The harmonicity of slice regular functions

In this article we investigate harmonicity, Laplacians, mean value theorems and related topics in the context of quaternionic analysis. We observe that a Mean Value Formula for slice regular functions holds true and it is a consequence of the well known Representation Formula for slice regular functions over $\mathbb{H}$. Motivated by this observation, we have constructed three order-two differential operators in the kernel of which slice regular functions are, answering positively to the question: is a slice regular function over $\mathbb{H}$ (analogous to an holomorphic function over $\mathbb{C}$)"harmonic"in some sense, i.e. is it in the kernel of some order-two differential operator over $\mathbb{H}$ ? Finally, some applications are deduced, such as a Poisson Formula for slice regular functions over $\mathbb{H}$ and a Jensen's Formula for semi-regular ones.


Introduction
In [19] and [20], Gentili and Struppa gave the following definition of slice regular function over the quaternions: Let Ω be a domain in H. A real differentiable function f : Ω → H is said to be slice regular if, ∀ I ∈ S = {q ∈ H, ℜe q = 0 : |q| = 1}, its restriction f I to the complex line C I = R + RI passing through the origin and containing 1 and I is holomorphic on Ω ∩ C I , which is equivalent to require that, ∀ I ∈ S, on Ω ∩ C I .
Later the notion of "slice regularity" was generalized to algebras other than H ( [15,22]). Definition 1.2. Let A be a finite-dimensional real alternative algebra with 1 and let S = {x ∈ A : x · x = −1}. Assume that for each element x ∈ A \ R there is a unique representation in the form x = α + βI with α ∈ R, β ∈ R >0 , I ∈ S.
Regularity of a function f : A → A is defined in the same way as for quaternions, i.e., f is regular iff (1) holds. 1 Let D ⊂ C be any symmetric set with respect to the real axis. A function F = F 1 + F 2 ı : D → A ⊗ C such that F (z) = F (z) is said to be a stem function.
If a stem function F induces the slice function f , we will write f = I(F ). Proposition 1.3. Let A, S and D be as in the above definition. Assume that D intersects R.
Then a slice function f : Ω D → A is regular if and only if its stem function F : D → A ⊗ C is holomorphic.
Classically, mean value theorems are closely related to harmonicity. We investigate mean value properties for quaternionic functions. We prove (Proposition 4.1) that a slice regular function f fulfills f (a + bI) = 1 2π S 2π 0 (1 − IJ)f (a + bJ + re Jθ )dθdµ(J), ∀a, b ∈ R, I ∈ S where µ is a probability measure on S which is invariant under the involution J → −J.
Conversely, we show that every continuous function f : H → H with this mean value property must be the sum of a regular and an antiregular function (Theorem 4.4).
We also show that for any point p ∈ H and every 3-sphere S containing p in its interior, there exists a H-valued measure on S such that f (p) = S f (q)dµ(q) for every regular function f (Theorem 7.1).
Over the field of complex numbers, the mean value property is equivalent to harmonicity. Therefore it is natural ask ourselves if slice regular functions were in the kernel of some order-two differential operator over H: in section 8 we answer positively to this question constructing three order-two differential operators in the kernel of which slice regular functions are. The first one is ∆ * : Let Ω D be a circular domain and let f : Ω D → H be a C 2 function. For every I ∈ S and every q ∈ C I \ R we set If f is a slice function, ∆ * f extends through the real axis. Therefore ∆ * is well defined as an operator on C 2 -slice functions. The second order-two differential operator is ∆ ′ : Here R w is an averaging operator which we define based on rotations, cf. section 6. We observe that (∆ ′ f )(q) = 1 2 ∆ I (T r(f ))(q) = 1 2 ∆ I (f + f c )(q) for all I ∈ S and every slice function f . The third order-two differential operator is ∆ ′′ : (∆ ′′ f )(q) = (∆ I N (f ))(q) = (∆ I (f · f c ))(q) for all I ∈ S. On one side ∆ * and ∆ ′ are R−linear operator, on the other hand ∆ ′′ is not a linear operator: but for ∆ ′′ a sort of Leibnitz rule for (f * g) holds true.
