Caratheodory operator of differential forms

Abstract

This article is devoted to extensions of some existing results about the Caratheodory operator from the function sense to the differential form situation. Similarly as the function sense, we obtain the convergence of sequences of differential forms defined by the Caratheodory operator. The main result in this article is the continuity and mapping property from one space of differential forms to another under some dominated conditions.

1 Introduction

It is well known that differential forms are generalizations of differentiable functions in R^Nand have been applied to many fields, such as potential theory, partial differential equations, quasi-conformal mappings, nonlinear analysis, electromagnetism, and control theory [1–12].

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 R^N, 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 × R → R by

$$T\omega \left(x\right)=f\left(x,\omega \left(x\right)\right).$$

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 Tω _n (s) = f(s, ω _n (s))(n = 1, 2, ...) convergence in measure to Tω(s) on Ω.

Theorem 1.1. The Caratheodory operator T maps $L^{p_1}\left(\Omega \right)$ into $L^{p_2}\left(\Omega \right)$ if and only if there exists a real number b > 0, and a function $a\left(x\right)\ge 0,a\left(x\right)\in L^{p_2}\left(\Omega \right)$ satisfies the following inequality

$$\left|f\left(x,\omega \right)\right|\le a\left(x\right)+b|\omega {|}^{\frac{p_1}{p_2}}\left(x\in \Omega ,\omega \in \left(-\infty ,+\infty \right)\right).$$

This article is to extend the above results to the space of differential forms.

2 Some preliminaries about differential forms

First, we introduce some notations and preliminaries about differential forms. Let Ω denote an open subset of R^N, N ≥ 2 and R = R¹, and the n-dimensional Lebesgue measure of a set Ω ⊂ R^Nis denoted by mes(Ω). Let {e₁, e₂, ..., e _n } denote the standard orthogonal basis of R^N. Λ^l(R^N) is the linear space of l-covectors, generated by the exterior products $e_I=e_{i_1}\wedge e_{i_2}\wedge \cdots \wedge e_{i_{\iota }}$, 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 $\wedge =\wedge \left(R^N\right)={\oplus }_{\iota =0}^N{\wedge }^{\iota }$ 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 ∧(R^N). Let D'(Ω, ∧^l) denote the space of all differential l-forms, and L^p(Ω, ∧^l) denote the space composed by the l-forms

$$\omega \left(x\right)=\sum _I{\omega }_I\left(x\right)d{x}_I=\sum _I{\omega }_{i_1{i}_2\dots i_{\iota }}\left(x\right)d{x}_{i_1}\wedge d{x}_{i_2}\wedge \cdots \wedge d{x}_{i_{\iota }},$$

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

$$\parallel \omega {\parallel }_{p,\Omega }={\left({\int }_{\Omega }|\omega \left(x\right){|}^{p}dx\right)}^{1∕p}={\left({\int }_{\Omega }{\left(\sum _I|{\omega }_I\left(x\right){|}^2\right)}^{p∕2}dx\right)}^{1∕p}<+\infty .$$

We see that

$$\left\{d{x}_{i_1}\wedge d{x}_{i_2}\wedge \dots \wedge d{x}_{i_{\iota }},1\le i_1<i_2<\cdots <i_{\iota }\le N\right\}$$

is a basis of the space ∧^l, then $dim\left({\wedge }^{\iota }\right)=dim\left({\wedge }^{\iota }\left(R^N\right)\right)=\sum _{\iota =0}^N\underset{N}{\overset{\iota }{C}}$ and

$$dim\left(\wedge \right)=\sum _{l=0}^{N}dim\left({\wedge }^l\left(R^N\right)\right)=\sum _{l=0}^N{C}_N^l=2^N.$$

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

Definition 2.1. For a mapping f : Ω × ∧^l→ ∧^l, where Ω is an open set in R^N, 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 × ∧^l→ R 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 ⊂ R^Nis measurable set(0 < mes Ω ≤ +∞), and f : Ω × ∧^l→ ∧^l. We define the Caratheodory operator T : ∧^l→ ∧^lfor differential forms by

$$T\omega \left(s\right)=f\left(s,\omega \left(s\right)\right).$$

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 $mes{F}_n>mes\Omega -\frac{1}{n}\left(n=1,2,\dots \right)$, such that f(x, ω) is continuous on F _n × ∧^l. Let $F={\cup }_{m=1}^{+\infty }{F}_n\subset \Omega$, 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

$$\left\{x\in F|f_I\left(x,\omega \right)\ge a\right\}=\bigcup _{n=1}^{+\infty }\left\{x\in F_n|f_I\left(x,\omega \right)\ge a\right\}$$

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 $mes{F}_n>mes\Omega -\frac{\eta }{2^n}$ and δ _n > 0 (n = 1, 2, ...) such that for any x₁, x₂ ∈ F _n with dist(x₁, x₂) < δ _n , ω_1,I, ω_2,I∈ [-n, n], |ω_1,I- ω₂, _I | < δ _n , where ω _i = ∑ _I ω _i,I dx _I , ω _i,I ∈ R, i = 1, 2, we have

$$|f\left(x_1,{\omega }_1\right)-f\left(x_2,{\omega }_2\right)|<\frac{1}{n}\left(n=1,2,...\right).$$

