2002-Fez conference on Partial Differential Equations, Electronic Journal of Differential Equations, Conference 09, 2002, pp 77–91. http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

A polyharmonic analogue of a Lelong theorem and polyhedric harmonicity cells ^{∗}

Mohamed Boutaleb

Abstract

We prove a polyharmonic analogue of a Lelong theorem using the topological method presented by Siciak for harmonic functions. Then we establish the harmonicity cells of a union, intersection, and limit of domains of ℝ^{n}. We also determine explicitly all the extremal points and support hyperplanes of polyhedric harmonicity cells in ℂ^{2}.

1 Introduction

Throughout this paper, D denotes a domain (a connected open) in ℝ^{n} with n ≥ 2, where D and ∂D are not empty. Since 1936, p-polyharmonic functions in D have been used in elasticity calculus [14]. These functions are C^{∞}-solutions of the partial differential equation

Δ^{p}f(x) = ∑_{|α|=p} \frac{p!}{α!} \frac{∂^{2|α|}f(x)}{∂x_{1}^{2α_{1}} … ∂x_{n}^{2α_{n}}} = 0, p ∈ N^{∗}, x ∈ D.

To study the singularities of these functions in D, Aronzajn [1, 2] considered the connected component H(D), containing D, of the open set ℂ^{n} \ ∪_{t∈∂D}Γ(t), where Γ(t) = {w ∈ ℂ^{n} : ∑_{j=1}^{n}(w_{j} − t_{j})^{2} = 0}. H(D) is called the harmonicity cell of D. Lelong [12, 13] proved that H(D) coincides with the set of points w ∈ ℂ^{n} such that there exists a path γ satisfying: γ(0) = w, γ(1) ∈ D and T [γ(τ)] ⊂ D for every τ in [0, 1], where T is the Lelong transformation, mapping points w = x + iy ∈ ℂ^{n} to (n − 2)-spheres 𝕊^{n−2}(x, ∥y∥) of the hyperplane of ℝ^{n} defined by: ⟨t − x, y⟩ = 0. This work can be divided into three sections: the first one treats a result on polyharmonic functions, the second some general properties on H(D), and the last one deals with a geometrical description of polyhedric harmonicity cells in ℂ^{2}.

Pierre Lelong [11] proved in addition that for every bounded domain D of ℝ^{n}, there exists a harmonic function f in D such that its domain of holomorphy (X_{f}, Φ) over ℂ^{n}satisfies Φ(X_{f}) = H(D), see also [4]. A concise proof of this result is given in Siciak’s paper [16] in the case of the Euclidean ball B_{n}^{r} = {x ∈ ℝ^{n}; ∥x∥ < 1}. In [5], we established that the former method can be applied to arbitrary domains. Also, V.Avanissian noted in [4] that the equality: (X_{f}, Φ) = (H(D), Id) holds in the following cases: D is starshaped with respect to some point x_{0} of D, or D is a C-domain ( that is D contains the convex hull of any (n − 2) dimensional-sphere included in D), or D ⊂ ℝ^{n} with n even and n ≥ 4. The object of Section 2 is to use a topological argument [16] to prove an analogous result for polyharmonic functions in D. As a consequence of this generalization we shall get

∗Mathematics Subject Classifications: 31A30, 31B30, 35J30.

Key words: Harmonicity cells, polyharmonic functions, extremal points,

Lelong transformation.

\begin{matrix}c \\ ◯ \end{matrix}2002 Southwest Texas State University.

Published December 28, 2002.

For every integer 1 ≤ p ≤ [\frac{n}{2}] and suitable domain D (say D is a C-domain, or in particular a convex domain), the harmonicity cell H(D) is nothing else but the greatest (in the inclusion sense) domain of ℂ^{n} whose trace on ℝ^{n} is D and to which all p-polyharmonic functions in D extends holomorphically.

In Section 3, we establish the harmonicity cell of an intersection, a union, and a limit of domains of ℝ^{n}, n ≥ 2. We give next in Section 4 some results about plane domains, prove the existence of polyhedric harmonicity cells in ℂ^{2}, and we calculate all extremal points of the harmonicity cell of a regular polygon. For an arbitrary convex polygon P_{n}, with n edges, we show that H(P_{n}) has exactly 2n faces in ℝ^{4} completely determined by means of the n support lines of P_{n}. It is well known by [10] that if we are given a complex analytic homeomorphism f : D_{1} → D_{2}, where D_{1}, D_{2} are domains of ℝ^{2}, D_{1}, D_{2} not equal to ℝ^{2} and ℝ^{2} ≃ ℂ, then H(D_{1}) and H(D_{2}) are analytically homeomorphic in ℂ^{2}. The holomorphic map Jf : H(D_{1}) → H(D_{2}) defined by w ↦ w′ with:

w′_{1} = \frac{f(w_{1} + iw_{2}) + \overline{f(\overline{w}_{1} + i\overline{w_{2}})}}{2}, w′_{2} = \frac{f(w_{1} + iw_{2}) − \overline{f(\overline{w}_{1} + i\overline{w_{2}})}}{2i}

realizes this homeomorphism.

In proposition 4.4, we show the continuity, according to the compact uniform topology, of the above Jarnicki extension f ↦ Jf and estimate ∥(Jf)(w)∥, w ∈ H(D) by means of sup_{z∈D} |f(z)|. As applications, we find the harmonicity cells of half strips and arbitrary convex plane polygonal domains (owing to an explicit calculation of their support function).

2 A polyharmonic analogue of Lelong theorem

Recall that any polyharmonic function u in D, being in particular analytic in D, has a holomorphic continuation ũ in a corresponding domain D^{u} of ℂ^{n} whose trace with ℝ^{n} is D. Therefore, given any integer p (0 < p < +∞) and any domain D of ℝ^{n}, one can associate a domain N H(D) of ℂ^{n} (depending on D only) such that the whole class H^{p}(D), of all p-polyharmonic functions in D, extends holomorphically to N H(D). This last complex domain, called the kernel of H(D), coincides with the set of all z ∈ H(D) satisfying C_{h}[T (z)] ⊂ D, where C_{h}[T (z)] denotes the convex hull of T (z). For more details see [4].

Making use of a topological argument appearing in [16], we will show now the following theorem.

Theorem 2.1 Let D be a bounded domain of ℝ^{n}, n ≥ 2, D ≠ ∅, ∂D ≠ ∅, and H(D) its harmonicity cell. Then for all integer 1 ≤ p ≤ [\frac{n}{2}], ([\frac{n}{2}] is the integer part of \frac{n}{2}) and all domains Ω̃ ⊃ H(D) the problem for the 2p-order linear partial differential operator Δ^{p}

Δ^{p}u = 0 in D
\overline{D_{1}}ũ = ⋯ = \overline{D_{n}}ũ = 0 in H(D)

has a solution h ∈ H^{p}(D) which cannot be holomorphically continued in Ω̃. Here Δ = Δ_{x} = ∂_{x_{1}x_{1}} + ⋯ + ∂_{x_{n}x_{n}} is the usual Laplacian of ℝ^{n}, \overline{D_{j}} = \frac{∂}{∂\overline{z_{j}}} , j = 1, 2, …, n.

Proof Let ξ ∈ ∂H(D) and 1 ≤ p ≤ [\frac{n}{2}]. Firstly, we will construct explicitly a p-polyharmonic function h_{ξ} in D whose holomorphic continuation h̃_{ξ} in 𝔑H(D) extends to the whole of H(D); however h̃_{ξ} cannot be holomorphically continued in a neighborhood of ξ. Next, we will deduce by a topological reasoning the existence of a p-polyharmonic function h in D such that H(D) is the domain of holomorphy of h̃.

