E extracta mathematicae Vol. 33, Núm. 1, 67 – 108 (2018) Partial Differential Equations and Strictly Plurisubharmonic Functions in Several Variables Jamel Abidi Department of Mathematics, Faculty of Sciences of Tunis 1060 Tunis, Tunisia abidijamel1@gmail.com Presented by Manuel Maestre Received September 22, 2014 Abstract: Using algebraic methods, we prove that there exists a fundamental relation be- tween partial differential equations and strictly plurisubharmonic functions over domains of Cn (n ≥ 1). Key words: Analytic convex and plurisubharmonic functions, harmonic function, maximal plurisubharmonic, differential equation, analysis, inequalities. AMS Subject Class. (2010): Primary 32A10, 32A60, 32A70, 32U05; Secondary 32W50. 1. Introduction We investigate in this paper the relation between strictly plurisubharmonic functions and partial differential equations in domains of Cn, (n ≥ 1). Various related results are obtained in this context. Several papers developed by Lelong [14, 15, 16, 17, 18], Sadullaev [20], Oka [19], Bremermann [4, 5],Siciak [21], Abidi [1], Cegrell [6] and others studied plurisubharmonic functions and related topics are of particular importance in this context. For example we can state that strictly plurisubharmonic functions and analytic subsets are related in domains of Cn as follows. Let A = {z ∈ C : f(z) = 0} and B = {z ∈ C : g(z) = 0} two analytic subsets of C, where f,g : C → C be 2 analytic functions, fg ̸= 0. Put f1 and g1 some analytic primitives of f and g respectively over C. Then A ∩ B = ∅ if and only if the function u, u(z,w) = |w − f1(z)|2 + |w − g1(z)|2 , (z,w) ∈ C2 , is strictly psh in C2. Some good references for the study of convex functions are [11, 13, 3]. For the study of analytic functions we cite the references [12, 10, 13]. For the study of the extension problem of analytic and plurisubharmonic functions we cite the references [7, 9, 6, 18, 8, 22, 23]. 67 68 j. abidi As usual, N := {1,2, . . .}, R and C are the sets of all natural, real and complex numbers, respectively. Let U be a domain of Rd, (d ≥ 2); md is the Lebesgue measure on Rd. Let f : U → C be a function; |f| is the modulus of f, Re(f) and Im(f) are respectively the real and imaginary parts of f. Let g : D → C be an analytic function, D is a domain of C. We denote by g(0) = g, g(1) = g′ is the holomorphic derivative of g over D. g(2) = g′′, g(3) = g′′′. In general g(m) = ∂mg ∂zm is the derivative of g of order m for all m ∈ N. Let z ∈ Cn, z = (z1, . . . ,zn), n ≥ 2. For j ∈ {1, . . . ,n}, we write z = (zj,Zj) = (z1, . . . ,zj−1,zj,zj+1, . . . ,zn) where Zj = (z1, . . . ,zj−1,zj+1, . . . ,zn) ∈ Cn−1. Ck(U) = {φ : U → C : φ is of class Ck in U}, k ∈ N ∪ {∞}. Let φ : U → C be a function of class C2. ∆(φ) is the Laplacian of φ. Let D be a domain of Cn, (n ≥ 1); psh(D) and prh(D) are respectively the class of plurisubharmonic and pluriharmonic functions on D. For all a ∈ C, |a| is the modulus of a; Re(a) and Im(a) are the real and the imaginary parts of a respectively. 