Superposition operator problems of Hölder-Lipschitz spaces

Yeli Niu; Heping Wang

doi:10.1515/dema-2024-0043

Artikel Open Access

Superposition operator problems of Hölder-Lipschitz spaces

Yeli Niu und Heping Wang

Veröffentlicht/Copyright: 27. November 2024

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Abstract

Let f be a function defined on the real line, and T f be the corresponding superposition operator which maps h to T f ( h ) , i.e., T f ( h ) = f ∘ h . In this article, the sufficient and necessary conditions such that T f maps periodic Hölder-Lipschitz spaces H p α into itself with 0 < α < 1 p and 1 p < α < 1 , where α is the smoothness index, are shown. Our result in the case 0 < α < 1 p may be the first result about the superposition operator problems of smooth function space containing unbounded functions.

Keywords: Hölder-Lipschitz spaces; superposition operator problem; locally Lipschitz function; Lipschitz function

MSC 2010: 26A15; 26A16; 47H30

1 Introduction

Given a real-valued function f defined on the real line R , the (autonomous) superposition operator T f is defined by

(1.1) T f h ( x ) = ( f ∘ h ) ( x ) = f ( h ( x ) ) , x ∈ [ a , b ] ,

where h is a real-valued function on [ a , b ] .

Let H ≡ H [ a , b ] be a Banach space of real-valued functions h : [ a , b ] ↦ R equipped with some norm ‖ ⋅ ‖ H . The superposition operator problem (SOP) refers to determining conditions on a function f on R , possibly both necessary and sufficient, under which the superposition operator T f maps H into itself. This problem is well known in many fields of nonlinear analysis and has its applications to operator theory, ordinary or partial differential equations, integral or integro-differential equations, considerations conducted in nonlinear functional analysis, and even physics and engineering [1,2]. The solution to SOP for a given H is sometimes very easy, sometimes highly nontrivial.

For many common Banach spaces H , the SOPs have been solved ([3,4] and references therein). For example, denote by WBV p ( [ a , b ] ) ( 1 ≤ p < ∞ ) the space of all functions of bounded p -variation in Wiener’s sense with finite norm

‖ h ‖ WBV p = ∣ h ( a ) ∣ + Var p W ( h ; [ a , b ] ) ,

where

Var p W ( h ; [ a , b ] ) = sup { I n } ∑ n = 1 ∞ ∣ h ( I n ) ∣ p 1 ⁄ p ,

and the supremum is taken over all sequences { I n } = { [ a n , b n ] } of nonoverlapping intervals in [ a , b ] , h ( I n ) = h ( b n ) − h ( a n ) . Denote by RBV p ( [ a , b ] ) ( 1 ≤ p < ∞ ) the space of all functions of bounded p -variation in Riesz’s sense with finite norm

‖ h ‖ RBV p = ∣ h ( a ) ∣ + Var p R ( h ; [ a , b ] ) ,

where

Var p R ( h ; [ a , b ] ) = sup { I n } ∑ n = 1 ∞ ∣ h ( I n ) ∣ p ∣ b n − a n ∣ p − 1 1 ⁄ p .

Particularly, when p = 1 , the spaces WBV p ( [ a , b ] ) and RBV p ( [ a , b ] ) recede to the well-known space B V ( [ a , b ] ) of functions of bounded variation.

Denote by Lip ( α , C [ a , b ] ) ( 0 < α ≤ 1 ) the space of all Lipschitz continuous functions of α -order with finite norm

‖ h ‖ Lip α = ∣ h ( a ) ∣ + sup x ≠ y , x , y ∈ [ a , b ] ∣ h ( x ) − h ( y ) ∣ ∣ x − y ∣ α .

We remark that when H = WBV p ( [ a , b ] ) [5,6], RBV p ( [ a , b ] ) , 1 ≤ p < ∞ [7], Lip ( α , C [ a , b ] ) [8], and A C [ a , b ] consisting of absolutely continuous functions on [ a , b ] [9], the superposition operator T f maps H into itself if and only if f is locally Lipschitz, i.e., there exists a positive constant K ( r ) depending only on r > 0 such that

(1.2) ∣ f ( x ) − f ( y ) ∣ ≤ K ( r ) ∣ x − y ∣ , ∣ x ∣ , ∣ y ∣ ≤ r .

We denote such f by f ∈ Lip loc . If K ( r ) in (1.2) is replaced by a constant K independent of r , then f is said to be Lipschitz and is denoted by f ∈ Lip .

Note that all the above spaces can be continuously embedded in the space B [ a , b ] consisting of bounded functions. For SOPs of smooth function spaces H [ a , b ] containing unbounded functions, we have few results.

Denote by Lip ( α , L p ( A ) ) the Hölder-Lipschitz spaces consisting of all functions h ∈ L p ( A ) for which

‖ △ t h ‖ p = ∫ A t ∣ h ( x + t ) − h ( x ) ∣ p d x 1 ⁄ p ≤ M t α , t > 0 ,

where A t ≔ [ a , b − t ] , if A = [ a , b ] , 0 < t ≤ b − a , and A t = A , if A is the torus T = { e 2 π i x , x ∈ R } . We remark that the spaces Lip ( α , L p ( A ) ) contain unbounded functions if 0 < α < 1 ⁄ p .

When α = 1 , Lip ( α , L p [ a , b ] ) = W p 1 [ a , b ] = RBV p ( [ a , b ] ) if 1 < p < ∞ ([10, Theorem 9.3] and [11, Chapter IX §4, Theorem 7]), Lip ( α , L p [ a , b ] ) = B V [ a , b ] if p = 1 ([10, Theorem 9.3]), where W p 1 [ a , b ] = { f ∣ f ∈ A C [ a , b ] and f ′ ∈ L p [ a , b ] } . It follows that for α = 1 , the superposition operator T f maps Lip ( α , L p [ a , b ] ) ( 1 ≤ p < ∞ ) into itself if and only if f is locally Lipschitz.

