Notions of Regularity for Functions of a Split-Quaternionic Variable
aa r X i v : . [ m a t h . C V ] J un Notions of Regularity for Functions of aSplit-Quaternionic Variable
John A. EmanuelloFlorida State University [email protected]
Craig A. NolderFlorida State University [email protected]
July 12, 2018
Abstract
Notions of a “holomorphic” function theory for functions of a split-quaternionic variablehave been of recent interest. We describe two found in the literature and show that one notionencompasses a small class of functions, while the other gives a richer collection. In the secondinstance, we describe a simple subclass of functions and give two examples of an analogue ofthe Cauchy-Kowalewski extension in this context.
One need look no further than a text on complex analysis, such as [1] or especially [5], to know thatalgebraic properties of C play a major role in the analysis and geometry of the plane. The simplefact that i = − ∆ z → f ( z + ∆ z ) − f ( z )∆ z means that the limit is the same whether ∆ z = ∆ x or ∆ z = i ∆ y . That is, ∂u∂x + i ∂v∂x = 1 i ∂u∂y + ∂v∂y , and the C-R equations are obtained: ∂u∂x = ∂v∂y and ∂v∂x = − ∂u∂y . The minus sign in the second equation occurs because i = − i , which is a direct consequence of i = −
1. Thus, when a function of a complex variable with C components is holomorphic if andonly if the C-R equations are satisfied. 1ne may also consider functions of a complex variable which are annihilated by the operator ∂ ¯ z := 12 (cid:18) ∂∂x + i ∂∂y (cid:19) . Indeed, a C function is holomorphic if and only if it is annihilated by ∂ ¯ z and its complex derivativeis given by ∂ z f , where ∂ z := 12 (cid:18) ∂∂x − i ∂∂y (cid:19) . Unlike in the complex case, when we consider functions of a split-quaternionic variable andexplore the two analogous ways of defining a holomorphic function, we find that they are not equiv-alent. Thus, two different theories of holomorphic functions can be studied, as in [8, 9]. However,the one in [8] stands out as the more natural analogue because it gives rise to a (relatively) largeclass of functions to be studied. Indeed, for the analogue defined in [9] we show (by adopting a proofof an analogous statement in Sudbery’s paper [13]) that only affine functions, which is a (relatively)small class of functions, satisfy the given conditions.
The split-quaternions are the real Clifford algebra Cℓ , := { Z = x + x i + x j + x ij : x , x , x , x ∈ R } . Functions of a split-quaternionic variable and notions of regularity have been the subject ofinterest in the literature [8, 9]. It is worth noting that the split-quaternions contain both thecomplex and split-complex numbers as subalgebras.In a manner similar to the split-complex case [4,6], we may obtain the indefinite quadratic form Q , by ZZ = x + x − x − x . Hence we shall identify the split-quaternions with R , .There are a number of ways to express the split-quaternions as 2 × R and C . Lemma 1.1.
As algebras, the split-quaternions and real × matrices are isomorphic.Proof. If we identify 1 ∼ " , i ∼ " −
11 0 , j ∼ " , then we may map Cℓ , to the real 2 × x + x i + x j + x ij " x + x − x + x x + x x − x . " x + x − x + x x + x x − x = x + x − x − x , which is the form Q , . It’s easy to check that this gives an algebra homomorphism. Further, " y y y y
12 [( y + y ) + ( y − y ) i + ( y + y ) j + ( y − y ) ij ]gives a two-sided inverse, so that the above is an algebra isomorphism. (cid:3) The functions we are concerned with are f : U ⊆ R , → Cℓ , , where U is open (in the euclidean sense). As higher dimensional analogues of functions of a complexvariable, we are interested in obtaining an analogous definition for holomorphic . As we shall see,there are various ways of doing this in the literature.The first and most interesting way is through split quaternionic valued differential operators [8].The second is more recent and less interesting and is obtained by considering a difference quotient [9]. Recall that in complex analysis, one considers the Dirac operators ∂ z and ∂ z , whose product (ineither order) gives the Laplacian for R , usually denoted by ∆. Of course, f is called holomorphicif ∂ z f = 0 and its (complex) derivative is given by ∂ z f . Additionally, the real and imaginary partsof f are harmonic functions, and the Dirichlet problem is well-posed.The question asked in the literature is: Can we define operators valued in Cℓ , which resemble ∂ z and ∂ z ? This question has been answered in the affirmative, although with little mention of thedifferential geometry which lies just below the surface.However the question we are really asking is: can we factorize the Laplacian in R , withlinear first order operators over Cℓ , ? In this semi-Riemannian manifold, the Laplacian, which isunderstood to be the derivative of the gradient, is given by [11]:∆ , = ∂ ∂x + ∂ ∂x − ∂ ∂x − ∂ ∂x . It is easy to check that the linear operators ∂ := ∂∂x + i ∂∂x − j ∂∂x − ij ∂∂x and ∂ := ∂∂x − i ∂∂x + j ∂∂x + ij ∂∂x , . Due to the non-commutativity of Cℓ , , these operators may be applied tofunctions on either the left or right and with different results, in general. Remark 2.1.
