RESEARC H Open Access Caratheodory operator of differential forms Zhaoyang Tang * and Jianmin Zhu * Correspondence: tzymath@gmail. com Department of Mathematics and System Science, National University of Defense Technology, Changsha, PR China 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. Keywords: differential forms, Caratheodory operator, continuity of operator 1 Introduction It is well known that differential forms are generalizations of differentiable functions in R N and have been applied to many fields, such as potential theory, partial differential equations, quasi-conformal mappings, nonlinear analysis, electromagnetism, and con- trol theory [1-12]. Oneoftheimportantworkinthefieldofdifferential forms is to develop various kinds of estimates and inequalities for differe ntial forms under some conditions. These results have wide applications in the A-harmonic equation, which implies more ver- sions 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 Car- atheodory operator, which form classic examples to discuss boundedness and continu- ity 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 Î G,-∞ <ω <+∞) satisfies the Caratheodory conditions, if 1. for almost all x Î 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 Tω ( x ) = f ( x, ω ( x )) . There are some essential results for Caratheodory operator as follows. Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 © 2011 Tang and Zhu; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribu tion License (h ttp://creativecommons.org/licenses/by/2.0), which permits unrestr icted use, distribution, and reproduction in any medium, provided the original work is properly cite d. Lemma 1.1. Suppose that mesΩ <+∞. Then f(x, ω) satisfies the Caratheodory condi- tion if and only if for any h > 0, there exists a bounded open set F ⊂ Ω with mesF >mesΩ - h, such that f(x, ω) is continuous on F × R. Lemma 1.2. Suppose that mesΩ < ∞.Ifω n (s)(n =1,2, )convergenceinmeasure to ω(s)onΩ,thenTω 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 ( ) into L p 2 ( ) if and only if there exists a real number b >0,andafunction a ( x ) ≥ 0, a ( x ) ∈ L p 2 ( ) satisfies the following inequality |f ( x, ω ) |≤a ( x ) + b|ω| p 1 p 2 ( x ∈ , ω ∈ ( −∞,+∞ )) . 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 = R 1 , and the n-dim ensional Lebesgue mea- sure of a set Ω ⊂ R N is denoted by mes(Ω). Let {e 1 , e 2 , ,e n } denote the standard orthogonal basis of R N . Λ l (R N )isthelinearspaceofl-covectors, generated by the exterior products e I = e i 1 ∧ e i 2 ∧···∧e i ι , corresponding to all ordered l-tuples I =(i 1 , i 2 , , i l ), 1 ≤ i 1 <i 2 <···<i l ≤ N, l =0,1, ,N. The Grassman algebra ∧ = ∧ R N = ⊕ N ι=0 ∧ ι is a graded algebra with respect to the exterior products. A differential l-form ω on Ω is a Schwartz distribut ion on Ω with values in ∧( R N ). Let D’ (Ω, ∧ l ) denote the space of all differential l-forms, and L p (Ω, ∧ l )denotethe space composed by the l-forms ω ( x ) = I ω I ( x ) dx I = I ω i 1 i 2 i ι ( x ) dx i 1 ∧ dx i 2 ∧···∧dx i ι , where ω I Î L p ( Ω, R) for all ordered l-tuples I.ThenL p ( Ω, ∧ l ), p ≥ 1 i s a Banach space with norm ω p, = |ω ( x ) | p dx 1 / p = ⎛ ⎝ I |ω I ( x ) | 2 p / 2 dx ⎞ ⎠ 1 / p < +∞ . We see that dx i 1 ∧ dx i 2 ∧ ∧ dx i ι ,1 ≤ i 1 < i 2 < ··· < i ι ≤ N is a basis of the space ∧ l , then dim ∧ ι = dim ∧ ι R N = N ι=0 C ι N and dim ( ∧ ) = N l = 0 dim ∧ l R N = N l = 0 C l N =2 N . Then we define the Caratheodory conditions and Caratheodory operator for differen- tial forms. Definition 2.1. For a mapping f : Ω × ∧ l ® ∧ l , where Ω is an open set in R N ,wesay that f satisfies Caratheodory conditions if 1. for all most s Î 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 Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 2 of 14 f J : Ω × ∧ l ® Randf J (s, ω) is continuous about ω for all most s Î Ω; and 2. for any fixed ω = Σ I ωdx I Î ∧ 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 Î ∧ 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) Î Ω × ∧ l . Definition 2.2. Suppose that Ω ⊂ R N is measurable set(0 <mesΩ ≤ +∞),andf: Ω × ∧ l ® ∧ l . We define the Caratheodory operator T : ∧ l ® ∧ l for 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 mesΩ <+∞. Then f(x, ω) satisfies the Caratheodory condi- tion, if and only if, for any h > 0, there exists a bounded closed set F ⊂ Ω,withmesF >mesΩ - h, 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 mesF n > mes − 1 n ( n =1,2, ) , such that f(x, ω)iscontinuousonF n × ∧ l .Let F = ∪ +∞ m =1 F n ⊂ ,thenmesF = mesΩ,andwhenx Î F, f(x, ω) is continuous on ∧ l . Hence, the first one of Caratheodory conditions is satisfied. For fixed ω Î ∧ l ,{x Î F n |f I (x, ω) ≥ a}(a Î 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,thenit is measurable on Ω. Hence, the second one of Caratheodory conditions is satisfied. Proof of necessity. For a given h > 0, we only need to prove the following result: there exists a bounded closed set F n ⊂ Ω with mesF n > mes − η 2 n and δ n >0(n = 1, 2, ) such that for any x 1 , x 2 Î F n with dist(x 1 , x 2 )<δ n , ω 1,I , ω 2,I Î [-n, n], |ω 1,I - ω 2 , I |<δ n ,whereω i = ∑ I ω i, I dx I , ω i,I Î 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 ⊂ is bounded and closed and satisfies mes(\F)=mes( +∞ n =1 (\F n ) ≤ +∞ n =1 mes(\F n ) < +∞ n =1 η 2 n = η . Then we can prove f(x, ω) is continuous on F × ∧ l as follows. For any given (x, ω) Î F × ∧ l , and ε > 0, we select n 0 with 1 n 0 <ε and |ω 1,I |<n 0 - 1 for all I. When (x 2 , ω 2 )in F × ∧ l satisfies dist(x 1 , x 2 ) <δ= min{δ n 0 ,1}, |ω 1,I − ω 2,I | < δ ,wehave 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 Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 3 of 14 |f (x 1 , ω 1 ) − f (x 2 , ω 2 )| < 1 n 0 <ε . Thus, f( x, ω) is continuous on F × ∧ l . Set 0 = {x ∈ |f ( x, ω ) is continuous on −∞<ω I < +∞ as function about ω} . According to the first one of Caratheodory conditions, we have mesΩ 0 = mesΩ. Let m,n = {x ∈ 0 |ω 1,I , ω 2,I ∈ [−n, n], and for all I|ω 1,I − ω 2,I | < 1 m contains that|f (x, ω 1 ) − f (x, ω 2 )|≤ 1 3n } (n = 1, 