This work is devoted to studying the SOPs of the Hölder-Lipschitz spaces H p α ≡ Lip ( α , L p ( T ) ) , 1 ≤ p < ∞ , 0 < α < 1 . For 1 ⁄ p < α < 1 , 1 < p < ∞ , the space H p α can be embedded into the continuous function space C ( T ) and any function in H p α is bounded. However, for 0 < α ≤ 1 ⁄ p , 1 ≤ p < ∞ , α < 1 , the space H p α contains unbounded functions [12]. The investigation in these two cases is completely different.

Our main results can be formulated as follows.

Theorem 1.1

Let 1 p < α < 1 , 1 < p < ∞ , and let T f be the superposition operator defined by (1.1). Then, T f maps H p α into itself if and only if f is locally Lipschitz.

Theorem 1.2

Let 0 < α < 1 p , 1 ≤ p < ∞ . Then, T f maps H p α into itself if and only if f is Lipschitz.

Theorem 1.3

Let α = 1 p , 1 < p < ∞ . Then, the sufficient condition for which T f maps H p α into itself is that f is Lipschitz, and the necessary but not sufficient condition is that f is locally Lipschitz.

The proofs of the sufficiency parts of Theorems 1.1–1.3 are standard and easy. For the proof of necessity of Theorem 1.1, we first show the continuity of f , and then the locally Lipschitz continuity of f . Our proof of Theorem 1.1 makes essential use of Terekhin’s estimates of L p -moduli of continuity in terms of p -moduli of continuity of a function in H p α . Such idea is proposed and used in [12,13].

For the proof of the necessity part of Theorem 1.2, apart from the one of locally Lipschitz continuity of f , we also show Lipschitz continuity of f in the end. We give and use estimates of L p -moduli of continuity of a step function to prove Theorem 1.2.

Remark 1.4

Theorem 1.2 may be the first result about the SOPs of smooth function space containing unbounded functions.

Remark 1.5

We do not know what is the sufficient and necessary condition for which T f maps H p 1 ⁄ p ( 1 < p < ∞ ) into itself. It is open. We conjecture that the sufficient and necessary condition for which T f maps H p 1 ⁄ p ( 1 < p < ∞ ) into itself is that f is Lipschitz.

2 Proof of Theorem 1.1

Let g be a 1-periodic function on the real line and 1 < p < ∞ . Following Terekhin [14], we define the modulus of p -continuity of g by

(2.1) ω 1 − 1 ⁄ p ( g ; δ ) = sup ‖ ℐ ‖ ≤ δ ∑ n = 1 ∞ ∣ g ( I n ) ∣ p 1 ⁄ p ( 0 < δ ≤ 1 ) ,

where the supremum takes over all sequences ℐ = { I n } = { [ a n , b n ] } of nonoverlapping intervals contained in a period with ‖ ℐ ‖ = sup n ∣ I n ∣ = sup n ∣ b n − a n ∣ ≤ δ . It follows from Hölder’s inequality that if f is absolutely continuous and f ′ ∈ L p , then

(2.2) ω 1 − 1 ⁄ p ( f ; δ ) ≤ δ 1 ⁄ p ′ ‖ f ′ ‖ p ( 0 < δ ≤ 1 ) ,

where 1 p + 1 p ′ = 1 . It is proved in [14, Corollary 1] that f ∈ H p α , 1 ⁄ p < α < 1 if and only if f can be modified on a set of zero measures in such a way that f is continuous and

(2.3) ω 1 − 1 ⁄ p ( f ; δ ) ≪ δ α − 1 ⁄ p ,

where A ≍ B means A ≪ B and A ≫ B , A ≪ B means that there exists a nonessential constant c > 0 such that A ≤ c B , and A ≫ B means B ≪ A . Hence, we assume that every function in H p α , α ∈ ( 1 ⁄ p , 1 ) is continuous.

Figure 1

The graph of g .

Proof of Theorem 1.1

Sufficiency.

Assume that f is locally Lipschitz. Let g be an arbitrary function in H p α , 1 ⁄ p < α < 1 , 1 < p < ∞ . Then, g is continuous and bounded. So there exists a constant r > 0 such that ‖ g ‖ ∞ = sup x ∈ T ∣ g ( x ) ∣ ≤ r . Since the function f is locally Lipschitz, there exists a positive constant k ( r ) such that for any x , y ∈ [ − r , r ] ,

∣ f ( x ) − f ( y ) ∣ ≤ k ( r ) ∣ x − y ∣ .

It follows that

‖ △ t ( f ∘ g ) ( ⋅ ) ‖ p = ‖ f ( g ( ⋅ + t ) ) − f ( g ( ⋅ ) ) ‖ p ≤ k ( r ) ‖ g ( ⋅ + t ) − g ( ⋅ ) ‖ p ≪ ∣ t ∣ α ,

which means that f ∘ g ∈ H p α . Hence, T f maps H p α into itself if f is locally Lipschitz.

Necessity.

Let α ∈ ( 1 ⁄ p , 1 ) , 1 < p < ∞ . Suppose that T f ( h ) = f ∘ h ∈ H p α for any h ∈ H p α . We want to show f ∈ Lip loc . We recall that any function h in H p α ( α > 1 ⁄ p ) is continuous on T . First, we prove that f is continuous on R . Otherwise, f is not continuous at some point x ∈ R . Without loss of generality we may assume that x = 0 , f ( 0 ) = 0 , and there exists a decreasing positive sequence { s n } satisfying s n < 2 − n and f ( s n ) > 1 .