There are other factorizations of ∆ , inside Cℓ , . Our choice of ∂ is deliberate–it is the gradient inside the semi-Riemannian manifold R , . For alternative interpretation of thisidea, see [10]. Definition 2.2.
Let U ⊂ Cℓ , ∼ = R , and let F : U → Cℓ , be C ( U ) . We say F is left regularif ∂F = 0 for every Z ∈ U . Similarly, we say F is right regular if F ∂ = 0 for every Z ∈ U . We have adopted the above definition from [8], which contains a proof of a Cauchy-like integralformula for left-regular functions.By multiplying arbitrary F with ∂ we obtain the following conditions which make it easier tocheck left and right regularity. Proposition 2.3.
Let F : U → Cℓ , be C ( U ) . Then F is left regular if and only if it satisfiesthe system of PDEs: ∂f ∂x − ∂f ∂x − ∂f ∂x − ∂f ∂x = 0 ∂f ∂x + ∂f ∂x + ∂f ∂x − ∂f ∂x = 0 ∂f ∂x − ∂f ∂x − ∂f ∂x − ∂f ∂x = 0 ∂f ∂x + ∂f ∂x + ∂f ∂x − ∂f ∂x = 0 . Proof.
The proof follows directly from the definition. Simply multiply in the proper order, collectlike components together, and equate them to zero to obtain the desired system. (cid:3)
Proposition 2.4.
Let F : U → Cℓ , be C ( U ) . Then F is right regular if and only if it satisfies he system of PDEs: ∂f ∂x − ∂f ∂x − ∂f ∂x − ∂f ∂x = 0 ∂f ∂x + ∂f ∂x − ∂f ∂x + ∂f ∂x = 0 ∂f ∂x + ∂f ∂x − ∂f ∂x + ∂f ∂x = 0 ∂f ∂x − ∂f ∂x − ∂f ∂x − ∂f ∂x = 0 . However, this notion of regularity is also some what unsatisfying, for simple analogues of holo-morphic functions in the complex plane are not regular.
Example 2.5.
Let A = a + ib + jc + ijd ∈ Cℓ , . Then AZ =( ax − bx + cx + dx ) + i ( bx + ax + dx − cx )+ j ( cx + dx + ax − bx ) + ij ( dx − cx + bx + ax ) . Thus, ∂ ( AZ ) = ( a + ib + jc + dij ) + i ( − b + ai + dj − cij ) − j ( c + id + aj + bij ) − ij ( d − ic − bj + aij )= − a + i b + j c + ij d = − A, A similar calculation shows that ( AZ ) ∂ = − A. Other calculations show that the function ZA is also neither left-regular nor right-regular. We obtain similar systems of PDEs if we consider the equations ∂F = 0 and F ∂ = 0: ∂f ∂x + ∂f ∂x + ∂f ∂x + ∂f ∂x = 0 ∂f ∂x − ∂f ∂x − ∂f ∂x + ∂f ∂x = 0 ∂f ∂x + ∂f ∂x − ∂f ∂x + ∂f ∂x = 0 ∂f ∂x − ∂f ∂x − ∂f ∂x + ∂f ∂x = 05nd ∂f ∂x + ∂f ∂x + ∂f ∂x + ∂f ∂x = 0 ∂f ∂x − ∂f ∂x + ∂f ∂x − ∂f ∂x = 0 ∂f ∂x − ∂f ∂x + ∂f ∂x − ∂f ∂x = 0 ∂f ∂x + ∂f ∂x + ∂f ∂x + ∂f ∂x = 0 . These also produce unsatisfying analogues of holomorphic since linear functions, again, fail theseconditions.
