# Caratheodory operator of differential forms

## 1 Introduction

It is well known that differential forms are generalizations of differentiable functions in RNand have been applied to many fields, such as potential theory, partial differential equations, quasi-conformal mappings, nonlinear analysis, electromagnetism, and control theory .

One of the important work in the field of differential forms is to develop various kinds of estimates and inequalities for differential forms under some conditions. These results have wide applications in the A-harmonic equation, which implies more versions of harmonic equations for functions [1, 5, 6].

The Caratheodory operator arose from the extension of Peano theorem about the existence of solutions to a first-order ordinary differential equation, which says that this kind equation has a solution under relatively mild conditions. It is very interesting to characterize equivalently the Caratheodory's conditions and the continuity of Caratheodory operator, which form classic examples to discuss boundedness and continuity of nonlinear operators and play an important part in advanced functional analysis.

For general function space, we define the Caratheodory operator as in [13, 14].

Definition 1.1. Suppose that G is measurable in RN, and 0 < mesG ≤ +∞. We say that function f(x, ω)(x G, -∞ < ω < +∞) satisfies the Caratheodory conditions, if

1. for almost all x G, f(x, ω) is continuous with respect to ω; and

2. for any ω, f(x, ω) is measurable about x on G.

For the function f(x, ω) with Caratheodory conditions, we define the Caratheodory operator T : G × RR by

$Tω x =f x , ω x .$

There are some essential results for Caratheodory operator as follows.

Lemma 1.1. Suppose that mes Ω < +∞. Then f(x, ω) satisfies the Caratheodory condition if and only if for any η > 0, there exists a bounded open set F Ω with mesF > mes Ω - η, such that f(x, ω) is continuous on F × R.

Lemma 1.2. Suppose that mes Ω < ∞. If ω n (s) (n = 1, 2, ...) convergence in measure to ω(s) on Ω, then n (s) = f(s, ω n (s))(n = 1, 2, ...) convergence in measure to (s) on Ω.

Theorem 1.1. The Caratheodory operator T maps $L p 1 Ω$ into $L p 2 Ω$ if and only if there exists a real number b > 0, and a function $a x ≥ 0 , a x ∈ L p 2 Ω$ satisfies the following inequality

$| f x , ω | ≤ a x + b | ω | p 1 p 2 x ∈ Ω , ω ∈ - ∞ , + ∞ .$

## 2 Some preliminaries about differential forms

First, we introduce some notations and preliminaries about differential forms. Let Ω denote an open subset of RN, N ≥ 2 and R = R1, and the n-dimensional Lebesgue measure of a set Ω RNis denoted by mes(Ω). Let {e1, e2, ..., e n } denote the standard orthogonal basis of RN. Λl(RN) is the linear space of l-covectors, generated by the exterior products $e I = e i 1 ∧ e i 2 ∧⋯∧ e i ι$, corresponding to all ordered l-tuples I = (i1, i2, ..., i l ), 1 ≤ i 1 < i2 < ··· < i l N, l = 0,1, ..., N. The Grassman algebra $∧=∧ R N = ⊕ ι = 0 N ∧ ι$ is a graded algebra with respect to the exterior products.

A differential l-form ω on Ω is a Schwartz distribution on Ω with values in (RN). Let D'(Ω, l) denote the space of all differential l-forms, and Lp(Ω, l) denote the space composed by the l-forms

$ω x = ∑ I ω I x d x I = ∑ I ω i 1 i 2 … i ι x d x i 1 ∧ d x i 2 ∧ ⋯ ∧ d x i ι ,$

where ω I Lp(Ω, R) for all ordered l-tuples I. Then Lp(Ω, l), p ≥ 1 is a Banach space with norm

$∥ ω ∥ p , Ω = ∫ Ω | ω x | p d x 1 ∕ p = ∫ Ω ∑ I | ω I x | 2 p ∕ 2 d x 1 ∕ p < + ∞ .$

We see that

$d x i 1 ∧ d x i 2 ∧ … ∧ d x i ι , 1 ≤ i 1 < i 2 < ⋯ < i ι ≤ N$

is a basis of the space l, then $d i m ∧ ι = d i m ∧ ι R N = ∑ ι = 0 N C N ι$ and

$d i m ∧ = ∑ l = 0 N d i m ∧ l R N = ∑ l = 0 N C N l = 2 N .$

Then we define the Caratheodory conditions and Caratheodory operator for differential forms.

Definition 2.1. For a mapping f : Ω × ll, where Ω is an open set in RN, we say that f satisfies Caratheodory conditions if 1. for all most s Ω, f(s, ω) is continuous with respect to ω, which means that f can be expanded as f(s, ω) = Σ J f J (s, ω)dx J where f J : Ω × lR and f J (s, ω) is continuous about ω for all most s Ω; and 2. for any fixed ω = Σ I ωdx I l, f(s, ω) is measurable about s, which means that each coefficient function f J (s, ω) is measurable about s for any fixed ω l.