2, ). With the density of the rational number, we have Ω 0 \Ω m , n ={x Î Ω 0 | there exists ω 1,I , ω 2,I Î [-n, n], such that |ω 1,I − ω 2,I | < 1 m , |f (x, ω 1 ) − f (x, ω 2 )| > 1 3n } = {x ∈ 0 | there exists rational number ω i,I Î [-n, n], i = 1, 2, such that | ω 1,I − ω 2,I | < 1 m , |f (x, ω 1 ) − f (x, ω 2 )| > 1 3n } . . With the second one of Caratheodory condition, for fixed ω 1 , ω 2 , {x ∈ 0 ||f (x, ω 1 ) − f (x, ω 2 )| > 1 3n } is measurable. Thus Ω 0 \ Ω m,n (as the countable union of such able sets) is measurable, too. So Ω m,n is measurable. Obviously, for fixed n we have Ω 1,n ⊂ Ω 2,n ⊂ ···. Let E n = +∞ m =1 m,n ⊂ 0 .Wewill prove that E n = Ω 0 . Actually, if E n ≠ Ω 0 , then there exists x 0 ∈ 0 \E n = +∞ m =1 ( 0 \ m,n ) . Thus, there exist ω ( m ) i = I ω ( m ) i , I dx I , i =1, 2 with ω (m) 1 , I , ω (m) 2 , I ∈ [−n, n ] , such that | ω (m) 1,I − ω (m) 2,I | < 1 m , and | f (x 0 , ω (m) 1 ) − f (x 0 , ω (m) 2 )| > 1 3n (m = 1, 2, ) . This obviously contradicts the unifo rm continuity of function f(x 0 , ω), where ω = ∑ I ω I dx I with -n ≤ ω I ≤ n for any I. Hence, we have proved E n = Ω 0 (n = 1, 2, ). Then we have l im m →+∞ mes m,n = mes 0 ( n = 1,2, ). For given n, we select m 0 such that mes m 0 ,n > mes 0 − 1 2 n η 3 . For any I,wedivideinterval[-n, n] into s =2nm 0 subintervals, and the endpoints of these subintervals are −n = ω ( 0 ) I <ω ( 1 ) I <ω ( 2 ) I < ··· <ω ( s ) I = n , where I =(i 1 , i 2 , , i l )andthenumberofallI is t = C l n .UsingLuzintheorem,there exist bounded closed sets D j ⊂ Ω 0 , such that Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 4 of 14 mesD j > mes 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 (SI) dx I stands for ω ( SI ) I selected from ω ( 0 ) I , ω ( 1 ) I , , ω ( s ) I for different I. Then ω (j) = I ω ( SI ) I dx I ∈∧ ι , and we know the total amount of these ω (j) Sis ( s +1 ) N r . Let D = r j =1 D j . According to the uniform continuity, there exists δ > 0 such that | f (x 1 , ω (j) ) − f (x 2 , ω (j) )| < 1 3n , for all x 1 , x 2 ∈ D and dist ( x 1 , x 2 ) <δfor j = 1, 2, , t . Now we select closed set F n ⊂ Ω m0,n ∩ D such that mesF n > mes( m 0 ,n ∩ D) − 1 2 n η 3 , and δ n satisfies 0 <δ n < min δ, 1 m 0 . We shall prove that F n and δ n are those that we need. Actually, 0 \( m 0 ,n ∩ D)=(\ m 0 ,n ) ∪ ( 0 \D) =(\ m 0 ,n ) ∪ ⎛ ⎝ r j=1 ( 0 \D i ) ⎞ ⎠ , then we have mes( 0 \( m 0 ,n ∩ D)) ≤ mes( 0 \ m 0 ,n )+ r j=1 mes( 0 \D j ) < 1 2 n η 3 + r j =1 η 3r2 n = 1 2 n 2η 3 . This leads to mes( m 0 ,n ∩ D) > mes 0 − 1 2 n 2η 3 . Thus, mesF n > mes 0 − η 2 n = mes − η 2 n . Suppose that x 1 , x 2 Î F n , d(x 1 , x 2 )<δ n , ω i,I Î [-n, n](i =1,2)|ω 1,I - ω 2,I |<δ n ,and ω (i+1) I − 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 (SI) I dx I such that for any I, we have | ω 1,I − ω (SI) 2,I | < 1 m 0 . Then, we obtain Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 5 of 14 |f (x 1 , ω 1 ) − f (x 1 , ω 2 )|≤|f (x 2 , ω 2 ) − f (x 2 , ω (j) )| + |f(x 2 , ω (j) ) − f (x 1 , ω (j) ) | + |f (x 1 , ω (j) ) − f (x 1 , ω 1 )| ≤ 1 3n + 1 3n + 1 3n = 1 n . 