Denote I n = [ 2 − n , 2 − n + 1 ] , n = 1 , 2 , … . Set g ( 0 ) = g ( 2 − n ) = 0 . On each interval I n , we define g ( x ) as a tent function such that g 2 − n + 2 − n + 1 2 = g ( 2 − n ⋅ 3 2 ) = s n . The graph of g is shown in Figure 1.

For any x , y ∈ [ 0 , 1 ] , we assume that x ∈ I n , y ∈ I m . Then, we have ∣ g ( x ) − g ( y ) ∣ ∣ x − y ∣ ≤ max { s n 2 − n ⋅ 1 2 , s m 2 − m ⋅ 1 2 } ≤ 2 . Hence, g ∈ Lip ( 1 , C ( T ) ) ⊆ H p α . However, we also have f ∘ g 2 − n ⋅ 3 2 = f ( s n ) > 1 and f ∘ g ( 0 ) = 0 , which imply that f ∘ g is not continuous at x = 0 . This leads to a contradiction. Hence, f is continuous.

We now prove that f is locally Lipschitz. Assume that f is not locally Lipschitz. Then, there exists a real number r > 0 and a sequence of intervals { [ a n , b n ] } which are subsets of [ − r , r ] and satisfies

(2.4) ∣ f ( b n ) − f ( a n ) ∣ ≥ k n ∣ b n − a n ∣ = 2 n + 2 ∣ b n − a n ∣ .

(We take k n = 2 n + 2 .) From the continuity of f and compactness of [ − r , r ] , we may assume that

max x , y ∈ [ − r , r ] ∣ f ( x ) − f ( y ) ∣ ≤ 1 .

Hence, we have

∣ b n − a n ∣ ≤ 1 k n .

Furthermore, there are a subsequence of { [ a n , b n ] } (without loss of generality, we still denote it as { [ a n , b n ] } ) and a constant ξ satisfying lim n → ∞ a n = lim n → ∞ b n = ξ . If a n < ξ < b n ,

k n ≤ ∣ f ( b n ) − f ( a n ) ∣ ∣ b n − a n ∣ ≤ max ∣ f ( ξ ) − f ( a n ) ∣ ∣ ξ − a n ∣ , ∣ f ( ξ ) − f ( b n ) ∣ ∣ ξ − b n ∣ .

Thus, we can assume that ξ < a n < b n . Considering subsequence if necessary, we may assume that

ξ < a n < b n < a n − 1 < b n − 1 ,

(2.5) ∣ a n − ξ ∣ ≤ 1 k n = 2 − n − 2 ,

and

b n − a n < b n − 1 − a n − 1 < 1 k n − 1 .

Set

(2.6) S n = ( b n − a n ) 1 − 1 α , l n = 1 k n ( b n − a n ) S n ≍ 1 k n ( b n − a n ) 1 ⁄ α ,

and

(2.7) δ n = A b n − a n 2 S n ≍ ( b n − a n ) 1 α , l 0 = 0 , δ 0 = 1 , a 0 = a 1 ,

where [ x ] denotes the largest integer not exceeding x , A is a positive constant satisfying

∑ n = 1 ∞ 2 l n δ n = ∑ n = 1 ∞ A ( b n − a n ) 1 k n ( b n − a n ) = 1 ⁄ 4 .

(The above equality is required just to ensure that the function g 1 , which is defined below, belongs to Lip.) The existence of A is guaranteed by the following inequality:

∑ n = 1 ∞ ( b n − a n ) 1 k n ( b n − a n ) ≤ ∑ n = 1 ∞ 1 k n = 1 ⁄ 4 .

We denote

J n 1 = ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) , ∑ j = 0 n ( 2 l j δ j + a j − a j + 1 ) ,

J n , 1 1 = ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) , ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) + 2 l n δ n ,

and

J n , 2 1 = ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) + 2 l n δ n , ∑ j = 0 n ( 2 l j δ j + a j − a j + 1 ) .

We define β ( x ) as follows:

β ( x ) = a n , x = ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) + 2 k δ n , 0 ≤ k ≤ l n , b n , x = ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) + ( 2 k + 1 ) δ n , 0 ≤ k ≤ l n − 1 , ξ , x = ∑ j = 0 ∞ ( 2 l j δ j + a j − a j + 1 ) = 1 ⁄ 4 + a 1 − ξ , a 1 , x = 1 linear , otherwise .

The graph of β is shown in Figure 2.

Let

β 2 ( x ) = ∑ n = 1 ∞ ( β ( x ) − a n ) χ J n , 1 1 ( x ) , β 1 ( x ) = β ( x ) − β 2 ( x ) ,

where χ A denotes the characteristic function of a set A . Then,

β 1 ( x ) = a n , x ∈ J n , 1 1 , ξ , x = 1 ⁄ 4 + a 1 − ξ , a 1 , x = 1 linear , otherwise .

The graph of β 1 is shown in Figure 3.

Since a 1 − ξ 1 − ( a 1 − ξ + 1 ⁄ 4 ) ≤ 1 (by (2.5)) and the length of J n , 2 1 is a n − a n + 1 , we obtain that for any x , y ∈ [ 0 , 1 ] ,

∣ β 1 ( x ) − β 1 ( y ) ∣ ∣ x − y ∣ ≤ 1 .

It follows that

β 1 ∈ Lip ( 1 , C ( T ) ) ⊆ H p α .

Now we show that β 2 ∈ H p α . We note that

β 2 ( x ) = 0 , x = ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) + 2 k δ n , 0 ≤ k ≤ l n , b n − a n , x = ∑ j = 0 n − 1 ( 2 l j δ j + a j − a j + 1 ) + ( 2 k + 1 ) δ n , 0 ≤ k ≤ l n − 1 , 0 , x ∈ J n , 2 1 ∪ [ 1 ⁄ 4 + a 1 − ξ , 1 ] , linear , otherwise .

