k w ( a , b ) , w(a,x_k)=2^k w(a,b), w ( a , x k ) = 2 k w ( a , b ) , and hence x k → a x_k \to a x k → a k → − ∞ k \to-\infty k → − ∞ k ≤ − 1 k \le-1 k ≤ − 1
I k = ( x k − 1 , x k ) , E = { x ∈ J : w ( x ) ≤ κ λ } , E k = E ∩ ( x k , x k + 1 ) . I_k=(x_{k-1},x_k), \qquad E=\{x\in J:w(x)\le\kappa\lambda\}, \qquad E_k=E\cap(x_k,x_{k+1}). I k = ( x k − 1 , x k ) , E = { x ∈ J : w ( x ) ≤ κλ } , E k = E ∩ ( x k , x k + 1 ) . The construction gives
w ( a , x k + 1 ) = 4 w ( I k ) . w(a,x_{k+1})=4w(I_k). w ( a , x k + 1 ) = 4 w ( I k ) . By the hypothesis on the averages, also at x = b x=b x = b
λ ( x k + 1 − a ) ≤ w ( a , x k + 1 ) . \lambda(x_{k+1}-a)\le w(a,x_{k+1}). λ ( x k + 1 − a ) ≤ w ( a , x k + 1 ) . Consequently,
w ( E k ) ≤ κ λ ∣ E k ∣ ≤ κ λ ( x k + 1 − a ) ≤ 4 κ w ( I k ) = β 1 / 4 w ( I k ) . w(E_k)\le\kappa\lambda\lvert E_k\rvert\le\kappa\lambda(x_{k+1}-a)\le4\kappa w(I_k)=\beta^{1/4}w(I_k). w ( E k ) ≤ κλ ∣ E k ∣ ≤ κλ ( x k + 1 − a ) ≤ 4 κ w ( I k ) = β 1/4 w ( I k ) . Applying (13) to x k − 1 < x k < x k + 1 x_{k-1}<x_k<x_{k+1} x k − 1 < x k < x k + 1
∣ E k ∣ < 1 4 ( x k + 1 − x k − 1 ) . \lvert E_k\rvert<\frac{1}{4}(x_{k+1}-x_{k-1}). ∣ E k ∣ < 4 1 ( x k + 1 − x k − 1 ) . All these triples lie in [ a , b ] ⊂ [ u , v ] [a,b]\subset[u,v] [ a , b ] ⊂ [ u , v ] ( x k , x k + 1 ) (x_k,x_{k+1}) ( x k , x k + 1 ) k ≤ − 1 k\le-1 k ≤ − 1 J J J
∣ E ∣ = ∑ k = − ∞ − 1 ∣ E k ∣ ≤ 1 4 ∑ k = − ∞ − 1 ( x k + 1 − x k − 1 ) ≤ 1 2 ∣ J ∣ . \lvert E\rvert=\sum_{k=-\infty}^{-1}\lvert E_k\rvert\le\frac{1}{4}\sum_{k=-\infty}^{-1}(x_{k+1}-x_{k-1})\le\frac{1}{2}\lvert J\rvert. ∣ E ∣ = k = − ∞ ∑ − 1 ∣ E k ∣ ≤ 4 1 k = − ∞ ∑ − 1 ( x k + 1 − x k − 1 ) ≤ 2 1 ∣ J ∣ . This proves (15).
Fix now J = ( a , b ) ⊂ U J=(a,b)\subset U J = ( a , b ) ⊂ U w χ J w\chi_J w χ J J J J
λ 0 = M − ( w χ J ) ( b ) . \lambda_0=M^-(w\chi_J)(b). λ 0 = M − ( w χ J ) ( b ) . For λ > λ 0 \lambda>\lambda_0 λ > λ 0
{ x ∈ R : M − ( w χ J ) ( x ) > λ } = ⋃ j J j , \{x\in\mathbb{R}:M^-(w\chi_J)(x)>\lambda\}=\bigcup_j J_j, { x ∈ R : M − ( w χ J ) ( x ) > λ } = j ⋃ J j , where J j = ( a j , b j ) J_j=(a_j,b_j) J j = ( a j , b j ) ( e ) ⇒ ( f ) (e)\Rightarrow(f) ( e ) ⇒ ( f ) J J J
w ( a j , b j ) b j − a j = λ , w ( a j , x ) x − a j ≥ λ ( a j < x < b j ) . \frac{w(a_j,b_j)}{b_j-a_j}=\lambda,\qquad\frac{w(a_j,x)}{x-a_j}\ge\lambda\qquad(a_j<x<b_j). b j − a j w ( a j , b j ) = λ , x − a j w ( a j , x ) ≥ λ ( a j < x < b j ) . Thus (15) applies to every J j J_j J j
w ( { x ∈ J : w ( x ) > λ } ) ≤ ∑ j w ( J j ) = λ ∑ j ∣ J j ∣ ≤ 2 λ ∑ j ∣ { x ∈ J j : w ( x ) > κ λ } ∣ ≤ 2 λ ∣ { x ∈ J : w ( x ) > κ λ } ∣ . w(\{x \in J : w(x) > \lambda\}) \le\sum_j w(J_j)
= \lambda\sum_j |J_j|
\le2\lambda\sum_j |\{x \in J_j : w(x) > \kappa\lambda\}|
\le2\lambda|\{x \in J : w(x) > \kappa\lambda\}|. w ({ x ∈ J : w ( x ) > λ }) ≤ j ∑ w ( J j ) = λ j ∑ ∣ J j ∣ ≤ 2 λ j ∑ ∣ { x ∈ J j : w ( x ) > κλ } ∣ ≤ 2 λ ∣ { x ∈ J : w ( x ) > κλ } ∣. Multiply this inequality by λ δ − 1 \lambda^{\delta-1} λ δ − 1 δ > 0 \delta> 0 δ > 0 ( λ 0 , ∞ ) (\lambda_0,\infty) ( λ 0 , ∞ )
1 δ ∫ { x ∈ J : w ( x ) > λ 0 } w ( x ) ( w ( x ) δ − λ 0 δ ) d x ≤ 2 ( 1 + δ ) κ 1 + δ ∫ J w ( x ) 1 + δ d x . \frac{1}{\delta}\int_{\{x\in J:w(x)>\lambda_0\}}w(x)(w(x)^\delta-\lambda_0^\delta)\,dx
\le\frac{2}{(1+\delta)\kappa^{1+\delta}}\int_Jw(x)^{1+\delta}\,dx. δ 1 ∫ { x ∈ J : w ( x ) > λ 0 } w ( x ) ( w ( x ) δ − λ 0 δ ) d x ≤ ( 1 + δ ) κ 1 + δ 2 ∫ J w ( x ) 1 + δ d x . The integrand w ( x ) ( w ( x ) δ − λ 0 δ ) w(x)(w(x)^\delta-\lambda_0^\delta) w ( x ) ( w ( x ) δ − λ 0 δ ) w ( x ) ≤ λ 0 w(x) \le\lambda_0 w ( x ) ≤ λ 0
1 δ ( ∫ J w ( x ) 1 + δ d x − λ 0 δ w ( J ) ) ≤ 2 ( 1 + δ ) κ 1 + δ ∫ J w ( x ) 1 + δ d x . \frac{1}{\delta}\left(\int_Jw(x)^{1+\delta}\,dx-\lambda_0^\delta w(J)\right)
\le\frac{2}{(1+\delta)\kappa^{1+\delta}}\int_Jw(x)^{1+\delta}\,dx. δ 1 ( ∫ J w ( x ) 1 + δ d x − λ 0 δ w ( J ) ) ≤ ( 1 + δ ) κ 1 + δ 2 ∫ J w ( x ) 1 + δ d x . All these integrals are finite because w ∈ L ∞ ( U ) w \in L^\infty(U) w ∈ L ∞ ( U ) U U U δ > 0 \delta> 0 δ > 0
2 δ ( 1 + δ ) κ 1 + δ ≤ 1 2 . \frac{2\delta}{(1+\delta)\kappa^{1+\delta}}\le\frac{1}{2}. ( 1 + δ ) κ 1 + δ 2 δ ≤ 2 1 . Setting r = 1 + δ r=1+\delta r = 1 + δ
∫ J w ( x ) r d x ≤ 2 [ M − ( w χ J ) ( b ) ] r − 1 w ( J ) (16) \int_Jw(x)^r\,dx\le2[M^-(w\chi_J)(b)]^{r-1}w(J) \tag*{(16)} ∫ J w ( x ) r d x ≤ 2 [ M − ( w χ J ) ( b ) ] r − 1 w ( J ) (16) for every J = ( a , b ) ⊂ U J=(a,b)\subset U J = ( a , b ) ⊂ U r r r β 1 / 4 \beta_{1/4} β 1/4
It remains to obtain (14) . Fix h > 0 h>0 h > 0 a a a
u ≤ a − h < a < a + h ≤ v . u\le a-h<a<a+h\le v. u ≤ a − h < a < a + h ≤ v . Write
ℓ = a − h , m = a , z = a + h , \ell=a-h,\qquad m=a,\qquad z=a+h, ℓ = a − h , m = a , z = a + h , and set
A = ∫ ℓ m w ( x ) r d x , W = w ( ℓ , z ) . A=\int_\ell^m w(x)^r\,dx,\qquad W=w(\ell,z). A = ∫ ℓ m w ( x ) r d x , W = w ( ℓ , z ) . For every x ∈ ( m , z ) x\in(m,z) x ∈ ( m , z ) (16) to ( ℓ , x ) (\ell,x) ( ℓ , x )
A ≤ ∫ ℓ x w ( y ) r d y ≤ 2 [ M − ( w χ ( ℓ , x ) ) ( x ) ] r − 1 w ( ℓ , x ) ≤ 2 [ M − ( w χ ( ℓ , x ) ) ( x ) ] r − 1 W . A\le\int_\ell^xw(y)^r\,dy
\le2[M^-(w\chi_{(\ell,x)})(x)]^{r-1}w(\ell,x)
\le2[M^-(w\chi_{(\ell,x)})(x)]^{r-1}W. A ≤ ∫ ℓ x w ( y ) r d y ≤ 2 [ M − ( w χ ( ℓ , x ) ) ( x ) ] r − 1 w ( ℓ , x ) ≤ 2 [ M − ( w χ ( ℓ , x ) ) ( x ) ] r − 1 W . Hence
M − ( w χ ( ℓ , z ) ) ( x ) ≥ ( A 2 W ) 1 / ( r − 1 ) ( m < x < z ) . M^-(w\chi_{(\ell,z)})(x)\ge\left(\frac{A}{2W}\right)^{1/(r-1)}
\qquad(m<x<z). M − ( w χ ( ℓ , z ) ) ( x ) ≥ ( 2 W A ) 1/ ( r − 1 ) ( m < x < z ) . The weak type ( 1 , 1 ) (1,1) ( 1 , 1 ) M − M^{-} M −
h ≤ C W ( 2 W A ) 1 / ( r − 1 ) . h \le CW\left(\frac{2W}{A}\right)^{1/(r-1)}. h ≤ C W ( A 2 W ) 1/ ( r − 1 ) . Rearranging gives
A ≤ C r h 1 − r W r . (17) A \le C_r h^{1-r}W^r. \tag*{(17)} A ≤ C r h 1 − r W r . (17) Finally, apply (13) with the triple ℓ < m < z \ell< m < z ℓ < m < z E = ( m , z ) E=(m,z) E = ( m , z ) η = 1 / 2 \eta=1/2 η = 1/2
∣ E ∣ z − ℓ = 1 2 , \frac{|E|}{z-\ell}=\frac{1}{2}, z − ℓ ∣ E ∣ = 2 1 , the conclusion of that implication is false. Consequently w ( m , z ) > β 1 / 2 w ( ℓ , m ) w(m,z)>\beta_{1/2}w(\ell,m) w ( m , z ) > β 1/2 w ( ℓ , m )
W ≤ ( 1 + β 1 / 2 − 1 ) w ( m , z ) . W\leq(1+\beta_{1/2}^{-1})w(m,z). W ≤ ( 1 + β 1/2 − 1 ) w ( m , z ) . Combining this estimate with (17) , we obtain
( 1 h ∫ ℓ m w ( x ) r d x ) 1 / r ≤ C 1 h ∫ m z w ( x ) d x , \left(\frac{1}{h}\int_{\ell}^{m}w(x)^r\,dx\right)^{1/r}\leq C\frac{1}{h}\int_{m}^{z}w(x)\,dx, ( h 1 ∫ ℓ m w ( x ) r d x ) 1/ r ≤ C h 1 ∫ m z w ( x ) d x , which is (14) . The constants depend only on β 1 / 4 \beta_{1/4} β 1/4 β 1 / 2 \beta_{1/2} β 1/2
The next lemma is the one-sided analogue of [3 Lemma 4.56] . In the two-sided setting, the proof first derives a uniform A ∞ A_{\infty} A ∞
w t ( x ) = t p ( x ) w_t(x)=t^{p(x)} w t ( x ) = t p ( x ) from the modular estimate of [3 Lemma 4.55] , and then applies a reverse Hölder self-improvement argument.
