# On Cauchy–Pompeiu’s Operator in Octant Ring

Authors: Hasan Baddour; Anas Kar
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this study, we have determined Cauchy–Pompeiu representation formula in octant unit ringR4 by modulation the main Cauchy–Pompeiu representation; this modulation was obtained by applying the parqueting-reflection method. Moreover, the boundedness of the modulated Cauchy–Pompeiu’s operator in R4 is proved by applying Schmitz’s inequality.


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## Body

## 1. Introduction

Many studies have investigated the modulation of Cauchy–Pompeiu representation formula in different particular domains, e.g., the ring [1], half-disk and half-ring [2], quarter-ring [3], half-hexagon [4], and lens and lune [5], to obtain an explicit solution to the Schwartz problem [6]:(1)∂z¯w=fonD,w=fon∂D.The Cauchy–Pompeiu representation formula is the fundamental tool for solving Schwartz problem which just has to be properly modified.The main Cauchy–Pompeiu representation formula is given as follows.Theorem 1 (Cauchy–Pompeiu representation [1]).
Any functionw∈C1D;ℂ∩CD¯;ℂ for a regular complex domain D⊂ℂ can be represented as follows:(2)wz=12πi∫∂Dwζdζζ−z−1π∫Dwζ¯ζdξdηζ−z∗,Orwz=−12πi∫∂Dwζdζ¯ζ−z¯−1π∫Dwζζdξdηζ−z¯∗∗.The area integral appearing in the complex Cauchy–Pompeiu representation∗ defines a weakly singular integral operator T; we call it as “the Cauchy–Pompeiu’s operator” [1]:(3)Tfz=−1π∫Dfζdξdηζ−z,f∈L1D;ℂ.Its properties have been studied by Vekua [7]. The operator T helps in solving Schwartz problem because it holds(4)∂z¯Tf=f,in a weak sense (the derivative in Sobolev space [6]).When Cauchy–Pompeiu representation formula is modulated, the operatorT is also modulated.In this study, we modulate the Cauchy–Pompeiu representation formula in octant unit ring by using the parqueting-reflection method; then, we define the modulatedT operator in R4, and we prove the boundedness of this operator.


## 2. The Octant Unit Ring

We define the octant unit ring as follows (Figure1):(5)R4=z∈ℂ;0<r<z<1,0<argz<π4.Figure 1