The graph of β 2 is shown in Figure 4.

For any 0 < δ < 1 , take n 0 such that δ n 0 + 1 < δ ≤ δ n 0 . We set

β 2 1 ( x ) = β 2 ( x ) χ [ 0 , ∑ j = 0 n 0 ( 2 l j δ j + a j − a j + 1 ) ] , β 2 2 ( x ) = β 2 ( x ) − β 2 1 ( x ) .

We recall from (2.6) and (2.7) that

δ n ≍ ( b n − a n ) 1 ⁄ α , l n ≍ 1 k n ( b n − a n ) 1 ⁄ α , and l n δ n ≍ 1 ⁄ k n ≍ 2 − n .

Noting that ( b n − a n ) − 1 ≍ δ n − α ≤ δ − α for n ≤ n 0 and 1 − 1 α < 0 , by (2.2) we obtain

ω 1 − 1 p ( β 2 1 ; δ ) ≤ δ 1 − 1 p ‖ β 2 1 ′ ‖ p = δ 1 − 1 p ∑ n = 1 n 0 b n − a n δ n p δ n ⋅ 2 l n 1 ⁄ p ≍ δ 1 − 1 p ∑ n = 1 n 0 ( b n − a n ) 1 − 1 α p 1 k n 1 p ≤ δ 1 − 1 p δ α − 1 ∑ n = 1 n 0 1 k n 1 p ≪ δ α − 1 p .

For any n > n 0 and p > 1 α , by (2.7) and (2.1) we have ( b n − a n ) ≍ δ n α ≤ δ α and

ω 1 − 1 p ( β 2 2 ; δ ) = sup ‖ ℐ ‖ ≤ δ ∑ n = 1 ∞ ∣ β 2 2 ( I n ) ∣ p 1 p = ∑ n = n 0 + 1 ∞ ( b n − a n ) p ⋅ 2 l n 1 p ≪ ∑ n = n 0 ∞ ( b n − a n ) p − 1 α 1 k n 1 p ≪ δ α − 1 p .

Hence,

ω 1 − 1 p ( β 2 ; δ ) ≤ ω 1 − 1 p ( β 2 1 ; δ ) + ω 1 − 1 p ( β 2 2 ; δ ) ≪ δ α − 1 p ,

which combining with (2.3) implies that β 2 ∈ H p α . Hence, β = β 1 + β 2 ∈ H p α . However, by (2.4) we have

ω 1 − 1 p ( f ∘ β ; δ n ) ≥ ∣ f ( b n ) − f ( a n ) ∣ ⋅ ( 2 l n ) 1 p ≥ δ n α − 1 p k n 1 − 1 p .

Since k n 1 − 1 p → ∞ as n → ∞ , by (2.3) we obtain T f ( β ) = f ∘ β ∉ H p α , which contradicts our assumption.

Theorem 1.1 is proved.□

$Figure 2 The graph of β \beta .$

Figure 2

The graph of β .

$Figure 3 The graph of β 1 {\beta }_{1} .$

Figure 3

The graph of β 1 .

$Figure 4 The graph of β 2 {\beta }_{2} .$

Figure 4

The graph of β 2 .

$Figure 5 The graph of ϕ \phi .$

Figure 5

The graph of ϕ .

$Figure 6 The graph of ψ \psi .$

Figure 6

The graph of ψ .

$Figure 7 The graph of γ \gamma .$

Figure 7

The graph of γ .

3 Proofs of Theorems 1.2 and 1.3

Proof of Theorem 1.2

Sufficiency.

Assume that f is Lipschitz. Then, there exists a constant k > 0 such that for any x , y ∈ R , ∣ f ( x ) − f ( y ) ∣ ≤ k ∣ x − y ∣ . Thus, for any function g ∈ H p α , 0 < α ≤ 1 ⁄ p , 1 ≤ p < ∞ , and t > 0 , we have

‖ △ t ( f ∘ g ) ( ⋅ ) ‖ p = ‖ f ∘ g ( ⋅ + t ) − f ∘ g ( ⋅ ) ‖ p ≤ k ‖ g ( ⋅ + t ) − g ( ⋅ ) ‖ p ≪ t α ,

which means that f ∘ g ∈ H p α . Hence, T f maps H p α , 0 < α ≤ 1 ⁄ p , 1 ≤ p < ∞ into itself if f is Lipschitz.

Necessity.

Let α ∈ ( 0 , 1 ⁄ p ] , 1 ≤ p < ∞ , α < 1 . Suppose that T f ( h ) = f ∘ h ∈ H p α for any h ∈ H p α . We want to show that f ∈ Lip loc if 0 < α ≤ 1 ⁄ p and f ∈ Lip if 0 < α < 1 ⁄ p .

First, we show that f is continuous. Otherwise, f is not continuous at some point x . Without loss of generality, we may assume that x = 0 , f ( 0 ) = 0 , and there exists a nonincreasing sequence { s n } satisfying 0 < s n < 2 − n and f ( s n ) > 1 . We want to construct a function ϕ such that ϕ ∈ H p α but f ∘ ϕ ∉ H p α .

Choose two positive numbers r and η such that

0 < r < p , 0 < η < α r p < p .

Set

(3.1) l n = [ s n − ( 1 − α p ) r − η ] + 1 , t n = s n r A ,

where A is a positive constant such that

∑ n = 1 ∞ 2 l n t n = 1 .