Construction of h_{ξ}. 1) D ⊂ ℂ ≃ ℝ^{2}: by [12], the boundary point ξ belongs to some isotropic cone of vertex a t ∈ ∂D , i.e. ξ ∈ Γ(t), or t ∈ T (ξ) = \{ξ_{1} + iξ_{2}, \overline{ξ_{1}} + i\overline{ξ_{2}}\}.

a) If t = ξ_{1} + iξ_{2} , we consider the function

h̃_{ξ}(z) = Ln \{[(ξ_{1} + iξ_{2}) − (z_{1} + iz_{2})][\overline{(ξ_{1} + iξ_{2}) − (\overline{z_{1}} + i\overline{z_{2}})}]\},

where the branch is chosen such that h̃_{ξ} is real in D. Note that h̃_{ξ} is holomorphic in H(D), its restriction ˜(h_{ξ}|D)(x) = 2 Ln ∥x−t∥, where x = x_{1}+ix_{2} is harmonic in D, and lim_{z→ξ} |h̃_{ξ}(z)| = ∞. Hence h̃_{ξ} cannot be holomorphically continued beyond ξ.

b) If t = \overline{ξ_{1}} + i\overline{ξ_{2}} , the function

h̃_{ξ}(z) = Ln \{[(\overline{ξ_{1}} + i\overline{ξ_{2}}) − (z_{1} + iz_{2})][\overline{(\overline{ξ_{1}} + i\overline{ξ_{2}}) − (\overline{z_{1}} + i\overline{z_{2}})}]\},

satisfies the same requirements of (a). 2) D ⊂ ℝ^{n}, n ≥ 3:

a) Suppose n even ≥ 4. There exists by [12] a point t ∈ ∂D such that ∑_{j=1}^{n}(ξ_{j} −

t_{j})^{2} = 0. Consider then h̃_{ξ}^{p} : H(D) → ℂ, z = (z_{1}, …, z_{n}) ↦ h̃_{ξ}^{p}(z) with

h̃_{ξ}^{p}(z) = \begin{cases}\frac{1}{[(z_{1}−t_{1})^{2}+...(z_{n}−t_{n})^{2}]^{\frac{n}{2} −p}} & when 1 ≤ p ≤ \frac{n}{2} − 1 \\ Ln ∑_{j=1}^{n}(z_{j} − t_{j})^{2} & when p = \frac{n}{2} \end{cases}

(The branch is chosen in the complex logarithm such that (h̃_{ξ}|D) is real in D). Since H(D is the connected component containing D of ℂ^{n}\ ∪_{t∈∂D} {z ∈ ℂ^{n} ∑_{j=1}^{n}(z_{j} −t_{j})^{2} = 0}, we see that h̃_{ξ}^{p} is defined and holomorphic in H(D) and

that lim_{z→ξ} |h̃_{ξ}^{p}(z)| = ∞. It remains thus to prove that the restriction h̃_{ξ}^{p}|D is actually p-polyharmonic in D. 2.a.i: 1 ≤ p ≤ \frac{n}{2} − 1. Since for every x ∈ D : (h̃_{ξ}^{p}|D)(x) = 1/(r^{n−2p}) depends only on r = ∥x − t∥ the proof can be carried out directly. Indeed, it is simplest to introduce polar coordinates with t as origin and to use

Δ^{p} = (\frac{∂^{2}}{∂r^{2}} + \frac{n − 1}{r} \frac{∂}{∂r} + \frac{1}{r^{2}} B)^{p},

where B is the Beltrami operator containing only derivatives with respect to the angles variables. Now by induction on q = 1, 2, . . .. we find after some calculus that for an arbitrary (complex) α,

Δ^{q}(r^{α}) = α(α + n − 2)(α − 2)(α + n − 4) … (α − 2q + 2)(α + n − 2q)r^{α−2q}.

Observe that if α = 2p − n we obtain

Δ^{q}(r^{2p−n}) = (2p − n)(2p − 2) … (2p − n − 2q + 2)(2p − 2q)r^{2p−n−2q},

which gives respectively for q = p and q = p − 1:

Δ^{p}(r^{2p−n}) =0,
Δ^{p−1}(r^{2p−n}) =(2p − n)(2p − 2)(2p − n − 2)(2p − 4) … (4 − n)2r^{2−n}.

Note that Δ^{p−1}(r^{2p−n}) ≠ 0 if n is even and greater than or equal to 6; in addition, the case n = 4 involves p = 1, and so the last equality holds since Δ(\frac{1}{r^{2}} ) = 0 , Δ^{0}(\frac{1}{r^{2}} ) = 1. 2.a.ii: p = \frac{n}{2}: Since for every x ∈ D, the restriction h̃_{ξ}^{p}|D : x ↦ 2 Ln r, where r = ∥x − t∥, is a radial function, we use the same process to verify that Ln r is a \frac{n}{2}-polyharmonic function in ℝ^{n} − \{0\} for all n = 2p ≥ 4. As Δ(Ln r) = (2p − 2)r^{−2} , and Δ^{q}(Ln r) = (2q − 2)Δ^{q−1}(r^{−2}), we can make use of the corresponding formula of 2.a.(i) with α = −2 to have

Δ^{q}(Ln r) = (−1)^{q−1}2^{q}(q − 1)[(q − 1)!](n − 4)(n − 6) … (n − 2q)\frac{1}{r^{2q}} .

The last equality holds actually for all n ≥ 2 and q ≥ 1 since by the case (1) above this result is true for n = 2. Observe that if n = 2p ≥ 4 one obtains

Δ^{q}(Ln r) = (−1)^{q−1}2^{q}(q − 1)[(q − 1)!](2p − 4) … (2p − 2q)\frac{1}{r^{2q}} in ℝ^{2p} − \{0\}.

Thus Δ^{q}(Ln r) ≠ 0 for q = 1, 2, …, p − 1, and Δ^{q}(Ln r) = 0 if q = p.

b) Suppose n is odd, n ≥ 3. We consider again

h̃_{ξ}^{p}(z) = \frac{1}{[[ξ_{1} − z_{1})^{2} + ⋯ + [ξ_{n} − z_{n})^{2}]^{\frac{n}{2} −p}}

with 1 ≤ p ≤ [\frac{n}{2}] − 1, where the chosen branch is such that h̃_{ξ}^{p}|D, (x) > 0 in D.

Note that h̃_{ξ}^{p}(z) is holomorphic in H(D) and infinite in any neighborhood of ξ. By a similar calculus, we find for every x ∈ D,

Δ^{p}[h̃_{ξ}^{p}|D(x)] = 0
Δ^{p−1}[h̃_{ξ}^{p}|D(x)] ≠ 0 .

Existence of h: In the following we shall make use of the lemma.

Lemma 2.2 Let O[H(D)] denote the Fréchet space of all holomorphic functions on H(D), if it is endowed with the topology (τ) of uniform convergence on compact subsets of H(D). Then for all integer p = 1, 2, …, the set

O^{p}[H(D)] = {F ∈ O[H(D)]; F |D ∈ H^{p}(D)}

is a close subspace of O[H(D)], and therefore it is itself a Fréchet space.