Throughout this article we assume that f(s, ω) satisfies the Caratheodory condition(C-condition). Similarly, we can define the continuity of f(s, ω) about (s, ω) Ω × l.

Definition 2.2. Suppose that Ω RNis measurable set(0 < mes Ω ≤ +∞), and f : Ω × ll. We define the Caratheodory operator T : llfor differential forms by

$Tω s =f s , ω s .$

## 3 Main results and proofs

There is a necessary and sufficient condition of the Caratheodory conditions:

Lemma 3.1. Suppose that mes Ω < +∞. Then f(x, ω) satisfies the Caratheodory condition, if and only if, for any η > 0, there exists a bounded closed set F Ω, with mesF > mes Ω - η, such that f(x, ω) is continuous on F × l.

Proof. Proof of sufficiency.

According to the hypothesis, there exists a bounded open set F n Ω, with $mes F n >mesΩ- 1 n n = 1 , 2 , …$, such that f(x, ω) is continuous on F n × l. Let $F = ∪ m = 1 + ∞ F n ⊂ Ω$, then mesF = mes Ω, and when x F, f(x, ω) is continuous on l. Hence, the first one of Caratheodory conditions is satisfied. For fixed ω l, {x F n |f I (x, ω) ≥ a}(a R) is bounded and closed. Then

${ x ∈ F | f I ( x , ω ) ≥ a } = ⋃ n = 1 + ∞ { x ∈ F n | f I ( x , ω ) ≥ a }$

is measurable, so that f(x, ω) = ∑ I f(x, ω)dx I is measurable on F respect to x, then it is measurable on Ω. Hence, the second one of Caratheodory conditions is satisfied.

Proof of necessity.

For a given η > 0, we only need to prove the following result: there exists a bounded closed set F n Ω with $mes F n >mesΩ- η 2 n$ and δ n > 0 (n = 1, 2, ...) such that for any x1, x2 F n with dist(x1, x2) < δ n , ω1,I, ω2,I [-n, n], |ω1,I- ω2, I | < δ n , where ω i = ∑ I ω i,I dx I , ω i,I R, i = 1, 2, we have

$| f ( x 1 , ω 1 ) - f ( x 2 , ω 2 ) | < 1 n ( n = 1 , 2 , . . . ) .$

Actually, if we have proved this conclusion, then $F= ⋂ n = 1 + ∞ F n ⊂Ω$ is bounded and closed and satisfies

$m e s ( Ω \ F ) = m e s ( ∪ n = 1 + ∞ ( Ω \ F n ) ≤ ∑ n = 1 + ∞ m e s ( Ω \ F n ) < ∑ n = 1 + ∞ η 2 n = η .$

Then we can prove f(x, ω) is continuous on F × las follows. For any given (x, ω) F × l, and ε > 0, we select n0 with $1 n 0 < ε$ and |ω1,I| < n0 - 1 for all I. When (x2, ω2) in F × lsatisfies $dist ( x 1 , x 2 ) <δ=min { δ n 0 , 1 } ,| ω 1 , I - ω 2 , I |<δ$, we have $x 1 , x 2 ∈F⊂ F n 0 ,dist ( x 1 , x 2 ) < δ n 0 , | ω 2 , I |<δ+ n 0 -1≤ n 0$, and

$| f ( x 1 , ω 1 ) - f ( x 2 , ω 2 ) | < 1 n 0 < ε .$

Thus, f(x, ω) is continuous on F × l. Set

$Ω 0 = { x ∈ Ω | f ( x , ω ) i s c o n t i n u o u s o n - ∞ < ω I < + ∞ a s f u n c t i o n a b o u t ω } .$

According to the first one of Caratheodory conditions, we have mes Ω0 = mes Ω. Let

$Ω m , n = { x ∈ Ω 0 | ω 1 , I , ω 2 , I ∈ [ − n , n ] , a n d f o r a l l I | ω 1 , I − ω 2 , I | < 1 m c o n t a i n s t h a t | f ( x , ω 1 ) − f ( x , ω 2 ) | ≤ 1 3 n } ( n = 1 , 2 , ... ) .$

With the density of the rational number, we have Ω0 m , n = {x Ω0| there exists ω 1,I , ω 2,I [-n, n], such that $| ω 1 , I − ω 2 , I | < 1 m , | f ( x , ω 1 ) − f ( x , ω 2 ) | > 1 3 n } = { x ∈ Ω 0 |$there exists rational number ω i,I [-n, n], i = 1, 2, such that $| ω 1 , I − ω 2 , I | < 1 m , | f ( x , ω 1 ) − f ( x , ω 2 ) | > 1 3 n } .$.