The existence of A is guaranteed by the following inequality:

∑ n = 1 ∞ 2 ( [ s n − ( 1 − α p ) r − η ] + 1 ) s n r ≪ ∑ n = 1 ∞ s n α r p − η ≤ ∑ n = 1 ∞ 2 − n ( α r p − η ) < ∞ .

Set

J n 2 = 1 − ∑ j = 1 n 2 l j t j , 1 − ∑ j = 1 n − 1 2 l j t j .

We divide J n equally into 2 l n subintervals and denote every part by

J n , m 2 = 1 − ∑ j = 1 n 2 l j t j + ( m − 1 ) t n , 1 − ∑ j = 1 n 2 l j t j + m t n , m = 1 , … , 2 l n .

Now, we define ϕ ( x ) as

ϕ ( x ) = 0 , x ∈ J n , 2 j − 1 2 , s n , x ∈ J n , 2 j 2 , n = 1 , 2 , … , j = 1 , … , l n .

Denote ϕ n ( x ) = ϕ ( x ) χ J n ( x ) . The graph of ϕ is shown in Figure 5.

For fixed n ≥ 1 , we estimate ‖ △ t ϕ n ‖ p p , 0 < t < 1 . Noting that t n ≍ s n r , by (3.1) we have for 0 < t < t n ,

(3.2) ‖ △ t ϕ n ‖ p p = ∫ [ 0 , 1 ] ∣ ϕ n ( x + t ) − ϕ n ( x ) ∣ p d x = s n p ⋅ ∣ { x : ∣ ϕ n ( x + t ) − ϕ n ( x ) ∣ = s n } ∣ = s n p ⋅ ( 2 t l n ) ≍ t s n ( p − r ) + ( α p r − η ) ≤ t α p s n ( p − r ) + ( α p r − η ) ,

where ∣ A ∣ denotes the Lebesgue measure of a set A . Similarly for t ≥ t n we have

(3.3) ‖ △ t ϕ n ‖ p p = s n p ⋅ ∣ { x : ∣ ϕ n ( x + t ) − ϕ n ( x ) ∣ = s n } ∣ ≤ s n p ⋅ ( 2 l n t n ) ≍ s n α r p − η + p ≪ t α p s n p − η .

For any t ∈ ( 0 , 1 ) , it follows from (3.2) and (3.3) that

‖ △ t ϕ ‖ p ≤ ∑ n = 1 ∞ ‖ Δ t ϕ n ‖ p = ∑ t n > t ‖ △ t ϕ n ‖ p + ∑ t n ≤ t ‖ △ t ϕ n ‖ p ≪ ∑ t n > t t α s n ( 1 − r p ) + α p r − η p + ∑ t n ≤ t t α s n − η p + 1 ≪ t α ∑ t n > t 2 − n ( 1 − r p ) + α p r − η p + ∑ t n ≤ t 2 − n ( − η p + 1 ) ≪ t α ,

which means that ϕ ∈ H p α .

However, we have

‖ △ t n ( f ∘ ϕ ) ‖ p ≥ ∫ ∪ j = 1 l n J n 2 j − 1 ∣ f ∘ ϕ ( x + t n ) − f ∘ ϕ ( x ) ∣ p d x 1 p = f ( s n ) ( l n t n ) 1 p ≥ ( l n t n ) 1 p ≫ s n α r − η p ≍ t n α s n − η p .

Since s n − η p > 2 n η p → ∞ as n → ∞ , we obtain f ∘ ϕ ∉ H p α . This contradicts our assumption. So f is continuous on R .

Second, we show that f ∈ Lip loc if 0 < α ≤ 1 ⁄ p , 1 ≤ p < ∞ , α < 1 . Assume that f is not locally Lipschitz. As in the proof of Theorem 1.1, we can assume that there exist sequences { a n } , { b n } , and ξ satisfying

(3.4) max { ∣ a n − ξ ∣ , ∣ b n − a n ∣ } ≤ 1 k n = 2 − n ,

ξ < a n < b n < a n − 1 < b n − 1 ,

and

(3.5) 1 ≥ ∣ f ( b n ) − f ( a n ) ∣ ≥ k n ( b n − a n ) = 2 n ( b n − a n ) .

(We take k n = 2 n .) Choose two positive numbers r and η such that

r α > 1 , ( r α − 1 ) p < η < p .

For example, we may take r = 3 2 α , η = 3 p 4 . Set

(3.6) t n = ( b n − a n ) r A , l n = ( b n − a n ) r α p − r − p k n η + 1 , t 0 = l 0 = 0 ,

where A is a positive constant satisfying

∑ n = 1 ∞ 2 l n t n = 1 .

We remark by (3.4) that for n ≥ 1 ,

l n ≥ ( b n − a n ) r ( α p − 1 ) − p k n η ≥ k n r ( − α p + 1 ) + ( p − η ) > 1 .

The existence of A is derived from

∑ n = 1 ∞ 2 ( b n − a n ) r α p − r − p k n η + 1 ( b n − a n ) r ≍ ∑ n = 1 ∞ ( b n − a n ) r α p − p k n η ≤ ∑ n = 1 ∞ k n p ( − r α + 1 ) − η < ∞ .

Denote J n 3 = ∑ j = 0 n − 1 2 l j t j , ∑ j = 0 n 2 l j t j and

J n , k 3 = ∑ j = 0 n − 1 2 l j t j + ( k − 1 ) t n , ∑ j = 0 n − 1 2 l j t j + k t n , ( k = 1 , … , 2 l n , n ≥ 1 ) .

We define ψ ( x ) as follows:

ψ ( x ) = a n , x ∈ J n , 2 j − 1 3 , b n , x ∈ J n , 2 j 3 , n = 1 , 2 , … , j = 1 , … , l n .

The graph of ψ is shown in Figure 6.