Example 2.6.
Let A = a + ib + jc + ijd ∈ Cℓ , . Then, ∂ ( AZ ) = ( a + ib + jc + dij ) − i ( − b + ai + dj − cij )+ j ( c + id + aj + bij ) + ij ( d − ic − bj + aij )= 4 a. A similar calculation shows that ( AZ ) ∂ = 4 A. Other calculations show that the function ZA is not annihilated by ∂ on either side. Recall, another (and probably primary) way to define holomorphic functions is via the limit of adifference quotient: lim ∆ z → f ( z + ∆ z ) − f ( z )∆ z . One obtains the Cauchy-Riemann equations by allowing ∆ z to approach 0 along the real axis andagain along the imaginary axis and then setting the results equal to each other.In Masouri et. al., a similar method is used to produce another analogue of holomorphic [9].However, since the split-quaternions are not commutative, so there are two ways to construct ananalogue of the difference quotient. In Masouri the “quotient” is defined bylim ∆ Z → ( f ( Z + ∆ Z ) − f ( Z )) (∆ Z ) − . When this limit exists, such functions are called right Cℓ , -differentiable. By setting ∆ Z equalto ∆ x , i ∆ x , j ∆ x , and ij ∆ x , taking the limit in each instance, we get four ways to take the“derivative” [9]. That is, 6im ∆ x → ( f ( ζ + ∆ x ) − f ( ζ )) (∆ x ) − = ∂f ∂x + i ∂f ∂x + j ∂f ∂x + ij ∂f ∂x , lim i ∆ x → ( f ( ζ + ∆ x ) − f ( ζ )) ( i ∆ x ) − = − i ∂f ∂x + ∂f ∂x + ij ∂f ∂x − j ∂f ∂x , lim j ∆ x → ( f ( ζ + ∆ x ) − f ( ζ )) ( j ∆ x ) − = j ∂f ∂x + ij ∂f ∂x + ∂f ∂x + i ∂f ∂x , and lim ij ∆ x → ( f ( ζ + ∆ x ) − f ( ζ )) ( ij ∆ x ) − = ij ∂f ∂x − j ∂f ∂x − i ∂f ∂x + ∂f ∂x . Equating the four results, we obtain the system of PDEs [9]: ∂f ∂x = ∂f ∂x = ∂f ∂x = ∂f ∂x ∂f ∂x = − ∂f ∂x = ∂f ∂x = − ∂f ∂x ∂f ∂x = − ∂f ∂x = ∂f ∂x = − ∂f ∂x ∂f ∂x = ∂f ∂x = ∂f ∂x = ∂f ∂x . Although the work which introduces this notion of differentiability, [9], does not mention anyspecific examples of functions of right Cℓ , -differentiable functions, an entire class of functions canbe easily shown to have this property. Example 2.7.
Recall that AZ + K =( ax − bx + cx + dx + k ) + i ( bx + ax + dx − cx + ℓ )+ j ( cx + dx + ax − bx + m ) + ij ( dx − cx + bx + ax + n ) . Notice f ( Z ) = AZ + K is right Cℓ , -differentiable. Indeed, the “derivative” is lim ∆ Z → ( A ( Z + ∆ Z ) + K − AZ − K ) (∆ Z ) − = lim ∆ Z → ( A ∆ Z ) (∆ Z ) − = A. Theorem 2.8.