With the second one of Caratheodory condition, for fixed $ω 1 , ω 2 , { x ∈ Ω 0 | | f ( x , ω 1 ) - f ( x , ω 2 ) | > 1 3 n }$ is measurable. Thus Ω0 \ Ωm,n(as the countable union of such able sets) is measurable, too. So Ωm,nis measurable. Obviously, for fixed n we have Ω1,n Ω2,n ···. Let $E n = ⋃ m = 1 + ∞ Ω m , n ⊂ Ω 0$. We will prove that E n = Ω0. Actually, if E n ≠ Ω0, then there exists $x 0 ∈ Ω 0 \ E n = ⋂ m = 1 + ∞ ( Ω 0 \ Ω m , n )$. Thus, there exist $ω i ( m ) = ∑ I ω i , I ( m ) d x I ,i=1,2$ with $ω 1 , I ( m ) , ω 2 , I ( m ) ∈ [ - n , n ]$, such that

$| ω 1 , I ( m ) - ω 2 , I ( m ) | < 1 m ,$

and

$| f ( x 0 , ω 1 ( m ) ) - f ( x 0 , ω 2 ( m ) ) | > 1 3 n ( m = 1 , 2 , . . . ) .$

This obviously contradicts the uniform continuity of function f(x0, ω), where ω = ∑ I ω I dx I with -nω I n for any I. Hence, we have proved E n = Ω0 (n = 1, 2, ...).

Then we have

$lim m → + ∞ m e s Ω m , n = m e s Ω 0 ( n = 1 , 2 , ... ) .$

For given n, we select m0 such that

$m e s Ω m 0 , n > m e s Ω 0 - 1 2 n η 3 .$

For any I, we divide interval [-n, n] into s = 2nm0 subintervals, and the endpoints of these subintervals are

$- n = ω I ( 0 ) < ω I ( 1 ) < ω I ( 2 ) < ⋯ < ω I ( s ) = n ,$

where I = (i1, i2, ..., i l ) and the number of all I is $t= C n l$. Using Luzin theorem, there exist bounded closed sets D j Ω0, such that

$m e s D j > m e s Ω 0 - η 3 ( s + 1 ) N 2 n ,$

and f I (x, ω(j)) is continuous on D j with respect to x (thereby uniform continuous), where $ω ( j ) = ∑ I ω ( S I ) d x I$ stands for $ω I ( S I )$ selected from $ω I ( 0 ) , ω I ( 1 ) ,..., ω I ( s )$ for different I. Then $ω ( j ) = ∑ I ω I ( S I ) d x I ∈ ∧ ι$, and we know the total amount of these ω(j)S is $( s + 1 ) N ≜r$.

Let $D= ⋂ j = 1 r D j$. According to the uniform continuity, there exists δ > 0 such that

$| f ( x 1 , ω ( j ) ) - f ( x 2 , ω ( j ) ) | < 1 3 n , f o r a l l x 1 , x 2 ∈ D$

and

$d i s t ( x 1 , x 2 ) < δ f o r j = 1 , 2 , . . . , t .$

Now we select closed set F n Ωm 0,nD such that

$m e s F n > m e s ( Ω m 0 , n ∩ D ) - 1 2 n η 3 ,$

and δ n satisfies $0< δ n . We shall prove that F n and δ n are those that we need.

Actually,

$Ω 0 \ ( Ω m 0 , n ∩ D ) = ( Ω \ Ω m 0 , n ) ∪ ( Ω 0 \ D ) (1) = ( Ω \ Ω m 0 , n ) ∪ ⋃ j = 1 r ( Ω 0 \ D i ) , (2) (3)$

then we have

$m e s ( Ω 0 \ ( Ω m 0 , n ∩ D ) ) ≤ m e s ( Ω 0 \ Ω m 0 , n ) + ∑ j = 1 r m e s ( Ω 0 \ D j ) (1) < 1 2 n η 3 + ∑ j = 1 r η 3 r 2 n = 1 2 n 2 η 3 . (2) (3)$

$m e s ( Ω m 0 , n ∩ D ) > m e s Ω 0 - 1 2 n 2 η 3 .$

Thus,

$m e s F n > m e s Ω 0 - η 2 n = m e s Ω - η 2 n .$

Suppose that x1, x2 F n , d(x1, x2) < δ n , ω i,I [-n, n] (i = 1, 2) |ω1,I- ω2,I| < δ n , and $ω I ( i + 1 ) - ω I ( i ) = 1 m 0 δ n < 1 m 0 , i = 0 , 1 , 2 , . . . , s - 1$, and there exist some $ω ( j ) = ∑ I ω I ( S I ) d x I$ such that for any I, we have $| ω 1 , I - ω 2 , I ( S I ) |< 1 m 0$. Then, we obtain

The proof of necessity is finished.