Set

ψ 1 ( x ) = ξ + ∑ n = 1 ∞ ( a n − ξ ) χ J n 3 ( x ) , ψ 2 ( x ) = ψ ( x ) − ψ 1 ( x )

and

ψ 1 , n ( x ) = ( a n − ξ ) χ J n 3 ( x ) , ψ 2 , n ( x ) = ψ 2 ( x ) χ J n 3 ( x ) .

We note that for t ∈ ( 0 , 1 ) ,

‖ △ t χ J n 3 ‖ p = ∣ { x : ∣ χ J n 3 ( x + t ) − χ J n 3 ( x ) ∣ = 1 } ∣ 1 p ≤ min { ( 2 t ) 1 ⁄ p , ( 4 l n t n ) 1 ⁄ p } .

It follows that for t ∈ ( 0 , 1 ) ,

‖ △ t ψ 1 ‖ p ≤ ∑ n = 1 ∞ ‖ △ t ψ 1 , n ‖ p ≪ ∑ 2 l n t n > t ( a n − ξ ) t 1 p + ∑ 2 l n t n ≤ t ( a n − ξ ) ( l n t n ) 1 p ≪ ∑ n = 1 ∞ 1 k n t 1 p = t 1 ⁄ p ≪ t α ,

which gives ψ 1 ∈ H p α .

We now prove that ψ 2 ∈ H p α . Similar to the proofs of (3.2) and (3.3), we can show that

‖ △ t ψ 2 , n ‖ p = ( ( b n − a n ) p ∣ { x : ∣ ψ 2 , n ( x + t ) − ψ 2 , n ( x ) ∣ = b n − a n } ∣ ) 1 p ≪ ( ( b n − a n ) p min { l n t , l n t n } ) 1 p .

Noting that t n ≍ ( b n − a n ) r and 0 < α ≤ 1 ⁄ p , by (3.6) we have for 0 < t < t n ,

( b n − a n ) ( l n t ) 1 ⁄ p ≪ t α t n 1 ⁄ p − α ( b n − a n ) r ( α − 1 p ) k n η p ≪ t α k n η p ,

and for 1 > t ≥ t n ,

( b n − a n ) ( l n t n ) 1 ⁄ p ≤ t α , ( b n − a n ) ( l n ) 1 ⁄ p ( t n ) 1 ⁄ p − α ≪ t α k n η p .

Hence, we have

‖ △ t ψ 2 ‖ p ≤ ∑ n = 1 ∞ ‖ △ t ψ 2 , n ‖ p ≤ ∑ t n > t ‖ △ t ψ 2 , n ‖ p + ∑ t n ≤ t ‖ △ t ψ 2 , n ‖ p ≪ ∑ t n > t t α k n η p + ∑ t n ≤ t t α k n η p ≪ t α ,

which implies ψ 2 ∈ H p α . Finally, we obtain ψ = ψ 1 + ψ 2 ∈ H p α .

However, we have

‖ △ t n ( f ∘ ψ ) ‖ p ≥ ∫ ∪ j = 1 l n J n , 2 j − 1 3 ∣ f ∘ ψ ( x + t n ) − f ∘ ψ ( x ) ∣ p d x 1 p ≫ k n ( b n − a n ) ( l n t n ) 1 p ≍ k n 1 − η p t n α .

Since k n 1 − η p → ∞ as n → ∞ , we obtain f ∘ ψ ∉ H p α , which contradicts our assumption. So f is locally Lipschitz.

It remains to show that f is Lipschitz if 0 < α < 1 ⁄ p , 1 ≤ p < ∞ . Assume that f is locally Lipschitz and is not Lipschitz. Then, f ∉ Lip ( − ∞ , − 2 ) or f ∉ Lip ( 2 , ∞ ) , where Lip I = { f ∣ sup x , y ∈ I ∣ f ( x ) − f ( y ) ∣ ∣ x − y ∣ < + ∞ } for a given interval I . There is no loss of generality in assuming that f ∉ Lip ( 2 , ∞ ) . Set k n = 2 n . There exist 2 < a 1 < b 1 such that

∣ f ( b 1 ) − f ( a 1 ) ∣ ≥ k 1 ∣ b 1 − a 1 ∣ .

Since f is not Lipschitz, we have f ∉ Lip ( max { b 1 , k 2 } , ∞ ) . So there exist max { b 1 , k 2 } < a 2 < b 2 such that

∣ f ( b 2 ) − f ( a 2 ) ∣ ≥ k 2 ∣ b 2 − a 2 ∣ .

In the same manner, we can obtain that there exists a sequence { [ a n , b n ] } of intervals satisfying

∣ f ( b n ) − f ( a n ) ∣ ≥ k n ∣ b n − a n ∣ , 2 < a 1 < b 1 < … < a n < b n < … , k n < a n < b n .

From the continuity of f and the compactness of [ a n , b n ] , we know that f is uniformly continuous on [ a n , b n ] . Hence, there is δ n > 0 such that for any x , y ∈ [ a n , b n ] , ∣ x − y ∣ < δ n , we have

∣ f ( x ) − f ( y ) ∣ < 1 .

We equally divide [ a n , b n ] into v n = b n − a n δ n + 1 parts and denote as d n , j = a n + j b n − a n v n ( 0 ≤ j ≤ v n ) , thus a n = d n , 0 < d n , 1 < … < d n , v n − 1 < d n , v n = b n and ∣ f ( d n , j ) − f ( d n , j + 1 ) ∣ < 1 , j = 0 , 1 , … , v n − 1 . We obtain

k n ≤ ∣ f ( b n ) − f ( a n ) ∣ ∣ b n − a n ∣ ≤ max 0 ≤ j ≤ v n − 1 ∣ f ( d n , j ) − f ( d n , j + 1 ) ∣ ∣ d n , j − d n , j + 1 ∣ = ∣ f ( d n , j n ) − f ( d n , j n + 1 ) ∣ ∣ d n , j n − d n , j n + 1 ∣ ,

for some j n , 0 ≤ j n ≤ v n − 1 . We replace [ a n , b n ] by [ d n , j n , d n , j n + 1 ] which is also denoted by [ a n , b n ] . Thus, we have

k n ∣ b n − a n ∣ ≤ ∣ f ( b n ) − f ( a n ) ∣ < 1 .

