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Привалов定理的拓广
引用本文:陆启铿,钟同德.Привалов定理的拓广[J].数学学报,1957,7(1):144-165.
作者姓名:陆启铿  钟同德
作者单位:中国科学院数学研究所 (陆启铿),厦门大学数学系(钟同德)
摘    要:<正> 设Ω是 m 个实变数 u_1,…,u_m 空间中的-p 维可定向流形Ω:(?)Ω称为属于 C~e 类(e是非负的整数),如果实函数 f_1,…,f_(m-p)皆有e次连续偏微商.Ω称为平滑的,如Ω属于 C~1 类并且矩阵

收稿时间:1956-6-26

AN EXTENSION OF PRIVALOF THEOREM
Institution:LOOK C.H(Institute of Mathematics,Academia Sinica)CHUNG T.D(Amoy University)
Abstract:Let m=2n(n≥2),z_α=u_α+iu_(n+α)(α=1,…,n),and(?)be a domainof the 2n-dimensional space of u_1,…,u_(2n),and its boundary Ω be a(2n—1)-dimensional smooth orientable manifold of olass C~2 defined byF(u_1,…u_(2n))=0.(1)Let K_(2n-1)(z,(?))be a complex exterior differential form of degree 2n-1(?)whereγ(z—ξ)denotes the euclidean distance of z=(z_1,…,z_n)andξ==(ξ_1,…,ξ),namely(?)(3)andσ=(n-2)1/2πi_n.It is known that if f(z)is regular in the domain(?)and on its boun-dary,then we have the Cauchy formula~(1])f(w)=σ∫_πf(z)K_(2n-1)(z,w),x∈(?)Since the exterior differential operatorD=sum from k=1 to 2n(?)du~k,if we set(?) then it is obviousD=d+(?)(5)First we prove the following two lemmas:Lemma 1.K_(2n-1)(z,ξ)is homologous to zero at those points z≠ξ.Lemma 2.K_(2n-1)(z,ξ)is invariant under the unitary transformation.We then obtainTheorem 1.If z_0 is a point on Ω,then the principle value of theintegral of K(2n-1,)(z, z_0) on Ω exists,andV.P.∫_π K_(2n-1)(z,z_0)=1/2.Corollary:If f(z)is defined on Ω and satisfies the H(?)lder conditionand z_0∈Ω,then the principle value of the integral(Ⅰ)∫_Ωf(z)K_(2n-1)(z, z_0)exists.Theorem 2.If Ω is a smooth orientable manifold belongs to class C~2and f(z)is a continuous fumction of complex value defined on Ω,whichsatisfies H(?)lder condition and difines a function F(w)in(?)such thatF(w)=∫_Ωf(z)K_(2n-1)(z,w)and if W_0 is an arbitrary point on Ω,then we have(Ⅱ)F_i(W_0)=V.P∫_Ωf(z)K_(2n-1)(z, w_0)+1/2f(w_0),(Ⅲ)F_e,(W_0)=V.P∫_Ωf(z)K_(2n-1)(Z, W_0)-1/2f(W_0)where F_i(W_0)and F_e(W_0)devote the limit values of F(w)when w approachesw_0 from the inner part and the outer part of the domain(?)respectively.Theorem 3.If f(z)is regular in(?)and on its boundary,thenF_i(z_0)=f(z_0).
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