Proof Let us consider F_{1}, F_{2}, . . .. a sequence in O^{p}[H(D)] ⊂ O[H(D)] converging to a function F , uniformly on every compact K′ of H(D). It is well known by a theorem of Weierstrass that F is also holomorphic in H(D), it remains thus to verify that Δ^{p}(F |D) = 0, p = 1, 2, . . .. By [7], page 161, for all multi-index β = (β_{1}, …, β_{n}) ∈ ℕ^{n}: D^{β}F_{j} → D^{β}F, uniformly on every compact K′ of H(D); in particular we also have (D^{β}F_{j})|D → (D^{β}F)|D uniformly on any compact K ⊂ D since we may treat all K′ ∩ ℝ^{n} ≠ ∅ as compact subsets of the real subspace in the complex (z_{1}, …, z_{n})−space. Now, note that

(D^{β}F_{j})|D = (D_{z}^{β}F_{j})|D

= (\frac{∂^{|β|}F_{j}}{∂z_{1}^{β_{1}} … ∂z_{n}^{β_{n}}})|D = \frac{∂^{|β|}}{∂x_{1}^{β_{1}} … ∂x_{n}^{β_{n}}} (F_{j}|D) = D_{x}^{β}(F_{j}|D),

where z_{j} = x_{j} + iy_{j} , j = 1, …, n. Then for q = 1, 2, …, p − 1, the sequence

(Δ_{z}^{q}F_{j})|D = [(∑_{j=1}^{n} \frac{∂^{2}}{∂z_{j}^{2}} )^{q}F_{j}]|D = ( ∑_{|α|=q} \frac{q!}{α!}D_{z}^{2α}F_{j})|D

= ∑_{|α|=q} \frac{q!}{α!}D_{x}^{2α}(F_{j}|D) = Δ_{x}^{q}(F_{j}|D),

being a finite sum of derivatives (D^{β}F_{j})|D, we have Δ_{x}^{q}(F_{j}|D) → Δ_{x}^{q}(F |D), uniformly on every compact K of D. Putting F_{j}|D = f_{j} and F |D = f, we have also for every x ∈ D: lim_{j→∞}[Δ^{q}f_{j}(x)] = Δ^{q}f(x), q = 1, 2, …, p − 1. Since each f_{j} is supposed p-polyharmonic in D for 1 ≤ p ≤ [\frac{n}{2}], we have f_{j} ∈ C_{ℝ}^{2p}(D) and f_{j} satisfies the appropriate mean value property, see [4]:

λ(f_{j}, x, R) = f_{j}(x) + ∑_{q=1}^{p−1} a_{q}R^{2q}Δ^{q}f_{j}(x)  (2.1)

for all x ∈ D, and R > 0 so small that B_{n}^{r}(x, R) = {y ∈ ℝ^{n}; ∥y − x∥ < R} ⊂ D, where λ(f_{j}, x, R) denotes the integral mean values over the surface ∂B_{n}^{r}(x, R):

λ(f_{j}, x, R) = \frac{Γ(\frac{n}{2})}{2\sqrt{π^{n}}} ∫_{∥a∥=1} f_{j}(x + Ra)dσ(a) ,

with dσ(a) an element of surface differential on the sphere S^{n−1}(O, 1) and a_{q} = Γ(\frac{n}{2})/(2^{2q}q!Γ(q + \frac{n}{2})). As f_{j} converges to f uniformly on the compact set S^{n−1}(x, R), the limit process applied to (2.1) yields of course (2.1) for f:

λ(f, x, R) = f(x) + ∑_{q=1}^{p−1} a_{q}R^{2q}Δ^{q}f(x);

that is, f = lim_{j→∞} f_{j} has the mean value property (2.1) in D. Thus f is p-polyharmonic in D.

To prove the existence of the aforesaid function h, let z^{(1)}, …, z^{(j)}. be a denumerable dense subset of the compact set ∂H(D). For every (j, k) ∈ ℕ^{∗^{2}}, let B_{n}^{c}(z^{(j)}, \frac{1}{k}) = \{w ∈ ℂ^{n}; ∥w − z^{(j)}∥ < \frac{1}{k} \} denote the hermitian ball of ℂ^{n} centered at z^{(j)} and of radius\frac{1}{k}, and put

H_{j,k}(D) = H(D) ∪ B_{n}^{c}(z^{(j)}, \frac{1}{k}).

Due to the density of {z^{(j)}}_{j∈N^{∗}} in ∂H(D), it is enough to prove the existence of a function belonging to O^{p}[H(D)] which cannot be holomorphically continued beyond H(D). This amounts to show the existence of a functions belonging to O^{p}[H(D)] which cannot be holomorphically continued to any domain H_{j,k}(D). Thanks to the construction step, for all j, k ∈ ℕ^{∗} and every p = 1, 2, …, [\frac{n}{2}], we have:

O^{p}[H(D)] \ R_{j,k}{O^{p}[H_{j,k}(D)]} ≠ ∅,

where R_{j,k} denotes the restriction mapping from O^{p}[H_{j,k}(D)] to O^{p}[H(D)], and O^{p} [H_{j,k}(D)] the space of all holomorphic functions in H_{j,k}(D) whose trace on ℝ^{n} is p-polyharmonic in D. The spaces O^{p}[H(D)] and O^{p}[H_{j,k}(D)] being Fréchet spaces by the Lemma above, and the linear and continuous mapping R_{j,k} being not onto, we deduce owing to a Banach Theorem [14] that the range of R_{j,k} is a subset of the first category of O^{p}[H(D)]; that is,

R_{j,k}{O^{p}[H_{j,k}(D)]} = ∪_{m=1}^{∞}X_{j,k}^{m},

where X_{j,k}^{m}, m = 1, 2, … are subsets of O^{p}[H(D)] satisfying (\overline{X_{j,k}^{m}})^{0} = ∅, with respect to the topology (τ). Observe that

∪_{j,k=1}^{∞}R_{j,k}{O^{p}[H_{j,k}(D)]} = ∪_{j,k=1}^{∞}(∪_{m=1}^{∞}X_{j,k}^{m}) = ∪_{j,k,m=1}^{∞}X_{j,k}^{m}

is also of the first category in O^{p}[H(D)]. Since O^{p}[H(D)] is in particular a Baire space, we have, of course,

O^{p}[H(D)] \ ∪_{j,k,m=1}^{∞}X_{j,k}^{m} = O^{p}[H(D)] \ ∪_{j,k=1}^{∞}R_{j,k}{O^{p}[H_{j,k}(D)]} ≠ ∅,

so we can pick up an element h of O^{p}[H(D)] which cannot be continued holomorphically through ∂H(D). □

Corollary 2.3 Let D be a C-domain of ℝ^{n}(n ≥ 2), or D be a convex domain of ℝ^{n}. Then for every integer p, (1 ≤ p ≤ [\frac{n}{2}], the integer part of \frac{n}{2}), the harmonicity cell of D satisfies

H(D) = [∩_{u∈H^{p}(D)}D^{u}]^{0}  (2.2)

where H^{p}(D) = {u ∈ C^{∞}(D); Δ^{p}u = 0 in D} and D^{u} is the complex domain of ℂ^{n} to which a polyharmonic function u extends holomorphically.

Proof By [4] Lemma 1.1.2, each p-polyharmonic function u in D, p ∈ ℕ^{∗}, is the restriction of a holomorphic function ũ in D^{u} ⊂ ℂ^{n} such that D^{u} ∩ ℝ^{n} = D. The former property is actually a consequence of the analyticity of u. In addition, the p-polyharmonicity of u implies more precisely that the kernel of H(D) is included in D^{u}(see [4] Theorem 5.2.6). If u wanders through the whole class H^{p}(D), we obtain: N H(D) ⊂ [∩_{u∈H^{p}(D)}D^{u}]^{0} (note here that the kernel of a harmonicity cell is a connected open of ℂ^{n}).

Inversely, due to Theorem 2.1 above, we can associate to every 1 ≤ p_{0} ≤ [\frac{n}{2}] a function f_{p_{0}} ∈ H^{p_{0}}(D) satisfying D^{f_{p_{0}}} = H(D). So, if 1 ≤ p ≤ [\frac{n}{2}] we get obviously the inclusion: [∩_{u∈H^{p}(D)}D^{u}]^{0} ⊂ D^{f_{p_{0}}} . Hence, one deduces N H(D) ⊂ [∩_{u∈H^{p}(D)}D^{u}]^{0} ⊂ H(D). Now the assumption on D guarantees that N H(D) = H(D); the desired equality follows. □

Observe that it is an unexpected result that the right-hand side of Equality (2.2) does not depend on the choice of p. This allows us thus to give the following result.

Corollary 2.4 For every bounded domain D of ℝ^{n}, n ≥ 2, with D ≠ ∅, and ∂D ≠ ∅, we have

H(D) = ∩_{1≤p≤[ \frac{n}{2} ]}[∩_{u∈H^{p}(D)}D^{u}]^{0}.