The main properties of ∆ * and ∆ ′ are the followings: Theorem 1.4 (Theorem 6.21). Let f : Ω D → H be a C 2 slice function. Assume that D is simply-connected. ∆ * f is vanishing identically if and only if f can be written as a sum of a regular function g and an anti-regular function h. Proposition 1.5 (Proposition 6.13). Let f : H → H be a slice function induced by a stem function F : C → H ⊗ C. Then ∆F is again a stem function, and ∆ * f is the slice function associated to ∆F . Proposition 1.6 (Proposition 6.28). Let h : H → R be a slice function with ∆ ′ h = 0. Then there exists a slice-preserving regular function f such that h = ℜe (f ). Proposition 1.7 (Proposition 6.31). Let u : H → R be a C 2 -function such that ∆ ′ u = 0 outside R.
Then u admits no isolated zero in any real point a ∈ R.
Finally, we provide a Jensen's formula. Then: We hope that this paper can provide new ideas for studying slice regular functions and their "harmonic properties" on slice regular quaternionic manifolds recently introduced by Bisi-Gentili in [7] and Angella-Bisi in [5].

Prerequisites about quaternionic functions
In this section we will overview the main notions and results needed for our aims. First of all, let us denote by H the real algebra of quaternions. An element x ∈ H is usually written as x = x 0 + ix 1 + jx 2 + kx 3 , where i 2 = j 2 = k 2 = −1 and ijk = −1. Given a quaternion x we introduce a conjugation in H (the usual one), as x c = x 0 − ix 1 − jx 2 − kx 3 ; with this conjugation we define the real part of x as ℜe(x) := (x + x c )/2 and the imaginary part as ℑm(x) := (x − x c )/2. With the notion of conjugation just defined we can write the euclidean square norm of a quaternion x as |x| 2 = xx c . The subalgebra of real numbers will be identified, of course, with the set R := {x ∈ H | ℑm(x) = 0}. Now, if x ∈ H is such that ℜe(x) = 0, then the imaginary part of x is such that (ℑm(x)/| ℑm(x)|) 2 = −1. More precisely, any imaginary quaternion I = ix 1 + jx 2 + kx 3 , such that x 2 1 + x 2 2 + x 2 3 = 1 is an imaginary unit. The set of imaginary units is then a real 2−sphere and it will be conveniently denoted as follows: S := {x ∈ H | x 2 = −1} = {x ∈ H | ℜe(x) = 0, |x| = 1}. With the previous notation, any x ∈ H can be written as x = α + Iβ, where α, β ∈ R and I ∈ S. Given any I ∈ S we will denote the real subspace of H generated by 1 and I as: Sets of the previous kind will be called slices.
We denote the 2−sphere with center α ∈ R and radius |β| (passing through α + Iβ ∈ H), as: 2.1. Slice functions and regularity. In this part we will recall the main definitions and features of slice functions. The theory of slice functions was introduced in [22] as a tool to generalize the theory of quaternionic regular functions defined on particular domains introduced in [19,20], to more general domains and to alternative * −algebras. The complexification of H is defined to be the real tensor product between H itself and C: (Here ı = 1 ⊗ i.) Note that H ⊗ C has a natural structure of an associative algebra induced by the algebra structures of H and C. Explicitly, the product on H ⊗ C is given as follows: if p 1 + q 1 ı, p 2 + q 2 ı belong to H ⊗ C, then, The usual complex conjugation p + qı = p − qı commutes with the following involution (the quaternionic conjugation) (p + qı) c = p c + q c ı.
We introduce now the class of subsets of H where our functions will be defined.
Definition 2.1. Given any set D ⊆ C, we define its circularization as the subset in H defined as follows: Such subsets of H are called circular sets. If D ⊂ C is an open connected subset such that D ∩ R = ∅, then Ω D (which is again open and connected and intersects the real line R) is called a slice domain (see [18]).
From now on, Ω D ⊂ H will always denote a circular domain arising as circularization of a symmetric domain D ⊂ C.
Definition 2.2. Let D ⊂ C be any symmetric set with respect to the real axis. A function for all α + iβ ∈ D and all I ∈ S. If a stem function F induces the slice function f , we will write f = I(F ). The set of slice functions defined on a certain circular domain Ω D will be denoted by S(Ω D ).
3. Let f : Ω D → H be a function defined on a circular domain Ω D . If there exists a function F = F 1 + F 2 ı : D → H ⊗ C such that equation (3) holds, then F is a stem function, i.e., F 1 (z) = F 1 (z) and F 2 (z) = −F 2 (z).
Proof. We observe that for all z = α + iβ ∈ D and all I ∈ S we have Examples of (left) slice functions are polynomials and functions given by power series in the variable x ∈ H with all coefficients on the right, i.e., a power series if convergent, defines a slice function.