Take l 0 = t 0 = 0 and

(3.7) l n = b n b n − a n 1 α + 1 , t n = A k n − η b n − 1 α p ( b n − a n ) 1 − α p α p p 1 − α p , n ≥ 1 ,

where 0 < η < 1 , and A is a positive constant satisfying

∑ n = 1 ∞ 2 l n t n = 1 .

The existence of A follows from the following inequality:

∑ n = 1 ∞ 2 b n b n − a n 1 α + 1 k n − η b n − 1 α p ( b n − a n ) 1 − α p α p p 1 − α p ≪ ∑ n = 1 ∞ b n − p 1 − α p k n − η p 1 − α p ≤ ∑ n = 1 ∞ k n − η p − p 1 − α p < ∞ .

Set J n 4 = ∑ j = 0 n − 1 2 l j t j , ∑ j = 0 n 2 l j t j , n = 1 , 2 , … and

J n , m 4 = ∑ j = 0 n − 1 2 l j t j + ( m − 1 ) t n , ∑ j = 0 n − 1 2 l j t j + m t n , m = 1 , … , 2 l n , n = 1 , 2 , … .

Similar to the construction as we did in the proof of locally Lipschitz continuity of f , we define γ ( x ) as follows:

γ ( x ) = a n , x ∈ J n , 2 k − 1 4 , 1 ≤ k ≤ l n , b n , x ∈ J n , 2 k 4 , 1 ≤ k ≤ l n , n = 1 , 2 , … .

We assert that γ ∈ L p , since

‖ γ ‖ p p ≤ ∑ n = 1 ∞ b n p 2 l n t n ≪ ∑ n = 1 ∞ k n − η p 1 − α p b n − α p 2 1 − α p ≪ ∑ n = 1 ∞ k n − η p − α p 2 1 − α p < ∞ .

The graph of γ is shown in Figure 7.

Write

γ 1 ( x ) = ∑ n = 1 ∞ a n χ J n 4 ( x ) , γ 2 ( x ) = γ ( x ) − γ 1 ( x ) = ∑ n = 1 ∞ ( b n − a n ) ∑ j = 1 l n χ J n , 2 j 4 ( x )

and

γ n , 1 ( x ) = a n χ J n 4 ( x ) , γ n , 2 ( x ) = ( b n − a n ) ∑ j = 1 l n χ J n , 2 j 4 ( x ) .

We now prove that γ ∈ H p α . The proof is similar to the one as in the proof of f ∈ Lip loc . For 0 < t < 1 we have

‖ △ t γ 1 ‖ p ≤ ∑ n = 1 ∞ ‖ △ t γ n , 1 ‖ p = ∑ n = 1 ∞ ( a n p ∣ { x : γ n , 1 ( x + t ) − γ n , 1 ( x ) = a n } ∣ ) 1 p ≤ ∑ 2 l n t n > t a n ( 2 t ) 1 p + ∑ 2 l n t n ≤ t a n ( 4 l n t n ) 1 p ≪ t α ∑ 2 l n t n > t b n ( l n t n ) 1 p − α + t α ∑ 2 l n t n ≤ t b n ( l n t n ) 1 p − α ≍ t α ∑ n = 1 ∞ k n − η ≪ t α

and

‖ △ t γ 2 ‖ p ≤ ∑ n = 1 ∞ ‖ △ t γ n , 2 ‖ p = ∑ n = 1 ∞ ( ( b n − a n ) p ∣ { x : γ n , 2 ( x + t ) − γ n , 2 ( x ) = b n − a n } ∣ ) 1 p ≪ ∑ t n > t ( b n − a n ) ( l n t ) 1 p + ∑ t n ≤ t ( b n − a n ) ( l n t n ) 1 p ≪ t α ∑ t n > t ( b n − a n ) l n 1 p t n 1 p − α + t α ∑ t n ≤ t ( b n − a n ) l n 1 p t n 1 p − α ≍ t α ∑ n = 1 ∞ k n − η ≪ t α ,

from which we deduce that γ = γ 1 + γ 2 ∈ H p α . However, we obtain

‖ △ t n ( f ∘ γ ) ‖ p ≥ k n ∣ b n − a n ∣ ( l n t n ) 1 p ≫ k n 1 − η t n α .

Since k n 1 − η → ∞ as n → ∞ , we have ( f ∘ γ ) ∉ H p α , which contradicts our assumption. So f is Lipschitz.

The proof of Theorem 1.2 is complete.□

Proof of Theorem 1.3

The proof of the sufficient condition and the necessary condition for which T f maps H p 1 ⁄ p ( 1 < p < ∞ ) into itself was given in the proof of Theorem 1.2. So it suffices to find out a locally Lipschitz function f for which T f does not map H p 1 ⁄ p ( 1 < p < ∞ ) into itself.

Let 1 < p < ∞ . We set

(3.8) ζ ( x ) = ln 1 ∣ x ∣ , x ∈ [ − 1 ⁄ 2 , 1 ⁄ 2 ] \ { 0 } 0 , x = 0

and f ( x ) = ∣ x ∣ a , a > 1 , x ∈ R . Then, f is locally Lipschitz. We want to prove that ζ ∈ H p 1 ⁄ p and f ∘ ζ ∉ H p 1 ⁄ p . First, we prove that ln 1 + 1 u p and ln 1 u − 1 p are Lebesgue integrable on [ 0 , 1 ] .