Remark 2.5 Putting p = 1 in Corollary 2.3, we find again an Avanissian s’ result (cf. [4] p.67): Let A(D) (Ha(D)) be the class of all real analytic (harmonic) functions on D ⊂ ℝ^{n}. For f ∈ A(D), we denote f̃: D^{f} → ℂ the holomorphic extension of f to the maximal domain D^{f} of ℂ^{n} (in the inclusion meaning).

Then the sets: A = ∩_{f∈A(D)}D^{f} and B = ∩_{f∈Ha(D)}D^{f} satisfy A^{∘}= ∅, B^{∘}= H(D).

3 Some properties of harmonicity cells

In [4], Avanissian established the following general results about the operation D ↦ H(D); see also [13].

Proposition 3.1 The harmonicity cells of domains of ℝ^{n}, n ≥ 2, satisfy

a) If D_{1} ∩ D_{2} = ∅, H(D_{1}) ∩ H(D_{2}) = ∅; if D_{1} ⊂ D_{2}, H(D_{1}) ⊂ H(D_{2}). b) H(∪_{ν∈J}D_{ν}) = ∪_{ν∈J}H(D_{ν}) for every exhaustive increasing family of do-

mains D_{ν}.

c) H(D)∩ℝ^{n} = D; H(D) is symmetric with respect to ℝ^{n}; and if D is convex then so is H(D).

d) If D is starshaped at a_{0}, then H(D) is starshaped at a_{0} and H(D) = {z ∈ ℂ^{n}; T (z) ⊂ D}.

e) ∂D ⊂ ∂H(D) ; if z ∈ \overline{H(D)} , T (z) ⊂ \overline{D} ; and if z ∈ ∂H(D) , T (z)∩∂D ≠ ∅.

f) δ[H(D)] ≤ 2[\frac{n}{2n+2}]^{\frac{1}{2}} δ(D), where δ(D) denotes the diameter of D.

g) H(D) may be explicitly obtained when D is a ball, a cube, or a difference of two balls.

h) If n = 2 and ℝ^{2} ≃ ℂ, H(D) = \{z ∈ ℂ^{2}; z_{1} + iz_{2} ∈ D, \overline{z_{1}} + i\overline{z_{2}} ∈ D\}.

In the following, we establish supplementary results. Let 𝔇^{n} denote henceforth the family of all domains D of ℝ^{n}, D ≠ ∅, ∂D ≠ ∅, and ℭ_{s}^{n} the family of all domains of ℂ^{n} which are symmetric with respect to ℝ^{n} = {x + iy ∈ ℂ^{n}; y = 0}.

Proposition 3.2 The mapping H : D ∈ 𝔇^{n} ↦ H(D) ∈ ℭ_{s}^{n} satisfies:

a) H is injective ; H(D) is bounded if and only if D is bounded.

b) For every compact set K ⊂ H(D), there exists a domain D_{1} ∈ 𝔇^{n} such that D_{1} is relatively compact in D and K ⊂ H(D_{1}).

c) If D_{1}, D_{2} ∈ 𝔇^{2} are such that D_{1} ∩ D_{2} is connected then H(D_{1} ∩ D_{2}) = H(D_{1})∩H(D_{2}). If (D_{j})_{j∈J} is a family of starshaped domains in ℝ^{n}(n ≥ 2) such that ∩_{j∈J}D_{j} is a starshaped domain, or if (D_{j})_{j∈J} is a family of convex domains in ℝ^{n} such that ∩_{j∈J}D_{j} is open, then H(∩_{j∈J}D_{j}) = ∩_{j∈J}H(D_{j}).

d) If D_{1}, D_{2} ∈ 𝔇^{2} with D_{1} ∩D_{2} ≠ ∅ then H(D_{1} ∪D_{2}) ⊃ H(D_{1})∪H(D_{2}), and the equality holds if and only if D_{1} ⊂ D_{2} or D_{2} ⊂ D_{1}. More generally, if (D_{j})_{j∈J} ⊂ 𝔇^{n}(n ≥ 2) is such that D_{i} ∩ D_{j} ≠ ∅ for all i, j ∈ J then H(∪_{j∈J}D_{j}) ⊃ ∪_{j∈J}H(D_{j}). The equality holds if ∪_{j∈J}D_{j} = D_{j_{0}} for a certain j_{0} ∈ J.

Proof a) By Proposition 3.1, H is well defined on 𝔇^{n} with values in ℭ_{s}^{n}; and if two harmonicity cells H(D) and H(D′) coincide in ℂ^{n}, then their traces on ℝ^{n} coincide also, that is D = D′. Besides, suppose that for some R > 0, D ⊂ B_{n}^{r}(0, R) = {x ∈ ℝ^{n}; ∥x∥ < R}. Then H(D) ⊂ H[B_{n}^{r}(0, R)] = LB(0, R) = {z ∈ ℂ^{n}; L(z) < R} (the Lie ball of ℂ^{n}), see [3, 5, 9], where

L(z) = [∥z∥^{2} + \sqrt{∥z∥^{4} − |∑_{j=1}^{n}z_{j}^{2}|^{2}} ]^{1/2}.

Since ∥z∥ ≤ L(z), we have LB(0, R) ⊂ B_{n}^{c}(0, R) = {z ∈ ℂ^{n}; ∥z∥ < R} and H(D) is bounded in ℂ^{n}. The converse is obvious since D ⊂ H(D).

b) Let us consider an increasing exhaustive family (D_{ν})_{ν∈J} (J is a fixed indices set) of bounded domains D_{ν} ∈ 𝔇^{n} such that D = ∪_{ν∈J}D_{ν}. Due to 3.1.b, the family (H(D_{ν}))_{ν∈J} of harmonicity cells of (D_{ν})_{ν∈J} is also increasing, exhaustive and satisfies H(D) = ∪_{ν∈J}H(D_{ν}). We then have: K ⊂ ∪_{ν∈J}H(D_{ν}). Since K is a compact set, we can extract from this open covering of K, a finite subcovering of K: K ⊂ ∪_{k=1}^{n}H(D_{ν_{k}}). Afterwards we have by 3.1.a: ∪_{k=1}^{n}H(D_{ν_{k}}) ⊂ H(∪_{k=1}^{n}D_{ν_{k}}) and K ⊂ H(∪_{k=1}^{n}D_{ν_{k}}). Seeing that D′ is a relatively compact domain in D and taking D′ = ∪_{k=1}^{n}D_{ν_{k}}, we obtain the desired result.

c) The inclusion H(D_{i} ∩ D_{j}) ⊂ H(D_{i}) ∩ H(D_{j}) is obvious from D_{i} ∩ D_{j} ⊂ D_{i}, D_{i} ∩ D_{j} ⊂ D_{j}. If w ∈ H(D_{i}) ∩ H(D_{j}), T (w) ⊂ D_{i} ∩ D_{j}, that is w ∈ H(D_{i} ∩ D_{j}). By similar arguments we obtain the general case.

d) D_{1} ∩ D_{2} ≠ ∅ guarantees that D_{1} ∪ D_{2} ∈ 𝔇^{n}. Since D_{i} ⊂ D_{1} ∪ D_{2}, i = 1, 2, H(D_{1}) ∪ H(D_{2}) ⊂ H(D_{1} ∪ D_{2}). Suppose now that D_{1} is neither included in D_{2}, nor D_{2} in D_{1}. If a and b are arbitrarily chosen in D_{1} \ D_{2} and D_{2} \ D_{1} respectively, and if ℝ^{2} ≃ ℂ, then the point w = (\frac{a+\overline{b}}{2}, \frac{a−\overline{b}}{2i}) of ℂ^{2} satisfies T (w) = {a, b} ⊂ D_{1} ∪ D_{2}. Now, the last hypothesis on D_{1} and D_{2} involves that w ∉ H(D_{1}) ∪ H(D_{2}). Besides, as D_{i} ∩ D_{j} ≠ ∅ we have ∪_{j∈J}D_{j} ∈ 𝔇^{n} and thus this union does possess a harmonicity cell in ℂ^{n}. The given inclusion is evident since D_{i} ⊂ ∪_{j∈J}D_{j}. Suppose in addition that D_{j_{0}} = ∪_{j∈J}D_{j}. From H(D_{j_{0}}) ⊂ ∪_{j∈J}H(D_{j}) and H(∪_{j∈J}D_{j}) ⊃ ∪_{j∈J}H(D_{j}), we deduce the equality H(∪_{j∈J}D_{j}) = ∪_{j∈J}H(D_{j}). □