Let F : U ⊆ R , → Cℓ , . Then F is right Cℓ , -differentiable if and only F ( Z ) = AZ + K , where A, K ∈ Cℓ , . That is, the right Cℓ , -differentiable functions must beaffine mappings. roof. As similar fact is true for functions of a quaternionic variable and so we follow a similarproof from Sudbery’s paper [13].First notice that Z = ( x + x i ) + ( x + x i ) j = z + wj . As such we may write f ( Z ) = g ( z, w ) + h ( z, w ) j , where g ( z, w ) = f ( z, w ) + if ( z, w ) and h ( z, w ) = f ( z, w ) + if ( z, w ).Now, the above system of PDEs gives us that g is holomorphic with respect to the complexvariables z and w . Similarly, h is holomorphic with respect to the complex variables w and z .Additionally, ∂g∂z = ∂f ∂x + i ∂f ∂x = ∂f ∂x + i ∂f ∂x = ∂h∂w ,∂g∂w = − ∂f ∂x + i ∂f ∂x = − ∂f ∂x + i ∂f ∂x = ∂h∂z . Now, g and h have continuous partial derivatives of all orders. Thus, we must have ∂ g∂z = ∂∂z (cid:18) ∂h∂w (cid:19) = ∂∂w (cid:18) ∂h∂z (cid:19) = 0 ,∂ h∂w = ∂∂w (cid:18) ∂g∂z (cid:19) = ∂∂z (cid:18) ∂g∂w (cid:19) = 0 ,∂ g∂w = ∂∂w (cid:18) ∂h∂z (cid:19) = ∂∂z (cid:18) ∂h∂w (cid:19) = 0 ,∂ h∂z = ∂∂z (cid:18) ∂g∂w (cid:19) = ∂∂w (cid:18) ∂g∂z (cid:19) = 0 . W.L.O.G. we may assume that U is connected and convex (since each connected component may becovered by convex sets, which overlap pair-wise on convex sets). Thus integrating on line segmentsallows us to conclude that g and h are linear: g ( z, w ) = α + βz + γw + δzw,h ( z, w ) = ǫ + ηz + θw + νzw. Since ∂g∂z = ∂h∂w , we must have that β = θ and δ = ν = 0. Also since ∂g∂w = ∂h∂z , it is the case that γ = η . Thus, f ( Z ) = g ( z, w ) + h ( z, w ) j = ( α + βz + γw ) + ( ǫ + γz + βw ) j = ( β + γj )( z + wj ) + ( α + ǫj )= AZ + K, as required. (cid:3) Remark 2.9.
The above theorem proves that right Cℓ , -differentiable functions are not left orright regular and conversely (except for when A = 0 ). Indeed, they are also not annihilated by ∂ oneither side. We are very grateful to Professor Uwe K¨ahler of University of Aveiro for bringing this paper to our attention.
8s an alternative to the definition found in [9], one may reverse the multiplication in the differ-ence quotient to obtain lim ∆ Z → (∆ Z ) − ( f ( Z + ∆ Z ) − f ( Z )) . When this limit exists, such functions are called left Cℓ , -differentiable. Proceeding as above, aslightly different system of PDEs than the one found in [9] is obtained: ∂f ∂x = ∂f ∂x = ∂f ∂x = ∂f ∂x ∂f ∂x = − ∂f ∂x = − ∂f ∂x = ∂f ∂x ∂f ∂x = ∂f ∂x = ∂f ∂x = ∂f ∂x ∂f ∂x = − ∂f ∂x = − ∂f ∂x = ∂f ∂x . Example 2.10.
Recall that AZ =( ax − bx + cx + dx ) + i ( bx + ax + dx − cx )+ j ( cx + dx + ax − bx ) + ij ( dx − cx + bx + ax ) . Notice f ( Z ) = AZ + K is not left Cℓ , -differentiable.However, the map F ( Z ) = ZA + K is left Cℓ , -differentiable. Indeed, the “derivative” is lim ∆ Z → (∆ Z ) − (( Z + ∆ Z ) A + K − ZA − K ) = lim ∆ Z → (∆ Z ) − (∆ ZA ) = A. Theorem 2.11.
Let F : U ⊆ R , → Cℓ , . Then F is left Cℓ , -differentiable if and only F ( Z ) = ZA + K , where A, K ∈ Cℓ , . That is, the left Cℓ , -differentiable functions must be affinemappings.Proof. We can make a few adjustments to the proof of the right Cℓ , -differentiable case.First note that if we write F ( Z ) = g ( z, w ) + jh ( z, w ), where g ( z, w ) = f ( z, w ) + if ( z, w ), h ( z, w ) = f ( z, w ) − if ( z, w ), and the complex variables z, w as above.The system of PDEs above assures that g is holomorphic with respect to z and w , while h isholomorphic with respect to z and w . Additionally, we get that ∂g∂z = ∂h∂w∂g∂w = ∂h∂z .