A function f : Ω D → H is a slice function if and only if it obeys the following "representation formula": (see [18,22] They are again stem functions. Let f be a slice function induced by a stem function F (i.e. f = I(F ) and let q ∈ H, q ∈ C I , I ∈ S. Then These derivatives are also called "Cullen derivatives".
We are now in the position to define slice regular functions (see Definition 8 in [22]). is (left) regular if its stem function F is holomorphic. The set of regular functions will be denoted by Equivalently, a slice function f ∈ S 1 (Ω D ) is regular if the following equation holds: ∂ c f (α + Jβ) = 0, ∀ α + Jβ ∈ Ω D . The set of regular functions is a real vector space and a right H-module. In the case in which Ω D is a slice domain, the definition of regularity is equivalent to the one given in [18].
We recall a key lemma of this theory that will be useful later on, [18]. Lemma 2.6 (Splitting). Let f be a regular function defined on an open set Ω of H. Then, for any I ∈ S and for any J ∈ S with J ⊥ I, there exist two holomorphic functions g I , h I : Ω ∩ C I → C I such that, ∀z = x + yI, it is : 2.1.2. Product of slice functions and their zero set. In general, the pointwise product of slice functions is not a slice function. However there is some product called "star product" which does turn slice functions into slice functions.
The following notion is of great importance in the theory. For the following basic facts on this "star product" see [22] and [18]. Explicitly, if F = F 1 + F 2 ı and G = G 1 + G 2 ı are stem functions, then Slice preserving functions satisfy the following characterization. Since real numbers commute with all quaternions, this has the following consequence: Let f, g ∈ S(Ω D ). If f is slice preserving, then .
If f and g are both slice-preserving, then f g = f * g = g * f = gf . As stated in [18], if f is a regular function defined on B ρ , the ball of center 0 and radius ρ, then it is slice preserving if and only if f can be expressed as a power series of the form x n a n , with a n real numbers.
The following definitions are taken from [18,22]. Another observation is that, if f is slice preserving, then f c = f and so f s = f 2 .
Frequently, the sum f + f c is denoted by T r(f ).

2.1.3.
Zeros of regular functions. We are going now to recall some key facts about the zeros of a slice function. Let f : Ω D → H be any slice function with zero locus . Let x ∈ Z(f ). There are the following three possibilities: x is a real zero; • x a punctual (non-real) zero, i.e., x / ∈ R and S x ∩ Z(f ) = {x}; • x a spherical zero, i.e., x / ∈ R and S x ⊂ Z(f ).
The inclusion holds for any two slice functions f, g : Ω D → H, while in general Z(g) ⊂ Z(f * g).
What is true in general is the following equality: Proof. Under the assumptions of the corollary, the symmetrization f s = f c * f vanishes identically on each Σ n . Hence (Ω D ∩C J )∩Z(f s ) contains an accumulation point for any J ∈ S. Consequently f must vanish identically.
2.1.5. Multiplicities of zeros. . Let f ∈ SR(Ω D ) such that f s does not vanish identically. Given n ∈ N and q ∈ Z(f ), we say that x is a zero of f of total multiplicity n, and we will denote it by Then there existsp ∈ S p and a regular function g such that f (q) = g(q) * (q −p).
Proof. There is an element a ∈ S p such that f c (a) = 0, implying that there exists a regular function h with f c (q) = (q−a) * h(q). It follows that f (q) = h c (q) * (q−ā).
2.1.6. Semi−regular functions and their poles. We will recall now some concept of "semi-regular functions" which are the quaternionic analog of meromorphic functions. Here our main references are [18] and [24].
The regularity of f − * on Ω D \ Z(f s ) just defined follows thanks to the last equality.
Let Ω D be a slice domain, q ∈ Ω D and f a regular function on Ω D \ S q . Then f has a pole of order d in q if (x − q) * d * f extends as a regular function through q Such a semi-regular function may be regarded as a map from Ω D toĤ = H∪{∞} by assigning ∞ as function value at each pole.
Proof. Due to the representation formula we have Adding both above equalities yields the assertion of the lemma.

Divisors
In complex analysis, the divisor of a holomorphic function is the formal sum of its zeroes, counted with the respective multiplicities. We propose that for a quaternionic regular function defined on Ω D the divisor should be defined as a formal sum of points in the closed upper plane intersected with D, i.e., on {z ∈ D : ℑm(z) ≥ 0}. Then the "(slice) divisor" div(f ) of f is defined as the formal Z-linear combination z∈D + m z (F ){z} where for z = x + yi the multiplicity m z (f ) is defined as follows: m z (f ) = m if in a neighbourhood of S x+yI = {x + yJ : J ∈ S} the function f can be written as with a i ∈ S x+yI and g being a regular function without zeros on S x+yI .