Let A = ∫ 0 1 ln 1 + 1 u p d u = ∫ 0 1 ln 1 + 1 u p d u and

B = ∫ 0 1 ln 1 u − 1 p d u = ∫ 0 1 2 ln 1 u − 1 p d u + ∫ 1 2 1 ln 1 u − 1 p d u = D + C .

For C , making the change in variable x = u 1 − u yields:

C = ∫ 1 2 1 ln 1 u − 1 p d u = ∫ 1 2 1 ln u 1 − u p d u = ∫ 1 + ∞ ( ln x ) p ( x + 1 ) 2 d x .

By the comparison theorem, and the convergence of integrals ∫ 0 1 u − 1 2 d u and ∫ 1 + ∞ u − 3 2 d u , as well as:

lim x → + ∞ ( ln x ) p ( x + 1 ) 2 x − 3 2 = 0 , lim u → 0 + ln 1 u − 1 p u − 1 2 = 0 , lim u → 0 + ln 1 + 1 u p u − 1 2 = 0 .

we obtain that ln 1 + 1 u p and ln 1 u − 1 p are Rieman integrable on [ 0 , 1 ] . Furthermore, by [15, Theorem 2.28] and the monotone convergence theorem [15, Theorem 2.14], we deduce that ln 1 + 1 u p and ln 1 u − 1 p are Lebesgue integrable on [ 0 , 1 ] .

Next we prove ζ ∈ H p 1 ⁄ p . Without loss of generality we can assume that 0 < t < 1 ⁄ 8 . We have

‖ △ t ζ ‖ p p = ∫ − 1 2 − 2 t + ∫ − 2 t − t + ∫ − t 0 + ∫ 0 t + ∫ t 1 2 − t + ∫ 1 2 − t 1 2 ∣ △ t ζ ∣ p d x = I + I I + I I I + I V + V + V I .

Using the transformation x = − u in I and the inequality ∣ ln ( 1 − s ) ∣ ≤ s 1 − s , 0 < s < 1 , we obtain

I = ∫ − 1 2 − 2 t ln x + t x p d x = ∫ 2 t 1 2 ln 1 − t u p d u ≤ ∫ 2 t 1 2 t u − t p d u ≪ t .

Applying the transformation x = − ( u + 1 ) t in I I , x = − u t in I I I , x = u t in I V , and noting that the functions ln 1 + 1 u p and ln 1 u − 1 p are integrable on [ 0 , 1 ] , we have

I I = ∫ − 2 t − t ln x + t x p d x = t ∫ 0 1 ln 1 + 1 u p d u ≪ t ,

I I I = ∫ − t 0 ln x + t x p d x = t ∫ 0 1 ln − 1 + 1 u p d u ≪ t ,

and

I V = ∫ 0 t ln x + t x p d x = t ∫ 0 1 ln 1 + 1 u p d u ≪ t .

Using the inequality ln ( 1 + x ) ≤ x , x > 0 in V , we obtain

V = ∫ t 1 2 − t ln x + t x p d x = ∫ t 1 2 − t ln 1 + t x p d x ≤ ∫ t 1 2 − t t x p d x ≪ t .

Finally, we note that x , ∣ x + t − 1 ∣ ∈ [ 1 ⁄ 4 , 1 ⁄ 2 ] if x ∈ [ 1 ⁄ 2 − t , 1 ⁄ 2 ] . This gives that

V I = ∫ 1 2 − t 1 2 ln x + t − 1 x p d x ≤ ( ln 2 ) p t ≪ t .

Hence, we have

‖ △ t ζ ‖ p = ( I + I I + I I I + I V + V + V I ) 1 p ≪ t 1 p ,

which implies that ζ ∈ H p 1 ⁄ p .

Next we show that f ∘ ζ ∉ H p 1 ⁄ p . We have for t ∈ ( 0 , 1 ⁄ ( 3 e ) ) ,

‖ △ t ( f ∘ ζ ) ‖ p p ≥ ∫ t 2 t ∫ 0 t ( f ∘ ζ ) ′ ( u + x ) d u p d x = ∫ t 2 t ( − a ) ∫ 0 t ln 1 u + x a − 1 1 x + u d u p d x ≥ ∫ t 2 t a ∫ 0 t ln 1 3 t a − 1 1 3 t d u p d x = a 3 p t ln 1 3 t p ( a − 1 ) .

Since ln 1 3 t a − 1 → ∞ as t → 0 + , we obtain f ∘ ζ ∉ H p 1 ⁄ p .

The proof of Theorem 1.3 is complete.□

Acknowledgements

The authors would like to express their gratitude to the referees for carefully reading this article and providing valuable suggestions and comments.

Funding information: Yeli Niu was supported by National Natural Science Foundation of China (Project No. 12201412). Part of the work was done during the Yeli Niu’s study in Capital Normal University.
Author contributions: All authors have agreed to take responsibility for the complete content of this manuscript. They have also approved its submission to the journal, reviewed all results, and given their approval for the final version of the manuscript. Heping Wang proposed the topic of the article, and Yeli Niu, together with Heping Wang, resolved and extend the topic together with Heping Wang.
Conflict of interest: The authors state no conflict of interest.

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Received: 2023-12-04

Revised: 2024-06-07

Accepted: 2024-06-11

Published Online: 2024-11-27

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https://doi.org/10.1515/dema-2024-0043

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Hölder-Lipschitz spaces; superposition operator problem; locally Lipschitz function; Lipschitz function

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