Actually,  gave a proof of the condition that the u in f(x, u) is a normal l- dimensional vector. With this lemma, we have the following result.

Lemma 3.2. Suppose that mes Ω < +∞. If for ω(x) = Σ I ω I (x)dx I , ω I (x) is measurable on Ω, then (x) = f(x, ω(x)) is measurable on Ω.

Proof. According to Lemma 3.1, there exists a closed set F n Ω with $mes F n >mesΩ- 1 n n = 1 , 2 , . . .$, such that f(x, ω) is continuous on F n × l. Suppose that F n Fn+1(n = 1, 2, ...), otherwise let $⋃ k = 1 n F k$ be the new F n . From Luzin Theorem, there exists a closed set $D n , I ⊂ F n ,mes D n , I >mes F n - 1 n t ( t = C N l )$ such that ω I (x) is continuous on Dn,I. Similarly, supposing Dn,I Dn+1,I(n = 1, 2, ...), and D n = ∩ I D n,I F n , we have $mes ( F n \ D n ) ≤ ∑ I ( m e s F n - m e s D n , I ) = ∑ I 1 n t = 1 n$, which deduces that for any I, ω I (x) is continuous on D n . We suppose D n D n+1 (n = 1, 2, ...) just as F n . Let $D= ⋃ n = 1 ∞ D n , H=Ω\D$, then

$m e s D = lim n → ∞ m e s D n = m e s Ω .$

Hence, we have mesH = 0. For any rational number a,

${ x | x ∈ D n , f I ( x , ω ( x ) ) ≥ a }$

is closed and can be expressed as

${ x | x ∈ D f I ( x , ω ( x ) ) ≥ a } = ⋃ n = 1 ∞ { x | x ∈ D n f I ( x , ω ( x ) ) ≥ a } .$

Thus, {x|x D n , f I (x, ω(x)) ≥ a} is measurable, which implies that

${ x | x ∈ D , f I ( x , ω ( x ) ) ≥ a }$

is measurable. Hence, f I (x, ω(x)) is measurable on Ω. Then, f(x, ω(x)) is measurable. This ends the proof.

In fact, this lemma is also true when mes Ω = +∞. For the Caratheodory operator T, we first prove that the operator maintains the convergence in measure for sequences of differential forms.

Theorem 3.1. Suppose that mes Ω < ∞. If ω n (s) (n = 1, 2, ...) converge in measure to ω(s) on Ω, then n (s) = f(s, ω n (s))(n = 1, 2, ...) converge in measure to (s) on Ω.

Proof.

For any σ > 0, let F n = {s|s Ω, |f(s, ω n (s)) - f(s, ω(s))| ≥ σ}. We need to prove

$lim n → ∞ m e s F n = 0 ,$

that is

$lim n → ∞ m e s D n = m e s Ω ,$

where

$D n = Ω \ F n = { s | s ∈ Ω , | f ( s , ω n ( s ) ) - f ( s , ω ( s ) ) | < σ } .$

Let

$Ω k = { s | s ∈ Ω satisfy that for any ω ′ ; if | ω ( s ) − ω ′ | < 1 k , then we have | f ( s , ω ( s ) ) − f ( s , ω ′ ) | < σ } ( k = 1 , 2 , ... ) .$

Clearly

$Ω 1 ⊂ Ω 2 ⊂ Ω 3 ⊂ ⋯ .$

Let $H= ⋃ k = 1 ∞ Ω k$. If s0 Ω \ H, then s0 Ω k (k = 1, 2, ...). So there exists $ω k ′$ such that

$| ω ( s 0 ) - ω k ′ | < 1 k .$

But

$| f ( s 0 , ω ( s 0 ) ) - f ( s 0 , ω k ′ ) | ≥ σ ( k = 1 , 2 , . . . ) .$

Thus f(s0, ω) is not continuous at ω = ω0 = ω(s0). Hence, because of f satisfying C-conditions, we know mes(Ω \ H) = 0, that is

$lim k → ∞ m e s Ω k = m e s H = m e s Ω .$

For all ε > 0, we can choose sufficiently large k0 such that

$m e s Ω k 0 > m e s Ω - ε 2 .$
(3.1)

Let

$Q n = { s | s ∈ Ω , | ω n ( s ) - ω ( s ) | ≥ 1 k 0 } R n = Ω \ Q n = { s | s ∈ Ω , | ω n ( s ) - ω ( s ) | < 1 k 0 } .$

As ω n (s) converge in measure to ω(s), we have

$lim n → ∞ m e s Q n = 0 ,$

that is

$lim n → ∞ m e s R n = m e s Ω .$

Thus, there exists a positive integer N such that

$m e s R n > m e s Ω - ε 2 i f n > N .$
(3.2)