Standard facts on zeros of regular functions (see 2.1.3) guarantee us the following properties: • are the isolated zeros with multiplicity n k and S c k +Jd k are the spherical zeros with multiplicity m k , then Then f has a zero only at I while g(q) = (q − J) * (q − I) has a zero only at J, but the divisor is the same: This notion of a divisor is easily extended to semi-regular functions f = g − * * h (g, h regular) by posing div(f ) = div(h) − div(g).

A mean value theorem
as well as and therefore Proof. This follows from combining the complex mean value theorem with the formulae relating slice and stem functions. Proof. This a special case of Proposition 4.1 with b = 0.
Remark 4.3. Note that in Corollary 4.2 we integrate over the sphere with radius r and center a, but not with respect to the euclidean volume element dV on the 3-sphere. This is crucial.
Then f is harmonic (in the sense of being the sum of a regular and an antiregular function) if and only if Proof. We assume that (6) holds. The function f is a slice function if and only if it satisfies the representation formula. Hence f is a slice function iff for all a, b ∈ R, H, I ∈ S and p = a + bI. This can be verified by explicit calculation: Thus f is a slice function induced by some stem function F . This stem function can be easily determined as Thus F 1 satisfies the ordinary mean value property for functions defined on C and therefore must be harmonic. Similar arguments apply to F 2 . As a result we see that F is the sum of a holomorphic and an antiholomorphic function from C to H ⊗ R C and consequently f : H → H is the sum of a regular and an antiregular function.
For the opposite direction, assume that f is the sum of a regular function and an antiregular function. Then (6) follows immediately from Proposition 4.1.

Generalized Representation Formula
This formula already appeared in [13]: see Theorem 3.2. Here we give a new proof and we deduce some consequences.
Proposition 5.1. Let f be a regular function and let I, J, H ∈ S (not necessarily orthogonal). Assume that J = I, H = I. Then the following equality holds: Proof. We have A linear combination of both equations yields: Proof. Since I, J are purely imaginary, we have JI = IJ. Therefore denotes the vector part of IJ. Define r = ℜe (IJ). Observe that r ∈] − 1, +1]. Using |IJ| = 1 we know that the vector part of IJ has norm √ 1 − r 2 . Therefore We observe that the map r → −1 + 2 1 + r is evidently an injective map from (−1, +1] to R + . As In a similar way one proves lim J→I M 2 (J, H) = 0 and the analog statement for H → I.

Rotations
For every w ∈ H * let S w : H → H denote the map given by S w (q) = w −1 qw. This is an orthogonal transformation of R 4 which fixes R pointwise. Observe that S −1 w = S w −1 . Proof. The group is SO(3, R). For each k ∈ N, let Σ k denote the set of all S w1 • . . . • S w 2k . Then Σ = ∪ k Σ k is the group generated by all the S w . (Σ is evidently a semigroup and in fact a group, because (S w ) −1 = S w −1 .) Σ is connected, because each Σ k is connected and Σ k ⊆ Σ k+1 . On the other hand, it is not commutative, since e.g. S I and S (I+J)/ Then Let v ∈ S. Then S v is an orthogonal transformation. Due to the invariance of the measure µ, we have: (Here S * v denotes the pull-back by the map S v .) It follows that We observe that w S w (q)dµ(w) = q, ∀q ∈ R, because S w (q) = q, ∀q ∈ R, w ∈ S. Now let q be in the orthogonal complement of R, i.e., in the real vector subspace V of H spanned by I, J, K. Since the integral is linear, and V is stabilized by every S w (w ∈ S) it follows that Combined with the fact H(q) ∈ R, ∀q ∈ H, we obtain for every q ∈ R and every q ∈ V . By R-linearity of the map H, it follows that this equality holds for all q ∈ H. (1) f is rotationally invariant.
(2) f (q) = f (S w q) for all q ∈ H, w ∈ S. Lemma 6.6. Let f denote a regular function, w ∈ H * . Then we can define another regular function R w f as Lemma 6.7. If f : H → H is induced by a stem function F : C → H ⊗ C, then R w f is induced by S w −1 (F ) with S w −1 acting via the first factor of the tensor product H ⊗ C.