9e also have that g and h have partial derivatives of all orders and similarly to the “right” casethe second partials vanish: ∂ g∂z = ∂ h∂w = ∂ g∂w = ∂ h∂z = 0 . Thus, by the same argument for the right Cℓ , -differentiable proof, we conclude that g and h are linear: g ( z, w ) = α + βz + γw + δzw,h ( z, w ) = ǫ + ηz + θw + νzw. Since ∂g∂z = ∂h∂w and ∂g∂w = ∂h∂z , we must have that β = θ , γ = η , and δ = ν = 0. Thus, f ( Z ) = g ( z, w ) + jh ( z, w )= ( α + βz + γw ) + j ( ǫ + γz + βw )= ( z + wj )( β + γj ) + ( α + ǫj )= ZA + K, as required. (cid:3) Remark 2.12.
Thus, right Cℓ , -differentiability is perhaps not a good analogue of holomorphic.Even though these are equivalent notions in the complex setting, in the split quaternionic settingthere are more directions in which to take the limit and this requires much stronger conditions.For this reason we are justified in studying functions in the kernels of the operators, and not the Cℓ , -differentiable functions. Given a F : U → Cℓ , whose components are at least C and which satisfies at least one of thefollowing: ∂F = 0 , F ∂ = 0 , ∂F = 0 , or F ∂ = 0 , must have components which satisfy John’s equation [8]:∆ , u = 0 . Such functions are said to be ultra-hyperbolic.In fact, we can use ultra-hyperbolic functions to build regular functions.
Theorem 2.13.
Let f : U → R be ultra-hyperbolic, then ∂f is both left and right regular.Proof. Write F = ∂f . Then clearly ∂F = ∆ , f = 0 = ( ∂f ) ∂ = F ∂. (cid:3)
10t turns out that left and right differentiable functions also have components which are ultra-hyperbolic.
Theorem 2.14.
Let F : U → Cℓ , , with components which are at least C , be left-differentiableor right-differentiable. Then the components of F are ultra-hyperbolic.Proof. Suppose F ( x , x , x , x ) = f + f i + f j + f ij is right-differentiable. Then notice that ∂ f ∂x + ∂ f ∂x − ∂ f ∂x − ∂ f ∂x = ∂∂x (cid:18) ∂f ∂x (cid:19) + ∂∂x (cid:18) − ∂f ∂x (cid:19) − ∂∂x (cid:18) − ∂f ∂x (cid:19) − ∂∂x (cid:18) ∂f ∂x (cid:19) = 0 . A similar argument works for the other f i and for the left-differentiable case. (cid:3) With all of these notions of holomorphic functions, it becomes necessary to choose one and deemit the “canonical” one. Since the difference quotients do not yield an extensive class of functions,we believe use of an operator to be the be the best place to start. Given the association between Cℓ , and R , , it seems ∂ is the ideal operator for our purposes, since it is also the gradient in R , (and since it is an analogue of ∂ ¯ z , which is times the gradient of R ). Given the overwhelmingconvention of applying operators on the left of functions, we choose left-regular to be the canonicalnotion of holomorphic.Indeed, this is the one chosen by Libine [8]. In his work, he shows that left-regular functionssatisfy a Cauchy-like integral formula. Theorem 3.1 (Libine’s Integral Formula) . Let U ⊆ Cℓ , be a bounded open (in the Euclideantopology) region with smooth boundary ∂U . Let f : U → Cℓ , be a function which extends to areal-differentiable function on an open neighborhood V ⊆ Cℓ , of U such that ∂f = 0 . Then for any Z ∈ Cℓ , such that the boundary of U intersects the cone C = n Z ∈ Cℓ , : ( Z − Z )( Z − Z ) = 0 o transversally, we have lim ǫ → − π Z ∂U ( Z − Z )( Z − Z )( Z − Z ) + iǫ k Z − Z k · dZ · f ( Z )= ( f ( Z ) if Z ∈ U else , where the three form dZ is given by dZ = dx ∧ dx ∧ dx − ( dx ∧ dx ∧ dx ) i + ( dx ∧ dx ∧ dx ) j − ( dx ∧ dx ∧ dx ) ij. Given this interesting property has been proven, it is some what surprising that a more detaileddescription of left-regular functions has not been given in the literature. So we conclude by showingthat some left regular functions have a simple description.11 .1 A Class of Left Regular Functions
To date, the author has not been able to find a description for left regular functions in a mannersimilar to the split-complex case [4,7]. It may be the case that no such description exists in general.However, it is possible to give a large class of left-regular functions a simple description.