Obviously, $Ω k 0 ∩ R n ⊂ D n$, so

$Ω \ D n ⊂ Ω \ ( Ω k 0 ∩ R n ) = ( Ω \ Ω k 0 ) ∪ ( Ω \ R n ) .$

Then, with (3.1) and (3.2), and mes Ω < ∞, we know that for n > N,

$0 ≤ m e s Ω - m e s D n = m e s ( Ω \ D n ) (1) ≤ m e s ( Ω \ Ω k 0 ) + m e s ( Ω \ R n ) (2) = ( m e s Ω - m e s Ω k 0 ) + ( m e s Ω - m e s R n ) (3) < ε 2 + ε 2 = ε . (4) (5)$

That is

$lim n → ∞ m e s D n = m e s Ω ,$

and we have

$lim n → ∞ m e s F n = 0 .$

Lemmas 3.3, 3.4 will be used in the following proof of the main theorem.

Lemma 3.3. In normed space, we have,

$| | x | - | y | | p ≤ | | x | p - | y | p | .$

where x, y is any element of this space.

The proof is easy to obtain and therefore omitted.

Lemma 3.4. Suppose ω n (s) Lp(Ω, l) and p ≥ 1. If ||ω n - ω0|| → 0, then there exists a subsequence $ω n k ( s )$ of ω n (s) such that $ω n k ( s ) → ω 0 ( s )$a.e.

Proof. For ω n (s) = Σ I ω n,I (s)ds I Lp(Ω, l), if

$| | ω n - ω 0 | | p → 0 ,$

using

$| | ω n , I - ω 0 , I | | p ≤ | | ω n - ω 0 | | p = | | ( ∑ ( ω n , I - ω 0 , I ) 2 ) p ∕ 2 | | p ,$

we have

$| | ω n , I - ω 0 , I | | p → 0 , for any I .$

As we know, for the first index I1, we have subsequence $ω n k , I 1 ( s ) → ω o , I 1 ( s )$a.e.. And for I2, we also have $|| ω n , I 2 - ω 0 , I 2 ||→0$. By fixing the index I2, there exists a subsequence $ω n k j , I 2 ( s )$ of the sequence $ω n k , I 2$ satisfying $ω n k j , I 2 ( s ) → ω 0 , I 2 ( s )$a.e..

By repeating the above procedure, we can find subsequence ω n,I (s) → ω0,I(s) a.e., for any I.

Hence, there exists a subsequence $ω n , I ( s ) → ω 0 , I ( s )$a.e. Thus, we complete the proof.

Theorem 3.2. Suppose mes Ω < ∞, p1, p2 ≥ 1. If f satisfies

$| f ( s , ω ) | ≤ a ( s ) + b | ω | p 1 ∕ p 2 s ∈ Ω , ω ∈ L p 1 ( Ω , ∧ ι ) ,$

where $a ( s ) ∈ L p 2 ( Ω )$, and b > 0 is a constant, then C-operator T maps $L p 1 ( Ω , ∧ ι )$ into $L p 2 ( Ω , ∧ ι )$ and simultaneously is bounded and continuous.

Proof. If $ω ( s ) ∈ L p 1 ( Ω , ∧ t ) ,$, we have $a ( s ) + b | ω ( s ) | p 1 ∕ p 2 ∈ L p 2 ( Ω )$, which implies that $Tω ( s ) =f ( s , ω ( s ) ) ∈ L p 2 ( Ω , ∧ ι )$ and

$T : L p 1 ( Ω , ∧ ι ) → L p 2 ( Ω , ∧ ι ) .$

With Minkowski inequality, for any $ω ( s ) ∈ L p 1 ( Ω , ∧ ι )$, we have

$| | T ω | | p 2 ≤ | | a ( s ) + b | ω ( s ) | p 1 ∕ p 2 | | p 2 (1) ≤ | | a ( s ) | | p 2 + b | | ω ( s ) | | p 1 p 1 ∕ p 2 . (2) (3)$

So T is bounded. Next we prove the continuity of T.