Proof. We have is induced by the stem function R(F ) where R denotes the real part in the first factor of the tensor product, i.e., R(a ⊗ b) = (ℜe a) ⊗ b.
Proof. By the Lemma 6.7, g is induced by w S w −1 F dµ(w). Aided by Lemma 2.3, this implies the assertion, because w S w −1 (q)dµ(w) = ℜe (q) for every q ∈ H (Lemma 6.3). Lemma 6.9. For a regular function f = +∞ k=0 q k a k we have We observe that R vw = R v • R w and S vw = S w • S v for v, w ∈ H * . Definition 6.10. Let I ∈ S. Then the differential operator ∆ I is defined as Lemma 6.11. Let I ∈ S. Then ∆ I (f ) vanishes along C I for regular functions f (and anti-regular functions f ).
Proof. This is clear, because the restriction of a regular function f to a complex line C I can be written as f (z) = f 1 (z) + f 2 (z)J where f i : C I → C I , for i = 1, 2, are entire functions with respect to the complex structure on C I and J orthogonal to I, by the splitting Lemma 2.6.
Then for every function f we have We define a differential operator ∆ * for slice functions as follows: Definition 6.12. Let Ω D be a circular domain and let f : Ω D → H be a slice function.
For every I ∈ S and every q ∈ C I we set We have to show that ∆ * is well-defined, i.e., that (∆ I f )(q) = (∆ J f )(q) for all I, J ∈ S, q ∈ R. This will be deduced from the proposition below. Proposition 6.13. Let f : H → H be a C 2 -function which is induced by a stem function F : C → H ⊗ C. Then F is likewise C 2 . Moreover, G = ∆F is again a stem function, i.e., τ (G(z)) = G(z) for τ (a ⊗ b) = a ⊗b, and Proof. This follows using the representation formula.
Corollary 6.14. ∆ * is well-defined. Proof. In the notation of the proposition we have (  Proof. Let F be the stem function of f . Then ∆ * f ≡ 0 implies that ∆F ≡ 0, i.e., F is harmonic. As a consquence, F and therefore also f must be real analytic. Proof. ∆ * f ≡ 0 implies the harmonicity of the stem function F . Now boundedness of f implies boundedness of F which leads to a contradiction unless F (ant therefore also f ) is constant.
After introducing ∆ * we now define the related notions ∂ * and∂ * : Definition 6.18. Let Ω D be a circular domain and let f : Ω D → H be a slice function. Let q ∈ Ω D , q = x + yI, x, y ∈ R, I ∈ S.
We define One may check that for slice functions these operators are well-defined even if q ∈ R: Proposition 6.19. ∂ * and∂ * are well-defined.
Proof. Let q = x ∈ R. Since f is a slice function, there is a stem function F = F 1 + F 2 ı such that and To show that (∂ * f )(x) is well-defined, we have to show that (∂ x f )(x) and I(∂ y f )(x) are independent of I, i.e., we have to verify that (∂ x F 2 )(x) = 0 and (∂ y F 1 )(x) = 0.
To deduce these two equalities, we recall that as a stem function F must fulfill F 1 (z) = F 1 (z) and F 2 (z) = −F 2 (z) for z ∈ C. This easily implies the assertions. Similarily one verifies that∂ * f is well-defined.
Proof. Let F be the stem function inducing f . Then ∆ * f is a slice function induced by the stem function ∆F (see Corollary 6.15). Now F is a map from D to the complex vector space H ⊗ C. Hence ∆F vanishes iff F is harmonic iff F = G + H for a holomorphic function G : D → H ⊗ C and an anti-holomorphic function We have to verify that G, H may be taken to be stem functions. To state it more precisely: We have to show: If F : D → H ⊗ C is a map such that F (z) = F (z) and such that F = G + H for a holomorphic map G and an anti-holomorphic map H, then G and H may be chosen in such a way that G(z) = G(z) and H(z) = H(z). We observe that c is totally imaginary and that Thus, by replacing G with G − c/2 we may turn G into a stem function. Correspondingly we replace H by H + c/2. Finally we let g, resp. h, be the slice functions induced by the stem functions G, resp. H, and observe that g is regular, because G is holomorphic and h is anti-regular, because H is anti-holomorphic. Now we demonstrate (2). First we observe that We already know that f is written as f = g + h where g is regular and h is anti-regular. By replacing g with g − g(0) and h with h + g(0) we may assume that g(0) = 0. Then we obtain g(q) = k≥0 q k a k with a 0 = 0 and h(q) = k≥0q and Since both power series assume only real values on R, all their coefficients have to be real-valued. Hence a k + b k and ka k are all real-valued. Combined with a 0 = 0 it follows that a k , b k ∈ R for all k ∈ N ∪ {0}.