In the present one-sided setting, we combine Lemma 5.1 with the classical theory of the classes A ∞ + A_{\infty}^{+} A ∞ + R H s + RH_{s}^{+} R H s + [18 , 2 , 22 ] .
Lemma 5.3. Let p ( ⋅ ) ∈ K 0 + ( R ) p(\cdot)\in K_{0}^{+}(\mathbb{R}) p ( ⋅ ) ∈ K 0 + ( R ) 1 < p − ≤ p + < ∞ 1<p_{-}\leq p_{+}<\infty 1 < p − ≤ p + < ∞ b 1 , b 2 > 0 b_1,b_2>0 b 1 , b 2 > 0
I 0 − = ( a 0 − h 0 , a 0 ) , I 0 + = ( a 0 , a 0 + h 0 ) , I_{0}^{-}=(a_0-h_0,a_0),\qquad I_{0}^{+}=(a_0,a_0+h_0), I 0 − = ( a 0 − h 0 , a 0 ) , I 0 + = ( a 0 , a 0 + h 0 ) , where h 0 > 0 h_0>0 h 0 > 0 U = ( a 0 − h 0 , a 0 + h 0 ) U=(a_0-h_0,a_0+h_0) U = ( a 0 − h 0 , a 0 + h 0 ) b 1 ≤ t ≤ b 2 ∥ χ U ∥ p ( ⋅ ) b_1\leq t\leq\frac{b_2}{\lVert\chi_U\rVert_{p(\cdot)}} b 1 ≤ t ≤ ∥ χ U ∥ p ( ⋅ ) b 2 s 0 > 1 s_0>1 s 0 > 1 C 0 > 0 C_0>0 C 0 > 0 p − , p + , b 1 , b 2 p_{-},p_{+},b_1,b_2 p − , p + , b 1 , b 2 [ p ] K 0 + [p]_{K_{0}^{+}} [ p ] K 0 +
\left(\fint_{I_{0}^{-}}t^{sp(x)}\,dx\right)^{1/s}\leq C_0\fint_{I_{0}^{+}}t^{p(x)}\,dx \tag*{(18)}
for every 1 < s < s 0 1<s<s_0 1 < s < s 0
Proof. Set w ( x ) = t p ( x ) w(x)=t^{p(x)} w ( x ) = t p ( x ) (13) of Lemma 5.2, with constants independent of t t t U U U η ∈ ( 0 , 1 ) \eta\in(0,1) η ∈ ( 0 , 1 ) β η ∈ ( 0 , 1 ) \beta_{\eta}\in(0,1) β η ∈ ( 0 , 1 ) t t t U U U
w ( E ) ≤ β η w ( a , b ) ⟹ ∣ E ∣ c − a < η (19) w(E)\leq\beta_{\eta}w(a,b)\quad\Longrightarrow\quad\frac{|E|}{c-a}<\eta \tag*{(19)} w ( E ) ≤ β η w ( a , b ) ⟹ c − a ∣ E ∣ < η (19) whenever a 0 − h 0 ≤ a < b < c ≤ a 0 + h 0 a_0-h_0\leq a<b<c\leq a_0+h_0 a 0 − h 0 ≤ a < b < c ≤ a 0 + h 0 E ⊂ ( b , c ) E\subset(b,c) E ⊂ ( b , c )
Indeed, since ( a , b ) ⊂ U (a,b)\subset U ( a , b ) ⊂ U t t t
∥ t χ ( a , b ) ∥ p ( ⋅ ) ≤ b 2 . \lVert t\chi_{(a,b)}\rVert_{p(\cdot)}\leq b_2. ∥ t χ ( a , b ) ∥ p ( ⋅ ) ≤ b 2 . Consequently w ( a , b ) ≤ B w(a,b)\leq B w ( a , b ) ≤ B B : = max { b 2 p − , b 2 p + } B:=\max\{b_2^{p_{-}},b_2^{p_{+}}\} B := max { b 2 p − , b 2 p + } 0 < β ≤ 1 0 < \beta\le1 0 < β ≤ 1 w ( E ) ≤ β w ( a , b ) w(E) \le\beta w(a,b) w ( E ) ≤ β w ( a , b ) ρ p ( ⋅ ) ( t χ E ) = w ( E ) ≤ B \rho_{p(\cdot)}(t\chi_E) = w(E) \le B ρ p ( ⋅ ) ( t χ E ) = w ( E ) ≤ B
∥ t χ E ∥ p ( ⋅ ) ≤ c 2 , c 2 : = max { 1 , B 1 / p − } . \|t\chi_E\|_{p(\cdot)} \le c_2,\qquad c_2 := \max\{1,B^{1/p_-}\}. ∥ t χ E ∥ p ( ⋅ ) ≤ c 2 , c 2 := max { 1 , B 1/ p − } . Suppose, on the contrary, that ∣ E ∣ ≥ η ( c − a ) |E| \ge\eta(c-a) ∣ E ∣ ≥ η ( c − a )
1 c − a ∫ b c t χ E ( y ) d y ≥ b 1 η . \frac{1}{c-a}\int_b^c t\chi_E(y)\,dy \ge b_1\eta. c − a 1 ∫ b c t χ E ( y ) d y ≥ b 1 η . Lemma 5.1, applied to t χ E t\chi_E t χ E
∫ a b ( t ∣ E ∣ c − a ) p ( x ) d x ≤ C η w ( E ) , \int_a^b \left(t\frac{|E|}{c-a}\right)^{p(x)}\,dx \le C_\eta w(E), ∫ a b ( t c − a ∣ E ∣ ) p ( x ) d x ≤ C η w ( E ) , where C η C_\eta C η t t t U U U a a a b b b c c c E E E
∫ a b ( t ∣ E ∣ c − a ) p ( x ) d x ≥ η p + w ( a , b ) . \int_a^b \left(t\frac{|E|}{c-a}\right)^{p(x)}\,dx \ge\eta^{p_+}w(a,b). ∫ a b ( t c − a ∣ E ∣ ) p ( x ) d x ≥ η p + w ( a , b ) . Choosing
β η = min { 1 2 , η p + 2 C η } \beta_\eta=\min\left\{\frac{1}{2},\frac{\eta^{p_+}}{2C_\eta}\right\} β η = min { 2 1 , 2 C η η p + } gives a contradiction. This proves (19).
Since t > 0 t>0 t > 0 p + < ∞ p_+<\infty p + < ∞ w ( x ) = t p ( x ) w(x)=t^{p(x)} w ( x ) = t p ( x ) L ∞ ( U ) L^\infty(U) L ∞ ( U ) t t t U U U r > 1 r>1 r > 1 C > 0 C>0 C > 0 t t t U U U
( ∫ I 0 − t r p ( x ) d x ) 1 / r ≤ C ∫ I 0 + t p ( x ) d x . \left(\int_{I_0^-}t^{rp(x)}\,dx\right)^{1/r}\le C\int_{I_0^+}t^{p(x)}\,dx. ( ∫ I 0 − t r p ( x ) d x ) 1/ r ≤ C ∫ I 0 + t p ( x ) d x . For every 1 < s < r 1<s<r 1 < s < r
( ∫ I 0 − t s p ( x ) d x ) 1 / s ≤ ( ∫ I 0 − t r p ( x ) d x ) 1 / r . \left(\int_{I_0^-}t^{sp(x)}\,dx\right)^{1/s}\le\left(\int_{I_0^-}t^{rp(x)}\,dx\right)^{1/r}. ( ∫ I 0 − t s p ( x ) d x ) 1/ s ≤ ( ∫ I 0 − t r p ( x ) d x ) 1/ r . Taking s 0 = r s_0=r s 0 = r s s s
The following lemma is a consequence of the boundedness criterion of Lerner and Ombrosi [15 Theorem 1.2] . We include the proof to explain its application to the right-sided maximal operator.