Octant unit disk.First, we will find the reflections of the domainR4 at its boundaries.Let∂1R4 be the segment r,1, ∂2R4 be the circular arc 1,eπ/4i:τ⟶eτi, where τ∈0,π/4, ∂3R4 be the segment eπ/4i,r.eπ/4i, and ∂4R4 be the circular arc r.eπ/4i,r:τ⟶r.eτi, where it is τ∈π/4,0.Letz∈R4; then,(1) The pointz is reflected at the segment ∂3R4 onto iz¯.(2) The pointsz,iz¯ are reflected at y-axis onto −z¯,iz, respectively.(3) The pointsz,iz¯,iz,−z¯ are reflected at x-axis onto z¯,−iz,−iz¯,−z, respectively.We obtain the following eight sets:(6)R4,0=R4,R4,1=z∈ℂ;0<r<z<1,π4<argz<π2=iz¯;z∈R4,R4,2=z∈ℂ;0<r<z<1,π2<argz<3π4=iz;z∈R4,R4,3=z∈ℂ;0<r<z<1,3π4<argz<π=−z¯;z∈R4,R4,4=z∈ℂ;0<r<z<1,π<argz<5π4=−z;z∈R4,R4,5=z∈ℂ;0<r<z<1,5π4<argz<3π2=−iz¯;z∈R4,R4,6=z∈ℂ;0<r<z<1,3π2<argz<7π4=−iz;z∈R4,R4,7=z∈ℂ;0<r<z<1,7π4<argz<2π=z¯;z∈R4.We can easily notice that (see Figure2)(7)∪k=07R4,k¯=R¯=z∈ℂ;0<r≤z≤1.Figure 2
Unit ring.The pointsz,iz¯,iz,−z¯,−z,−iz¯,−iz,z¯ are reflected at ∂D onto the points:(8)1z¯,iz,iz¯,−1z,−1z¯,−iz,−iz¯,1z.The pointsz,iz¯,iz,−z¯,−z,−iz¯,−iz,z¯ are reflected at ∂Dr onto the points:(9)r2z¯,ir2z,ir2z¯,−r2z,−r2z¯,−ir2z,−ir2z¯,r2z,where Dr=z∈ℂ;z<r<1Point (7) is reflected at ∂Dr onto the points:(10)r2z,ir2z¯,ir2z,−r2z¯,−r2z,−ir2z¯,−ir2z,r2z¯.Point (10) is reflected at ∂D onto(11)1r2z¯,ir2z,ir2z¯,−1r2z,−1r2z¯,−ir2z,−ir2z¯,1r2z.Point (11) is reflected at ∂Dr onto(12)r4z,ir4z¯,ir4z,−r4z¯,−r4z,−ir4z¯,−ir4z,r4z¯.When continuing that way, we obtain(13)±r2kz,±r2kz¯,±ir2kz,±ir2kz¯,±1r2kz,±1r2kz¯,±ir2kz,±ir2kz¯,where k∈ℕ.Point (9) is reflected at ∂D onto(14)zr2,iz¯r2,izr2,−z¯r2,−zr2,−iz¯r2,−izr2,z¯r2.Points (14) are reflected at ∂Dr onto(15)r4z¯,ir4z,ir4z¯,−r4z,−r4z¯,−ir4z,−ir4z¯,r4z.When continuing that way, we obtain(16)±zr2k,±z¯r2k,±izr2k,±iz¯r2k,±r2kz,±r2kz¯,±ir2kz,±ir2kz¯,where k∈ℕEach point of sets (13) and (16) belong to a different domain Dm; this domains parquets the complex plane:(17)ℂ=∪m∈IDm¯,Iis countable set.


## 3. Main Result

In this section, we have determined the Cauchy–Pompeiu representation formula in octant unit ringR4 by applying Theorem 1 at points (13) and (16).Theorem 2.
Any functionw∈C1R4;ℂ∩CR4¯;ℂ can be represented as follows:(18)wz=2πi∫ζ=1Reζ,Imζ>0Rewζζ4+z4ζ4−z4−ζ¯4+z4ζ¯4−z4+2∑k=1∞r8kζ4r8kζ4−z4−z4r8kz4−ζ4+z4r8kz4−ζ¯4−ζ¯4r8kζ¯4−z4.dζζ−2πi∫ζ=rReζ,Imζ>0Rewζζ4+z4ζ4−z4−ζ¯4+z4ζ¯4−z4+2∑k=1∞r8kζ4r8kζ4−z4−z4r8kz4−ζ4+z4r8kz4−ζ¯4−ζ¯4r8kζ¯4−z4.dζζ+4πi∫r1Rewζt3t4−z4−t3z41−t4z4+∑k=1∞r8kt3r8kt4−z4−z4tr8kz4−t4+t3z4r8kt4z4−1−1tr8k−t4z4dt+4πi∫1rRewteπ/4it3t4+z4+t3z4t4z4+1+∑k=1∞r8kt3r8kt4+z4−z4r8kz4+t4+t3z4r8kt4z4−1−1r8k+t4z4dt+4π∫ζ=1Reζ,Imζ>0Imwζ.dζζ−4π∫R4wζ¯ζζ3ζ4−z4+ζ3z4ζ4z4−1+∑k=1∞r8kζ3r8kζ4−z4−z4ζr8kz4−ζ4+ζ3z4r8kζ4z4−1−1ζr8k−ζ4z4−wζ¯ζ¯ζ¯3ζ¯4−z4+ζ¯3z4ζ¯4z4−1+∑k=1∞r8kζ¯3r8kζ¯4−z4−z4ζ¯r8kz4−ζ¯4+ζ¯3z4r8kζ¯4z4−1−1ζ¯r8k−ζ¯4z4dξdη,where ζ=ξ+iη, and z∈R4.Proof.