If T is discontinuous in $L p 1 ( Ω , ∧ ι )$, that is to say, there exist ${ ω n } ⊂ L p 1 ( Ω , ∧ ι ) ( n = 0 , 1 , 2 . . . )$ and ε0 > 0, such that

$| | ω n ( s ) - ω 0 ( s ) | | p 1 → 0 ,$

but

$| | T ω n ( s ) - T ω 0 ( s ) | | p 2 ≥ ε 0 .$

Let

$f n ( s ) = T ω n ( s ) = f ( s , ω n ( s ) ) , g n ( s ) = a ( s ) + b | ω n ( s ) | p 1 ∕ p 2 .$

Then

$| f n ( s ) | ≤ g n ( s ) .$

According to Minkowski inequality, we have

$- | | ω n - ω 0 | | p 1 ≤ | | ω n | | p 1 - | | ω 0 | | p 1 ≤ | | ω n - ω 0 | | p 1 ,$

So

$| | | ω n | | - | | ω 0 | | | ≤ | | ω n - ω 0 | | p 1 .$

and

$| | ω n - ω 0 | | p 1 → 0 ,$

so

$| | ω n | | p 1 → | | ω 0 | | p 1 .$

With the first one of C-conditions and Lemma 3.4, there exists a subsequence (suppose that {ω n } is this subsequence) such that

$f n ( s ) → f 0 ( s ) , g n ( s ) → g 0 ( s ) a . e .$

Obviously, $| | ω n | p 1 - | ω 0 | p 1 | ≤ | ω n | p 1 + | ω 0 p 1 |$. According to Lemma Fatou, we have

$∫ Ω lim n → ∞ ( | ω n | p 1 + | ω 0 | p 1 - | | ω n | p 1 - | ω 0 | p 1 | ) d s ≤ lim n → ∞ ∫ Ω ( | ω n | p 1 + | ω 0 | p 1 - | | ω n | p 1 - | ω 0 | p 1 | ) d s .$

Then

$2 | | ω 0 | | p 1 p 1 ≤ 2 | | ω 0 | | p 1 p 1 - lim n → ∞ ∫ Ω ( | | ω n | p 1 - | ω 0 | p 1 | ) d s .$

Hence, there exists a subsequence ${ ω n k ( s ) }$ such that

$lim k → ∞ ∫ Ω | | ω n k | p 1 - | ω 0 | p 1 | d s = 0 .$

Suppose that $lim n → ∞ ∫ Ω || ω n | p 1 -| ω 0 | p 1 |ds=0$

and with Lemma 3.3, we have

$∫ Ω | g n ( s ) - g 0 ( s ) | p 2 d s = b p 2 ∫ Ω | | ω n | p 1 ∕ p 2 - | ω 0 | p 1 ∕ p 2 | p 2 d s (1) ≤ b p 2 ∫ Ω | | ω n | p 1 - | ω 0 | p 1 | | d s → 0 ( n → ∞ ) . (2) (3)$

As

$lim n → ∞ ∫ Ω | | ω n | p 1 - | ω 0 | p 1 | d s = 0 ,$

we have

$lim n → ∞ ∫ Ω | g n ( s ) - g 0 ( s ) | p 2 d s = 0 .$

Then

$| | g n ( s ) | | p 2 → | | g 0 ( s ) | | p 2 ( n → ∞ ) .$

And

$| f n ( s ) - f 0 ( s ) | p 2 ≤ 2 p 2 ( | g n ( s ) | p 2 + | g 0 ( s ) | p 2 ) .$

Applying Lemma Fatou to ${ 2 p 2 | g n ( s ) | p 2 + | g 0 ( s ) | p 2 - | f n ( s ) - f 0 ( s ) | p 2 }$, we have

$lim n → ∞ ∫ Ω | f n ( s ) - f 0 ( s ) | p 2 d s = 0 .$

According to Lemma 3.4, there exists a subsequence ${ f n k }$ such that

$lim k → ∞ ∫ Ω | f n k ( s ) - f 0 ( s ) | p 2 d s = lim n → ∞ ∫ Ω | f n ( s ) - f 0 ( s ) | p 2 d s = 0 ,$

that is

$| | T ω n k - T ω 0 | | → 0 ( k → ∞ ) .$

This is contradictory to $||T ω n ( s ) -T ω 0 ( s ) | | p 2 ≥ ε 0$. Hence, T is continuous. This ends the proof.

Actually, we can prove that the condition in Theorem 3.2 is a necessary and sufficient condition.

Theorem 3.3. The Caratheodory operator T maps continuously and boundedly $L p 1 ( Ω , ∧ ι )$ into $L p 2 ( Ω , ∧ ι )$, if and only if, there exists $b>0,a ( x ) ≥0,a ( x ) ∈ L p 2 ( Ω )$ satisfying the following inequality,

$| f ( x , ω ) | ≤ a ( x ) + b | ω | p 1 p 2 ( x ∈ Ω , ω ∈ ∧ ι ) .$
(3.3)

Proof. The proof of sufficiency has been proved in Theorem 3.2.