Lemma 5.4. Let p ( ⋅ ) ∈ P ( R ) p(\cdot)\in\mathcal{P}(\mathbb{R}) p ( ⋅ ) ∈ P ( R ) 1 < p − ≤ p + < ∞ 1<p_-\le p_+<\infty 1 < p − ≤ p + < ∞ s ∗ > 1 s_*>1 s ∗ > 1 C > 0 C>0 C > 0
∥ M + ∥ L s p ( ⋅ ) ( R ) → L s p ( ⋅ ) ( R ) ≤ C ( s − 1 ) − 1 / ( s p − ) ( 1 < s < s ∗ ) , (20) \|M^+\|_{L^{sp(\cdot)}(\mathbb{R})\to L^{sp(\cdot)}(\mathbb{R})}\le C(s-1)^{-1/(sp_-)}\qquad(1<s<s_*), \tag*{(20)} ∥ M + ∥ L s p ( ⋅ ) ( R ) → L s p ( ⋅ ) ( R ) ≤ C ( s − 1 ) − 1/ ( s p − ) ( 1 < s < s ∗ ) , (20) where C C C s s s M + M^+ M + L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R )
Proof. For 0 < r < 1 0<r<1 0 < r < 1
M r + f = ( M + ( ∣ f ∣ r ) ) 1 / r . M_r^+f=(M^+(|f|^r))^{1/r}. M r + f = ( M + ( ∣ f ∣ r ) ) 1/ r . The basis { ( x , x + h ) } h > 0 , x ∈ R \{(x,x+h)\}_{h>0},x\in\mathbb{R} {( x , x + h ) } h > 0 , x ∈ R M + M^+ M + [15 Definition 1.1] ; this was already observed by Sawyer [24 ] , as noted in [15 p. 55] . Moreover, L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R ) [15 Theorem 1.2, (i) $\Rightarrow$ (ii)] , it suffices to prove that
lim ε → 0 + ε ∥ M 1 − ε + ∥ L p ( ⋅ ) → L p ( ⋅ ) = 0. (21) \lim_{\varepsilon\to0^+}\varepsilon\left\|M^+_{1-\varepsilon}\right\|_{L^{p(\cdot)}\to L^{p(\cdot)}}=0. \tag*{(21)} ε → 0 + lim ε M 1 − ε + L p ( ⋅ ) → L p ( ⋅ ) = 0. (21) Set r = 1 − ε r=1-\varepsilon r = 1 − ε s = 1 / r s=1/r s = 1/ r
∥ M r + f ∥ p ( ⋅ ) = ∥ M + ( ∣ f ∣ r ) ∥ s p ( ⋅ ) , ∥ ∣ f ∣ r ∥ s p ( ⋅ ) = ∥ f ∥ p ( ⋅ ) r . \left\|M^+_r f\right\|_{p(\cdot)}=\left\|M^+\left(|f|^r\right)\right\|_{sp(\cdot)},\qquad\left\||f|^r\right\|_{sp(\cdot)}=\|f\|^r_{p(\cdot)}. M r + f p ( ⋅ ) = M + ( ∣ f ∣ r ) s p ( ⋅ ) , ∥ ∣ f ∣ r ∥ s p ( ⋅ ) = ∥ f ∥ p ( ⋅ ) r . Since every non-negative function in L s p ( ⋅ ) L^{sp(\cdot)} L s p ( ⋅ ) ∣ f ∣ r |f|^r ∣ f ∣ r f ∈ L p ( ⋅ ) f\in L^{p(\cdot)} f ∈ L p ( ⋅ )
∥ M r + ∥ L p ( ⋅ ) → L p ( ⋅ ) = ( ∥ M + ∥ L s p ( ⋅ ) → L s p ( ⋅ ) ) s . (22) \left\|M^+_r\right\|_{L^{p(\cdot)}\to L^{p(\cdot)}}=\left(\left\|M^+\right\|_{L^{sp(\cdot)}\to L^{sp(\cdot)}}\right)^s. \tag*{(22)} M r + L p ( ⋅ ) → L p ( ⋅ ) = ( M + L s p ( ⋅ ) → L s p ( ⋅ ) ) s . (22) Using ε = ( s − 1 ) / s \varepsilon=(s-1)/s ε = ( s − 1 ) / s
ε ∥ M 1 − ε + ∥ L p ( ⋅ ) → L p ( ⋅ ) ≤ C s s ( s − 1 ) 1 − 1 / p − . \varepsilon\left\|M^+_{1-\varepsilon}\right\|_{L^{p(\cdot)}\to L^{p(\cdot)}}\leq\frac{C^s}{s}(s-1)^{1-1/p_-}. ε M 1 − ε + L p ( ⋅ ) → L p ( ⋅ ) ≤ s C s ( s − 1 ) 1 − 1/ p − . The right-hand side tends to zero as s → 1 + s\to1^+ s → 1 + p − > 1 p_->1 p − > 1
Proof of Theorem 1.1. We are now ready to prove Theorem 1.1 . The argument follows the general strategy of [3 Theorem 4.52] . In place of the usual Calderón–Zygmund decomposition, we use the one-sided level-set decomposition for M + M^+ M + [16 p. 525, proof of Theorem 2, equation (5.1)] , with g ≡ 1 g\equiv1 g ≡ 1
We shall work with 1 < s < s 0 1<s<s_0 1 < s < s 0 s 0 > 1 s_0>1 s 0 > 1 b 1 = 1 / 2 b_1=1/2 b 1 = 1/2 b 2 b_2 b 2 s s s s 0 s_0 s 0 min { s 0 , 2 } \min\{s_0,2\} min { s 0 , 2 } s 0 ≤ 2 s_0\leq2 s 0 ≤ 2 [3 Theorem 4.52] , we may reduce to the case in which f f f
∥ f ∥ s p ( ⋅ ) = 1. \|f\|_{sp(\cdot)}=1. ∥ f ∥ s p ( ⋅ ) = 1. We decompose f = f 1 + f 2 f=f_1+f_2 f = f 1 + f 2
f 1 = f χ { f > 1 } , f 2 = f χ { f ≤ 1 } . f_1=f\chi_{\{f>1\}},\qquad f_2=f\chi_{\{f\leq1\}}. f 1 = f χ { f > 1 } , f 2 = f χ { f ≤ 1 } . Since f 1 > 1 f_1>1 f 1 > 1 s > 1 s>1 s > 1
∥ f 1 ∥ p ( ⋅ ) ≤ ∥ f 1 ∥ s p ( ⋅ ) ≤ 1. \|f_1\|_{p(\cdot)}\leq\|f_1\|_{sp(\cdot)}\leq1. ∥ f 1 ∥ p ( ⋅ ) ≤ ∥ f 1 ∥ s p ( ⋅ ) ≤ 1. We first estimate the large part f 1 f_1 f 1 A > 1 A>1 A > 1 k ≥ 0 k\geq0 k ≥ 0
Ω k + = { x ∈ R : M + f 1 ( x ) > A k } . \Omega_k^+=\{x\in\mathbb{R}:M^+f_1(x)>A^k\}. Ω k + = { x ∈ R : M + f 1 ( x ) > A k } . By this one-sided level-set decomposition,
Ω k + = ⋃ j I j k , I j k = ( a j k , b j k ) , \Omega_k^+=\bigcup_j I_j^k,\qquad I_j^k=(a_j^k,b_j^k), Ω k + = j ⋃ I j k , I j k = ( a j k , b j k ) , where the intervals I j k I_j^k I j k
1 ∣ I j k ∣ ∫ I j k f 1 ( y ) d y = A k . (23) \frac{1}{|I_j^k|}\int_{I_j^k}f_1(y)\,\mathrm{d}y=A^k. \tag*{(23)} ∣ I j k ∣ 1 ∫ I j k f 1 ( y ) d y = A k . (23) Moreover, if x ∈ I j k x\in I_j^k x ∈ I j k
1 b j k − x ∫ x b j k f 1 ( y ) d y > A k . (24) \frac{1}{b_j^k-x}\int_x^{b_j^k}f_1(y)\,\mathrm{d}y>A^k. \tag*{(24)} b j k − x 1 ∫ x b j k f 1 ( y ) d y > A k . (24) This is the one-dimensional one-sided substitute for the Calderón–Zygmund cubes used in the proof of the two-sided result; see, for instance, [24 , 16 , 17 ] . We shall also use the level pieces
E j k = I j k ∩ { M + f 1 ≤ A k + 1 } . E_j^k=I_j^k\cap\{M^+f_1\le A^{k+1}\}. E j k = I j k ∩ { M + f 1 ≤ A k + 1 } . Since
Ω k + ∖ Ω k + 1 + = ⋃ j E j k , \Omega_k^+\setminus\Omega_{k+1}^+=\bigcup_j E_j^k, Ω k + ∖ Ω k + 1 + = j ⋃ E j k , the sets E j k E_j^k E j k j j j k k k
∣ Ω k + 1 + ∩ I j k ∣ ≤ 1 A ∣ I j k ∣ . (25) \left|\Omega_{k+1}^+\cap I_j^k\right|\le\frac{1}{A}|I_j^k|. \tag*{(25)} Ω k + 1 + ∩ I j k ≤ A 1 ∣ I j k ∣. (25) Indeed,
A k + 1 ∣ Ω k + 1 + ∩ I j k ∣ = ∑ I i k + 1 ⊂ I j k A k + 1 ∣ I i k + 1 ∣ = ∑ I i k + 1 ⊂ I j k ∫ I i k + 1 f 1 ( y ) d y ≤ ∫ I j k f 1 ( y ) d y = A k ∣ I j k ∣ . \begin{aligned}
A^{k+1}\left|\Omega_{k+1}^+\cap I_j^k\right|
=\sum_{I_i^{k+1}\subset I_j^k}A^{k+1}|I_i^{k+1}| \\
&=\sum_{I_i^{k+1}\subset I_j^k}\int_{I_i^{k+1}}f_1(y)\,dy \\
&\le\int_{I_j^k}f_1(y)\,dy=A^k|I_j^k|.
\end{aligned} A k + 1 Ω k + 1 + ∩ I j k = I i k + 1 ⊂ I j k ∑ A k + 1 ∣ I i k + 1 ∣ = I i k + 1 ⊂ I j k ∑ ∫ I i k + 1 f 1 ( y ) d y ≤ ∫ I j k f 1 ( y ) d y = A k ∣ I j k ∣. Consequently,
∣ E j k ∣ ≥ ( 1 − 1 A ) ∣ I j k ∣ . (26) |E_j^k|\ge\left(1-\frac{1}{A}\right)|I_j^k|. \tag*{(26)} ∣ E j k ∣ ≥ ( 1 − A 1 ) ∣ I j k ∣. (26) This is the one-sided analogue of the large-subset property used in the classical proof.
We now work inside a fixed component
I = I j k = ( a , b ) , λ = A k . I=I_j^k=(a,b),\qquad\lambda=A^k. I = I j k = ( a , b ) , λ = A k . The following geometric partition is chosen so that Lemmas 5.3 and 5.1 can be applied successively. Let
ℓ 0 = b − a 3 , ℓ n + 1 = 2 3 ℓ n , n ≥ 0 , \ell_0=\frac{b-a}{3},\qquad\ell_{n+1}=\frac{2}{3}\ell_n,\qquad n\ge0, ℓ 0 = 3 b − a , ℓ n + 1 = 3 2 ℓ n , n ≥ 0 , and define
L n = ( b − 3 ℓ n , b − 2 ℓ n ) , M n = ( b − 2 ℓ n , b − ℓ n ) , R n = ( b − ℓ n , b ) . L_n=(b-3\ell_n,b-2\ell_n),\qquad M_n=(b-2\ell_n,b-\ell_n),\qquad R_n=(b-\ell_n,b). L n = ( b − 3 ℓ n , b − 2 ℓ n ) , M n = ( b − 2 ℓ n , b − ℓ n ) , R n = ( b − ℓ n , b ) . For each n n n ℓ n \ell_n ℓ n { L n } n ≥ 0 \{L_n\}_{n\ge0} { L n } n ≥ 0
I = ⋃ n = 0 ∞ L n I=\bigcup_{n=0}^{\infty}L_n I = n = 0 ⋃ ∞ L n up to a set of measure zero.