Corollary 3.3 If (D_{j})_{j≥1} is a monotonous sequence in 𝔇^{n}, so is (H(D_{j}))_{j≥1} in ℭ_{s}^{n}; and writing D = lim_{j→∞} D_{j}, we have lim_{n→∞} H(D_{n}) = H(D) under the assumptions that: ∪_{j≥1}D_{j} ≠ ℝ^{n} if the sequence (D_{j})_{j≥1} is increasing, and that ∩_{j∈J}D_{j} ∈ 𝔇^{n} in the decreasing case.

Proof If the sequence is increasing then lim inf_{n→∞} D_{n} = ∪_{n≥1}(∩_{k≥n}D_{k}) = ∪_{n≥1}D_{n}, lim sup_{n→∞} D_{n} = ∩_{n≥1}(∪_{k≥n}D_{k}) = ∩_{n≥1}(∪_{k≥n}D_{k}) = ∪_{k≥1}D_{k}, so lim_{n→∞} D_{n} = ∪_{n≥1}D_{n}. Since ∪_{n≥1}D_{n} ≠ ∅ and ∪_{n≥1}D_{n} ≠ ℝ^{n}, we deduce that lim_{n≥1} D_{n} is an element of 𝔇^{n}.

Next, (H(D_{n}))_{n≥1} being also increasing, lim_{n→∞} H(D_{n}) = ∪_{n≥1}H(D_{n}). Now by 3.2.d: ∪_{n≥1}H(D_{n}) ⊂ H(∪_{n≥1}D_{n}). Moreover, if w_{0} ∈ H(∪_{n≥1}D_{n}) one has T (w_{0}) ⊂ ∪_{n≥1}D_{n}; then by 3.2.b and the fact that T (z) is a compact set for every z ∈ ℂ^{n}, there exists n_{0} ≥ 0 such that T (w_{0}) ⊂ D_{n_{0}} i.e. w_{0} ∈ H(D_{n_{0}}) ⊂ ∪_{n≥1}H(D_{n}). Thus H(lim_{n→∞} D_{n}) ⊂ lim_{n→∞} H(D_{n}), which involves the aforesaid equality. In case of a decreasing sequence (D_{n})_{n≥1} one has lim inf_{n→∞} D_{n} = ∪_{n≥1}(∩_{k≥n}D_{k}) = ∪_{n≥1}(∩_{k≥1}D_{k}) = ∩_{k≥1}D_{k}, and lim sup_{n→∞} D_{n} = ∩_{n≥1}(∪_{k≥n}D_{k}) = ∩_{n≥1}D_{n}. So lim_{n→∞} D_{n} = ∩_{n≥1}D_{n}, which is in 𝔇^{n} by hypothesis. Now, (H(D_{n}))_{n≥1} being decreasing one also has: lim_{n→∞} H(D_{n}) = H(lim_{n→∞} D_{n}). □

Corollary 3.4 The mapping D ↦ H(D) is not a surjective operator: The unit hermitian ball B_{n}^{c} = {z ∈ ℂ^{n}; ∥z∥ < 1} does not represent a harmonicity cell in ℂ^{n}.

Proof. Since B_{n}^{c} ∩ ℝ^{n} = B_{n}^{r} = {x ∈ ℝ^{n}; ∥x∥ < 1} is a convex domain of ℝ^{n} (according to the induced topology), we have to find a point w_{0} ∈ B_{n}^{c} for which the Lelong sphere T (w_{0}) is not contained into B_{n}^{r}. For, put w_{0} = ρ (i, 1, …, 1) ∈ ℂ^{n} where ρ > 0 is small enough for w_{0} to belong at B_{n}^{c} and for T (w_{0}) to contain a certain ξ_{0} ∈ ℝ^{n} with ∥ξ_{0}∥ ≥ 1. Taking [n + 2 \sqrt{n − 1}]^{−1/2} < ρ < 1/n and writing w_{0} = x_{0} + i y_{0} we see that a ξ_{0} satisfying

[⟨ξ_{0} − x_{0}, y_{0}⟩ = 0, ∥ξ_{0} − x_{0}∥ = ∥y_{0}∥, and ∥ξ_{0}∥ ≥ 1];

that is,

ρξ_{1} = 0, ξ_{1}^{2} + (ξ_{2} − ρ)^{2} + … (ξ_{n} − ρ)^{2} = ρ^{2} and ξ_{1}^{2} + ⋯ + ξ_{n}^{2} ≥ 1

is given by: ξ_{0} = ρ[1 + (n − 1)^{\frac{−1}{2}} ](0, 1, …, 1).

Remark 3.5 Due to propositions 3.1 and 3.2 above, the definition of a harmonicity cell may be naturally extended to arbitrary open sets of ℝ^{n} for n ≥ 1 as follows H(∅) = ∅, H(ℝ^{n}) = ℂ^{n}, H(]a, b[) = ℂ for ]a, b[⊂ ℝ, and H(O) = ∪_{i∈I}H(O_{i}), where O is an open set of ℝ^{n}, (O_{i})_{i∈I} the family of the connected components of O.

Remark 3.6 Some properties are not always preserved by D ↦ H(D); this is especially the case if:

(i) D is simply connected in ℝ^{n} with n ≥ 3. Indeed, the two domains D = ℝ^{n} − {0} and H(D) = ℂ^{n} − {z ∈ ℂ^{n}; z_{1}^{2} + ⋯ + z_{n}^{2} = 0}, having 0 and ℤ respectively as fundamental groups, they offer then an example of a not simply connected harmonicity cell corresponding to a real simply connected domain; for π_{1}[H(D)] = ℤ, see [6].

(ii) D is strictly convex in ℝ^{n}with n ≥ 2. An example is given by the harmonicity cell of the unit ball B_{n}^{r} of ℝ^{n}. If E(\overline{V}) denotes the set of all extremal points of a convex V we have E(\overline{B_{n}^{r}}) = ∂B_{n}^{r} since these two sets coincide with the unit Euclidean sphere S^{n−1} of ℝ^{n}. Nevertheless, by [9]: E(\overline{H(B_{n}^{r})}) = ∂^{∨}[H(B_{n}^{r})] = \{w = xe^{iθ} ∈ ℂ^{n}; x ∈ S^{n−1}, θ ∈ ℝ\}, where ∂^{∨}U

denotes the Ŝilov boundary of U ⊂ ℂ^{n}; thus: E(\overline{H(B_{n}^{r})}) ⊂^{≠} ∂[H(B_{n}^{r})].

(iii) D is partially - circled in ℂ^{n} ≃ ℝ^{2n}, n ≥ 2, that is (for instance): z ∈ D ⇒ (z_{1}, …, z_{n−1}, e^{iθ}z_{n}) ∈ D, for all θ ∈ ℝ. Indeed if D = B_{n}^{c} = {z ∈ ℂ^{n}; ∥z∥ < 1}, H(B_{n}^{c}) is not partially - circled in ℂ^{2n} with respect to w_{2n} since w_{0} = \sqrt{1 + 2n}(1, …, 1) ∈ ℂ^{2n} satisfies L(w_{0}) = \sqrt{2n/(1 + 2n)} < 1, but L[(2n + 1)^{\frac{−1}{2}} , …, (2n + 1)^{\frac{−1}{2}} , i(2n + 1)^{\frac{−1}{2}} ] = [2n + 2\sqrt{2n − 2}]^{\frac{1}{2}} (2n + 1)^{\frac{−1}{2}} > 1. On the other hand, B_{n}^{c} is even circled (at the origin).