Using Theorem1 for the points ±r2kz,±z/r2k, for every k∈ℕ, we find(19)wz=12πi∫∂R4wζdζζ−z−1π∫R4wζ¯ζdξdηζ−z,(20)0=12πi∫∂R4wζdζζ+z−1π∫R4wζ¯ζdξdηζ+z,(21)0=12πi∫∂R4wζ∑k=1∞1ζ−r2kzdζ−1π∫R4wζ¯ζ∑k=1∞1ζ−r2kzdξdη,0=12πi∫∂R4wζ∑k=1∞1ζ+r2kzdζ−1π∫R4wζ¯ζ∑k=1∞1ζ+r2kzdξdη,0=12πi∫∂R4wζ∑k=1∞1ζ−z/r2kdζ−1π∫R4wζ¯ζ∑k=1∞1ζ−z/r2kdξdη,(22)0=12πi∫∂R4wζ∑k=1∞1ζ+z/r2kdζ−1π∫R4wζ¯ζ∑k=1∞1ζ+z/r2kdξdη.
When adding (19)–(22) and fixing the terms, we obtain(23)wz=12πi∫∂R4wζ2ζζ2−z2+∑k=1∞2ζζ2−r4kz2+2ζr4kr4kζ2−z2dζ−1π∫R4wζ¯ζ2ζζ2−z2+∑k=1∞2ζζ2−r4kz2+2ζr4kr4kζ2−z2dξdη.
Similarly, when using Theorem1 for the points, ±ir2kz,±iz/r2k, for k∈ℕ, and fixing the terms, we find(24)0=12πi∫∂R4wζ2ζζ2+z2+∑k=1∞2ζζ2+r4kz2+2ζr4kr4kζ2+z2dζ−1π∫R4wζ¯ζ2ζζ2+z2+∑k=1∞2ζζ2+r4kz2+2ζr4kr4kζ2+z2dξdη.
When using Theorem1 for the points, ±1/r2kz,±r2k/z,±i/r2kz,±ir2k/z, for k∈ℕ, and fixing the terms we also find(25)0=12πi∫∂R4wζ2ζz2ζ2z2−1+∑k=1∞2ζz2r4kr4kζ2z2−1+2ζz2ζ2z2−r4kdζ−1π∫R4wζ¯ζ2ζz2ζ2z2−1+∑k=1∞2ζz2r4kr4kζ2z2−1+2ζz2ζ2z2−r4kdξdη,(26)0=12πi∫∂R4wζ2ζz2ζ2z2+1+∑k=1∞2ζz2r4kr4kζ2z2+1+2ζz2ζ2z2+r4kdζ−1π∫R4wζ¯ζ2ζz2ζ2z2+1+∑k=1∞2ζz2r4kr4kζ2z2+1+2ζz2ζ2z2+r4kdξdη.
Finally, when using Theorem1 for the points, we obtain(27)±r2kz¯,±z¯r2k,±ir2kz¯,±iz¯r2k,±1r2kz¯,±r2kz¯,±ir2kz¯,±ir2kz¯.
Fork∈ℕ, we find(28)0=12πi∫∂R4wζ2ζζ2−z¯2+∑k=1∞2ζζ2−r4kz¯2+2ζr4kr4kζ2−z¯2dζ−1π∫R4wζ¯ζ2ζζ2−z¯2+∑k=1∞2ζζ2−r4kz¯2+2ζr4kr4kζ2−z¯2dξdη,(29)0=12πi∫∂R4wζ2ζζ2+z¯2+∑k=1∞2ζζ2+r4kz¯2+2ζr4kr4kζ2+z¯2dζ−1π∫R4wζ¯ζ2ζζ2+z¯2+∑k=1∞2ζζ2+r4kz¯2+2ζr4kr4kζ2+z¯2dξdη,(30)0=12πi∫∂R4wζ2ζz¯2ζ2z¯2−1+∑k=1∞2ζz¯2r4kr4kζ2z¯2−1+2ζz¯2ζ2z¯2−r4kdζ−1π∫R4wζ¯ζ2ζz¯2ζ2z¯2−1+∑k=1∞2ζz¯2r4kr4kζ2z¯2−1+2ζz¯2ζ2z¯2−r4kdξdη,(31)0=12πi∫∂R4wζ2ζz¯2ζ2z¯2+1+∑k=1∞2ζz¯2r4kr4kζ2z¯2+1+2ζz¯2ζ2z¯2+r4kdζ−1π∫R4wζ¯ζ2ζz¯2ζ2z¯2+1+∑k=1∞2ζz¯2r4kr4kζ2z¯2+1+2ζz¯2ζ2z¯2+r4kdξdη.