Set
t n = 1 2 ℓ n ∫ R n f 1 ( y ) d y . (27) t_n=\frac{1}{2\ell_n}\int_{R_n}f_1(y)\,dy. \tag*{(27)} t n = 2 ℓ n 1 ∫ R n f 1 ( y ) d y . (27) By (24) , applied at the left endpoint of R n R_n R n
1 ℓ n ∫ R n f 1 ( y ) d y > λ , \frac{1}{\ell_n}\int_{R_n} f_1(y)\,\mathrm{d}y > \lambda, ℓ n 1 ∫ R n f 1 ( y ) d y > λ , and therefore
t n > λ 2 ≥ 1 2 . (28) t_n > \frac{\lambda}{2} \ge\frac{1}{2}. \tag*{(28)} t n > 2 λ ≥ 2 1 . (28) We next verify the upper bound required in Lemma 5.3. By Hölder’s inequality and ∥ f 1 ∥ p ( ⋅ ) ≤ 1 \lVert f_1\rVert_{p(\cdot)} \le1 ∥ f 1 ∥ p ( ⋅ ) ≤ 1
t n = 1 2 ℓ n ∫ R n f 1 ( y ) d y ≤ C 2 ℓ n ∥ χ R n ∥ p ′ ( ⋅ ) . t_n = \frac{1}{2\ell_n}\int_{R_n} f_1(y)\,\mathrm{d}y \le\frac{C}{2\ell_n}\lVert\chi_{R_n}\rVert_{p'(\cdot)}. t n = 2 ℓ n 1 ∫ R n f 1 ( y ) d y ≤ 2 ℓ n C ∥ χ R n ∥ p ′ ( ⋅ ) . Apply the K 0 + K_0^+ K 0 +
b − 3 ℓ n < b − ℓ n < b . b-3\ell_n < b-\ell_n < b. b − 3 ℓ n < b − ℓ n < b . The corresponding left interval agrees, up to a set of measure zero, with L n ∪ M n L_n\cup M_n L n ∪ M n R n R_n R n
∥ χ L n ∪ M n ∥ p ( ⋅ ) ∥ χ R n ∥ p ′ ( ⋅ ) ≤ 3 [ p ] K 0 + ℓ n . \lVert\chi_{L_n\cup M_n}\rVert_{p(\cdot)}\lVert\chi_{R_n}\rVert_{p'(\cdot)} \le3[p]K_0^+\ell_n. ∥ χ L n ∪ M n ∥ p ( ⋅ ) ∥ χ R n ∥ p ′ ( ⋅ ) ≤ 3 [ p ] K 0 + ℓ n . Consequently,
t n ≤ C ∥ χ L n ∪ M n ∥ p ( ⋅ ) . (29) t_n \le\frac{C}{\lVert\chi_{L_n\cup M_n}\rVert_{p(\cdot)}}. \tag*{(29)} t n ≤ ∥ χ L n ∪ M n ∥ p ( ⋅ ) C . (29) Together with (28) , this verifies the hypotheses of Lemma 5.3 for
I 0 − = L n , I 0 + = M n , I_0^- = L_n,\qquad I_0^+ = M_n, I 0 − = L n , I 0 + = M n , with b 1 = 1 / 2 b_1=1/2 b 1 = 1/2 b 2 > 0 b_2>0 b 2 > 0 j j j k k k n n n s s s L n L_n L n M n M_n M n
( ∫ L n t n s p ( x ) d x ) 1 / s ≤ C ∫ M n t n p ( x ) d x . (30) \left(\int_{L_n}t_n^{s p(x)}\,\mathrm{d}x\right)^{1/s}\le C\int_{M_n}t_n^{p(x)}\,\mathrm{d}x. \tag*{(30)} ( ∫ L n t n s p ( x ) d x ) 1/ s ≤ C ∫ M n t n p ( x ) d x . (30) On the other hand, from the definition of t n t_n t n
t n = 1 ∣ M n ∣ + ∣ R n ∣ ∫ R n f 1 ( y ) d y . t_n=\frac{1}{\lvert M_n\rvert+\lvert R_n\rvert}\int_{R_n}f_1(y)\,\mathrm{d}y. t n = ∣ M n ∣ + ∣ R n ∣ 1 ∫ R n f 1 ( y ) d y . Therefore Lemma 5.1 , applied to the triple
b − 2 ℓ n < b − ℓ n < b , b-2\ell_n < b-\ell_n < b, b − 2 ℓ n < b − ℓ n < b , gives
∫ M n t n p ( x ) d x ≤ C ∫ R n f 1 ( y ) p ( y ) d y . (31) \int_{M_n}t_n^{p(x)}\,\mathrm{d}x\le C\int_{R_n}f_1(y)^{p(y)}\,\mathrm{d}y. \tag*{(31)} ∫ M n t n p ( x ) d x ≤ C ∫ R n f 1 ( y ) p ( y ) d y . (31) Since
∣ L n ∣ = ∣ M n ∣ = ∣ R n ∣ = ℓ n , \lvert L_n\rvert=\lvert M_n\rvert=\lvert R_n\rvert=\ell_n, ∣ L n ∣ = ∣ M n ∣ = ∣ R n ∣ = ℓ n , (30) and (31) imply
∫ L n t n s p ( x ) d x ≤ C ℓ n ( ∫ R n f 1 ( y ) p ( y ) d y ) s . (32) \int_{L_n}t_n^{s p(x)}\,\mathrm{d}x\le C\ell_n\left(\int_{R_n}f_1(y)^{p(y)}\,\mathrm{d}y\right)^s. \tag*{(32)} ∫ L n t n s p ( x ) d x ≤ C ℓ n ( ∫ R n f 1 ( y ) p ( y ) d y ) s . (32) Using (28) , we conclude that
∫ L n λ s p ( x ) d x ≤ C ℓ n ( ∫ R n f 1 ( y ) p ( y ) d y ) s . (33) \int_{L_n}\lambda^{s p(x)}\,\mathrm{d}x\le C\ell_n\left(\int_{R_n}f_1(y)^{p(y)}\,\mathrm{d}y\right)^s. \tag*{(33)} ∫ L n λ s p ( x ) d x ≤ C ℓ n ( ∫ R n f 1 ( y ) p ( y ) d y ) s . (33) Thus (33) is the one-sided counterpart of the combined use of the modular and reverse Hölder estimates in the proof of [3 Theorem 4.52] .
It remains to sum these local inequalities over the level intervals. The intervals I j k I_j^k I j k k k k k k k
For each k ≥ 0 k \ge0 k ≥ 0
F k = { x ∈ R : M + f 1 ( x ) ≤ A k + 2 } . F_k = \{x \in\mathbb{R} : M^+f_1(x) \le A^{k+2}\}. F k = { x ∈ R : M + f 1 ( x ) ≤ A k + 2 } . Fix n n n E j k ∩ L n ≠ ∅ E_j^k \cap L_n \ne\varnothing E j k ∩ L n = ∅ x n ∗ ∈ E j k ∩ L n x_n^* \in E_j^k \cap L_n x n ∗ ∈ E j k ∩ L n M + f 1 ( x n ∗ ) ≤ A k + 1 M^+f_1(x_n^*) \le A^{k+1} M + f 1 ( x n ∗ ) ≤ A k + 1 v = f 1 v=f_1 v = f 1 x = x n ∗ x=x_n^* x = x n ∗ y = b − ℓ n y=b-\ell_n y = b − ℓ n λ 1 = A k + 1 \lambda_1=A^{k+1} λ 1 = A k + 1 λ 2 = A k + 2 \lambda_2=A^{k+2} λ 2 = A k + 2
We define
J n = ( x n ∗ , b − ℓ n ) . J_n = (x_n^*,b-\ell_n). J n = ( x n ∗ , b − ℓ n ) . Because x n ∗ ∈ L n x_n^* \in L_n x n ∗ ∈ L n ∣ J n ∣ ≥ ℓ n |J_n| \ge\ell_n ∣ J n ∣ ≥ ℓ n
∣ F k ∩ J n ∣ ≥ ( 1 − 1 A ) ∣ J n ∣ ≥ ( 1 − 1 A ) ℓ n . (34) |F_k \cap J_n| \ge\left(1-\frac{1}{A}\right)|J_n| \ge\left(1-\frac{1}{A}\right)\ell_n. \tag*{(34)} ∣ F k ∩ J n ∣ ≥ ( 1 − A 1 ) ∣ J n ∣ ≥ ( 1 − A 1 ) ℓ n . (34) Set g ( y ) = f 1 ( y ) p ( y ) g(y)=f_1(y)^{p(y)} g ( y ) = f 1 ( y ) p ( y ) z ∈ J n z \in J_n z ∈ J n ( z , b ) (z,b) ( z , b ) R n R_n R n b − z ≤ 3 ℓ n b-z \le3\ell_n b − z ≤ 3 ℓ n
M + g ( z ) ≥ 1 b − z ∫ R n g ( y ) d y ≥ 1 3 ∫ R n g ( y ) d y . M^+g(z) \ge\frac{1}{b-z}\int_{R_n}g(y)\,dy \ge\frac{1}{3}\int_{R_n}g(y)\,dy. M + g ( z ) ≥ b − z 1 ∫ R n g ( y ) d y ≥ 3 1 ∫ R n g ( y ) d y . Using (33) and (34), we obtain
∫ E j k ∩ L n λ s p ( x ) d x ≤ C ℓ n ( ∫ R n g ( y ) d y ) s ≤ C ∫ F k ∩ J n ( M + g ( z ) ) s d z . (35) \int_{E_j^k\cap L_n}\lambda^{sp(x)}\,dx
\le C\ell_n\left(\int_{R_n}g(y)\,dy\right)^s
\le C\int_{F_k\cap J_n}(M^+g(z))^s\,dz. \tag*{(35)} ∫ E j k ∩ L n λ s p ( x ) d x ≤ C ℓ n ( ∫ R n g ( y ) d y ) s ≤ C ∫ F k ∩ J n ( M + g ( z ) ) s d z . (35) The intervals J n J_n J n
J n ⊂ ( b − 3 ℓ n , b − ℓ n ) , J_n \subset(b-3\ell_n,b-\ell_n), J n ⊂ ( b − 3 ℓ n , b − ℓ n ) , and since ℓ n + 1 = 2 3 ℓ n \ell_{n+1}=\frac{2}{3}\ell_n ℓ n + 1 = 3 2 ℓ n
∑ n χ J n ≤ C . \sum_n \chi_{J_n}\le C. n ∑ χ J n ≤ C . Summing (35) in n n n
∫ E j k ( A k ) s p ( x ) d x ≤ C ∫ F k ∩ I j k ( M + g ( z ) ) s d z . (36) \int_{E_j^k}(A^k)^{sp(x)}\,dx
\le C\int_{F_k\cap I_j^k}(M^+g(z))^s\,dz. \tag*{(36)} ∫ E j k ( A k ) s p ( x ) d x ≤ C ∫ F k ∩ I j k ( M + g ( z ) ) s d z . (36) For fixed k k k I j k I_j^k I j k
∑ j ∫ E j k ( A k ) s p ( x ) d x ≤ C ∫ Ω k + ∩ F k ( M + g ( z ) ) s d z . \sum_j\int_{E_j^k}(A^k)^{sp(x)}\,dx
\le C\int_{\Omega_k^+\cap F_k}(M^+g(z))^s\,dz. j ∑ ∫ E j k ( A k ) s p ( x ) d x ≤ C ∫ Ω k + ∩ F k ( M + g ( z ) ) s d z . But
Ω k + ∩ F k ⊂ { A k < M + f 1 ≤ A k + 2 } . \Omega_k^+\cap F_k\subset\{A^k<M^+f_1\le A^{k+2}\}. Ω k + ∩ F k ⊂ { A k < M + f 1 ≤ A k + 2 } . These bands have uniformly bounded overlap as k k k
∑ k , j ∫ E j k ( A k ) s p ( x ) d x ≤ C ∫ R [ M + ( f 1 p ( ⋅ ) ) ( x ) ] s d x . (37) \sum_{k,j}\int_{E_j^k}(A^k)^{sp(x)}\,dx
\le C\int_{\mathbb{R}}\left[M^+\left(f_1^{p(\cdot)}\right)(x)\right]^s\,dx. \tag*{(37)} k , j ∑ ∫ E j k ( A k ) s p ( x ) d x ≤ C ∫ R [ M + ( f 1 p ( ⋅ ) ) ( x ) ] s d x . (37) We now use the classical L s L^{s} L s s s s 1 < s ≤ 2 1<s\le2 1 < s ≤ 2