4 Harmonicity cells of polygonal plane domains

The case n = 2 is rather special since the Lelong map T is given by: T (z) = \{z_{1} + iz_{2}, \overline{z_{1}} + i\overline{z_{2}}\}, where z ∈ ℂ^{2} and ℝ^{2} ≃ ℂ. So, in [5], we have determined explicitly the harmonicity cells of some plane domains and shed light on the close connection between the set E(\overline{D}), of all the extremal points of a convex domain D of ℝ^{2}, and the set E(\overline{H(D)}), see also [4]. We will give now some properties and constructions which are proper to the complex plane. More precisions on the Jarnicki extension given in Section 1 will also be established.

Proposition 4.1 The operator H : 𝔇^{2} → ℭ_{s}^{2} satisfies

a) If D is circled at z_{0} ∈ ℂ, balanced at z_{0} ∈ D, or simply connected, then so is H(D) respectively.

b) If P_{n}^{a} is an arbitrary convex polygon with n edges, then the harmonicity cell H(P_{n}^{a}) is of polyhedric form in ℂ^{2} with 2n faces and n^{2} vertices. Furthermore, identifying ℂ^{2} with ℝ^{4} by writing y = (x_{3}, x_{4}) and x + iy = (x_{1}, x_{2}, x_{3}, x_{4}), each support line of P_{n}^{a} defined, for a certain j = 1, …, n, by a_{j}x_{1}+b_{j}x_{2}−α_{j} = 0, (a_{j}, b_{j}, α_{j} ∈ ℝ), generates two support hyperplanes of H(P_{n}^{a}) of respective equations:

a_{j}x_{1}+b_{j}x_{2}+b_{j}x_{3}−a_{j}x_{4}−α_{j} = 0 and a_{j}x_{1}+b_{j}x_{2}−b_{j}x_{3}+a_{j}x_{4}−α_{j} = 0.

c) Let P_{n}^{r} denote the regular polygon which vertices are ω_{k} = e^{2ikπ/n}, k = 0, …, n − 1. Then

H(P_{n}^{r}) =\{w = x + iy ∈ ℂ^{2} : x_{1} cos(2k + 1)\frac{π}{n} + x_{2} sin(2k + 1)\frac{π}{n}

+ \sqrt{∥y∥^{2} − [y_{1} cos(2k + 1)\frac{π}{n} + y_{2} sin(2k + 1)\frac{π}{n}]^{2}} < cos \frac{π}{n},

k = 0, …, n − 1}.

d) The n^{2} vertices of \overline{H(P_{n}^{r})} are given by ω_{km} = x_{km} + iy_{km} and \overline{ω_{km}} = x_{km} − iy_{km}, (0 ≤ k ≤ m ≤ n − 1), where

x_{km} = \frac{1}{2}(cos \frac{2kπ}{n} + cos \frac{2mπ}{n} , sin \frac{2kπ}{n} + sin \frac{2mπ}{n}),

y_{kk} = 0, k = 0, …, n − 1,

y_{km} = \frac{sin π(m − k) /n}{\sqrt{2}[1 − cos 2π(m − k) /n]^{1/2}}

×(sin \frac{2πm}{n} − sin \frac{2πk}{n} , cos \frac{2πk}{n} − cos \frac{2πm}{n}).

Proof a) For θ ∈ ℝ, z_{0} = a + ib ∈ ℂ, and w = (w_{1}, w_{2}) ∈ H(D), we see that z_{0} + e^{iθ}w remains in H(D). Since T (z_{0} + e^{iθ}w) = {a + e^{iθ}w_{1} + i(b + e^{iθ}w_{2}), a+e^{−iθ}\overline{w_{1}}+i(b+e^{−iθ}\overline{w_{2}})\} = \{z_{0}+e^{iθ}(w_{1}+iw_{2}), z_{0}+e^{−iθ}(\overline{w_{1}}+i\overline{w_{2}})\}, and as D is circled with respect to z_{0}, we have T (z_{0}+e^{iθ}w) ⊂ D. If the above circled domain D is supposed starshaped at z_{0} too, then H(D) is also starshaped at z_{0} (by 3.1.d) that is, H(D) is balanced at z_{0}. Let D ∈ D^{2} be a simply connected domain and f a holomorphic one-one map sending D onto B = {z ∈ ℂ; |z| < 1}. By Jarnicki Theorem , f extends to a holomorphic homeomorphism Jf : H(D) → H(B). Now, by [4], H(B) is the unit disk of (ℂ^{2}, L), where L is the Lie norm; this means that H(B) is convex and in particular simply connected. Since Jf is a homeomorphism, H(D) is also simply connected.

b) Suppose that P_{n}^{a} is defined by:

P_{n}^{a} = {x = x_{1} + ix_{2} ∈ ℝ^{2}; ⟨x, V^{j}⟩ < α_{j}, j = 1, …, n},

with given vectors V^{j} = (a_{j}, b_{j}) ∈ ℝ^{2} and scalars α_{j} ∈ ℝ. By 3.1.d, one has w = x + iy ∈ H(P_{n}^{a}) ⟺ x + T (iy) ⊂ P_{n}^{a} ⟺ x + ξ ∈ P_{n}^{a}, ∀ξ ∈ T (iy) ⟺ ⟨x, V^{j}⟩ + max_{ξ∈T (iy)}⟨ξ, V^{j}⟩ < α_{j}, j = 1, …, n. Since T (iy) = {(−y_{2}, y_{1}), (y_{2}, −y_{1})}, we have

H(P_{n}^{a}) = {w = x + iy ∈ ℂ^{2}; ⟨w, U^{j}⟩ < α_{j} and ⟨w, W^{j}⟩ < α_{j}, j = 1, …, n},

where w = (x_{1}, x_{2}, x_{3}, x_{4}), y = (x_{3}, x_{4}), U^{j} = (a_{j}, b_{j}, −b_{j}, a_{j}), and W^{j} = (a_{j}, b_{j}, b_{j}, −a_{j}), while ⟨, ⟩ denotes the usual scalar product in ℝ^{4}. From the expression above, we deduce that the harmonicity cell of an arbitrary convex polygon (not necessarily bounded) with n edges is a polyhedron of ℂ^{2} ≃ ℝ^{4} having 2n faces and by [5], n^{2} vertices.

c) For the regular polygon P_{n}^{r}, we have also another expression of its harmonicity cell. Indeed, if ℂ ≃ ℝ^{2}, we put ω_{n} = ω_{0}, ω_{k} = (cos \frac{2kπ}{n} , sin \frac{2kπ}{n}), and V^{k} = ω_{k+1} − ω_{k} = (a_{k}, b_{k}), k = 0, …, n − 1. By (b) we have

H(P_{n}^{r}) = \{x ∈ ℝ^{2}; ⟨x, V^{k}⟩ + max_{ξ∈T (iy)}⟨ξ, V^{k}⟩ < cos \frac{π}{n}, k = 0, …, n − 1\}.

By the method of Lagrange multipliers [4], we find max_{ξ∈T (iy)}⟨ξ, V^{k}⟩ = [∥y∥^{2} − ⟨y, V^{k}⟩^{2}]^{1/2}; the announced expression of H(P_{n}^{r}) follows.

d) Applying the following two lemmas proved in [5], (see also [4]) we obtain all the extremal points of \overline{H(P_{n}^{r})} by means of those of \overline{P_{n}^{r}} □

Lemma 4.2 If D is a non empty convex domain of ℝ^{n}, n ≥ 2, ∂D ≠ ∅, then E(\overline{D)} ⊂ E(\overline{H(D)}).