Theorem 3.2.
Let F : U ⊆ R , → Cℓ , have the form F ( x , x , x , x ) = ( g ( x + x , x + x ) + g ( x − x , x − x ))+ ( g ( x − x , x − x ) + g ( x + x , x + x )) i + ( g ( x + x , x + x ) − g ( x − x , x − x )) j + ( g ( x − x , x − x ) − g ( x + x , x + x )) ij, where g i ∈ C ( U ) . Then ∂F = 0 .Proof. We can easily check that such an F satisfies the necessary system of PDEs. However, it isfar more enlightening to see how one can arrive at such a solution.Write F = f + f i + f j + f ij . Using an argument from [4], we have that ∂ = (cid:18) ∂∂x − j ∂∂x (cid:19) + i (cid:18) ∂∂x − j ∂∂x (cid:19) = 2 (cid:18) ∂∂v j + + ∂∂u j − (cid:19) + 2 i (cid:18) ∂∂v j + + ∂∂u j − (cid:19) := ∂ + i∂ , where u = x + x , v = x − x , u = x + x , u = x − x , j + = 1 + j j − = 1 − j j + and j − are idempotents and annihilate each other. Also, notice that ij + = j − i and ij − = j + i .Similarly, we may write F = ( F j + + F j − ) + i ( F j + + F j − ) . Now, one way in which ∂F = 0 is if ∂ ( F j + + F j − ) = ∂ ( F j + + F j − )= ∂ ( i ( F j + + F j − )) = ∂ ( i ( F j + + F j − )) = 0 . j + and j − , we see that the conditions implies that ∂F ∂v = ∂F ∂v = 0 ∂F ∂u = ∂F ∂u = 0 ∂F ∂u = ∂F ∂u = 0 ∂F ∂v = ∂F ∂v = 0This, of course, means that F = F ( u , u ) , F = F ( v , v ) , F = F ( v , v ) , F = F ( u , u ) . Translating back to the original coordinates, we see F has the desired form. (cid:3) The converse is not true, in general. Here is a simple counter-example.
Example 3.3.
Consider the Cℓ , -valued function f ( x , x , x , x ) = x x x − x x x i + x x x j + x x x ij. It is easy to check that f satisfies the necessary system of PDEs so that ∂f = 0 . However, noticethat if we write f as in the above proof, then f = (cid:0) u − v (cid:1) u j + − v j − ) + i (cid:0) u v (cid:1) − u j + + v j − ) . Now, ∂ " (cid:0) u − v (cid:1) u j + − v j − ) = − v u j + − u v . Thus, f is not of the form as prescribed in Theorem 3.2. In a manner similar to the Cℓ ,n case, we can also take a Cℓ , -valued function whose componentsare real analytic and generate a left regular function valued in Cℓ , . In fact, there are two ways todo this. The first borrows heavily from a result found in Brackx, Delanghe, and Sommen’s book [2].13 heorem 3.4. Let g ( x , x ) be a Cℓ , -valued function on U ⊆ R with real-analytic components.Then the function f ( Z ) = ∞ X k =0 ∂ " x k +10 + x k +11 (2 k + 1)! ! ∆ k g ( x , x ) , where ∆ is the Laplace operator in the x x -plane, is left-regular in an open neighborhood of { (0 , } × U in R , and f (0 , , x , x ) = g ( x , x ) − ig ( x x ) .Proof. We proceed by a similar proof found in [2].Let g ( x , x ) = g ( x , x ) + g ( x , x ) i + g ( x , x ) j + g ( x , x ) ij . Since g ℓ is analytic, then anapplication of Taylor’s theorem gives that on every compact set K ⊂ U there are constants c K and λ K , depending on K , such thatsup ( x ,x ) ∈ K (cid:12)(cid:12) ∆ k g ( x , x ) (cid:12)(cid:12) ≤ (2 k )! c K λ kK andsup ( x ,x ) ∈ K (cid:12)(cid:12)(cid:12)(cid:12) ∂∂x ℓ ∆ k g ( x , x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (2 k + 