∫ R ( M + g ( x ) ) s d x ≤ C s − 1 ∫ R ∣ g ( x ) ∣ s d x (38) \int_{\mathbb{R}}(M^{+}g(x))^{s}\,dx\le\frac{C}{s-1}\int_{\mathbb{R}}|g(x)|^{s}\,dx \tag*{(38)} ∫ R ( M + g ( x ) ) s d x ≤ s − 1 C ∫ R ∣ g ( x ) ∣ s d x (38) where C C C s s s
For completeness, this estimate follows directly from the weak type ( 1 , 1 ) (1,1) ( 1 , 1 ) λ > 0 \lambda>0 λ > 0
M + ( g χ { ∣ g ∣ ≤ λ / 2 } ) ≤ λ 2 , M^{+}(g\chi_{\{|g|\le\lambda/2\}})\le\frac{\lambda}{2}, M + ( g χ { ∣ g ∣ ≤ λ /2 } ) ≤ 2 λ , and hence
{ M + g > λ } ⊂ { M + ( g χ { ∣ g ∣ > λ / 2 } ) > λ 2 } . \{M^{+}g>\lambda\}\subset\left\{M^{+}(g\chi_{\{|g|>\lambda/2\}})>\frac{\lambda}{2}\right\}. { M + g > λ } ⊂ { M + ( g χ { ∣ g ∣ > λ /2 } ) > 2 λ } . The weak type ( 1 , 1 ) (1,1) ( 1 , 1 )
∣ { M + g > λ } ∣ ≤ C λ ∫ { ∣ g ∣ > λ / 2 } ∣ g ( y ) ∣ d y . |\{M^{+}g>\lambda\}|\le\frac{C}{\lambda}\int_{\{|g|>\lambda/2\}}|g(y)|\,dy. ∣ { M + g > λ } ∣ ≤ λ C ∫ { ∣ g ∣ > λ /2 } ∣ g ( y ) ∣ d y . Using the distribution-function formula and Fubini’s theorem,
∫ R ( M + g ( x ) ) s d x = s ∫ 0 ∞ λ s − 1 ∣ { M + g > λ } ∣ d λ ≤ C s ∫ R ∣ g ( y ) ∣ ∫ 0 2 ∣ g ( y ) ∣ λ s − 2 d λ d y ≤ C s − 1 ∫ R ∣ g ( y ) ∣ s d y , \begin{aligned}
\int_{\mathbb{R}}(M^{+}g(x))^{s}\,dx
&=s\int_{0}^{\infty}\lambda^{s-1}|\{M^{+}g>\lambda\}|\,d\lambda\\
&\le Cs\int_{\mathbb{R}}|g(y)|\int_{0}^{2|g(y)|}\lambda^{s-2}\,d\lambda\,dy\\
&\le\frac{C}{s-1}\int_{\mathbb{R}}|g(y)|^{s}\,dy,
\end{aligned} ∫ R ( M + g ( x ) ) s d x = s ∫ 0 ∞ λ s − 1 ∣ { M + g > λ } ∣ d λ ≤ C s ∫ R ∣ g ( y ) ∣ ∫ 0 2∣ g ( y ) ∣ λ s − 2 d λ d y ≤ s − 1 C ∫ R ∣ g ( y ) ∣ s d y , where in the last inequality we used 1 < s ≤ 2 1<s\le2 1 < s ≤ 2
Applying this estimate to
g = f 1 p ( ⋅ ) , g=f_{1}^{p(\cdot)}, g = f 1 p ( ⋅ ) , we obtain
∫ R [ M + ( f 1 p ( ⋅ ) ) ( x ) ] s d x ≤ C s − 1 ∫ R f 1 ( x ) s p ( x ) d x ≤ C s − 1 . \int_{\mathbb{R}}\left[M^{+}\left(f_{1}^{p(\cdot)}\right)(x)\right]^{s}\,dx\le\frac{C}{s-1}\int_{\mathbb{R}}f_{1}(x)^{sp(x)}\,dx\le\frac{C}{s-1}. ∫ R [ M + ( f 1 p ( ⋅ ) ) ( x ) ] s d x ≤ s − 1 C ∫ R f 1 ( x ) s p ( x ) d x ≤ s − 1 C . Since the sets E j k E_{j}^{k} E j k { M + f 1 > 1 } \{M^{+}f_{1}>1\} { M + f 1 > 1 } M + f 1 ( x ) ≤ A k + 1 M^{+}f_{1}(x)\le A^{k+1} M + f 1 ( x ) ≤ A k + 1 E j k E_{j}^{k} E j k
( M + f 1 ( x ) ) s p ( x ) ≤ A s p + ( A k ) s p ( x ) on E j k . (M^{+}f_{1}(x))^{sp(x)}\le A^{sp_{+}}(A^{k})^{sp(x)}\qquad\text{on }E_{j}^{k}. ( M + f 1 ( x ) ) s p ( x ) ≤ A s p + ( A k ) s p ( x ) on E j k . Since s 0 ≤ 2 s_{0}\le2 s 0 ≤ 2 A s p + A^{sp_{+}} A s p +
∫ { M + f 1 > 1 } ( M + f 1 ( x ) ) s p ( x ) d x ≤ C s − 1 . (39) \int_{\{M^{+}f_{1}>1\}}(M^{+}f_{1}(x))^{sp(x)}\,dx\le\frac{C}{s-1}. \tag*{(39)} ∫ { M + f 1 > 1 } ( M + f 1 ( x ) ) s p ( x ) d x ≤ s − 1 C . (39) The complementary region is simpler. On { M + f 1 ≤ 1 } \{M^{+}f_{1}\le1\} { M + f 1 ≤ 1 } p ( x ) ≥ p − p(x)\ge p_{-} p ( x ) ≥ p −
( M + f 1 ( x ) ) s p ( x ) ≤ ( M + f 1 ( x ) ) s p − . (M^{+}f_{1}(x))^{sp(x)}\le(M^{+}f_{1}(x))^{sp_{-}}. ( M + f 1 ( x ) ) s p ( x ) ≤ ( M + f 1 ( x ) ) s p − . Since s p − > 1 sp_{-}>1 s p − > 1 L s p − L^{sp_{-}} L s p − M + M^{+} M + f 1 > 1 f_{1}>1 f 1 > 1
∫ { M + f 1 ≤ 1 } ( M + f 1 ( x ) ) s p ( x ) d x ≤ C ∫ R f 1 ( x ) s p − d x ≤ C ∫ R f 1 ( x ) s p ( x ) d x ≤ C . \int_{\{M^{+}f_{1}\le1\}}(M^{+}f_{1}(x))^{sp(x)}\,dx\le C\int_{\mathbb{R}}f_{1}(x)^{sp_{-}}\,dx\le C\int_{\mathbb{R}}f_{1}(x)^{sp(x)}\,dx\le C. ∫ { M + f 1 ≤ 1 } ( M + f 1 ( x ) ) s p ( x ) d x ≤ C ∫ R f 1 ( x ) s p − d x ≤ C ∫ R f 1 ( x ) s p ( x ) d x ≤ C . Combining this estimate with (39) , we obtain
∫ R ( M + f 1 ( x ) ) s p ( x ) d x ≤ C s − 1 . (40) \int_{\mathbb{R}} \left(M^{+}f_1(x)\right)^{sp(x)}\,dx \le\frac{C}{s-1}. \tag*{(40)} ∫ R ( M + f 1 ( x ) ) s p ( x ) d x ≤ s − 1 C . (40) We next turn to the small part f 2 f_2 f 2 M + f 2 ≤ M f 2 M^{+}f_2 \le Mf_2 M + f 2 ≤ M f 2 ∥ M f 2 ∥ s p ( ⋅ ) \|Mf_2\|_{sp(\cdot)} ∥ M f 2 ∥ s p ( ⋅ ) s s s M M M L s p ( ⋅ ) ( R ) L^{sp(\cdot)}(\mathbb{R}) L s p ( ⋅ ) ( R ) p ( ⋅ ) p(\cdot) p ( ⋅ ) K 0 K_0 K 0 Example 6.3 ). However, the estimate of M f 2 Mf_2 M f 2 [3 Theorem 4.52] uses only the N ∞ N_\infty N ∞ [3 Theorem 3.16, pp. 104–106] , where, as observed in [3 pp. 131–132] , the decay of the exponent at infinity enters only through the embeddings of L ∞ L^\infty L ∞ p ( ⋅ ) p(\cdot) p ( ⋅ ) p ∞ p_\infty p ∞ N ∞ N_\infty N ∞ [20 ] .
Since s > 1 s>1 s > 1
∣ 1 s p ( x ) − 1 s p ∞ ∣ − 1 = s ∣ 1 p ( x ) − 1 p ∞ ∣ − 1 ≥ ∣ 1 p ( x ) − 1 p ∞ ∣ − 1 , \left|\frac{1}{sp(x)}-\frac{1}{sp_\infty}\right|^{-1}
= s\left|\frac{1}{p(x)}-\frac{1}{p_\infty}\right|^{-1}
\ge\left|\frac{1}{p(x)}-\frac{1}{p_\infty}\right|^{-1}, s p ( x ) 1 − s p ∞ 1 − 1 = s p ( x ) 1 − p ∞ 1 − 1 ≥ p ( x ) 1 − p ∞ 1 − 1 , so s p ( ⋅ ) ∈ N ∞ sp(\cdot)\in N_\infty s p ( ⋅ ) ∈ N ∞ s p ∞ sp_\infty s p ∞ Λ ∞ \Lambda_\infty Λ ∞ p ∞ ∈ [ p − , p + ] p_\infty\in[p_-,p_+] p ∞ ∈ [ p − , p + ] 1 < s < s 0 ≤ 2 1<s<s_0\le2 1 < s < s 0 ≤ 2 s p ∞ sp_\infty s p ∞ [ p − , 2 p + ] ⊂ ( 1 , ∞ ) [p_-,2p_+]\subset(1,\infty) [ p − , 2 p + ] ⊂ ( 1 , ∞ ) M M M L s p ∞ ( R ) L^{sp_\infty}(\mathbb{R}) L s p ∞ ( R ) s s s [3 p. 169] ,
∥ M + f 2 ∥ s p ( ⋅ ) ≤ ∥ M f 2 ∥ s p ( ⋅ ) ≤ C , (41) \|M^{+}f_2\|_{sp(\cdot)}\le\|Mf_2\|_{sp(\cdot)}\le C, \tag*{(41)} ∥ M + f 2 ∥ s p ( ⋅ ) ≤ ∥ M f 2 ∥ s p ( ⋅ ) ≤ C , (41) where C C C p ( ⋅ ) p(\cdot) p ( ⋅ ) s s s
We now combine the estimates for f 1 f_1 f 1 f 2 f_2 f 2 (40) and the modular–norm inequalities,
∥ M + f 1 ∥ s p ( ⋅ ) ≤ C ( s − 1 ) − 1 / ( s p − ) . \|M^{+}f_1\|_{sp(\cdot)}\le C(s-1)^{-1/(sp_-)}. ∥ M + f 1 ∥ s p ( ⋅ ) ≤ C ( s − 1 ) − 1/ ( s p − ) . Indeed, after enlarging C C C 1 < s < s ∗ ≤ 2 1<s<s_*\le2 1 < s < s ∗ ≤ 2 s p ( ⋅ ) sp(\cdot) s p ( ⋅ ) s p − sp_- s p − (41) gives
∥ M + f 2 ∥ s p ( ⋅ ) ≤ C . \|M^{+}f_2\|_{sp(\cdot)}\le C. ∥ M + f 2 ∥ s p ( ⋅ ) ≤ C . Here s ∗ > 1 s_*>1 s ∗ > 1 s s s
By sublinearity, we therefore have
∥ M + f ∥ s p ( ⋅ ) ≤ C ( s − 1 ) − 1 / ( s p − ) \|M^{+}f\|_{sp(\cdot)}\le C(s-1)^{-1/(sp_-)} ∥ M + f ∥ s p ( ⋅ ) ≤ C ( s − 1 ) − 1/ ( s p − ) for every non-negative, bounded, compactly supported function f f f ∥ f ∥ s p ( ⋅ ) = 1 \|f\|_{sp(\cdot)}=1 ∥ f ∥ s p ( ⋅ ) = 1
For an arbitrary f ∈ L s p ( ⋅ ) ( R ) f\in L^{sp(\cdot)}(\mathbb{R}) f ∈ L s p ( ⋅ ) ( R )