Proof of necessity. First, we suppose f(x, 0) ≡ 0. With the continuity and bound-edness of T, we know there exists b > 0 such that

$∫ Ω | ω ( x ) | p 1 d x ≤ 1 ⇒ ∫ Ω | f ( x , ω ( x ) ) | p 2 d x ≤ b p 2 .$
(3.4)

Then, we define a function on Ω × l:

$g ( x , ω ) = | f ( x , ω ) | - b | ω | p 1 p 2 , if | f ( x , ω ) | ≥ b | ω | p 1 p 2 ; 0 , if | f ( x , ω ) | ≥ b | ω | p 1 p 2 .$

Suppose that $ω ( x ) ∈ L p 1 ( Ω , ∧ ι )$. Let F = {x|x Ω, g(x, ω(x)) > 0} and $∫ F |ω ( x ) | p 1 dx=n+α$, where n is a nonnegative integer and 0 ≤ α < 1. With the absolute continuity of integral, we can divide F into n + 1 measurable set Ω1, Ω2, ..., Ωn+1which is mutually disjoint and

$∫ Ω i | ω ( x ) | p 1 d x ≤ 1 ( i = 1 , 2 , . . . , n + 1 ) .$

Hence, according to (3.4), we have

$∫ F | f ( x , ω ( x ) ) | p 2 d x = ∑ i = 1 n + 1 ∫ Ω i | f ( x , ω ( x ) ) | p 2 d x (1) ≤ ( n + 1 ) b p 2 (2) (3)$

Then

$∫ Ω [ g ( x , ω ( x ) ) ] p 2 d x = ∫ F [ g ( x , ω ( x ) ) ] p 2 d x = ∫ F [ f ( x , ω ( x ) ) | − b | ω ( x ) | p 1 p 2 ] p 2 d x ≤ ∫ F [ | f ( x , ω ( x ) ) | p 2 d x − b p 2 ∫ F | ω ( x ) | p 1 d x ≤ ( n + 1 ) b p 2 − ( n + α ) b p 2 ≤ b p 2 .$
(3.5)

We use the following inequality in the above statement,

$( u - υ ) r ≤ u r - υ r , ∀ u ≥ υ ≥ 0 , r ≥ 1 .$

As f(x, ω) satisfies the Caratheodory condition, g(x, ω) also satisfies it. Then there exist D Ω, mes(Ω\D) = 0, and f(x, ω) is continuous with respect to ω for any x D. Let $D= ⋃ k = 1 ∞ D k$, where mesD k < +∞ (k = 1, 2, ...) and D1 D2 D3 . Let ω k (x) = Σ I ωk,I(x)dx I on x D k , where for any I,

$- k ≤ ω I * ≤ k , a n d g ( x , ω * ) = m a x - k ≤ ω I ≤ k g ( x , ω ) .$

We choose ω k as follows. First, we sort all of the I with an order $ω 1 , ω 2 ,..., ω r r ≜ C N l$. For I1, let $ω I 1 *$ be the smallest value of which satisfies the above qualification; for I2, let $ω I 2 *$ be the smallest value of which satisfies the above qualification and under the condition that we have selected $ω I 1 *$; then we repeat this procedure. Hence, $ω k ( x ) = ∑ I ω k , I ( x ) d x I = ∑ I ω I * d x I$ on D k . If x Ω \ D k , let ωk,I= 0, then ω k (x) = 0.

Now we prove that ωk,I(x) is measurable on Ω for any I. Obviously we only need to prove this on D k . We suppose that any other ωk,I'is fixed where the maximum can be got; then g(x, ω) only has two variables x, ω I . Hence, let g(x, ω I ) stand for the function we are studying here, which actually is $g ( x , ω ) ,ω= ∑ I , ω I * ,d x I ,d x I ,+ ω k , I d x I$. Using Lemma (3.1), there exists bounded closed set $F k , n ⊂ D k ,mes F k , n >mes D k - 1 n ( n = 1 , 2 , . . . )$ such that g(x, ω I ) is continuous on Fk,n× (-∞, +∞). Clearly, we can suppose Fk,n Fk,n+1(n = 1, 2, ...). Let $G k = ⋃ n = 1 ∞ F k , n$. Then mes(D k \ G k ) = 0. For any real number a, we investigate the set Hk,n= {x|x Fk,n,ωk,I> a}. Suppose that x0 Hk,n, let η = ωk,I(x0) - a > 0, and we select δ > 0 such that δ < η and -k < ωk,I- δ. According to the definition of ωk,I, we know g(x0, ω I ) < g(x0, ωk,I(x0)), for all -kω < ωk,I(x0). Hence,

$2 δ = g ( x 0 , ω k , I ( x 0 ) ) - m a x - k ≤ ω ≤ ω k , I ( x 0 ) - δ g ( x 0 , ω I ) > 0 .$

With the uniform continuity of g(x, ω I ) on Fk,n× [-k, k], there exists $ρ x 0 >0$ such that if $|x- x 0 |< ρ x 0$ and x Fk,n, then we always have