It is also evident that if both (∂ * f )(x) = k x k−1 ka k and (∂ * f )(x) = k x k−1 kb k are reals over the reals, then also f (x) = k x k−1 k(a k + b k ) is real over the reals, which is the last equivalent condition of (2).

Definition 6.22.
( We observe that (∆ ′ f )(q) = 1 2 ∆ * (T r(f ))(q) = 1 2 ∆ * (f + f c )(q) for all I ∈ S. Analogously we can introduce another second order operator in the following way: for all I ∈ S. On one side ∆ ′ and ∆ * are linear operators, on the other hand, ∆ ′′ is not a linear operator.
By the way, for ∆ ′′ a product formula holds. Let f, g be slice functions. Then Remark 6.23. In [14] the following global first order differential operator was introduced: Whereas the operator ∆ * is defined everywhere only if applied to slice functions, G is everywhere defined for any C 1 -function. Still, we believe that the additional factor y 2 (which guarantees the applicability of the operator even to non-slice functions) is somewhat unnatural.
In particular, if we try to construct a second order differential operator, we see thatḠ G = y 4 ∆ * − 2Iy 3∂ * . ThusḠG even applied to slice functions on a complex slice C I is not merely the multiple of the ordinary complex Laplacian. Moreover,ḠG annihilates regular functions, but not antiregular functions. (1) f is rotationally equivariant.
(2) R I (f ) = f for all I ∈ S.
(3) S I (f (q)) = f (S I (q)) for all I ∈ S and q ∈ H.
(4) If g : q → g · q denotes the R-linear action of SO(3, R) on H which is trivial on R and which is the natural orthogonal transformations on I, J, K = R 3 , then f : H → H is equivariant for this action, i.e., g · f (q) = f (g · q) for all g ∈ SO(3, R), q ∈ H. (5) f is induced by a stem function F : C → H ⊗ C with F (C) ⊂ R ⊗ C, i.e., both F 1 and F 2 are real-valued for F = F 1 + F 2 ı.  (3) implies S I (f (q)) = f (S I (q)). For q ∈ C I we have S I (q) = q and therefore S I (f (q)) = f (q), which implies f (q) ∈ C I . Thus f (C I ) ⊂ C I . We define functions g, h : C → R such that Now, for every J ∈ S, there is an orthogonal transformation φ fixing R with φ(I) = J. Using (4), we have Thus f : H → H is induced by the stem function F given as F (x + yi) = g(x, y) ⊗ 1 + h(x, y) ⊗ i. (The fact F (z) = F (z) follows from Lemma 2.3.) This completes the proof of (3) ⇒ (5). (1) h is induced by a stem function.
(4) h is rotationally invariant. Moreover, in this case the stem function H can be defined as for all x, y ∈ R, I ∈ S.
Proposition 6.28. Let h : H → R be a slice function with ∆ ′ h = 0.
Then there exists a slice-preserving regular function f such that h = ℜe (f ).
Proof. Since h is real-valued and a slice function, it is also rotationally equivariant and rotationally invariant (due to Lemma 6.27). Being rotationally equivariant implies ∆ ′ h = ∆ * h. Being rotationally invariant implies h(x + yI) = h(x − yI) ∀x, y ∈ R, ∀I ∈ S which in turn implies that ∂ ∂y h vanishes on the real line. Hence all the assumptions of Theorem 6.21 (2) are fulfilled, and we may deduce that h = g + k where g is regular, k is anti-regular and both g and k are slice-preserving. The condition h(q) ∈ R is equivalent to h(q) − h(q) = 0. Therefore g(q) − k(q) = g(q) − k(q) ∀q ∈ H. The left hand side of this equation is a regular function, while the right hand side is anti-regular. Thus both must be constant. That the right hand side is constant, implies: k(q) = g(q) − g(0) + k(0) ∀q ∈ H. Hence h(q) = g(q) + k(q) = g(q) + g(q) − g(0) + k(0) = 2 ℜe (g(q)) − g(0) + k(0) ∀q ∈ H.