Actually, if we have proved this conclusion, then $F={\bigcap }_{n=1}^{+\infty }{F}_n\subset \Omega$ is bounded and closed and satisfies

$$mes(\Omega \backslash F)=mes({\displaystyle \underset{n=1}{\overset{+\infty }{\cup }}(\Omega \backslash F_n)}\le {\displaystyle \sum _{n=1}^{+\infty }mes(\Omega \backslash F_n)<{\displaystyle \sum _{n=1}^{+\infty }\frac{\eta }{2^n}=\eta .}}$$

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 $\frac{1}{n_0}<\epsilon$ and |ω_1,I| < n₀ - 1 for all I. When (x₂, ω₂) in F × ∧^lsatisfies $dist\left(x_1,x_2\right)<\delta =min\left\{{\delta }_{n_0},1\right\},|{\omega }_{1,I}-{\omega }_{2,I}|<\delta$, we have $x_1,x_2\in F\subset F_{n_0},dist\left(x_1,x_2\right)<{\delta }_{n_0,}\left|{\omega }_{2,I}\right|<\delta +n_0-1\le n_0$, and

$$|f\left(x_1,{\omega }_1\right)-f\left(x_2,{\omega }_2\right)|<\frac{1}{n_0}<\epsilon .$$

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

$${\Omega }_0=\left\{x\in \Omega |f\left(x,\omega \right)iscontinuouson-\infty <{\omega }_I<+\infty asfunctionabout\omega \right\}.$$

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

$$\begin{array}{l}{\Omega }_{m,n}=\{x\in {\Omega }_0|{\omega }_{1,I},{\omega }_{2,I}\in [-n,n],andforallI|{\omega }_{1,I}-{\omega }_{2,I}|<\frac{1}{m}\\ containsthat|f(x,{\omega }_1)-f(x,{\omega }_2)|\le \frac{1}{3n}\}(n=1,2,\mathrm{...}).\end{array}$$

With the density of the rational number, we have Ω₀\Ω _m , _n = {x ∈ Ω₀| there exists ω _1,I , ω _2,I ∈ [-n, n], such that $|{\omega }_{1,I}-{\omega }_{2,I}|<\frac{1}{m},|f(x,{\omega }_1)-f(x,{\omega }_2)|>\frac{1}{3n}\}=\{x\in {\Omega }_0|$there exists rational number ω _i,I ∈ [-n, n], i = 1, 2, such that $|{\omega }_{1,I}-{\omega }_{2,I}|<\frac{1}{m},|f(x,{\omega }_1)-f(x,{\omega }_2)|>\frac{1}{3n}\}.$.

With the second one of Caratheodory condition, for fixed ${\omega }_1,{\omega }_2,\left\{x\in {\Omega }_0\left|\right|f\left(x,{\omega }_1\right)-f\left(x,{\omega }_2\right)|>\frac{1}{3n}\right\}$ is measurable. Thus Ω₀ \ Ω_m,n(as the countable union of such able sets) is measurable, too. So Ω_m,nis measurable. Obviously, for fixed n we have Ω_1,n⊂ Ω_2,n⊂ ···. Let $E_n={\bigcup }_{m=1}^{+\infty }{\Omega }_{m,n}\subset {\Omega }_0$. We will prove that E _n = Ω₀. Actually, if E _n ≠ Ω₀, then there exists $x_0\in {\Omega }_0\backslash E_n={\bigcap }_{m=1}^{+\infty }\left({\Omega }_0\backslash {\Omega }_{m,n}\right)$. Thus, there exist ${\omega }_i^{\left(m\right)}={\sum }_I{\omega }_{i,I}^{\left(m\right)}d{x}_I,i=1,2$ with ${\omega }_{1,I}^{\left(m\right)},{\omega }_{2,I}^{\left(m\right)}\in \left[-n,n\right]$, such that

$$|{\omega }_{1,I}^{\left(m\right)}-{\omega }_{2,I}^{\left(m\right)}|<\frac{1}{m},$$

and

$$|f\left(x_0,{\omega }_1^{\left(m\right)}\right)-f\left(x_0,{\omega }_2^{\left(m\right)}\right)|>\frac{1}{3n}\left(m=1,2,...\right).$$

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

Then we have

$$\underset{m\to +\infty }{\text{lim}}mes{\Omega }_{m,n}=mes{\Omega }_0(n=1,2,\mathrm{...}).$$

For given n, we select m₀ such that

$$mes{\Omega }_{m_0,n}>mes{\Omega }_0-\frac{1}{2^n}\frac{\eta }{3}.$$

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

$$-n={\omega }_I^{\left(0\right)}<{\omega }_I^{\left(1\right)}<{\omega }_I^{\left(2\right)}<\cdots <{\omega }_I^{\left(s\right)}=n,$$

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

$$mes{D}_j>mes{\Omega }_0-\frac{\eta }{3{\left(s+1\right)}^N{2}^n},$$

and f _I (x, ω^(j)) is continuous on D _j with respect to x (thereby uniform continuous), where ${\omega }^{\left(j\right)}={\sum }_I{\omega }^{\left(SI\right)}d{x}_I$ stands for ${\omega }_I^{\left(SI\right)}$ selected from ${\omega }_I^{\left(0\right)},{\omega }_I^{\left(1\right)},...,{\omega }_I^{\left(s\right)}$ for different I. Then ${\omega }^{\left(j\right)}={\sum }_I{\omega }_I^{\left(SI\right)}d{x}_I\in {\wedge }^{\iota }$, and we know the total amount of these ω^(j)S is ${\left(s+1\right)}^N\triangleq r$.