Since0∉R4, we can rewrite (23)–(31) as follows:(32)wz=1πi∫∂R4wζζ2ζ2−z2+∑k=1∞r4kζ2r4kζ2−z2+z2ζ2−r4kz2dζζ−2π∫R4wζ¯ζζζ2−z2+∑k=1∞r4kζr4kζ2−z2+z2ζζ2−r4kz2dξdη,(33)0=1πi∫∂R4wζζ2ζ2+z2+∑k=1∞r4kζ2r4kζ2+z2−z2ζ2+r4kz2dζζ−2π∫R4wζ¯ζζζ2+z2+∑k=1∞r4kζr4kζ2−z2−z2ζζ2−r4kz2dξdη,(34)0=1πi∫∂R4wζζ2z2ζ2z2−1+∑k=1∞r4kζ2z2r4kζ2z2−1+1ζ2z2−r4kdζζ−2π∫R4wζ¯ζζz2ζ2z2−1+∑k=1∞r4kζz2r4kζ2z2−1+1ζζ2z2−r4kdξdη,(35)0=1πi∫∂R4wζζ2z2ζ2z2+1+∑k=1∞r4kζ2z2r4kζ2z2+1−1ζ2z2+r4kdζζ−2π∫R4wζ¯ζζz2ζ2z2+1+∑k=1∞r4kζz2r4kζ2z2+1−1ζζ2z2+r4kdξdη,(36)0=1πi∫∂R4wζζ2ζ2−z¯2+∑k=1∞r4kζ2r4kζ2−z¯2+z¯2ζ2−r4kz¯2dζζ−2π∫R4wζ¯ζζζ2−z¯2+∑k=1∞r4kζr4kζ2−z¯2+z¯2ζζ2−r4kz¯2dξdη,(37)0=1πi∫∂R4wζζ2ζ2+z¯2+∑k=1∞r4kζ2r4kζ2+z¯2−z¯2ζ2+r4kz¯2dζζ−2π∫R4wζ¯ζζζ2+z¯2+∑k=1∞r4kζr4kζ2+z¯2−z¯2ζζ2+r4kz¯2dξdη,(38)0=1πi∫∂R4wζζ2z¯2ζ2z¯2−1+∑k=1∞r4kζ2z¯2r4kζ2z¯2−1+1ζ2z¯2−r4kdζζ−2π∫R4wζ¯ζζz¯2ζ2z¯2−1+∑k=1∞r4kζz¯2r4kζ2z¯2−1+1ζζ2z¯2−r4kdξdη,(39)0=1πi∫∂R4wζζ2z¯2ζ2z¯2+1+∑k=1∞r4kζ2z¯2r4kζ2z¯2+1−1ζ2z¯2+r4kdζζ−2π∫R4wζ¯ζζz¯2ζ2z¯2+1+∑k=1∞r4kζz¯2r4kζ2z¯2+1−1ζζ2z¯2+r4kdξdη.
Adding (32) and (33) gives(40)wz=2πi∫∂R4wζζ4ζ4−z4+∑k=1∞r8kζ4r8kζ4−z4+z4ζ4−r8kz4dζζ−4π∫R4wζ¯ζζ3ζ4−z4+∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4dξdη.
Adding (36) and (37) gives(41)0=2πi∫∂R4wζζ4ζ4−z¯4+∑k=1∞r8kζ4r8kζ4−z¯4+z¯4ζ4−r8kz¯4dζζ−4π∫R4wζ¯ζζ3ζ4−z¯4+∑k=1∞r8kζ3r8kζ4−z¯4+z¯4ζζ4−r8kz¯4dξdη.
Adding (34) and (35) gives(42)0=2πi∫∂R4wζζ4z4z4ζ4−1+∑k=1∞r8kζ4z4r8kz4ζ4−1+1z4ζ4−r8kdζζ−4π∫R4wζ¯ζζ3z4ζ4z4−1+∑k=1∞r8kζ3z4r8kζ4z4−1+1ζζ4z4−r8kdξdη.