$| g ( x , ω I ) - g ( x 0 , ω I ) | < δ f o r a l l - k ≤ ω ≤ k .$

Hence, we know that

$g ( x , ω k , I ( x 0 ) ) > g ( x , ω I ) , f o r a l l - k ≤ ω I ≤ ω k , I ( x 0 ) - δ .$

Then ωk,I(x) > ωk,I(x0) - δ > ωk,I(x0) - η = a, for all $|x- x 0 |< ρ x 0 ,x∈ F k , n$. Let $S ( s 0 , ρ x 0 )$ stand for the open ball ${ x | x ∈ R N , | x - x 0 | < ρ x 0 }$. Then we have proved $S ( x 0 , ρ x 0 ) ∩ F k , n ⊂ H k , n$. Hence, we know that

$H k , n = ⋃ x ∈ H k , n S ( x , ρ x ) ∩ F k , n .$

As $⋃ x ∈ H k , n S ( x , ρ x )$ is open set and Fk,nis closed set, Hk,nis measurable. Thus, ${ x | x ∈ G k , ω k , I > a } = ⋃ n = 1 ∞ H k , n$ is also measurable. Then we know that ω k,I (x) is measurable on G k , so is on D k .

Now we prove that ωk,I(x) is measurable for any I. According to Lemma 3.2, f(x, ω k (x)) is measurable, therefore g(x, ω k (x)) is measurable, too.

Obviously,

$∫ Ω | ω ( x ) | p 1 d x = ∫ D k | ω k ( x ) | p 1 d x ≤ ( C n l ) p 1 k p 1 m e s D k < + ∞ .$

So $ω k ( x ) ∈ L p 1 ( Ω , ∧ ι )$. Using (3.5), we have

$∫ Ω [ g ( x , ω k ( x ) ) ] p 2 d x ≤ b p 2 ( k = 1 , 2 , . . . ) .$
(3.6)

Let $a ( x ) = s u p ω ∈ ∧ ι g ( x , ω ) ( x ∈ Ω ) .$.

Clearly, when x D, we have

$a ( x ) = l i m k → ∞ g ( x , ω k ( x ) ) ,$
(3.7)

and so a(x) is nonnegative measurable function. With (3.6) and (3.7), using Lemma Fatou, we have that

$∫ Ω [ a ( x ) ] p 2 d x ≤ lim k → ∞ ∫ Ω [ g ( x , ω k ( x ) ) ] p 2 d x ≤ b p 2 .$

Hence, $a ( x ) ∈ L p 2 ( Ω )$. As

$a ( x ) = s u p ω ∈ ∧ ι g ( x , ω ) ≥ s u p ω ∈ ∧ ι { | f ( x , ω ) | - b | ω | p 1 p 2 } ,$

thus

$| f ( x , ω ) | ≤ a ( x ) + b | ω | p 1 p 2 ( x ∈ Ω , ω ∈ ∧ ι ) .$

Hence, under the hypothesis f(x, 0) ≡ 0, we prove the necessity of the theorem. For general situation, let f1(x, ω) = f(x, ω) - f(x, 0). Then f1(x, 0) ≡ 0. Applying the above conclusion to f1, we know that there exist $a 1 ( x ) ≥ 0 , a 1 ( x ) ∈ L p 1 ( Ω )$ and b > 0, such that

$| f 1 ( x , ω ) | ≤ a ( x ) + b | ω | p 1 p 2 , ∀ x ∈ Ω , ω ∈ ∧ ι ,$

where $a ( x ) = a 1 ( x ) +|f ( x , 0 ) |≥0,a ( x ) ∈ L p 2 ( Ω )$. Then we know $f ( x , ω ) ∈ L p 2 ( Ω , ∧ ι ) .$. Hence, we complete the proof.

In the proofs above, when l = 0, Caratheodory operator T: Ω × R Ω × R degenerates to normal function space. We can also find that when l1l2, these conclusions are still true. Actually, we only need to redefine an appropriate norm. For $ω ( x ) = ∑ l = 0 n ω l ( x ) ∈ ∧$, let

$| | ω ( x ) | | Ω , ∧ , p = ∑ l = 0 n | | ω l ( x ) | | Ω , ∧ ι , p ,$

where ω l (x) = Σ I ωI,l(x)dx I . Then we can extend T to $L p 1 ( Ω × ∧ ) ↦ L p 2 ( Ω × ∧ )$ and the above conclusions still come into existence.

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## Acknowledgements

ZT was supported by the Fundamental Research Foundation of NUDT (NO. JC10-02-02), and by the NSF of Hunan Province (No. 11JJ3004).

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Correspondence to Zhaoyang Tang.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

JZ carries out the main idea and ZT gives the main proof.

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Tang, Z., Zhu, J. Caratheodory operator of differential forms. J Inequal Appl 2011, 88 (2011). https://doi.org/10.1186/1029-242X-2011-88 