Lemma 4.3 Let D be a non empty convex domain, ∂D ≠ ∅, in ℂ ≃ ℝ^{2}.

a) Every point w ∈ E(\overline{H(D)}) satisfies T (w) ⊂ E(\overline{D)}. b)Conversely, given arbitrary points a and b of E(\overline{D)}, there exists w ∈ E(\overline{H(D)}) such that T (w) = {a, b}.

Let U, V be two domains of ℂ^{n}, n ≥ 1. we denote hom(U, V) the set of all holomorphic homeomorphisms F : U → V, and hom_{r}(H(D), H(D′)) the set of all F ∈ hom(H(D), H(D′)) of which the restriction F |_{D} belongs to hom(D, D′), where D, D′ ∈ D^{2} and ℂ ≃ ℝ^{2}.

Proposition 4.4 Let D, D′ ⊂ ℂ be two non empty domains with D ≠ ℂ, D′ ≠ ℂ. The Jarnicki extension J is an injective continuous mapping from hom(D, D′) onto hom_{r}(H(D), H(D′)) according to the compact uniform topology (τ).

Furthermore, hom_{r}(H(D), H(D′)) ≃ hom(D, D′) (topologically homeomorphic); and for a holomorphic homeomorphism f : D → D′ we have the estimate

∥Jf(w)∥ ≤ sup_{z∈D} |f(z)|, for every w ∈ H(D).

Proof If f and f′ are such that Jf = Jf′ on H(D) then by [10], f = (Jf)|D = (Jf′)|D = f′ on D. Let (f_{n})_{n≥1} be a convergent sequence in (hom(D, D′), τ). By 3.2.b,to test (J f_{n})_{n≥1} for compact uniform convergence in the harmonicity cell of D it is not really necessary to check uniform convergence on every compact set K in H(D) - checking it on the closed harmonicity cells \overline{H(D_{0})} where D_{0} is an arbitrary relatively compact domain in D is enough. Now if w_{0} ∈ H(D_{0}) with w_{0} = (w_{1}^{0}, w_{2}^{0}):

∥Jf_{n}(w_{0}) − Jf(w_{0})∥^{2} = A_{n}^{2}(w) + B_{n}^{2}(w),

where f = lim_{n→∞} f_{n}, and

A_{n} = \frac{1}{2}|[f_{n}(w_{1}^{0} + iw_{2}^{0}) − f(w_{1}^{0} + iw_{2}^{0})] + [\overline{f_{n}(\overline{w_{1}^{0}} + i\overline{w_{2}^{0}})} − \overline{f(\overline{w_{1}^{0}} + i\overline{w_{2}^{0}})}]|,

B_{n=} \frac{1}{2}|[f_{n}(w_{1}^{0} + iw_{2}^{0}) − f(w_{1}^{0} + iw_{2}^{0})] − [\overline{f_{n}(\overline{w_{1}^{0}} + i\overline{w_{2}^{0}})} − \overline{f(\overline{w_{1}^{0}} + i\overline{w_{2}^{0}})}]|.

Both A_{n} and B_{n} are bounded above by \frac{1}{2} sup_{w∈H(D_{0})} |f_{n}(w_{1} + iw_{2}) − f(w_{1} + iw_{2})| + \frac{1}{2} sup_{w∈H(D_{0})} |f_{n}(\overline{w_{1}} + i\overline{w_{2}}) − f(\overline{w_{1}} + i\overline{w_{2}}). By 3.1.h: w ∈ H(D_{0}) if and only if w_{1} + iw_{2} ∈ D_{0} and \overline{w_{1}} + i\overline{w_{2}} ∈ D_{0}. Thus:

A_{n} ≤ sup_{z∈D_{0}} |f_{n}(z) − f(z)|, B_{n} ≤ sup_{z∈D_{0}} |f_{n}(z) − f(z)|,

sup_{w∈\overline{H(D_{0})}} ∥Jf_{n}(w) − Jf(w)∥ ≤ \sqrt{2} sup_{z∈\overline{D_{0}}} |f_{n}(z) − f(z)|.

Since lim_{n→∞} sup_{z∈\overline{D_{0}}} |f_{n}(z) − f(z)| = 0, we have Jf_{n} → Jf, according to (τ). The mapping J : hom(D, D′) → hom_{r}(H(D), H(D′)) is continuous and injective. To see that this mapping is onto, take F ∈ hom_{r}(H(D), H(D′)) and observe that (by [10]) J(F |D) and F are both holomorphic homeomorphisms from H(D) onto H(D′) having the same restriction on D : (J(F |D))|D = F |D. So by the uniqueness principle of analytic extension in ℂ^{n} : J(F |D) = F. Conversely, putting: R = J^{−1}and making use of 3.1.c, e and 3.2.b, we have for every D_{0} ⊂ D with \overline{D_{o}} compact: sup\overline{H(D_{0})} ∥F_{n} − F ∥ ≥ sup\overline{D_{0}} |RF_{n} − RF |, which implies that R is also continuous. Finally,we have

∥Jf(w)∥^{2} =\frac{1}{4}|f(w_{1} + iw_{2}) + \overline{f(\overline{w_{1}} + i\overline{w_{2}})}|^{2} + \frac{1}{4}|f(w_{1} + iw_{2}) − \overline{f(\overline{w_{1}} + i\overline{w_{2}})}|^{2}
=\frac{1}{2}[|f(w_{1} + iw_{2})|^{2} + |f(\overline{w_{1}} + i\overline{w_{2}})|^{2}]
≤\frac{1}{2}[(sup_{\overline{D}} |f|)^{2} + (sup_{\overline{D}} |f|)^{2}] = (sup_{\overline{D}} |f|)^{2}.
□

Remark 4.5 The notion of harmonicity cells has a functorial aspects; indeed let 𝔇^{2} still denote the category of all domains D of ℝ^{2} ≃ ℂ, D ≠ ∅, ∂D ≠ ∅ with arrows in hom(D_{1}, D_{2}), and ℭ_{s}^{2} the category of all domains U of ℂ^{2} which are symmetric with respect to ℝ^{2}, with arrows F in hom(U_{1}, U_{2}). Then, by the uniqueness theorem of holomorphic continuation in ℂ^{n}, to the composition:

D_{1} →^{f} D_{2} →^{g} D_{3} corresponds H(D_{1}) →^{Jf} H(D_{2}) →^{Jg} H(D_{3}) such that: J(g ∘ f) =

(Jg) ∘ (Jf); next f = Id in Jarnicki Theorem (Section 1) gives: J Id_{D} = Id_{H(D)}. This means that the operator: D ∈ 𝔇^{2} ↦ H(D) ∈ ℭ_{s}^{2} and f ∈ hom(D_{1}, D_{2}) ↦ H(f) = Jf ∈ hom[H(D_{1}), H(D_{2})] may be considered as a covariant functor between the said categories. The representability of this functor and its classifying object will be discussed in a further paper.

Example If V is an arbitrary half strip of ℝ^{2}, there exists an usual transformation f, mapping V onto V′ = {x ∈ ℝ^{2} : x_{1} > a, k_{1} < x_{2} < k_{2}}, for some a > 0, k_{1}, k_{2} ∈ ℝ. Now by [4, 7], we have for all convex domains U of ℝ^{n} (n ≥ 2):

H(U) = {w = x + iy ∈ ℂ^{n}; max_{t∈T (iy)}[ max_{ξ∈S^{n−1}}(⟨x + t, ξ⟩ − sup_{u∈U}⟨ξ, u⟩)] < 0}.