Adding (38) and (39) gives(43)0=2πi∫∂R4wζζ4z¯4z¯4ζ4−1+∑k=1∞r8kζ4z¯4r8kz¯4ζ4−1+1z¯4ζ4−r8kdζζ−4π∫R4wζ¯ζζ3z¯4ζ4z¯4−1+∑k=1∞r8kζ3z¯4r8kζ4z¯4−1+1ζζ4z¯4−r8kdξdη.
When taking the complex conjugate of (41) and (43), we can obtain(44)0=2πi∫ζ=1Reζ,Imζ>0wζ¯11−ζ4z4+∑k=1∞r8k1r8k−ζ4z4+z4ζ41−r8kz4dζζ+2πi∫ζ=rReζ,Imζ>0wζ¯r8r8−ζ4z4+∑k=1∞r8kr8r8k+1−ζ4z4+ζ4z4r8−r8kζ4z4dζζ−2πi∫r1wt¯t4t4−z4+∑k=1∞r8kt4r8kt4−z4+z4t4−r8kz4dtt−2πi∫1rwt.eπ/4i¯t4t4+z4+∑k=1∞r8kt4r8kt4+z4−z4t4+r8kz4dtt−4π∫R4wζ¯ζ¯ζ¯3ζ¯4−z4+∑k=1∞r8kζ¯3r8kζ¯4−z4+z4ζ¯ζ¯4−r8kz4dξdη,(45)0=2πi∫ζ=1Reζ,Imζ>0wζ¯z4z4−ζ4+∑k=1∞r8kz4r8kz4−ζ4+ζ4z4−r8kζ4dζζ+2πi∫ζ=rReζ,Imζ>0wζ¯r8z4r8z4−ζ4+∑k=1∞r8kr8z4r8k+1z4−ζ4+ζ4z4r8−r8kζ4dζζ−2πi∫r1wt¯t4z4z4t4−1+∑k=1∞r8kt4z4r8kz4t4−1+1z4t4−r8kdtt−2πi∫1rwt.eπ/4i¯t4z4z4t4+1+∑k=1∞r8kt4z4r8kz4t4+1−1z4t4+r8kdtt−4π∫R4wζ¯ζ¯ζ¯3z4ζ¯4z4−1+∑k=1∞r8kζ¯3z4r8kζ¯4z4−1+1ζ¯ζ¯4z4−r8kdξdη.
When adding (40)–(42) and then subtracting (44) and (45) from the result, we can obtain(46)wz=2πi∫ζ=1Reζ,Imζ>0wζζ4ζ4−z4+ζ4z4ζ4z4−1+∑k=1∞r8kζ4r8kζ4−z4+z4ζ4−r8kz4+ζ4z4r8kζ4z4−1+1z4ζ4−r8k−wζ¯11−ζ4z4+z4z4−ζ4+∑k=1∞r8kζ4z4−r8kζ4+z4r8kz4−ζ4+ζ4z41−r8kζ4z4+1r8k−z4ζ4dζζ+2πi∫ζ=rReζ,Imζ>0wζζ4ζ4−z4+ζ4z4ζ4z4−1+∑k=1∞r8kζ4r8kζ4−z4+z4ζ4−r8kz4+ζ4z4r8kζ4z4−1+1z4ζ4−r8k−wζ¯r8r8−ζ4z4+r8z4r8z4−ζ4+∑k=1∞r8kr8r8k+1−ζ4z4+z4ζ4r8−r8kz4ζ4+r8z4r8k+1z4−ζ4+ζ4r8z4−r8kζ4dζζ+2πi∫r1wtt4t4−z4+t4z4t4z4−1+∑k=1∞r8kt4r8kt4−z4+z4t4−r8kz4+t4z4r8kt4z4−1+1z4t4−r8k−wt¯t4z4t4z4−1+t4t4−z4+∑k=1∞r8kt4r8kt4−z4+z4t4−r8kz4+t4z4r8kt4z4−1+1t4z4−r8kdtt+2πi∫1rwt.eπ/4it4t4+z4+t4z4t4z4+1+∑k=1∞r8kt4r8kt4+z4−z4t4+r8kz4+t4z4r8kt4z4+1−1z4t4+r8k−wt.eπ/4i¯t4t4+z4+t4z4t4z4+1+∑k=1∞r8kt4r8kt4+z4−z4t4+r8kz4+t4z4r8kt4z4+1−1t4z4+r8kdtt−4π∫R4wζ¯ζζ3ζ4−z4+ζ3z4z4ζ4−1+∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4+ζ3z4r8kζ4z4−1+1ζz4ζ4−r8k−wζ¯ζ¯ζ¯3ζ¯4−z4+ζ¯3z4ζ¯4z4−1+∑k=1∞r8kζ¯3r8kζ¯4−z4+z4ζ¯ζ¯4−r8kz4+ζ¯3z4r8kζ¯4z4−1+1ζ¯ζ¯4z4−r8kdξdη.