Finally, h(H) ⊂ R implies −g(0) + k(0) ∈ R. This proves the assertion for f (q) = 2g(q) − g(0) + k(0). Proposition 6.29. Let u be real-valued and rotationally invariant (in the sense of Definition 6.4). Let f be regular. Then Proof. This is a consequence of the complex computation Proof. We start proving that, under the hypotheses of the proposition, Indeed ∀ w ∈ S: By (7), where g ′ := ∂ c g. Proposition 6.31. Let u : H → R be a C 2 -function such that ∆ ′ u ≡ 0 on H \ R. Then u admits no real isolated zero.
Proof. Assume the contrary, i.e., assume that u has an isolated zero in a point a ∈ R. Then there exists an open neighborhood W of a such that u has no zero on W \ {a}. For dimension reasons, W \ {a} is connected. Thus u is either everywhere positive or everywhere negative on W \ {a}. Without loss of generality, assume that u > 0 on W \ {a}.
For q sufficiently close to a (but q = a) we have S q ⊂ W \ {a}. For such q we have (R w u)(q) > 0 ∀w ∈ S and thereforeũ(q) > 0. On the other hand,ũ(a) = u(a), because a ∈ R. Thusũ has a strict local minimum in a. By construction ∆ ′ u = ∆ * ũ . Fix I ∈ S. Then ∆ Iũ = ∆ * ũ on C I \ R. Nowũ is a C 2 -function, because u is a C 2 -function. Thus the vanishing of ∆ Iũ on the dense open subset C I \ R implies that ∆ Iũ vanishes on all of C I , i.e.,ũ is a harmonic function on C I with a strict minimum in a. This is impossible. Theorem 7.1. Let p ∈ H and let S ⊂ H be a 3-dimensional sphere such that p is in its interior. Let Ω denote a circular domain containing both S and its interior.
Then there exists an H-valued measure µ on S which is absolutely continuous with respect to the euclidean measure such that for every regular function f defined on Ω.
Proof. We first discuss the special case where p ∈ R. In this case for each I ∈ S the restriction of f to C I is holomorphic, and C I and S intersect in a 1-dimensional sphere which contains p in its interior. We thus may construct dµ first taking the measure on S ∩ C I given by the classical Poisson formula, and then averaging over I ∈ S with respect to any probability measure of S. Now assume p ∈ R. Fix I ∈ S such that p ∈ C I . Let c = s + vJ (s, v ∈ R, v > 0, J ∈ S) denote the center of the sphere S and ρ its radius. DefineḠ = {t + yi ∈ C : t, y ∈ R, y > 0, ∃ H ∈ S : t + yH ∈ S}.
We observe that S is connected and that H → |yH − vJ| 2 (for fixed y, v > 0, J ∈ S) defines a continuous map which evidently takes its maximum in H = −J (with (y + v) 2 as its value) and its minimum in H = J (with value |y − v| 2 ).
From this we may deduce : Let us now fix t ∈ R and investigate for which y > 0 we have t + iy ∈Ḡ. We define K = ρ 2 − |t − s| 2 and obtain the following condition: It follows that the interior G ofḠ is simply-connected and therefore biholomorphic to the unit disc. Now p is in the interior of G. We fix such a biholomorphic map ψ : G → ∆ and recall that it extends continuously to the respective boundaries. We may and do require ψ(p) = 0.
By the classical mean value theorem for every holomorphic function F : C → C, every r > 0 and every probability measure σ on [0, 1].
Pulling-back with ψ yields a probability measure dξ on G such that for every holomorphic function F . We may assume dξ to be absolutely continuous. For each point t + yi ∈ G we have a 2-sphere t + yS in H. Its intersection with S is again a sphere. We let τ denote the involution defined by sending each element of Σ t,y = S ∩ (t + yS) to its antipodal element.
By the generalized transformation formula we obtain functions m i : Σ t,y → H, for i = 1, 2, such that for every regular function f .
In particular for every probability measure α on Σ t,y . Hence we may choose an absolutely continuous probability measure β t,y on Σ t,y such that We recall that regular functions restrict to holomorphic ones on C I . We may therefore combine the above constructions (see equations (8), (9)) to obtain f (p) = t+yi∈G q∈Σt,y f (q)dβ(q)dξ(t + yi) (with Σ t,y = S ∩ (t + yS)).

Poisson's formula.