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-dimen- sional vector. With this lemma, we have the following result. Lemma 3.2. Supp ose that mesΩ <+∞.Ifforω(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 mesF n > mes − 1 n ( n = 1,2, ) ,suchthatf(x, ω) i s continuous on F n × ∧ l .Suppose that F n ⊂ F n+1 (n=1, 2, ), otherwise let n k =1 F k be the new F n . From Luzin Theorem, there exists a closed set D n,I ⊂ F n , mesD n,I > mesF n − 1 nt (t = C l N ) such that ω I (x)is continuous on D n,I .Similarly,supposingD n,I ⊂ D n+1,I (n = 1, 2, ), and D n = ∩ I D n,I ⊂ F n ,wehave mes(F n \D n ) ≤ I (mesF n − mesD n,I )= I 1 nt = 1 n , which deduces that for any I, ω I (x)iscontinuousonD n .WesupposeD n ⊂ D n+1 (n = 1, 2, ) just as F n .Let D = ∞ n =1 D n, H = \ D , then mesD = lim n → ∞ mesD n = mes . 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 ∈ Df I (x, ω(x)) ≥ a} = ∞ n =1 {x|x ∈ D n f I (x, ω(x)) ≥ a} . Thus, {x| x Î 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 mesΩ =+∞. For the Caratheodory operator T, we first prove that the operator m aintains the converge nce in measure for sequences of differential forms. Theorem 3.1. Suppose that mesΩ < ∞.Ifω n (s)(n = 1, 2, ) converge in measure to ω(s)onΩ, then Tω n (s)=f(s, ω n (s))(n=1, 2, ) converge in measure to Tω(s)onΩ. Proof. For any s > 0, let F n ={s|s Î Ω,|f(s, ω n (s)) - f(s, ω(s))| ≥ s}. We need to prove lim n → ∞ mesF n =0 , Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 6 of 14 that is lim n → ∞ mesD n = mes , where D n = \F n = {s|s ∈ , |f ( s, ω n ( s )) − f ( s, ω ( s )) | <σ} . Let k = {s|s ∈ satisfy that for any ω ;if|ω(s) − ω | < 1 k ,thenwehav e |f ( s, ω ( s )) − f ( s, ω ) | <σ} ( k = 1,2, ) . Clearly 1 ⊂ 2 ⊂ 3 ⊂ ··· . Let H = ∞ k =1 k .Ifs 0 Î Ω \ H,thens 0 ∉ 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(s 0 , ω) is not continuous at ω = ω 0 = ω(s 0 ). Hence, because of f satisfying C- conditions, we know mes(Ω \ H) = 0, that is lim k → ∞ mes k = mesH = mes . For all ε > 0, we can choose sufficiently large k 0 such that mes k 0 > mes − ε 2 . (3:1) Let Q n = {s|s ∈ , |ω n (s) − ω(s)|≥ 1 k 0 } R n = \Q n = {s|s ∈ , |ω n (s) − ω(s)| < 1 k 0 } . As ω n (s) converge in measure to ω(s), we have lim n → ∞ mesQ n =0 , that is lim n → ∞ mesR n = mes . Thus, there exists a positive integer N such that mesR n > mes − ε 2 ifn > N . (3:2) Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 7 of 14 Obviously, k 0 ∩ R n ⊂ D n ,so \D n ⊂ \( k 0 ∩ R n )=(\ k 0 ) ∪ (\R n ) . Then, with (3.1) and (3.2), and mesΩ < ∞ , we know that for n >N, 0 ≤ mes − mesD n = mes(\D n ) ≤ mes( \ k 0 )+mes(\R n ) =(mes − mes k 0 )+(mes − mesR n ) < ε 2 + ε 2 = ε. That is lim n → ∞ mesD n = mes , and we have lim n → ∞ mesF 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) Î L p ( Ω, ∧ l )andp ≥ 1. If ||ω n - ω 0 || ® 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 Î L p (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 I 1 , we have subsequence ω n k ,I 1 (s) → ω o,I 1 (s ) a.e And for I 2 ,wealsohave ||ω n,I 2 − ω 0,I 2 || → 0 . By fixing the index I 2 ,thereexistsasubse- quence 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) ® ω 0,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 mesΩ < ∞, p 1 , p 2 ≥ 1. If f satisfies | f ( s, ω ) |≤a ( s ) + b|ω| p 1 /p 2 s ∈ , ω ∈ L p 1 ( , ∧ ι ), where a ( s ) ∈ L p 2 ( ) , and b > 0 is a constant, then C-operator T maps L p 1 ( , ∧ ι ) into L p 2 ( , ∧ ι ) and simultaneously is bounded and continuous. Proof.If ω ( s ) ∈ L p 1 ( , ∧ t ), ,wehave a ( s ) + b|ω ( s ) | p 1 /p 2 ∈ L p 2 ( ) , which implies that Tω ( s ) = f ( s, ω ( s )) ∈ L p 2 ( , ∧ ι ) and Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 8 of 14 T : L p 1 ( , ∧ ι ) → L p 2 ( , ∧ ι ). With Minkowski inequality, for any ω ( s ) ∈ L p 1 ( , ∧ ι ) , we have ||Tω|| p 2 ≤||a(s)+b|ω(s)| p 1 / p 2 || p 2 ≤||a(s)|| p 2 + b||ω(s)|| p 1 /p 2 p 1 . So T is bounded. Next we prove the continuity of T. If T is discontinuous in L p 1 ( , ∧ ι ) ,thatistosay,thereexist {ω n }⊂L p 1 ( , ∧ ι )( n = 0,1,2 ) and ε 0 > 0, such that ||ω n (s) − ω 0 (s)|| p 1 → 0 , but | |Tω n (s) − Tω 0 (s)|| p 2 ≥ ε 0 . Let f n ( s ) = Tω 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 firs t one of C-conditions and Lemma 3.4, there exists a subsequence (sup- pose 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 p 1 0 | . According to Lemma Fatou, we have lim n→∞ (|ω n | p 1 + |ω 0 | p 1 −||ω n | p 1 −|ω 0 | p 1 |)ds ≤ lim n→∞ (|ω n | p 1 + |ω 0 | p 1 −||ω n | p 1 −|ω 0 | p 1 |)ds . Then 2||ω 0 || p 1 p 1 ≤ 2||ω 0 || p 1 p 1 − lim n→∞ (||ω n | p 1 −|ω 0 | p 1 |)ds . Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 9 of 14 Hence, there exists a subsequence {ω n k (s) } such that lim k→∞ ||ω n k | p 1 −|ω 0 | p 1 |ds =0 . Suppose that lim n→∞ ||ω n | p 1 −|ω 0 | p 1 |ds = 0 and with Lemma 3.3, we have ∫ |g n (s) − g 0 (s)| p 2 ds = b p 2 ||ω n | p 1 /p 2 −|ω 0 | p 1 /p 2 | p 2 ds ≤ b p 2 ||ω n | p 1 −|ω 0 | p 1 ||ds → 0(n →∞) . As lim n→∞ ||ω n | p 1 −|ω 0 | p 1 |ds =0 , we have lim n→∞ |g n (s) − g 0 (s)| p 2 ds =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→∞ |f n (s) − f 0 (s)| p 2 ds =0 . According to Lemma 3.4, there exists a subsequence {f n k } such that lim k→∞ |f n k (s) − f 0 (s)| p 2 ds = lim n→∞ |f n (s) − f 0 (s)| p 2 ds =0, that is | |Tω n k − Tω 0 || → 0(k →∞) . This is contradictory to ||Tω n (s) − Tω 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 suffi- cient condition. Theorem 3.3. The Caratheodory operator T maps continuously and boundedly L p 1 ( , ∧ ι ) into L p 2 ( , ∧ ι ) , if and only if, there exists b > 0, a ( x ) ≥ 0, a ( x ) ∈ L p 2 ( ) satisfying the following inequality, | f ( x, ω ) |≤ a ( x ) + b|ω| p 1 p 2 ( x ∈ , ω ∈∧ ι ). (3:3) Proof. The proof of sufficiency has been proved in Theorem 3.2. Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 10 of 14 [...]...Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 11 of 14 Proof of necessity First, we suppose f(x, 0) ≡ 0 With the continuity and boundedness of T, we know there exists b > 0 such that |ω(x)|p1 dx ≤ 1 ⇒ |f (x, ω(x))|p2 dx ≤ bp2 (3:4)... value of which satisfies the above ∗ qualification; for I2, let ωI2 be the smallest value of which satisfies the above qualifica∗ tion and under the condition that we have selected ωI1; then we repeat this procedure Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Hence, ωk (x) = ωk,I (x)dxI = I I Page 12 of 14... relationship between closed functional and continuous functional J Longyan Teachers Coll (Sci Nat Ed) 8, 10–15 (1990) doi:10.1186/1029-242X-2011-88 Cite this article as: Tang and Zhu: Caratheodory operator of differential forms Journal of Inequalities and Applications 2011 2011:88 Submit your manuscript to a journal and benefit from: 7 Convenient online submission 7 Rigorous peer review 7 Immediate publication... Research Foundation of NUDT (NO JC10-02-02), and by the NSF of Hunan Province (No 11JJ3004) Authors’ contributions JZ carries out the main idea and ZT gives the main proof Competing interests The authors declare that they have no competing interests Received: 15 May 2011 Accepted: 18 October 2011 Published: 18 October 2011 References 1 Agarwal, PR, Ding, S, Nolder, CA: Inequalities for Differential Forms... (x) ∈ Lp1 ( , ∧ι ) Using (3.5), we have [g(x, ωk (x))]p2 dx ≤ bp2 Let a(x) = supω∈∧ι g (x, ω) (x ∈ (k = 1, 2, ) ) (3:6) Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88 Page 13 of 14 Clearly, when x Î D, we have a(x) = limk→∞ g(x, ωk (x)), (3:7) and so a(x) is nonnegative measurable function With (3.6) and (3.7),... Xing, Y, Li, R: Some Caccioppoli estimates for differential forms J Inequal Appl 2009, 11 (2009) Article ID 734528, 3 Aronsson, G, Lindqvist, P: On p-harmonic functions in the plane and their stream functions J Diff Eqns 74, 157–178 (1988) doi:10.1016/0022-0396(88)90022-8 Tang and Zhu Journal of Inequalities and Applications 2011, 2011:88 http://www.journalofinequalitiesandapplications.com/content/2011/1/88... 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 a1 (x) ≥ 0, a1 (x) ∈ Lp1 ( ) and b > 0, such that p1 |f1 (x, ω)| ≤ a(x) + b|ω| p2 , ∀x ∈ , ω ∈ ∧ι , where a(x) = a1 (x) + |f (x, 0)| ≥ 0, a(x) ∈ Lp2 ( ) Then we know f (x, ω) ∈ Lp2 ( , ∧ι ) Hence, we complete the proof In the proofs above,... 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 Î D Let D = ∞ Dk, where mesDk < +∞ (k = 1, 2, ) and D1 ⊂ D2 ⊂ D3 ⊂ ⋅⋅⋅ Let ωk(x) = k=1 ΣI ωk,I (x)dxI on x Î Dk, where for any I, ∗ −k ≤ ωI ≤ k, and g(x, ω∗) = max−k≤ωI ≤k g(x, ω) We choose ω k as follows First, we sort all of the I with an order... H k,n , let h = ω k,I (x 0) - a > 0, and we select δ > 0 such that δ . some dominated conditions. Keywords: differential forms, Caratheodory operator, continuity of operator 1 Introduction It is well known that differential forms are generalizations of differentiable functions. 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. Access Caratheodory operator of differential forms Zhaoyang Tang * and Jianmin Zhu * Correspondence: tzymath@gmail. com Department of Mathematics and System Science, National University of Defense