We notice(47)r8r8−ζ4z4+r8z4r8z4−ζ4+∑k=1∞r8kr8r8k+1−ζ4z4+z4ζ4r8−r8kz4ζ4+r8z4r8k+1z4−ζ4+ζ4r8z4−r8kζ4=z4ζ41−z4ζ4+ζ4z4−ζ4+∑k=1∞r8k1r8k−ζ4z4+z4ζ4r8−r8kz4ζ4+z4r8kz4−ζ4+ζ4r8z4−r8kζ4.
By puttingwζ=Rewζ+i.Imwζ on 3.26 and fixing the terms, we end up the proof.
The Cauchy–Pompeiu’s operator inR4 is the area integral in the modulated Cauchy–Pompeiu representation in R4; it is given by the following formula:(48)TR4fz=−4π∫R4wζ¯ζζ3ζ4−z4+ζ3z4z4ζ4−1+∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4+ζ3z4r8kζ4z4−1+1ζz4ζ4−r8k−wζ¯ζ¯ζ¯3ζ¯4−z4+ζ¯3z4ζ¯4z4−1+∑k=1∞r8kζ¯3r8kζ¯4−z4+z4ζ¯ζ¯4−r8kz4+ζ¯3z4r8kζ¯4z4−1+1ζ¯ζ¯4z4−r8kdξdη,f∈L1R4;ℂ.


## 4. The OperatorTR4

The Cauchy–Pompeiu’s operator inR4 includes an infinite series; therefore, we have to prove its convergence; this will be done through the following two lemmas.Lemma 1.
For everyz,ζ∈R4, with z≠ζ, we have the following:(49)ζ−z<ζ−zk,zk=±1r2kz,±ir2kz,±zr2k,±izr2k,ζ−r2z¯<ζ−zk,zk=±r2kz,±ir2kz,±r2kz,±ir2kz.where k∈ℕ.Proof.
Let=ξ+iη and z=x+iy; then,(50)ζ−1z¯2−ζ−z2=ζ−1z¯ζ¯−1z−ζ−zζ¯−z¯=ζ2−ζ¯z¯−ζz+1z2−ζ2+zζ¯+z¯ζ−z2=ζ¯z2−1z¯+ζz2−1z−z2−1z2+1z2=z2−1z2zζ¯+z¯ζ−z2−1=1x2+y2−1x2+y2−2ξx−2ηy+1≥1x2+y2−11−ξ2−η2>0⇒ζ−z<ζ−1z¯,where x−ξ2≥0⇒x2−2ξx≥−ξ2,y−η2≥0⇒y2−2yη≥−η2,(51)z2=x2+y2<1⇒1x2+y2−1>0,ζ2=ξ2+η2<1⇒1−ξ2−η2>0.
Sinceζ−1/z¯<ζ−1/z≤ζ−1/r2kz, for every k∈ℕ, we have(52)ζ−z<ζ−1z≤ζ−1rkz.
By the symmetry properties, we have(53)ζ−1z<ζ−iz≤ζ−ir2kz,ζ−1z<ζ+iz≤ζ+ir2kz,ζ−1z<ζ+1z≤ζ+1r2kz.
This leads to(54)ζ−z<ζ−zk,zk=±1r2kz,±ir2kz,for every k∈ℕ.
On the contrary, we have(55)ζ−z≤ζ−zr2k.
By the symmetry properties, we have(56)ζ−zr2k<ζ+zr2k,ζ−zr2k<ζ−izr2k,ζ−zr2k<ζ+izr2k.
This leads to(57)ζ−z<ζ−zk,zk=±zr2k,±izr2k,for every k∈ℕ.