f N = min { ∣ f ∣ , N } χ ( − N , N ) . f_N=\min\{|f|,N\}\chi_{(-N,N)}. f N = min { ∣ f ∣ , N } χ ( − N , N ) . Then f N ↑ ∣ f ∣ f_N\uparrow|f| f N ↑ ∣ f ∣ M + f N ↑ M + f M^{+}f_N\uparrow M^{+}f M + f N ↑ M + f ∥ f N ∥ s p ( ⋅ ) ≤ ∥ f ∥ s p ( ⋅ ) \|f_N\|_{sp(\cdot)}\le\|f\|_{sp(\cdot)} ∥ f N ∥ s p ( ⋅ ) ≤ ∥ f ∥ s p ( ⋅ ) [3 Theorem 2.61] yields
∥ M + ∥ L s p ( ⋅ ) → L s p ( ⋅ ) ≤ C ( s − 1 ) − 1 / ( s p − ) ( 1 < s < s ∗ ) . (42) \|M^{+}\|_{L^{sp(\cdot)}\to L^{sp(\cdot)}}\le C(s-1)^{-1/(sp_-)}
\qquad(1<s<s_*). \tag*{(42)} ∥ M + ∥ L s p ( ⋅ ) → L s p ( ⋅ ) ≤ C ( s − 1 ) − 1/ ( s p − ) ( 1 < s < s ∗ ) . (42) Lemma 5.4 now implies that M + M^{+} M + L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R ) Theorem 1.1 . □
Examples
We present three examples illustrating the relation between K 0 K_0 K 0 K 0 + K_0^+ K 0 + N ∞ N_\infty N ∞ K 0 K_0 K 0
[ p ] K 0 : = sup I ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ∣ I ∣ < ∞ , [p]_{K_0}:=\sup_I\frac{\|\chi_I\|_{p(\cdot)}\|\chi_I\|_{p'(\cdot)}}{|I|}<\infty, [ p ] K 0 := I sup ∣ I ∣ ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) < ∞ , where the supremum is taken over all bounded intervals. By monotonicity of the Luxemburg norm, whenever a < b < c a<b<c a < b < c
∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ ∥ χ ( a , c ) ∥ p ( ⋅ ) ∥ χ ( a , c ) ∥ p ′ ( ⋅ ) . \|\chi_{(a,b)}\|_{p(\cdot)}\|\chi_{(b,c)}\|_{p'(\cdot)}\leq\|\chi_{(a,c)}\|_{p(\cdot)}\|\chi_{(a,c)}\|_{p'(\cdot)}. ∥ χ ( a , b ) ∥ p ( ⋅ ) ∥ χ ( b , c ) ∥ p ′ ( ⋅ ) ≤ ∥ χ ( a , c ) ∥ p ( ⋅ ) ∥ χ ( a , c ) ∥ p ′ ( ⋅ ) . Consequently,
K 0 ⊂ K 0 + , [ p ] K 0 + ≤ [ p ] K 0 . (43) K_0\subset K_0^+,\qquad[p]_{K_0^+}\leq[p]_{K_0}. \tag*{(43)} K 0 ⊂ K 0 + , [ p ] K 0 + ≤ [ p ] K 0 . (43) We shall repeatedly use the elementary estimate
∥ χ E ∥ q ( ⋅ ) ≤ ∣ E ∣ 1 / q + ( E ) , 0 < ∣ E ∣ ≤ 1 , (44) \|\chi_E\|_{q(\cdot)}\leq|E|^{1/q^+(E)},\qquad0<|E|\leq1, \tag*{(44)} ∥ χ E ∥ q ( ⋅ ) ≤ ∣ E ∣ 1/ q + ( E ) , 0 < ∣ E ∣ ≤ 1 , (44) where q + ( E ) = ess sup E q q^+(E)=\operatorname{ess\,sup}_E q q + ( E ) = ess sup E q χ E / ∣ E ∣ 1 / q + ( E ) \chi_E/|E|^{1/q^+(E)} χ E /∣ E ∣ 1/ q + ( E ) I I I ∣ I ∣ ≤ 1 |I|\leq1 ∣ I ∣ ≤ 1
∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ∣ I ∣ ≤ ∣ I ∣ 1 / p + ( I ) − 1 / p − ( I ) . (45) \frac{\|\chi_I\|_{p(\cdot)}\|\chi_I\|_{p'(\cdot)}}{|I|}\leq|I|^{1/p^+(I)-1/p^-(I)}. \tag*{(45)} ∣ I ∣ ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ≤ ∣ I ∣ 1/ p + ( I ) − 1/ p − ( I ) . (45) Example 6.1 (The condition K 0 + K_0^+ K 0 + N ∞ N_\infty N ∞ φ ∈ C ∞ ( R ) \varphi\in C^\infty(\mathbb{R}) φ ∈ C ∞ ( R )
0 ≤ φ ≤ 1 , supp φ ⊂ [ − 1 / 2 , 1 / 2 ] , φ = 1 on [ − 1 / 4 , 1 / 4 ] . 0\leq\varphi\leq1,\qquad\operatorname{supp}\varphi\subset[-1/2,1/2],\qquad\varphi=1\ \text{on }[-1/4,1/4]. 0 ≤ φ ≤ 1 , supp φ ⊂ [ − 1/2 , 1/2 ] , φ = 1 on [ − 1/4 , 1/4 ] . Define
p ( x ) = 3 + ∑ k = 1 ∞ φ ( x − e k ) . p(x)=3+\sum_{k=1}^{\infty}\varphi(x-e^k). p ( x ) = 3 + k = 1 ∑ ∞ φ ( x − e k ) . Then
p ( ⋅ ) ∈ K 0 ⊂ K 0 + , p ( ⋅ ) ∉ N ∞ . p(\cdot)\in K_0\subset K_0^+,\qquad p(\cdot)\notin N_\infty. p ( ⋅ ) ∈ K 0 ⊂ K 0 + , p ( ⋅ ) ∈ / N ∞ . This is the exponent from [3 Example 4.51] .
Proof. The supports of the summands are pairwise disjoint. Thus 3 ≤ p ≤ 4 3\leq p\leq4 3 ≤ p ≤ 4 p p p L L L
Fix an interval I I I D = ∣ I ∣ D=|I| D = ∣ I ∣ D ≤ 1 D\leq1 D ≤ 1
p + ( I ) − p − ( I ) ≤ L D , 1 p − ( I ) − 1 p + ( I ) ≤ L D 9 . p_+(I)-p_-(I)\leq LD,\qquad\frac{1}{p_-(I)}-\frac{1}{p_+(I)}\leq\frac{LD}{9}. p + ( I ) − p − ( I ) ≤ L D , p − ( I ) 1 − p + ( I ) 1 ≤ 9 L D . Therefore, by (45),
∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) D ≤ D − L D / 9 ≤ exp ( L 9 e ) . \frac{\|\chi_I\|_{p(\cdot)}\|\chi_I\|_{p'(\cdot)}}{D}\leq D^{-LD/9}\leq\exp\left(\frac{L}{9e}\right). D ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ≤ D − L D /9 ≤ exp ( 9 e L ) . Suppose now that D ≥ 1 D\geq1 D ≥ 1
S = ⋃ k ≥ 1 [ e k − 1 / 2 , e k + 1 / 2 ] . S=\bigcup_{k\geq1}[e^k-1/2,e^k+1/2]. S = k ≥ 1 ⋃ [ e k − 1/2 , e k + 1/2 ] . There is an absolute constant C C C
∣ I ∩ S ∣ ≤ C log ( e + D ) . (46) |I\cap S|\leq C\log(e+D). \tag*{(46)} ∣ I ∩ S ∣ ≤ C log ( e + D ) . (46) Indeed, if the first and last intervals in this union meeting I I I m m m n n n e n − e m ≤ D + 1 e^n-e^m\leq D+1 e n − e m ≤ D + 1
n − m + 1 ≤ 1 + log ( 1 + D + 1 e ) . n-m+1\leq1+\log\left(1+\frac{D+1}{e}\right). n − m + 1 ≤ 1 + log ( 1 + e D + 1 ) . Each of these intervals has length one.
Since p ≥ 3 p\geq3 p ≥ 3 D ≥ 1 D\geq1 D ≥ 1
∫ I ( D − 1 / 3 ) p ( x ) d x ≤ 1 , ∥ χ I ∥ p ( ⋅ ) ≤ D 1 / 3 . \int_I\left(D^{-1/3}\right)^{p(x)}dx\leq1,\qquad\|\chi_I\|_{p(\cdot)}\leq D^{1/3}. ∫ I ( D − 1/3 ) p ( x ) d x ≤ 1 , ∥ χ I ∥ p ( ⋅ ) ≤ D 1/3 . Moreover, 4 / 3 ≤ p ′ ≤ 3 / 2 4/3\leq p'\leq3/2 4/3 ≤ p ′ ≤ 3/2 p ′ = 3 / 2 p'=3/2 p ′ = 3/2 S S S A ≥ 1 A\geq1 A ≥ 1 (46) gives
∫ I ( 1 A D 2 / 3 ) p ′ ( x ) d x ≤ A − 3 / 2 + C A − 4 / 3 D − 8 / 9 log ( e + D ) . \int_I\left(\frac{1}{AD^{2/3}}\right)^{p'(x)}dx\leq A^{-3/2}+CA^{-4/3}D^{-8/9}\log(e+D). ∫ I ( A D 2/3 1 ) p ′ ( x ) d x ≤ A − 3/2 + C A − 4/3 D − 8/9 log ( e + D ) . The function D − 8 / 9 log ( e + D ) D^{-8/9}\log(e+D) D − 8/9 log ( e + D ) D ≥ 1 D\geq1 D ≥ 1 A A A I I I
∥ χ I ∥ p ′ ( ⋅ ) ≤ A D 2 / 3 . \|\chi_I\|_{p'(\cdot)}\leq AD^{2/3}. ∥ χ I ∥ p ′ ( ⋅ ) ≤ A D 2/3 . This proves p ∈ K 0 p\in K_0 p ∈ K 0 p ∈ K 0 + p\in K_0^+ p ∈ K 0 +
Finally, suppose that p ∈ N ∞ p\in N_\infty p ∈ N ∞ p ∞ p_\infty p ∞ p = 3 p=3 p = 3 ( − ∞ , 0 ) (-\infty,0) ( − ∞ , 0 ) p ∞ = 3 p_\infty=3 p ∞ = 3 p = 4 p=4 p = 4
E = ⋃ k ≥ 1 [ e k − 1 / 4 , e k + 1 / 4 ] , ∣ E ∣ = ∞ . E=\bigcup_{k\geq1}[e^k-1/4,e^k+1/4],\qquad|E|=\infty. E = k ≥ 1 ⋃ [ e k − 1/4 , e k + 1/4 ] , ∣ E ∣ = ∞. On this set,
∣ 1 p ( x ) − 1 3 ∣ = 1 12 . \left|\frac{1}{p(x)}-\frac{1}{3}\right|=\frac{1}{12}. p ( x ) 1 − 3 1 = 12 1 . Thus, for every Λ > 0 \Lambda>0 Λ > 0
∫ E exp ( − Λ ∣ 1 p ( x ) − 1 3 ∣ − 1 ) d x = ∫ E e − 12 Λ d x = ∞ . \int_E\exp\left(-\Lambda\left|\frac{1}{p(x)}-\frac{1}{3}\right|^{-1}\right)dx=\int_E e^{-12\Lambda}dx=\infty. ∫ E exp ( − Λ p ( x ) 1 − 3 1 − 1 ) d x = ∫ E e − 12Λ d x = ∞. Consequently, p ∉ N ∞ p\notin N_\infty p ∈ / N ∞
Example 6.2 (An exponent without local log-Hölder continuity). Fix 0 < α < 1 0<\alpha<1 0 < α < 1
r α = exp ( − 3 1 / α ) . r_\alpha=\exp(-3^{1/\alpha}). r α = exp ( − 3 1/ α ) . Define
p α ( x ) = { ( 1 2 + ( log ( 1 / ∣ x ∣ ) ) − α ) − 1 , 0 < ∣ x ∣ < r α , 2 , x = 0 , 6 5 , ∣ x ∣ ≥ r α . p_\alpha(x)=
\begin{cases}
\left(\dfrac{1}{2}+(\log(1/|x|))^{-\alpha}\right)^{-1}, & 0<|x|<r_\alpha,\\
2, & x=0,\\
\dfrac{6}{5}, & |x|\geq r_\alpha.