1)! c K λ kK , where | · | denotes the euclidean norm in R .Thus, sup ( x ,x ) ∈ K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ " x k +10 + x k +11 (2 k + 1)! ! ∆ k g ( x , x ) = sup ( x ,x ) ∈ K (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) x k (2 k )! + i x k (2 k )! (cid:19) ∆ k g − x k +10 + x k +11 (2 k + 1)! (cid:18) j ∂∂x ∆ k g + ij ∂∂x ∆ k g (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ sup ( x ,x ) ∈ K (cid:20)(cid:12)(cid:12)(cid:12)(cid:12) x k (2 k )! (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12) ∆ k g (cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) x k (2 k )! (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12) ∆ k g (cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x k +10 + x k +11 (2 k + 1)! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ∂∂x ∆ k g (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ∂∂x ∆ k g (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) ≤ c K (cid:2) (1 + 2 | x | ) x k λ kK + (1 + 2 | x | ) x k λ kK (cid:3) , so that f converges uniformly on [ K ⊆ U (cid:20)(cid:18) − √ λ K , √ λ K (cid:19) × (cid:18) − √ λ K , √ λ K (cid:19) × ˚ K (cid:21) . ∂f = ∞ X k =0 ∆ , " x k +10 + x k +11 (2 k + 1)! ! ∆ k g ( x , x ) = ∞ X k =0 ∆ , x k +10 + x k +11 (2 k + 1)! ! ∆ k g ( x , x )+ ∞ X k =0 x k +10 + x k +11 (2 k + 1)! ! ∆ , (cid:0) ∆ k g ( x , x ) (cid:1) = ∞ X k =1 x k − + x k − (2 k − ! ∆ k g ( x , x ) − ∞ X k =0 x k +10 + x k +11 (2 k + 1)! ! ∆ k +1 g ( x , x )= 0 , as needed. (cid:3) Example 3.5.
Let g ( x , x ) = x x . Then ∆ g = 0 and the formula above gives f ( Z ) = ∂ [( x + x )( x x )]= x x − x x i + ( x x + x x ) j + ( x x + x x ) . A less trivial example demonstrates that the more complicated g is the more complicated f is. Example 3.6.
Let g ( x , x ) = x + x x . Thus, ∆ g = 12 x + 6 x x and ∆ g = 24 .Then from the formula, we get f ( Z ) = (cid:2) x + 3 x (2 x + x x ) + x + x x (cid:3) + (cid:2) x + 3 x (2 x + x x ) + x + x x (cid:3) i + (cid:20) ( x x ) + x ( x + x ) + x (cid:21) j + (cid:20) ( x x ) + x ( x + x ) + x (cid:21) ij. We can define a true extension of an analytic function which is left regular and closely resemblesthe Cauchy-Kowalewski extension found in [3,12]. Again, we are again grateful to Brackx et. al fortheir proof in the Cℓ ,n case, which again gives the convergence of the series. Theorem 3.7 (Cauchy-Kowalewski Extension in Cℓ , ) . Let g ( x , x , x ) be a Cℓ , -valued functionwhose components are real-analytic functions on U ⊆ R . Then the function f ( x , x , x , x ) = ∞ X k =0 ( − x ) k k ! D k g ( x , x , x ) , here D = ∂ − ∂∂x , is left-regular in an open neighborhood of { } × U , and f (0 , x , x , x ) = g ( x , x , x ) . The following lemma will be useful in demonstrating the convergence of f in an open neighbor-hood of U . Lemma 3.8.
Let g ( x , x , x ) be a Cℓ , -valued function whose components are real-analytic func-tions on U ⊆ R . Then on a compact set K , there are constants c K and λ K such that (cid:12)(cid:12) D k g ( x , x , x ) (cid:12)(cid:12) ≤ k c K ( k !) λ kK . Proof of the Lemma.