We have(58)r2z=r2z<r2,r2z¯=r2z>r2⇒r2z<r2<r2z¯<r<ζ.
Therefore,(59)ζ−r2z¯<ζ−r2z.
Since(60)ζ−r2z≤ζ−r2kz,k∈ℕ,we can obtain(61)ζ−r2z¯<ζ−r2kz,k∈ℕ.
By symmetry properties, we have(62)ζ−r2z¯<ζ−r2z≤ζ−r2kz,ζ−r2z¯<ζ−r2z<ζ+r2z≤ζ+r2kz,ζ−r2z¯<ζ−r2z<ζ−ir2z≤ζ−ir2kz,ζ−r2z¯<ζ−r2z<ζ+ir2z≤ζ+ir2kz,ζ−r2z¯<ζ−r2z<ζ+r2z≤ζ+r2kz,ζ−r2z¯<ζ−r2z<ζ−ir2z≤ζ−ir2kz,ζ−r2z¯<ζ−r2z<ζ+ir2z≤ζ+ir2kz,for every k∈ℕ.
This means(63)ζ−r2z¯<ζ−zk,zk=±r2kz,±ir2kz,±r2kz,±ir2kz.Lemma 2.
For everyz,ζ∈R4, with z≠ζ, 0<r<1, the series is as follows:(64)A=∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4+ζ3z4r8kζ4z4−1+1ζz4ζ4−r8k.
Is it uniformly converged.Proof.
First, we have(65)ζ3r8kζ4−z4<1r8kζ4−z4=14z3r2k.1ζ−z/r2k−1ζ+z/r2k+iζ−iz/r2k−iζ+iz/r2k<14r2k+31ζ−z/r2k+1ζ+z/r2k+1ζ−iz/r2k+1ζ+iz/r2k,ζ<1,r<z.
From Lemma1, we have(66)1ζ−zk<1ζ−z,zk=±zr2k,±izr2k,⇒ζ3r8kζ4−z4<1r2k+3.1ζ−z.
Second, we have(67)z4ζζ4−r8kz4=z4r6nζ1ζ−r2kz−1ζ+r2kz+iζ−ir2kz−iζ+ir2kz<14r6k+11ζ−r2kz+1ζ+r2kz+1ζ−ir2kz+1ζ+ir2kz.
From Lemma1, we have(68)1ζ−zk<1ζ−r2/z¯,zk=±r2kz,±ir2kz⇒z4ζζ4−r8kz4<1r6k+1.1ζ−r2/z¯.
Third, we have(69)ζ3z4r8kζ4z4−1<z4r8kζ4z4−1=14z3r2k.1ζ−1/r2kz−1ζ+1/r2kz+iζ−i/r2kz−iζ+i/r2kz.
From Lemma1, we have(70)1ζ−zk<1ζ−z,zk=±1r2kz±ir2kz,⇒ζ3z4r8kζ4z4−1<1r2k+3.1ζ−z.
Fourth, we have(71)1ζζ4z4−r8k=1ζzr6.1ζ−r2k/z−1ζ+r2k/z+iζ−ir2k/z−iζ+ir2k/z<14r6n+21ζ−r2k/z+1ζ+r2k/z+1ζ−ir2k/z+1ζ+ir2k/z.
From Lemma1, we have(72)1ζ−zk<1ζ−r2/z¯,zk=±r2kz,±ir2kz,⇒1ζζ4z4−r8k<1r6k+2.1ζ−r2/z¯.
When using (66), (68), (70), and (72), we can obtain(73)A≤∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4+ζ3z4r8kζ4z4−1+1ζz4ζ4−r8k<∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4+ζ3z4r8kζ4z4−1+1ζz4ζ4−r8k<∑k=1∞r8k1r2k+3ζ−z+1r6k+1ζ−r2/z¯+1r2k+3ζ−z+1r6k+2ζ−r2/z¯≤2ζ−z∑k=1∞r6k−3+2ζ−r2/z¯∑k=1∞r2k−2<+∞,r6<r2<1.
And, the seriesA is uniformly converged.


## 5. The Boundedness of Cauchy–Pompeiu’s Operator inR4

The boundedness of the operatorT is proved by Tutschke and Mshimba in [8] by applying the Schmitz inequality. In this section, we will use the same technique to prove the boundedness of the modulated Cauchy–Pompeiu’s operator in R4.Theorem 3.