\end{cases} p α ( x ) = ⎩ ⎨ ⎧ ( 2 1 + ( log ( 1/∣ x ∣ ) ) − α ) − 1 , 2 , 5 6 , 0 < ∣ x ∣ < r α , x = 0 , ∣ x ∣ ≥ r α . Then
p α ∈ K 0 ∩ N ∞ ⊂ K 0 + ∩ N ∞ , p_\alpha\in K_0\cap N_\infty\subset K_0^+\cap N_\infty, p α ∈ K 0 ∩ N ∞ ⊂ K 0 + ∩ N ∞ , but p α p_\alpha p α [3 Example 4.59] , with the constant extension matching the boundary value.
Proof. Write p = p α p=p_{\alpha} p = p α r = r α r=r_{\alpha} r = r α
6 5 ≤ p ( x ) ≤ 2. \frac{6}{5}\leq p(x)\leq2. 5 6 ≤ p ( x ) ≤ 2. Since p = 6 / 5 p=6/5 p = 6/5 ( − r , r ) (-r,r) ( − r , r ) N ∞ N_{\infty} N ∞ p ∞ = 6 / 5 p_{\infty}=6/5 p ∞ = 6/5 0 < ∣ x ∣ < r 0<|x|<r 0 < ∣ x ∣ < r
2 − p ( x ) = 4 ( log ( 1 / ∣ x ∣ ) ) − α 1 + 2 ( log ( 1 / ∣ x ∣ ) ) − α . 2-p(x)=\frac{4(\log(1/|x|))^{-\alpha}}{1+2(\log(1/|x|))^{-\alpha}}. 2 − p ( x ) = 1 + 2 ( log ( 1/∣ x ∣ ) ) − α 4 ( log ( 1/∣ x ∣ ) ) − α . Hence
∣ p ( x ) − p ( 0 ) ∣ log ( 1 / ∣ x ∣ ) ⟶ ∞ as x → 0. |p(x)-p(0)|\log(1/|x|)\longrightarrow\infty\qquad\text{as }x\to0. ∣ p ( x ) − p ( 0 ) ∣ log ( 1/∣ x ∣ ) ⟶ ∞ as x → 0. Thus local log-Hölder continuity fails. It remains to verify K 0 K_0 K 0 ( − r , r ) (-r,r) ( − r , r ) 0 < t < r 0<t<r 0 < t < r
a ( t ) = 1 p ( t ) = 1 2 + ( log ( 1 / t ) ) − α . a(t)=\frac{1}{p(t)}=\frac{1}{2}+(\log(1/t))^{-\alpha}. a ( t ) = p ( t ) 1 = 2 1 + ( log ( 1/ t ) ) − α . The function a a a
a ′ ( t ) = α t ( log ( 1 / t ) ) α + 1 , a'(t)=\frac{\alpha}{t(\log(1/t))^{\alpha+1}}, a ′ ( t ) = t ( log ( 1/ t ) ) α + 1 α , and
a ′ ′ ( t ) = − α ( log ( 1 / t ) − α − 1 ) t 2 ( log ( 1 / t ) ) α + 2 < 0. a''(t)=-\frac{\alpha(\log(1/t)-\alpha-1)}{t^2(\log(1/t))^{\alpha+2}}<0. a ′′ ( t ) = − t 2 ( log ( 1/ t ) ) α + 2 α ( log ( 1/ t ) − α − 1 ) < 0. Let I = ( x , y ) ⊂ ( 0 , r ) I=(x,y)\subset(0,r) I = ( x , y ) ⊂ ( 0 , r ) D = y − x D=y-x D = y − x D ≤ x D\leq x D ≤ x
a ( y ) − a ( x ) ≤ D a ′ ( x ) ≤ D a ′ ( D ) = α ( log ( 1 / D ) ) α + 1 . a(y)-a(x)\leq Da'(x)\leq Da'(D)=\frac{\alpha}{(\log(1/D))^{\alpha+1}}. a ( y ) − a ( x ) ≤ D a ′ ( x ) ≤ D a ′ ( D ) = ( log ( 1/ D ) ) α + 1 α . Since p p p ( 0 , r ) (0,r) ( 0 , r )
∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) D ≤ exp ( ( a ( y ) − a ( x ) ) log ( 1 / D ) ) ≤ e α . \frac{\|\chi_I\|_{p(\cdot)}\|\chi_I\|_{p'(\cdot)}}{D}\leq\exp((a(y)-a(x))\log(1/D))\leq e^{\alpha}. D ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ≤ exp (( a ( y ) − a ( x )) log ( 1/ D )) ≤ e α . To treat intervals close to the origin, we claim that
∥ χ ( 0 , y ) ∥ p ( ⋅ ) ≤ 3 y 1 / p ( y ) , 0 < y ≤ r . (47) \|\chi_{(0,y)}\|_{p(\cdot)}\leq3y^{1/p(y)},\qquad0<y\leq r. \tag*{(47)} ∥ χ ( 0 , y ) ∥ p ( ⋅ ) ≤ 3 y 1/ p ( y ) , 0 < y ≤ r . (47) Set k = log ( 1 / y ) k=\log(1/y) k = log ( 1/ y )
A ( u ) = 1 2 + u − α , λ = y 1 / p ( y ) = e − k A ( k ) . A(u)=\frac{1}{2}+u^{-\alpha},\qquad\lambda=y^{1/p(y)}=e^{-kA(k)}. A ( u ) = 2 1 + u − α , λ = y 1/ p ( y ) = e − k A ( k ) . The substitution t = e − u t=e^{-u} t = e − u
∫ 0 y λ − p ( t ) d t = ∫ k ∞ exp ( − u + k A ( k ) A ( u ) ) d u . \int_0^y \lambda^{-p(t)}\,dt=\int_k^\infty\exp\left(-u+\frac{kA(k)}{A(u)}\right)\,du. ∫ 0 y λ − p ( t ) d t = ∫ k ∞ exp ( − u + A ( u ) k A ( k ) ) d u . For u ≥ k u\geq k u ≥ k
A ( k ) − A ( u ) ≤ α k − α − 1 ( u − k ) , A ( u ) ≥ 1 2 . A(k)-A(u)\leq\alpha k^{-\alpha-1}(u-k),\qquad A(u)\geq\frac{1}{2}. A ( k ) − A ( u ) ≤ α k − α − 1 ( u − k ) , A ( u ) ≥ 2 1 . Since k − α ≤ 1 / 3 k^{-\alpha} \le1/3 k − α ≤ 1/3
− u + k A ( k ) A ( u ) = − ( u − k ) + k ( A ( k ) − A ( u ) ) A ( u ) ≤ − ( 1 − 2 α k − α ) ( u − k ) ≤ − 1 3 ( u − k ) . -u+\frac{kA(k)}{A(u)}=-(u-k)+\frac{k(A(k)-A(u))}{A(u)}
\le-(1-2\alpha k^{-\alpha})(u-k)
\le-\frac{1}{3}(u-k). − u + A ( u ) k A ( k ) = − ( u − k ) + A ( u ) k ( A ( k ) − A ( u )) ≤ − ( 1 − 2 α k − α ) ( u − k ) ≤ − 3 1 ( u − k ) . Consequently,
∫ 0 y λ − p ( t ) d t ≤ 3. \int_{0}^{y}\lambda^{-p(t)}\,dt\le3. ∫ 0 y λ − p ( t ) d t ≤ 3. Since p ≥ 1 p\ge1 p ≥ 1 λ \lambda λ 3 λ 3\lambda 3 λ p ′ p' p ′ ( 0 , r ) (0,r) ( 0 , r )
∥ χ ( 0 , y ) ∥ p ′ ( ⋅ ) ≤ y 1 / p ′ ( y ) . \left\|\chi_{(0,y)}\right\|_{p'(\cdot)}\le y^{1/p'(y)}. χ ( 0 , y ) p ′ ( ⋅ ) ≤ y 1/ p ′ ( y ) . Combining the last two estimates, we obtain
∥ χ ( 0 , y ) ∥ p ( ⋅ ) ∥ χ ( 0 , y ) ∥ p ′ ( ⋅ ) ≤ 3 y . (48) \left\|\chi_{(0,y)}\right\|_{p(\cdot)}\left\|\chi_{(0,y)}\right\|_{p'(\cdot)}\le3y. \tag*{(48)} χ ( 0 , y ) p ( ⋅ ) χ ( 0 , y ) p ′ ( ⋅ ) ≤ 3 y . (48) If I = ( x , y ) ⊂ ( 0 , r ) I=(x,y)\subset(0,r) I = ( x , y ) ⊂ ( 0 , r ) D > x D>x D > x y < 2 D y<2D y < 2 D I I I 6 D 6D 6 D
Evenness handles intervals in ( − r , 0 ) (-r,0) ( − r , 0 ) I I I y = max { ∣ inf I ∣ , ∣ sup I ∣ } y=\max\{|\inf I|,|\sup I|\} y = max { ∣ inf I ∣ , ∣ sup I ∣ } I ⊂ ( − y , y ) I\subset(-y,y) I ⊂ ( − y , y ) y ≤ ∣ I ∣ y\le|I| y ≤ ∣ I ∣
∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ≤ 4 ∥ χ ( 0 , y ) ∥ p ( ⋅ ) ∥ χ ( 0 , y ) ∥ p ′ ( ⋅ ) ≤ 12 ∣ I ∣ . \|\chi_I\|_{p(\cdot)}\|\chi_I\|_{p'(\cdot)}
\le4\|\chi_{(0,y)}\|_{p(\cdot)}\|\chi_{(0,y)}\|_{p'(\cdot)}
\le12|I|. ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ≤ 4∥ χ ( 0 , y ) ∥ p ( ⋅ ) ∥ χ ( 0 , y ) ∥ p ′ ( ⋅ ) ≤ 12∣ I ∣. This proves the required estimate for all intervals contained in ( − r , r ) (-r,r) ( − r , r )
We now check that the estimate persists under the constant extension. The function 1 / p 1/p 1/ p [ r / 2 , ∞ ) [r/2,\infty) [ r /2 , ∞ ) ( − ∞ , − r / 2 ] (-\infty,-r/2] ( − ∞ , − r /2 ] L α L_\alpha L α I I I ( − r , r ) (-r,r) ( − r , r ) D = ∣ I ∣ < r / 2 D=|I|<r/2 D = ∣ I ∣ < r /2
∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) D ≤ D − L α D ≤ e L α / e . \frac{\|\chi_I\|_{p(\cdot)}\|\chi_I\|_{p'(\cdot)}}{D}
\le D^{-L_\alpha D}\le e^{L_\alpha/e}. D ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ≤ D − L α D ≤ e L α / e . For r / 2 ≤ D ≤ 1 r/2\le D\le1 r /2 ≤ D ≤ 1 6 / 5 ≤ p ≤ 2 6/5\le p\le2 6/5 ≤ p ≤ 2
∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) D ≤ D − 1 / 3 ≤ ( r / 2 ) − 1 / 3 . \frac{\|\chi_I\|_{p(\cdot)}\|\chi_I\|_{p'(\cdot)}}{D}
\le D^{-1/3}\le(r/2)^{-1/3}. D ∥ χ I ∥ p ( ⋅ ) ∥ χ I ∥ p ′ ( ⋅ ) ≤ D − 1/3 ≤ ( r /2 ) − 1/3 . Finally, if D ≥ 1 D\ge1 D ≥ 1 p ≥ 6 / 5 p\ge6/5 p ≥ 6/5
∥ χ I ∥ p ( ⋅ ) ≤ D 5 / 6 . \|\chi_I\|_{p(\cdot)}\le D^{5/6}. ∥ χ I ∥ p ( ⋅ ) ≤ D 5/6 . Since p ′ = 6 p'=6 p ′ = 6 ( − r , r ) (-r,r) ( − r , r ) p ′ ≥ 2 p'\ge2 p ′ ≥ 2
∫ I ( 1 2 D 1 / 6 ) p ′ ( x ) d x ≤ 2 − 6 + 2 r 4 D 1 / 3 ≤ 2 − 6 + r 2 < 1. \int_I\left(\frac{1}{2D^{1/6}}\right)^{p'(x)}dx
\le2^{-6}+\frac{2r}{4D^{1/3}}
\le2^{-6}+\frac{r}{2}<1. ∫ I ( 2 D 1/6 1 ) p ′ ( x ) d x ≤ 2 − 6 + 4 D 1/3 2 r ≤ 2 − 6 + 2 r < 1. Thus
∥ χ I ∥ p ′ ( ⋅ ) ≤ 2 D 1 / 6 , \|\chi_I\|_{p'(\cdot)}\le2D^{1/6}, ∥ χ I ∥ p ′ ( ⋅ ) ≤ 2 D 1/6 , which completes the proof that p ∈ K 0 p\in K_0 p ∈ K 0 p ∈ K 0 + p\in K_0^+ p ∈ K 0 +
Example 6.3 (A one-sided exponent outside K 0 K_0 K 0 Define
p ( x ) = { 2 , x ≤ 0 , 3 , 0 < x ≤ 1 , 4 − x , 1 < x < 2 , 2 , x ≥ 2. p(x)=
\begin{cases}
2, & x\le0,\\
3, & 0<x\le1,\\
4-x, & 1<x<2,\\
2, & x\ge2.