Let $D={\bigcap }_{j=1}^r{D}_j$. According to the uniform continuity, there exists δ > 0 such that

$$|f\left(x_1,{\omega }^{\left(j\right)}\right)-f\left(x_2,{\omega }^{\left(j\right)}\right)|<\frac{1}{3n},forall{x}_1,x_2\in D$$

and

$$dist\left(x_1,x_2\right)<\delta forj=1,2,...,t.$$

Now we select closed set F _n ⊂ Ω_{m 0,n}∩ D such that

$$mes{F}_n>mes\left({\Omega }_{m_0,n}\cap D\right)-\frac{1}{2^n}\frac{\eta }{3},$$

and δ _n satisfies $0<{\delta }_n<min\left\{\delta ,\frac{1}{m_0}\right\}$. We shall prove that F _n and δ _n are those that we need.

Actually,

$$\begin{array}{lll}\hfill {\Omega }_0\backslash \left({\Omega }_{m_0,n}\cap D\right)& =\left(\Omega \backslash {\Omega }_{m_0,n}\right)\cup \left({\Omega }_0\backslash D\right)& \hfill \text{(1)}\\ =\left(\Omega \backslash {\Omega }_{m_0,n}\right)\cup \left(\bigcup _{j=1}^r\left({\Omega }_0\backslash D_i\right)\right),& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

then we have

$$\begin{array}{lll}\hfill mes\left({\Omega }_0\backslash \left({\Omega }_{m_0,n}\cap D\right)\right)& \le mes\left({\Omega }_0\backslash {\Omega }_{m_0,n}\right)+\sum _{j=1}^{r}mes\left({\Omega }_0\backslash D_j\right)& \hfill \text{(1)}\\ <\frac{1}{2^n}\frac{\eta }{3}+\sum _{j=1}^r\frac{\eta }{3r{2}^n}=\frac{1}{2^n}\frac{2\eta }{3}.& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

This leads to

$$mes\left({\Omega }_{m_0,n}\cap D\right)>mes{\Omega }_0-\frac{1}{2^n}\frac{2\eta }{3}.$$

Thus,

$$mes{F}_n>mes{\Omega }_0-\frac{\eta }{2^n}=mes\Omega -\frac{\eta }{2^n}.$$

Suppose that x₁, x₂ ∈ F _n , d(x₁, x₂) < δ _n , ω _i,I ∈ [-n, n] (i = 1, 2) |ω_1,I- ω_2,I| < δ _n , and ${\omega }_I^{\left(i+1\right)}-{\omega }_I^{\left(i\right)}=\frac{1}{m_0}\left({\delta }_n<\frac{1}{m_0}\right),i=0,1,2,...,s-1$, and there exist some ${\omega }^{\left(j\right)}={\sum }_I{\omega }_I^{\left(SI\right)}d{x}_I$ such that for any I, we have $|{\omega }_{1,I}-{\omega }_{2,I}^{\left(SI\right)}|<\frac{1}{m_0}$. Then, we obtain

$$\begin{array}{lll}\hfill |f\left(x_1,{\omega }_1\right)-f\left(x_1,{\omega }_2\right)|& \le |f\left(x_2,{\omega }_2\right)-f\left(x_2,{\omega }^{\left(j\right)}\right)|+|f\left(x_2,{\omega }^{\left(j\right)}\right)-f\left(x_1,{\omega }^{\left(j\right)}\right)|& \hfill \text{(1) }\\ +|f\left(x_1,{\omega }^{\left(j\right)}\right)-f\left(x_1,{\omega }_1\right)|& \hfill \text{(2) }\\ \le \frac{1}{3n}+\frac{1}{3n}+\frac{1}{3n}=\frac{1}{n}.& \hfill \text{(3) }\\ \hfill \text{(4) }\end{array}$$

The proof of necessity is finished.

Actually, [15] 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 Tω(x) = f(x, ω(x)) is measurable on Ω.