Proposition 7.2 (Poisson's Formula). Let u : B R → R be a rotationally invariant C 2 -function. Assume that ∆ ′ u ≡ 0. Let a ∈ R. Then the following formula holds: Proof. Due to Lemma 6.27 the function u is a slice function. Thus we may conclude from Proposition 6.28 that u is the real part of a slice-preserving regular function f . Therefore for each I ∈ S, the restriction of u to C I is the real part of a holomorphic function from C I to C I and the above formula follows from the complex Poisson formula. Remark 7.3. As in Proposition 4.1, it is possible to generalize the above Poisson Formula at any a ∈ H with an integration on the circularization of ∂∆(a, r) ∪ ∂∆(a, r) instead of an integration on ∂B R .

A Jensen's Formula
The goal of this section is to prove a quaternionic version of Jensen's formula. For this purpose we need Blaschke factors. 8.1. Quaternionic ρ-Blaschke factors. In this subsection we are going to reproduce some results proved in [2,3] for a modification of quaternionic Blaschke factors.
In particular, Remark 8.2. We observe that (B a,ρ ) − * has a zero of multiplicity one at a and no other zeros or poles in B ρ .
The following is a consequence of Theorem 5.5 of [2]. with equality if f is slice-preserving.
Proof. Fix an imaginary unit I. Choose another imaginary unit J such that IJ = −JI (i.e., I and J are supposed to be orthogonal). Thus, using the "Splitting Lemma" 2.6, there are two holomorphic functions g, h with values in C I defined on a neighborhood of∆ ρ = {z ∈ C I : |z| ≤ ρ} such that Then |f (q)| 2 = |g(q)| 2 + |h(q)| 2 . Now, log |g| 2 + |h| 2 is subharmonic for any two holomorphic functions g, h : C → C. Thus we have subharmonicity of log |f | 2 and consequently log |f (0)| ≤ 1 2π 2π 0 log |f (ρe It )|dt Finally, by integration over the sphere of imaginary units we obtain the assertion.
In order to deal with the general case (where the function f is allowed to have zeros or poles) we need some preparations. Assume that |g(q)| = 1 for all q ∈ ∂B ρ . Then |f (q)| = |(f * g)(q)| and |g − * (q)| = 1 for all q ∈ ∂B ρ .
Proof. The formula |p −1 qp| = |q|, ∀p ∈ H * , q ∈ H implies that f (q) −1 qf (q) ∈ ∂B ρ whenever q ∈ ∂B ρ . Combined with for q with f (q) = 0 and |g(q)| = 1 ∀q ∈ ∂B ρ we obtain If we apply this equation to f = g − * , we obtain Proposition 8.6. Let f be a semi-regular function on a neighborhood ofB ρ , with neither zeros nor poles on ∂B ρ . Then there exist "ρ-Blaschke factors" B 1 , . . . , B r and a regular function without zeros f 0 such that: Here a function B is called ρ-Blaschke factor, if B = B a,ρ or B = B − * a,ρ for an element a ∈ B ρ .
Proof. Let g, h be regular functions such that f = g − * * h. First we claim that there exist ρ-Blaschke factors B 1 , . . . , B s and a regular functionh without zeros such that f = g − * * h * B 1 * . . . * B s . We proceed recursively. If h admits a zero in a point a ∈ B ρ , then there exists an element b ∈ S a such that h = h 0 * (q − b). Using the language of divisors as explained in section 3 we may reformulate this as follows: Proof. If f has neither zeros, nor poles, this is Proposition 8.4. For the general case, we consider div(f ) = k m k {p k }. First of all, since B ρ ⊂ Ω, it follows from Corollary 2.14 that there are only finitely many k with m k = 0, hence the sums in the statement are finite and k |m k | < ∞.

Recall that
From Proposition 8. 6 we deduce that f may be represented in the form where f 0 is regular on a neighborhood ofB ρ with neither zeros nor poles and each B i equals B ǫi * pi,ρ for some p i ∈ B ρ and ǫ i ∈ {+1, −1}. Now log |f (0)| = log |f 0 (0)| + (with equality if f is slice-preserving) due to Proposition 8.4. Combining these facts, we obtain the assertion. However, there is no similar formula for T r(f ) = f + f c , because div(f + f c ) is not completely determined by div(f ) whereas div(f s ) = 2 · div(f ). Therefore the Jensen's inequality (10) implies: Thus (13) log M f (R) − log |f (0)| ≥ n f (r) (log R − log r) for all 0 < r < R such that f has no zeros on ∂B R . By continuity (in R) it follows that (13) holds without any assumption whether there are zeros on ∂B R or not. This yields the assertion.