\end{cases} p ( x ) = ⎩ ⎨ ⎧ 2 , 3 , 4 − x , 2 , x ≤ 0 , 0 < x ≤ 1 , 1 < x < 2 , x ≥ 2. Then
p ( ⋅ ) ∈ K 0 + ∩ N ∞ , p ( ⋅ ) ∉ K 0 . p(\cdot)\in K_0^+\cap N_\infty,\qquad p(\cdot)\notin K_0. p ( ⋅ ) ∈ K 0 + ∩ N ∞ , p ( ⋅ ) ∈ / K 0 . In particular,
K 0 ∩ N ∞ ⊊ K 0 + ∩ N ∞ . K_0\cap N_\infty\subsetneq K_0^+\cap N_\infty. K 0 ∩ N ∞ ⊊ K 0 + ∩ N ∞ . Proof. Clearly 2 ≤ p ( x ) ≤ 3 2\le p(x)\le3 2 ≤ p ( x ) ≤ 3 p = 2 p=2 p = 2 ( 0 , 2 ) (0,2) ( 0 , 2 ) N ∞ N_\infty N ∞ p ∞ = 2 p_\infty=2 p ∞ = 2
To prove K 0 + K_0^+ K 0 + x ↦ p ( x ) + x x\mapsto p(x)+x x ↦ p ( x ) + x
p ( x ) − p ( y ) ≤ y − x , x < y . (49) p(x)-p(y)\le y-x,\qquad x<y. \tag*{(49)} p ( x ) − p ( y ) ≤ y − x , x < y . (49) Fix a < b < c a<b<c a < b < c
I − = ( a , b ) , I + = ( b , c ) , D = c − a . I^-=(a,b),\qquad I^+=(b,c),\qquad D=c-a. I − = ( a , b ) , I + = ( b , c ) , D = c − a . Suppose first that D ≤ 1 D\le1 D ≤ 1
P = ess sup I − p , Q = ess inf I + p . P=\underset{I^-}{\operatorname{ess\,sup}}\,p,\qquad Q=\underset{I^+}{\operatorname{ess\,inf}}\,p. P = I − ess sup p , Q = I + ess inf p . By (49), P − Q ≤ D P-Q\le D P − Q ≤ D ∣ I ± ∣ ≤ D \lvert I^\pm\rvert\le D ∣ I ± ∣ ≤ D
∥ χ I − ∥ p ( ⋅ ) ≤ D 1 / P , ∥ χ I + ∥ p ′ ( ⋅ ) ≤ D 1 − 1 / Q . \lVert\chi_{I^-}\rVert_{p(\cdot)}\le D^{1/P},\qquad\lVert\chi_{I^+}\rVert_{p'(\cdot)}\le D^{1-1/Q}. ∥ χ I − ∥ p ( ⋅ ) ≤ D 1/ P , ∥ χ I + ∥ p ′ ( ⋅ ) ≤ D 1 − 1/ Q . If P ≤ Q P\le Q P ≤ Q D D D P > Q P>Q P > Q P Q ≥ 4 PQ\ge4 P Q ≥ 4
1 P − 1 Q = − P − Q P Q ≥ − D 4 . \frac{1}{P}-\frac{1}{Q}=-\frac{P-Q}{PQ}\ge-\frac{D}{4}. P 1 − Q 1 = − P Q P − Q ≥ − 4 D . Thus, in either case,
∥ χ I − ∥ p ( ⋅ ) ∥ χ I + ∥ p ′ ( ⋅ ) D ≤ D − D / 4 ≤ e 1 / ( 4 e ) . \frac{\lVert\chi_{I^-}\rVert_{p(\cdot)}\lVert\chi_{I^+}\rVert_{p'(\cdot)}}{D}\le D^{-D/4}\le e^{1/(4e)}. D ∥ χ I − ∥ p ( ⋅ ) ∥ χ I + ∥ p ′ ( ⋅ ) ≤ D − D /4 ≤ e 1/ ( 4 e ) . Suppose next that D ≥ 1 D\ge1 D ≥ 1 p ( ⋅ ) ≥ 2 p(\cdot)\ge2 p ( ⋅ ) ≥ 2
∫ I − ( 1 D ) p ( x ) d x ≤ ∣ I − ∣ D ≤ 1 , \int_{I^-}\left(\frac{1}{\sqrt{D}}\right)^{p(x)}\,dx\le\frac{\lvert I^-\rvert}{D}\le1, ∫ I − ( D 1 ) p ( x ) d x ≤ D ∣ I − ∣ ≤ 1 , and hence
∥ χ I − ∥ p ( ⋅ ) ≤ D . \lVert\chi_{I^-}\rVert_{p(\cdot)}\le\sqrt{D}. ∥ χ I − ∥ p ( ⋅ ) ≤ D . Moreover, p ′ ( ⋅ ) = 2 p'(\cdot)=2 p ′ ( ⋅ ) = 2 ( 0 , 2 ) (0,2) ( 0 , 2 ) p ′ ( ⋅ ) ≥ 3 / 2 p'(\cdot)\ge3/2 p ′ ( ⋅ ) ≥ 3/2
∫ I + ( 1 4 D ) p ′ ( x ) d x ≤ ∣ I + ∖ ( 0 , 2 ) ∣ 16 D + ∣ I + ∩ ( 0 , 2 ) ∣ ( 4 D ) 3 / 2 ≤ 1 16 + 1 4 < 1. \int_{I^+}\left(\frac{1}{4\sqrt{D}}\right)^{p'(x)}\,dx\le\frac{\lvert I^+\setminus(0,2)\rvert}{16D}+\frac{\lvert I^+\cap(0,2)\rvert}{(4\sqrt{D})^{3/2}}\le\frac{1}{16}+\frac{1}{4}<1. ∫ I + ( 4 D 1 ) p ′ ( x ) d x ≤ 16 D ∣ I + ∖ ( 0 , 2 )∣ + ( 4 D ) 3/2 ∣ I + ∩ ( 0 , 2 )∣ ≤ 16 1 + 4 1 < 1. It follows that
∥ χ I + ∥ p ′ ( ⋅ ) ≤ 4 D . \lVert\chi_{I^+}\rVert_{p'(\cdot)}\le4\sqrt{D}. ∥ χ I + ∥ p ′ ( ⋅ ) ≤ 4 D . Combining the two cases proves p ∈ K 0 + p\in K_0^+ p ∈ K 0 +
To see that p ∉ K 0 p \notin K_0 p ∈ / K 0 0 < h < 1 0<h<1 0 < h < 1 I h = ( − h , h ) I_h=(-h,h) I h = ( − h , h ) 3 3 3 ( 0 , h ) (0,h) ( 0 , h ) 2 2 2 ( − h , 0 ) (-h,0) ( − h , 0 )
∥ χ I h ∥ p ( ⋅ ) ≥ ∥ χ ( 0 , h ) ∥ p ( ⋅ ) = h 1 / 3 \|\chi_{I_h}\|_{p(\cdot)}\geq\|\chi_{(0,h)}\|_{p(\cdot)}=h^{1/3} ∥ χ I h ∥ p ( ⋅ ) ≥ ∥ χ ( 0 , h ) ∥ p ( ⋅ ) = h 1/3 and
∥ χ I h ∥ p ′ ( ⋅ ) ≥ ∥ χ ( − h , 0 ) ∥ p ′ ( ⋅ ) = h 1 / 2 . \|\chi_{I_h}\|_{p'(\cdot)}\geq\|\chi_{(-h,0)}\|_{p'(\cdot)}=h^{1/2}. ∥ χ I h ∥ p ′ ( ⋅ ) ≥ ∥ χ ( − h , 0 ) ∥ p ′ ( ⋅ ) = h 1/2 . Consequently,
∥ χ I h ∥ p ( ⋅ ) ∥ χ I h ∥ p ′ ( ⋅ ) ∣ I h ∣ ≥ 1 2 h − 1 / 6 ⟶ ∞ as h ↓ 0. \frac{\|\chi_{I_h}\|_{p(\cdot)}\|\chi_{I_h}\|_{p'(\cdot)}}{|I_h|}\geq\frac{1}{2}h^{-1/6}\longrightarrow\infty\qquad\text{as }h\downarrow0. ∣ I h ∣ ∥ χ I h ∥ p ( ⋅ ) ∥ χ I h ∥ p ′ ( ⋅ ) ≥ 2 1 h − 1/6 ⟶ ∞ as h ↓ 0. Thus p ( ⋅ ) ∉ K 0 p(\cdot)\notin K_0 p ( ⋅ ) ∈ / K 0
Remark 6.4. By Theorem 1.1, the right-sided maximal operator M + M^+ M + L p ( ⋅ ) ( R ) L^{p(\cdot)}(\mathbb{R}) L p ( ⋅ ) ( R ) K 0 K_0 K 0 N ∞ N_\infty N ∞
AI Disclosure
The original ideas underlying this work are the author’s own. During the preparation of this manuscript, the author used Claude and ChatGPT to discuss technical details of the proofs, assist in drafting and checking parts of the mathematical arguments, organize references, and improve the clarity and readability of the exposition. All mathematical statements and proofs in the final manuscript were written, reviewed, and verified by the author, who takes full responsibility for the content of this manuscript.
(L. A. Vallejos) FAMAF, Universidad Nacional de Córdoba, CIEM (CONICET), 5000 Córdoba, Argentina
Email address: lucas.vallejos@unc.edu.ar