Proof. According to Lemma 3.1, there exists a closed set F _n ⊂ Ω with $mes{F}_n>mes\Omega -\frac{1}{n}\left(n=1,2,...\right)$, such that f(x, ω) is continuous on F _n × ∧^l. Suppose that F _n ⊂ F_n+1(n = 1, 2, ...), otherwise let ${\bigcup }_{k=1}^n{F}_k$ be the new F _n . From Luzin Theorem, there exists a closed set $D_{n,I}\subset F_n,mes{D}_{n,I}>mes{F}_n-\frac{1}{nt}\left(t=C_N^l\right)$ such that ω _I (x) is continuous on D_n,I. Similarly, supposing D_n,I⊂ D_n+1,I(n = 1, 2, ...), and D _n = ∩ _I D _n,I ⊂ F _n , we have $mes\left(F_n\backslash D_n\right)\le {\sum }_I\left(mes{F}_n-mes{D}_{n,I}\right)={\sum }_I\frac{1}{nt}=\frac{1}{n}$, 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 $D={\bigcup }_{n=1}^{\infty }{D}_{n,}H=\Omega \backslash D$, then

$$mesD=\underset{n\to \infty }{lim}mes{D}_n=mes\Omega .$$

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

$$\left\{x|x\in D_n,f_I\left(x,\omega \left(x\right)\right)\ge a\right\}$$

is closed and can be expressed as

$$\left\{x|x\in D{f}_I\left(x,\omega \left(x\right)\right)\ge a\right\}=\bigcup _{n=1}^{\infty }\left\{x|x\in D_n{f}_I\left(x,\omega \left(x\right)\right)\ge a\right\}.$$

Thus, {x|x ∈ D _n , f _I (x, ω(x)) ≥ a} is measurable, which implies that

$$\left\{x|x\in D,f_I\left(x,\omega \left(x\right)\right)\ge a\right\}$$

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 Tω _n (s) = f(s, ω _n (s))(n = 1, 2, ...) converge in measure to Tω(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

$$\underset{n\to \infty }{lim}mes{F}_n=0,$$

that is

$$\underset{n\to \infty }{lim}mes{D}_n=mes\Omega ,$$

where

$$D_n=\Omega \backslash F_n=\left\{s|s\in \Omega ,|f\left(s,{\omega }_n\left(s\right)\right)-f\left(s,\omega \left(s\right)\right)|<\sigma \right\}.$$

Let

$$\begin{array}{l}{\Omega }_k=\left\{s\right|s\in \Omega \text{satisfy}\text{that}\text{for}\text{any}{\omega }^{\prime };\text{if}|\omega (s)-{\omega }^{\prime }|<\frac{1}{k},\text{then}\text{we}\text{have}\\ |f(s,\omega (s))-f(s,{\omega }^{\prime })|<\sigma \}(k=1,2,\mathrm{...}).\end{array}$$

Clearly

$${\Omega }_1\subset {\Omega }_2\subset {\Omega }_3\subset \cdots .$$

Let $H={\bigcup }_{k=1}^{\infty }{\Omega }_k$. If s₀ ∈ Ω \ H, then s₀ ∉ Ω _k (k = 1, 2, ...). So there exists ${\omega }_k^{\prime }$ such that

$$|\omega \left(s_0\right)-{\omega }_k^{\prime }|<\frac{1}{k}.$$

But

$$|f\left(s_0,\omega \left(s_0\right)\right)-f\left(s_0,{\omega }_k^{\prime }\right)|\ge \sigma \left(k=1,2,...\right).$$

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

$$\underset{k\to \infty }{lim}mes{\Omega }_k=mesH=mes\Omega .$$

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

$$mes{\Omega }_{k_0}>mes\Omega -\frac{\epsilon }{2}.$$

(3.1)

Let

$$\begin{array}{c}{Q}_n=\left\{s|s\in \Omega ,|{\omega }_n\left(s\right)-\omega \left(s\right)|\ge \frac{1}{k_0}\right\}\\ R_n=\Omega \backslash Q_n=\left\{s|s\in \Omega ,|{\omega }_n\left(s\right)-\omega \left(s\right)|<\frac{1}{k_0}\right\}.\end{array}$$

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

$$\underset{n\to \infty }{lim}mes{Q}_n=0,$$

that is

$$\underset{n\to \infty }{lim}mes{R}_n=mes\Omega .$$

Thus, there exists a positive integer N such that

$$mes{R}_n>mes\Omega -\frac{\epsilon }{2}ifn>N.$$

(3.2)

Obviously, ${\Omega }_{k_0}\cap R_n\subset D_n$, so

$$\Omega \backslash D_n\subset \Omega \backslash \left({\Omega }_{k_0}\cap R_n\right)=\left(\Omega \backslash {\Omega }_{k_0}\right)\cup \left(\Omega \backslash R_n\right).$$

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

$$\begin{array}{lll}\hfill 0& \le mes\Omega -mes{D}_n=mes\left(\Omega \backslash D_n\right)& \hfill \text{(1)}\\ \le mes\left(\Omega \backslash {\Omega }_{k_0}\right)+mes\left(\Omega \backslash R_n\right)& \hfill \text{(2)}\\ =\left(mes\Omega -mes{\Omega }_{k_0}\right)+\left(mes\Omega -mes{R}_n\right)& \hfill \text{(3)}\\ <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

That is

$$\underset{n\to \infty }{lim}mes{D}_n=mes\Omega ,$$

and we have

$$\underset{n\to \infty }{lim}mes{F}_n=0.$$

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

Lemma 3.3. In normed space, we have,

$$\left|\right|x|-|y|{|}^p\le ||x{|}^p-|y{|}^p|.$$

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) ∈ L^p(Ω, ∧^l) and p ≥ 1. If ||ω _n - ω₀|| → 0, then there exists a subsequence ${\omega }_{n_k}\left(s\right)$ of ω _n (s) such that ${\omega }_{n_k}\left(s\right)\to {\omega }_0\left(s\right)$a.e.