Taylor’s theorem, again, gives that on a compact set K , there are constants c K and λ K such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ k ∂x k ∂x k ∂x k g ( x , x , x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c K ( k !) λ kK . It is worth mentioning that we first saw the above inequality in [2].Notice that when k is even, D k is a scalar operator. So suppose k is even. Then by the trinomialtheorem, D k = X k + k + k = k k !( k )!( k )!( k )! ∂ k ∂x k ∂x k ∂x k . Then, (cid:12)(cid:12) D k g ( x , x , x ) (cid:12)(cid:12) ≤ X k + k + k = k k !( k )!( k )!( k )! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ k ∂x k ∂x k ∂x k g ( x , x , x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c K ( k !) λ K X k + k + k = k k !( k )!( k )!( k )!= 3 k c K ( k !) λ kK . Now suppose k is odd. Then we have D k = X k + k + k = k − ( k − k )!( k )!( k )! i ∂ k ∂x k +11 ∂x k ∂x k − j ∂ k ∂x k ∂x k +12 ∂x k − ij ∂ k ∂x k ∂x k ∂x k +13 ! . (cid:12)(cid:12) D k g ( x , x , x ) (cid:12)(cid:12) ≤ X k + k + k = k − ( k − k )!( k )!( k )! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i ∂ k g ( x , x , x ) ∂x k +11 ∂x k ∂x k − j ∂ k g ( x , x , x ) ∂x k ∂x k +12 ∂x k − ij ∂ k g ( x , x , x ) ∂x k ∂x k ∂x k +13 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X k + k + k = k − ( k − k )!( k )!( k )! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ k g ( x , x , x ) ∂x k +11 ∂x k ∂x k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ k g ( x , x , x ) ∂x k ∂x k +12 ∂x k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ k g ( x , x , x ) ∂x k ∂x k ∂x k +13 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)! = X k + k + k = k − ( k − k )!( k )!( k )! (cid:0) c K ( k !) λ kK (cid:1) = 3 k c K ( k !) λ kK , as required. (cid:3) Proof of the Theorem.
The above lemma gives us that on a compact set K ⊂ U there are constants c K and λ K , depending on K , such that (cid:12)(cid:12)(cid:12)(cid:12) ( − x ) k k ! D k g ( x , x , x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c K (3 λ K ) k x k . Thus, f converges uniformly on [ K ⊆ U (cid:20)(cid:18) − λ K , λ K (cid:19) × ˚ K (cid:21) . The essential calculation is ∂f = ∞ X k =0 ∂ (cid:18) ( − x ) k k ! D k g ( x , x , x ) (cid:19) = ∞ X k =0 ∂ (cid:18) ( − x ) k k ! (cid:19) D k g ( x , x , x ) + ∞ X k =0 ( − x ) k k ! ∂ (cid:0) D k g ( x , x , x ) (cid:1) = − ∞ X k =1 ( − x ) k − ( k − D k g ( x , x , x ) + ∞ X k =0 ( − x ) k k ! D k +1 g ( x , x , x )= 0 , as required. (cid:3) emark 3.9. We may think of the above extension as a solution to the boundary value problem: ( ∂f ( x , x , x , x ) = 0 f (0 , x , x , x ) = g ( x , x , x ) . Example 3.10.
Consider the homogeneous polynomial of degree 2 g ( x , x , x ) = x + x + x + x x + x x + x x . Now, Dg ( x , x , x ) = (2 x + x + x ) i − (2 x + x + x ) j − (2 x + x + x ) ijD g ( x , x , x ) = − D ( x , x , x ) = 0 . Thus, f ( Z ) = (cid:0) − x + x + x + x + x x + x x + x x (cid:1) − x (2 x + x + x ) i − x (2 x + x + x ) j − x (2 x + x + x ) ij. is the left- regular function obtained by the above theorem. Remark 3.11.
In both of these formulas, a polynomial g will be transformed to a Clifford valuedfunction where every component is a polynomial. This is the case because polynomials have partialderivatives of after a certain order. That is, D k g and ∆ k g will be uniformly for all k > M forsome finite M . References [1] Lars V. Ahlfors.
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