This formula gives H(U) by means of the support function of U : δ_{U}(ξ) = sup_{u∈U}⟨ξ, u⟩. Making use of the fact that the function u ↦ ξ_{1}u_{1} + ξ_{2}u_{2}, being harmonic in V′, attains its supremum at some point of ∂V′. We find by simple calculations that

δ_{V′}(ξ) = \begin{cases}+∞ & if ξ_{1} > 0 \\ aξ_{1} + k_{2}ξ_{2} & if ξ_{1} ≤ 0 and ξ_{2} ≥ 0 \\ aξ_{1} + k_{1}ξ_{2} & if ξ_{1} ≤ 0 and ξ_{2} ≤ 0 \end{cases}

where ξ ∈ Γ, the unit circle of ℂ. Next, to search the supremum on Γ of the function g(ξ_{1}, ξ_{2}) = ⟨x + t, ξ⟩ − δ_{V′}(ξ), we restrict the study to {ξ ∈ Γ : ξ_{1} ≤ 0}. Since g(ξ_{1}, ξ_{2}) = g(ξ_{1}, ±\sqrt{1 − ξ_{1}^{2}}), with ξ_{1} ∈ [−1, 0], we put

g_{1}(ξ_{1}) = g(ξ_{1}, \sqrt{1 − ξ_{1}^{2}}) = α_{1}ξ_{1} + α_{2}\sqrt{1 − ξ_{1}^{2}} and g_{2}(ξ_{1}) = α_{1}ξ_{1} − β\sqrt{1 − ξ_{1}^{2}}, where α_{1} = x_{1} + t_{1} − a, α_{2} = x_{2} − t_{2} − k_{2}, β = x_{2} − t_{2} − k_{1}. One obtains that g′_{1}(ξ_{1}) = 0 if ξ_{1} = ±α_{1}/\sqrt{α_{1}^{2} + α_{2}^{2}} (when α_{1} ≠ 0 or α_{2} ≠ 0). In addition, the study of variations of g_{1}(ξ_{1}), in −1 ≤ ξ_{1} ≤ 0, in each of the three cases: α_{1} ≤ 0, (α_{1} ≥ 0 and α_{2} ≤ 0), and (α_{1} ≥ 0 and α_{2} ≥ 0) leads to max_{−1≤ξ_{1}≤0} g_{1}(ξ_{1}) = max(−α_{1}, α_{2}). Obviously, this equality holds even if α_{1} = α_{2} = 0. A similar calculus for g_{2}(ξ_{1}) gives max_{−1≤ξ_{1}≤0} g_{2}(ξ_{1}) = max(−β, −α_{2}). Putting γ = max( −α_{1}, α_{2}), δ = − min(β, α_{2}), and as T (iy) = {(−y_{2}, y_{1}), (y_{2}, −y_{1})}, we obtain the equivalence

max(γ, δ) < 0 ⇔ \{ \begin{matrix}a − x_{1} + y_{2} < 0, x_{2} + y_{1} − k_{2} < 0, k_{1} − x_{2} − y_{1} < 0, \\ a − x_{1} − y_{2} < 0, x_{2} − y_{1} − k_{2} < 0, k_{1} − x_{2} + y_{1} < 0. \end{matrix}

At last, writing min(u, v) = \frac{1}{2}(u + v − |u − v|) , and by the Jarnicki extension f ↦ Jf = f̃ (see section 1), we deduce H(V) = (f̃)^{−1}[H(V′)] , where

H(V′) = \{w = x + iy ∈ ℂ^{2}; |y_{1}| < \frac{k_{2} − k_{1}}{2} − |x_{2} − \frac{k_{1} + k_{2}}{2} |, |y_{2}| < x_{1} − a\}.

Example The harmonicity cell of an arbitrary convex polygon P′_{n} may be explicited by means of the n vertices ω′_{0}, …, ω′_{n−1}. For, put α = \frac{ω′_{0}+ω′_{2}}{2} and consider the translation τ_{−α} : z ↦ z − α. The domain P_{n} = τ_{−α}(P′_{n}) is also a convex polygon, with O ∈ P_{n} and n vertices ω_{0}, …, ω_{n−1}, given by ω_{k} = ω′_{k} − α. Making use of (d) and (h) in Proposition 3.1, we find after calculus and simplifications:

H(P_{n}) = \{w = x + iy ∈ ℂ^{2} : sgn(Im \overline{ω_{k}}ω_{k+1}) Im \overline{x}(ω_{k+1} − ω_{k})
+ \sqrt{|y|^{2}|ω_{k+1} − ω_{k}|^{2} − Im^{2}\overline{y}(ω_{k+1} − ω_{k})}
< | Im \overline{ω_{k}}ω_{k+1}|, k = 0, 1, …, n − 1\}

with ℝ^{2} ≃ ℂ, Im z is the imaginary part of z, and sgn α is the sign of α. Note that P′_{n} = τ_{α}P_{n} means that [w′ ∈ H(P′_{n})] if and only if [w′ −α ∈ H(P_{n})]. If now P′_{n,r} is some regular polygon, it is enough to consider its circumscribed circle C(β, R), centered at β ∈ ℝ^{2}, with radius R > 0. Next, applying successively the translation τ_{−β}, the homothety h_{\frac{1}{R}} and a suitable rotation ρ_{θ}, we obtain P_{n}^{r} = ρ_{θ}h_{1/R}τ_{−β}P′_{n,r} which is studied in Proposition 4.1.c. Note that the same process applies to arbitrary regular polyhedrons in ℝ^{n}, n ≥ 3.

References

[1] N. Aronszajn: Sur les décompositions des fonctions analytiques uniformes et sur leurs applications, Acta. math. 65 (1935) 1-156.

[2] N. Aronszajn, M. C. Thomas, J. L.Leonard: Polyharmonic functions, Clarendon. Press. Oxford(1983).

[3] V. Avanissian: Sur les fonctions harmoniques d’ordre quelconque et leur prolongement analytique dans ℂ^{n}. Séminaire P. Lelong-H.Skoda, Lecture Notes in Math, n^{0}919, Springer-Verlag, Berlin (1981) 192-281.

[4] V. Avanissian: Cellule d’harmonicité et prolongement analytique complexe, Travaux en cours, Hermann, Paris (1985).

[5] M. Boutaleb: Sur la cellule d’harmonicité de la boule unité de ℝ^{n}- Doctorat de 3^{0}cycle, U.L.P. Strasbourg, France (1983).

[6] E. Brieskorn: Beispiele zur Differentialtopologie von Singularitäten, Inv.Math.2 (1966) 1-14.

[7] R.Coquereaux, A.Jadczyk: Conformal Theories, Curved phase spaces, Relativistic wavelets and the Geometry of complex domains, Centre de physique théorique, Section 2, Case 907. Luminy, 13288. Marseille, France (1990).

[8] M. Hervé: Les fonctions analytiques, Presses Universitaires de France (1982), Paris.

[9] L. K. Hua: Harmonic Analysis of Functions of Several Complex Variables in the Classical Domains. Transl of Math. Monographs 6, Amer. Math. Soc. Provid. R. I.(1963).

[10] M. Jarnicki: Analytic Continuation of harmonic functions, Zesz. Nauk. U J, Pr. Mat 17, (1975) 93-104.

[11] P. Lelong: Sur la définition des fonctions harmoniques d’ordre infini, C. R. Acad. Sci. Paris 223 (1946) 372-374.

[12] P. Lelong: Prolongement analytique et singularités complexes des fonctions harmoniques, Bull. Soc. Math. Belg. 7 (1954-55) 10-23.

[13] P. Lelong: Sur les singularités complexes d’une fonction harmonique, C. R. Acad. Sci. Paris 232 (1951) 1895-1897.

[14] M. Nicolesco: Les fonctions polyharmoniques, Hermann Paris(1936).

[15] W. Rudin: Functional Analysis, Mc Graw Hill (1973).

[16] J. Siciak, Holomorphic continuation of harmonic functions. Ann. Pol. Math. XXIX (1974) 67-73.1.

Mohamed Boutaleb Département de Mathématiques et Informatique Faculté des Sciences Dhar-Mahraz B. P. 1796 Atlas, Fès, Maroc e-mail: mboutalebmoh@yahoo.fr