Proof. For ω _n (s) = Σ _I ω _n,I (s)ds _I ∈ L^p(Ω, ∧^l), if

$$\left|\right|{\omega }_n-{\omega }_0|{|}_p\to 0,$$

using

$$\left|\right|{\omega }_{n,I}-{\omega }_{0,I}|{|}_p\le ||{\omega }_n-{\omega }_0|{|}_p=\left|\right|{\left(\sum {\left({\omega }_{n,I}-{\omega }_{0,I}\right)}^2\right)}^{p∕2}|{|}_p,$$

we have

$$\left|\right|{\omega }_{n,I}-{\omega }_{0,I}|{|}_p\to 0,\mathsf{\text{for}}\mathsf{\text{any}}I.$$

As we know, for the first index I₁, we have subsequence ${\omega }_{n_k,I_1}\left(s\right)\to {\omega }_{o,I_1}\left(s\right)$a.e.. And for I₂, we also have $\left|\right|{\omega }_{n,I_2}-{\omega }_{0,I_2}\left|\right|\to 0$. By fixing the index I₂, there exists a subsequence ${\omega }_{n_{k_j},I_2}\left(s\right)$ of the sequence ${\omega }_{n_k,I_2}$ satisfying ${\omega }_{n_{k_j},I_2}\left(s\right)\to {\omega }_{0,I_2}\left(s\right)$a.e..

By repeating the above procedure, we can find subsequence ω _n,I (s) → ω_0,I(s) a.e., for any I.

Hence, there exists a subsequence ${\omega }_{n,I}\left(s\right)\to {\omega }_{0,I}\left(s\right)$a.e. Thus, we complete the proof.

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

$$\left|f\left(s,\omega \right)\right|\le a\left(s\right)+b|\omega {|}^{p_1∕p_2}s\in \Omega ,\omega \in L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right),$$

where $a\left(s\right)\in L^{p_2}\left(\Omega \right)$, and b > 0 is a constant, then C-operator T maps $L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)$ into $L^{p_2}\left(\Omega ,{\wedge }^{\iota }\right)$ and simultaneously is bounded and continuous.

Proof. If $\omega \left(s\right)\in L^{p_1}\left(\Omega ,{\wedge }^t\right),$, we have $a\left(s\right)+b|\omega \left(s\right){|}^{p_1∕p_2}\in L^{p_2}\left(\Omega \right)$, which implies that $T\omega \left(s\right)=f\left(s,\omega \left(s\right)\right)\in L^{p_2}\left(\Omega ,{\wedge }^{\iota }\right)$ and

$$T:L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)\to L^{p_2}\left(\Omega ,{\wedge }^{\iota }\right).$$

With Minkowski inequality, for any $\omega \left(s\right)\in L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)$, we have

$$\begin{array}{lll}\hfill \left|\right|T\omega |{|}_{p_2}& \le \left|\right|a\left(s\right)+b\left|\omega \left(s\right){|}^{p_1∕p_2}\right|{|}_{p_2}& \hfill \text{(1)}\\ \le \left|\right|a\left(s\right)|{|}_{p_2}+b|\left|\omega \left(s\right)\right|{|}_{p_1}^{p_1∕p_2}.& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

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

If T is discontinuous in $L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)$, that is to say, there exist $\left\{{\omega }_n\right\}\subset L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)\left(n=0,1,2...\right)$ and ε₀ > 0, such that

$$\left|\right|{\omega }_n\left(s\right)-{\omega }_0\left(s\right)|{|}_{p_1}\to 0,$$

but

$$\left|\right|T{\omega }_n\left(s\right)-T{\omega }_0\left(s\right)|{|}_{p_2}\ge {\epsilon }_0.$$

Let

$$f_n\left(s\right)=T{\omega }_n\left(s\right)=f\left(s,{\omega }_n\left(s\right)\right),g_n\left(s\right)=a\left(s\right)+b|{\omega }_n\left(s\right){|}^{p_1∕p_2}.$$

Then

$$\left|f_n\left(s\right)\right|\le g_n\left(s\right).$$

According to Minkowski inequality, we have

$$-\left|\right|{\omega }_n-{\omega }_0|{|}_{p_1}\le |\left|{\omega }_n\right|{|}_{p_1}-\left|\right|{\omega }_0|{|}_{p_1}\le ||{\omega }_n-{\omega }_0|{|}_{p_1},$$

So

$$\left|\right|\left|{\omega }_n\right||-|\left|{\omega }_0\right|\left|\right|\le \left|\right|{\omega }_n-{\omega }_0|{|}_{p_1}.$$

and

$$\left|\right|{\omega }_n-{\omega }_0|{|}_{p_1}\to 0,$$

so

$$\left|\right|{\omega }_n|{|}_{p_1}\to |\left|{\omega }_0\right|{|}_{p_1}.$$

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

$$f_n\left(s\right)\to f_0\left(s\right),g_n\left(s\right)\to g_0\left(s\right)a.e.$$