(Schmitz’s inequality [8]). For any regular domain D⊂ℂ, we have(74)∫Ddξdηζ−zα≤2π2−αSDπ1−α/2,where ζ=ξ+iη, z∈D, α is a real number consist 0<α<2, and SD is the area of the domain D.Theorem 4.
LetD⊂ℂ be a bounded domain; then, for f∈LpD¯;ℂ, p>2, T is a completely linear operator from LpD¯;ℂ into Cαℂ with α=p−2/p.Theorem 5.
The Cauchy–Pompeiu’s operator in the octant ringR4 is(75)TR4:CPR4;ℂ⟶CαR4¯;ℂ,TR4fz=−4π,∫R4fζζ3ζ4−z4+ζ3z4z4ζ4−1+∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4+ζ3z4r8kζ4z4−1+1ζz4ζ4−r8k−fζ¯ζ¯3ζ¯4−z4+ζ¯3z4ζ¯4z4−1+∑k=1∞r8kζ¯3r8kζ¯4−z4+z4ζ¯ζ¯4−r8kz4+ζ¯3z4r8kζ¯4z4−1+1ζ¯ζ¯4z4−r8kdξdη,which is bounded, where p>2 and α=p−2/p.Proof.
First, we have(76)4ζ3ζ4−z4=1ζ−z+1ζ+z+1ζ−iz+1ζ+iz⇒4ζ3ζ4−z4<4ζ−z,4ζ3z4ζ4z4−1=1ζ−1/z+1ζ+1/z+1ζ−i/z+1ζ+i/z⇒4ζ3z4ζ4z4−1<4ζ−z.
From Lemmas1 and 2, we can write(77)TR4fz≤4π∫R4fζ.ζ3ζ4−z4+ζ3z4z4ζ4−1+∑k=1∞r8kζ3r8kζ4−z4+z4ζζ4−r8kz4+ζ3z4r8kζ4z4−1+1ζz4ζ4−r8k+|fζ.ζ¯3ζ¯4−z4+ζ¯3z4ζ¯4z4−1+∑k=1∞r8kζ¯3r8kζ¯4−z4+z4ζ¯ζ¯4−r8kz4+ζ¯3z4r8kζ¯4z4−1+1ζ¯ζ¯4z4−r8kdξdη,f∈LpR4;ℂ,p>2<8π∫R4fCp8ζ−z+∑k=1∞2r6k−3ζ−z+2r2k−2ζ−r2/z¯dξdη,where fCp=maxζ∈R4fζ, with f∈CpR4;ℂ⊂LpR4;ℂ,(78)TR4fz<8fCpπ∫R48ζ−z+2∑k=1∞r2k−3ζ−z+r2k−3ζ−r2/z¯dξdη≤16fCpπ∫R44ζ−z+11−r2r−1ζ−z+r−1ζ−r2/z¯dξdη≤16fCpπ∫R44+1r1−r21ζ−z+1r1−r2.1ζ−r2/z¯dξdη⇒TR4fz<16πfCpI1+I2,where(79)I1=4+1r1−r2∫R4dξdηζ−z.
By Schmitz’s inequality, we have(80)I1≤4+1r1−r2.2π2−1π/81−r2π1−1/2=4+1r1−r2.π1−r22.
On the contrary, we have(81)I2=1r1−r2∫R4dξdηζ−r2/z¯≤1r1−r2∫R4dξdηζ−r2/z≤∫r1∫0π/4tdtdθt−r2/z=π4∫r1tdtt−r2/z,where r<z<1⇒1<1/z<1/r⇒r2<r2/z<r⇒−r<−r2/z<−r2,r<t<1.
Therefore, we have(82)0<t−r2z<1−r2⇒t−r2z=t−r2zI2≤π4∫r11+r2/zt−r2/zdt=π4t+r2zlnt−r2zr1I2≤π41−r+r2rlnz−r2rz−r2<π41−r+rz−r2rz−r≤π41−r+z+r<π/2⇒I1+I2<4+1r1−r2.π1−r22+π2=λ⇒TR4fz<16λπfCp.
And, the operatorTR4 is bounded.



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*Source: 1816883-2022-12-15.xml*