Obviously, $\left|\right|{\omega }_n{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right|\le |{\omega }_n{|}^{p_1}+|{\omega }_0^{p_1}|$. According to Lemma Fatou, we have

$$\begin{array}{c}{\int }_{\Omega }{\underset{}{lim}}_{n\to \infty }\left(|{\omega }_n{|}^{p_1}+|{\omega }_0{|}^{p_1}-\left|\right|{\omega }_n{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right|\right)ds\\ \le {\underset{}{lim}}_{n\to \infty }{\int }_{\Omega }\left(|{\omega }_n{|}^{p_1}+|{\omega }_0{|}^{p_1}-\left|\right|{\omega }_n{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right|\right)ds.\end{array}$$

Then

$$2\left|\right|{\omega }_0|{|}_{p_1}^{p_1}\le 2|\left|{\omega }_0\right|{|}_{p_1}^{p_1}-{\underset{}{lim}}_{n\to \infty }{\int }_{\Omega }\left(\left|\right|{\omega }_n{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right|\right)ds.$$

Hence, there exists a subsequence $\left\{{\omega }_{n_k}\left(s\right)\right\}$ such that

$$\underset{k\to \infty }{lim}{\int }_{\Omega }\left|\right|{\omega }_{n_k}{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right|ds=0.$$

Suppose that $\underset{n\to \infty }{lim}{\int }_{\Omega }\left|\right|{\omega }_n{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right|ds=0$

and with Lemma 3.3, we have

$$\begin{array}{lll}\hfill {\int }_{\Omega }|g_n\left(s\right)-g_0\left(s\right){|}^{p_2}ds& =b^{p_2}{\int }_{\Omega }\left|\right|{\omega }_n{|}^{p_1∕p_2}-|{\omega }_0{|}^{p_1∕p_2}{|}^{p_2}ds& \hfill \text{(1)}\\ \le b^{p_2}{\int }_{\Omega }\left|\right|{\omega }_n{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right||ds\to 0\left(n\to \infty \right).& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

As

$$\underset{n\to \infty }{lim}{\int }_{\Omega }\left|\right|{\omega }_n{|}^{p_1}-\left|{\omega }_0{|}^{p_1}\right|ds=0,$$

we have

$$\underset{n\to \infty }{lim}{\int }_{\Omega }|g_n\left(s\right)-g_0\left(s\right){|}^{p_2}ds=0.$$

Then

$$\left|\right|g_n\left(s\right)|{|}_{p_2}\to |\left|g_0\left(s\right)\right|{|}_{p_2}\left(n\to \infty \right).$$

And

$$|f_n\left(s\right)-f_0\left(s\right){|}^{p_2}\le 2^{p_2}\left(|g_n\left(s\right){|}^{p_2}+|g_0\left(s\right){|}^{p_2}\right).$$

Applying Lemma Fatou to $\left\{2^{p_2}|g_n\left(s\right){|}^{p_2}+|g_0\left(s\right){|}^{p_2}-|f_n\left(s\right)-f_0\left(s\right){|}^{p_2}\right\}$, we have

$${\underset{}{lim}}_{n\to \infty }{\int }_{\Omega }|f_n\left(s\right)-f_0\left(s\right){|}^{p_2}ds=0.$$

According to Lemma 3.4, there exists a subsequence $\left\{f_{n_k}\right\}$ such that

$$\underset{k\to \infty }{lim}{\int }_{\Omega }|f_{n_k}\left(s\right)-f_0\left(s\right){|}^{p_2}ds={\underset{}{lim}}_{n\to \infty }{\int }_{\Omega }|f_n\left(s\right)-f_0\left(s\right){|}^{p_2}ds=0,$$

that is

$$\left|\right|T{\omega }_{n_k}-T{\omega }_0\left|\right|\to 0\left(k\to \infty \right).$$

This is contradictory to $\left|\right|T{\omega }_n\left(s\right)-T{\omega }_0\left(s\right)|{|}_{p_2}\ge {\epsilon }_0$. 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 $L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)$ into $L^{p_2}\left(\Omega ,{\wedge }^{\iota }\right)$, if and only if, there exists $b>0,a\left(x\right)\ge 0,a\left(x\right)\in L^{p_2}\left(\Omega \right)$ satisfying the following inequality,

$$\left|f\left(x,\omega \right)\right|\le a\left(x\right)+b|\omega {|}^{\frac{p_1}{p_2}}\left(x\in \Omega ,\omega \in {\wedge }^{\iota }\right).$$

(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

$${\int }_{\Omega }|\omega \left(x\right){|}^{p_1}dx\le 1\Rightarrow {\int }_{\Omega }|f\left(x,\omega \left(x\right)\right){|}^{p_2}dx\le b^{p_2}.$$

(3.4)

Then, we define a function on Ω × ∧^l:

$$g\left(x,\omega \right)=\left\{\begin{array}{cc}\hfill \left|f\left(x,\omega \right)\right|-b|\omega {|}^{\frac{p_1}{p_2}},\hfill & \hfill \mathsf{\text{if}}\left|f\left(x,\omega \right)\right|\ge b|\omega {|}^{\frac{p_1}{p_2}};\hfill \\ \hfill 0,\hfill & \hfill \mathsf{\text{if}}\left|f\left(x,\omega \right)\right|\ge b|\omega {|}^{\frac{p_1}{p_2}}.\hfill \end{array}\right.$$

Suppose that $\omega \left(x\right)\in L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)$. Let F = {x|x ∈ Ω, g(x, ω(x)) > 0} and ${\int }_F|\omega \left(x\right){|}^{p_1}dx=n+\alpha$, 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, ..., Ω_n+1which is mutually disjoint and

$${\int }_{{\Omega }_i}|\omega \left(x\right){|}^{p_1}dx\le 1\left(i=1,2,...,n+1\right).$$

Hence, according to (3.4), we have

$$\begin{array}{lll}\hfill {\int }_F|f\left(x,\omega \left(x\right)\right){|}^{p_2}dx& =\sum _{i=1}^{n+1}{\int }_{{\Omega }_i}|f\left(x,\omega \left(x\right)\right){|}^{p_2}dx& \hfill \text{(1)}\\ \le {\left(n+1\right)}^{b^{p_2}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

Then

$$\begin{array}{l}{\displaystyle {\int }_{\Omega }{[g(x,\omega (x))]}^{p_2}dx=}{\displaystyle {\int }_F{[g(x,\omega (x))]}^{p_2}dx}\\ ={\displaystyle {\int }_F{[f(x,\omega (x))|-b|\omega (x){|}^{\frac{p_1}{p_2}}]}^{p_2}dx}\\ \le {\displaystyle {\int }_F[|f(x,\omega (x)){|}^{p_2}dx}-b^{p_2}{\displaystyle {\int }_F|\omega (x){|}^{p_1}}dx\\ \le (n+1)b^{p_2}-(n+\alpha )b^{p_2}\\ \le b^{p_2}.\end{array}$$

(3.5)

We use the following inequality in the above statement,

$${\left(u-\upsilon \right)}^r\le u^r-{\upsilon }^r,\forall u\ge \upsilon \ge 0,r\ge 1.$$

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 $D={\bigcup }_{k=1}^{\infty }{D}_k$, where mesD _k < +∞ (k = 1, 2, ...) and D1 ⊂ D2 ⊂ D3 ⊂ ⋅⋅⋅. Let ω _k (x) = Σ _I ω_k,I(x)dx _I on x ∈ D _k , where for any I,

$$-k\le {\omega }_I^{*}\le k,andg\left(x,\omega *\right)=ma{x}_{-k\le {\omega }_I\le k}g\left(x,\omega \right).$$

We choose ω _k as follows. First, we sort all of the I with an order ${\omega }_1,{\omega }_2,...,{\omega }_r\left(r\triangleq C_N^l\right)$. For I₁, let ${\omega }_{I_1}^{*}$ be the smallest value of which satisfies the above qualification; for I₂, let ${\omega }_{I_2}^{*}$ be the smallest value of which satisfies the above qualification and under the condition that we have selected ${\omega }_{I_1}^{*}$; then we repeat this procedure. Hence, ${\omega }_k\left(x\right)=\sum _I{\omega }_{k,I}\left(x\right)d{x}_I=\sum _I{\omega }_I^{*}d{x}_I$ on D _k . If x ∈ Ω \ D _k , let ω_k,I= 0, then ω _k (x) = 0.

Now we prove that ω_k,I(x) is measurable on Ω for any I. Obviously we only need to prove this on D _k . We suppose that any other ω_k,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 $g\left(x,\omega \right),\omega ={\sum }_I,{\omega }_I^{*},d{x}_I,d{x}_I,+{\omega }_{k,I}d{x}_I$. Using Lemma (3.1), there exists bounded closed set $F_{k,n}\subset D_k,mes{F}_{k,n}>mes{D}_k-\frac{1}{n}\left(n=1,2,...\right)$ such that g(x, ω _I ) is continuous on F_k,n× (-∞, +∞). Clearly, we can suppose F_k,n⊂ F_k,n+1(n = 1, 2, ...). Let $G_k={\bigcup }_{n=1}^{\infty }{F}_{k,n}$. Then mes(D _k \ G _k ) = 0. For any real number a, we investigate the set H_k,n= {x|x ∈ F_k,n,ω_k,I> a}. Suppose that x₀ ∈ H_k,n, let η = ω_k,I(x₀) - a > 0, and we select δ > 0 such that δ < η and -k < ω_k,I- δ. According to the definition of ω_k,I, we know g(x₀, ω _I ) < g(x₀, ω_k,I(x₀)), for all -k ≤ ω < ω_k,I(x₀). Hence,

$$2\delta =g\left(x_0,{\omega }_{k,I}\left(x_0\right)\right)-ma{x}_{-k\le \omega \le {\omega }_{k,I}\left(x_0\right)-\delta }g\left(x_0,{\omega }_I\right)>0.$$

With the uniform continuity of g(x, ω _I ) on F_k,n× [-k, k], there exists ${\rho }_{x0}>0$ such that if $|x-x_0|<{\rho }_{x_0}$ and x ∈ F_k,n, then we always have

$$|g\left(x,{\omega }_I\right)-g\left(x_0,{\omega }_I\right)|<\delta forall-k\le \omega \le k.$$

Hence, we know that

$$g\left(x,{\omega }_{k,I}\left(x_0\right)\right)>g\left(x,{\omega }_I\right),forall-k\le {\omega }_I\le {\omega }_{k,I}\left(x_0\right)-\delta .$$

Then ω_k,I(x) > ω_k,I(x₀) - δ > ω_k,I(x₀) - η = a, for all $|x-x_0|<{\rho }_{x_0},x\in F_{k,n}$. Let $S\left(s_0,{\rho }_{x_0}\right)$ stand for the open ball $\left\{x|x\in R^N,|x-x_0|<{\rho }_{x_0}\right\}$. Then we have proved $S\left(x_0,{\rho }_{x_0}\right)\cap F_{k,n}\subset H_{k,n}$. Hence, we know that

$$H_{k,n}=\left(\bigcup _{x\in H_{k,n}}S\left(x,{\rho }_x\right)\right)\cap F_{k,n}.$$

As ${\bigcup }_{x\in H_{k,n}}S\left(x,{\rho }_x\right)$ is open set and F_k,nis closed set, H_k,nis measurable. Thus, $\left\{x|x\in G_k,{\omega }_{k,I}>a\right\}={\bigcup }_{n=1}^{\infty }{H}_{k,n}$ 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 ω_k,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,

$${\int }_{\Omega }|\omega \left(x\right){|}^{p_1}dx={\int }_{D_k}|{\omega }_k\left(x\right){|}^{p_1}dx\le {\left(C_n^l\right)}^{p_1}{k}^{p_1}mes{D}_k<+\infty .$$

So ${\omega }_k\left(x\right)\in L^{p_1}\left(\Omega ,{\wedge }^{\iota }\right)$. Using (3.5), we have

$${\int }_{\Omega }{\left[g\left(x,{\omega }_k\left(x\right)\right)\right]}^{p_2}dx\le b^{p_2}\left(k=1,2,...\right).$$

(3.6)

Let $a\left(x\right)=su{p}_{\omega \in {\wedge }^{\iota }g}\left(x,\omega \right)\left(x\in \Omega \right).$.

Clearly, when x ∈ D, we have

$$a\left(x\right)=li{m}_{k\to \infty }g\left(x,{\omega }_k\left(x\right)\right),$$

(3.7)

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

$${\int }_{\Omega }{\left[a\left(x\right)\right]}^{p_2}dx\le {\underset{}{lim}}_{k\to \infty }{\int }_{\Omega }{\left[g\left(x,{\omega }_k\left(x\right)\right)\right]}^{p_2}dx\le b^{p_2}.$$

Hence, $a\left(x\right)\in L^{p_2}\left(\Omega \right)$. As

$$a\left(x\right)=su{p}_{\omega \in {\wedge }^{\iota }g}\left(x,\omega \right)\ge su{p}_{\omega \in {\wedge }^{\iota }}\left\{\left|f\left(x,\omega \right)\right|-b|\omega {|}^{\frac{p_1}{p_2}}\right\},$$

thus

$$\left|f\left(x,\omega \right)\right|\le a\left(x\right)+b|\omega {|}^{\frac{p_1}{p_2}}\left(x\in \Omega ,\omega \in {\wedge }^{\iota }\right).$$

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 $a_1\left(x\right)\ge 0,a_1\left(x\right)\in L^{p_1}\left(\Omega \right)$ and b > 0, such that

$$\left|f_1\left(x,\omega \right)\right|\le a\left(x\right)+b|\omega {|}^{\frac{p_1}{p_2}},\forall x\in \Omega ,\omega \in {\wedge }^{\iota },$$

where $a\left(x\right)=a_1\left(x\right)+\left|f\left(x,0\right)\right|\ge 0,a\left(x\right)\in L^{p_2}\left(\Omega \right)$. Then we know $f\left(x,\omega \right)\in L^{p_2}\left(\Omega ,{\wedge }^{\iota }\right).$. 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 l₁ ≠ l₂, these conclusions are still true. Actually, we only need to redefine an appropriate norm. For $\omega \left(x\right)={\sum }_{l=0}^n{\omega }_l\left(x\right)\in \wedge$, let

$$\left|\right|\omega \left(x\right)|{|}_{\Omega ,\wedge ,p}=\sum _{l=0}^n|\left|{\omega }_l\left(x\right)\right|{|}_{\Omega ,{\wedge }^{\iota },p},$$

where ω _l (x) = Σ _I ω_I,l(x)dx _I . Then we can extend T to $L^{p_1}\left(\Omega \times \wedge \right)\mapsto L^{p_2}\left(\Omega \times \wedge \right)$ and the above conclusions still come into existence.