2. Main results We begin this study by the next result. Theorem 2.1. Let h1, . . . ,hN : C → R be N harmonic functions, N ≥ 1. For the function u(z,w) = |w − h1(z)|2 + · · · + |w − hN(z)|2, (z,w) ∈ C2, the following conditions are equivalent: (a) u is strictly psh in C2; (b) { z ∈ C : ∂h1 ∂z (z) = 0 } ∩ · · · ∩ { z ∈ C : ∂hN ∂z (z) = 0 } = ∅. Proof. (a) ⇒ (b) Because |w − h1|2 + · · · + |w − hN|2 = N|w|2 + (h21 + · · · + h2N) − w(h1 + · · · + hN) − w(h1 + · · · + hN), u is a function of class C ∞ in C2. Now let (z,w) ∈ C2, ∂2u ∂z∂z (z,w) = 2 [∣∣∣∣∂h1∂z (z) ∣∣∣∣2 + · · · + ∣∣∣∣∂hN∂z (z) ∣∣∣∣2 ] , ∂2u ∂w∂w (z,w) = N , ∂2u ∂z∂w (z,w) = − ( ∂h1 ∂z (z) + · · · + ∂hN ∂z (z) ) . strictly plurisubharmonic functions 69 The Levi hermitian form associated to u is now L(u)(z,w)(α,β) = ∂2u ∂z∂z (z,w)αα + ∂2u ∂w∂w (z,w)ββ + 2 Re[ ∂2u ∂z∂w (z,w)αβ] = 2 [∣∣∣∣∂h1∂z (z) ∣∣∣∣2 + · · · + ∣∣∣∣∂hN∂z (z) ∣∣∣∣2 ] αα + Nββ + 2 Re [ − ( ∂h1 ∂z (z) + · · · + ∂hN ∂z (z) ) αβ ] > 0 , for all (z,w) ∈ C2 and (α,β) ∈ C2\{(0,0)}. Thus∣∣∣∣∂h1∂z (z) + · · · + ∂hN∂z (z) ∣∣∣∣2 < 2N [∣∣∣∣∂h1∂z (z) ∣∣∣∣2 + · · · + ∣∣∣∣∂hN∂z (z) ∣∣∣∣2 ] for each z ∈ C. Now we use the following lemma. Lemma 2.2. Let a1, . . . ,aN ∈ C and N ≥ 1. We have (i) N(|a1|2 + · · · + |aN|2) ≥ |a1 + · · · + aN|2 ; (ii) M(|a1|2 + · · · + |aN|2) > |a1 + · · · + aN|2 if M > N and there exists j0 such that aj0 ̸= 0. Proof.( N∑ j=1 aj )( N∑ k=1 ak ) = N∑ j,k=1 ajak ≤ N∑ j,k=1 |aj||ak| ≤ N∑ j,k=1 ( |aj|2 2 + |aj|2 2 ) = 2 N∑ j,k=1 |aj|2 2 = N∑ k=1 N∑ j=1 |aj|2 = N N∑ j=1 |aj|2. Now we complete the proof of the theorem. Put A = [∣∣∣∣∂h1∂z ∣∣∣∣2 + · · · + ∣∣∣∣∂hN∂z ∣∣∣∣2 ] + [ (2N − 1) (∣∣∣∣∂h1∂z ∣∣∣∣2 + · · · + ∣∣∣∣∂hN∂z ∣∣∣∣2 ) − ∣∣∣∣∂h1∂z + · · · + ∂hN∂z ∣∣∣∣2 ] ; 70 j. abidi A > 0 over C. A = B + C, where B ≥ 0, C ≥ 0. Then A = 0 if and only if B = C = 0. Thus if z ∈ C such that B(z) = ∣∣∣∂h1∂z (z)∣∣∣2 + · · · + ∣∣∣∂hN∂z (z)∣∣∣2 = 0, then ∂h1 ∂z (z) = · · · = ∂hN ∂z (z) = 0. Therefore C(z) = (2N −1) (∣∣∣∂h1∂z (z)∣∣∣2 +· · ·+∣∣∣∂hN∂z (z)∣∣∣2) − ∣∣∣∂h1∂z (z) + · · · + ∂hN∂z (z)∣∣∣2 = 0. The converse is also true. We conclude that A(z) > 0 if and only if B(z) > 0, for all z ∈ C. Then ∣∣∣∂h1∂z ∣∣∣2 + · · · + ∣∣∣∂hN∂z ∣∣∣2 > 0 over C if and only if u is strictly psh in C2. But ∣∣∣∂h1∂z ∣∣∣2 + · · · + ∣∣∣∂hN∂z ∣∣∣2 > 0 in C if and only if{ z ∈ C : ∂h1 ∂z (z) = 0 } ∩ · · · ∩ { z ∈ C : ∂hN ∂z (z) = 0 } = ∅ . By this proof we deduce also (b) ⇒ (a). For analytic functions, we have now. Theorem 2.3. Let g1, . . . ,gN : D → C be N analytic functions, N ∈ N\{1}, D is a domain of C. Put u(z,w) = |w − g1(z)|2 + · · · + |w − gN(z)|2, (z,w) ∈ D × C. Then u is strictly psh in D × C if and only if N∑ j,k=1 g′jg ′ kδjk > 0 over D, where δjk = (N − 1 if j = k and − 1 if j ̸= k), j,k ∈ {1, . . . ,N}. Proof. The function u = N|w|2 + ( |g1|2 + · · · + |gN|2 ) −w(g1 + · · · +gN) − w(g1 + · · · + gN) is of class C∞ over D × C. Let (z,w) ∈ C2, ∂2u ∂z∂z (z,w) = |g′1(z)| 2 + · · · + |g′N(z)| 2 , ∂2u ∂w∂w (z,w) = N , ∂2u ∂z∂w (z,w) = − ( g′1(z) + · · · + g ′ N(z) ) . Assume that u is strictly psh in D × C. The Levi hermitian form of u is now L(u)(z,w)(α,β) = [ |g′1(z)| 2 + · · · + |g′N(z)| 2 ] αα + Nββ + 2 Re [ − (g′1(z) + strictly plurisubharmonic functions 71 · · · + g′N(z))αβ ] > 0 for all (α,β) ∈ C2\{(0,0)}. Thus |g′1 + · · · + g ′ N| 2 < N[|g′1| 2 + · · · + |g′N| 2] over D. Then N∑ j,k=1(j ̸=k) g′jg ′ kδjk + (N − 1) [ |g′1| 2 + · · · + |g′N| 2 ] = N∑ j,k=1 g′jg ′ kδjk > 0 on D. Now assume that N∑ j,k=1 g′jg ′ kδjk > 0 on D. By the above proof, |g ′ 1 + · · · + g′N| 2 < N [ |g′1| 2 + · · · + |g′N| 2 ] . It follows that L(u)(z,w)(α,β) > 0 for each (α,β) ∈ C2\{0}. The theorem below gives a fundamental part of this paper and the study of the relation between partial differential equations and strictly plurisubhar- monic functions over domains of Cn, (n ≥ 1). Theorem 2.4. Let g : D → C be a function, D is a domain of C. Put v(z,w) = |w − g(z)|2, for (z,w) ∈ D × C. The following assertions are equiv- alent: (a) v is strictly psh in D × C ; (b) g is harmonic in D and { z ∈ D : ∂g ∂z (z) = 0 } = ∅. Proof. (a) ⇒ (b) v is strictly psh in D × C, then v is psh in D × C. Therefore g is harmonic in D by Abidi [1]. It follows that v is a function of class C∞ in D×C. Let (z,w) ∈ D×C. Write v(z,w) = |w|2+|g(z)|2−wg(z)−wg(z). We have ∂2v ∂z∂z (z,w)αα + ∂2v ∂w∂w (z,w)ββ + 2 Re [ ∂2v ∂z∂w (z,w)αβ ] > 0 for all (α,β) ∈ C2\{0}, and ∂2v ∂z∂z = ∣∣∣∣∂g∂z ∣∣∣∣2 + ∣∣∣∣∂g∂z ∣∣∣∣2 , ∂2v∂w∂w = 1 , ∂ 2v ∂z∂w = − ∂g ∂z . Therefore ∣∣ − ∂g ∂z ∣∣2 < ∣∣∂g ∂z ∣∣2 + |∂g ∂z ∣∣2, and consequently ∣∣∂g ∂z ∣∣2 > 0 over D. (b) ⇒ (a) Since g is harmonic in D, then v is a function of class C∞ in D × C. We have ∂2v ∂z∂z = ∣∣∣∣∂g∂z ∣∣∣∣2 + ∣∣∣∣∂g∂z ∣∣∣∣2 , ∂2v∂w∂w = 1 , ∂ 2v ∂z∂w = − ∂g ∂z . 72 j. abidi The Levi hermitian form of v is L(v)(z,w)(α,β) = ∂2v ∂z∂z (z,w)αα + ∂2v ∂w∂w (z,w)ββ + 2 Re [ ∂2v ∂z∂w (z,w)αβ ] = [∣∣∣∣∂g∂z(z) ∣∣∣∣2 + ∣∣∣∣∂g∂z(z) ∣∣∣∣2 ] αα + ββ + 2 Re [ − ∂g ∂z (z)αβ ] , for (z,w) ∈ D × C, (α,β) ∈ C2. Since ∣∣∂g ∂z (z) ∣∣2 > 0, for every z ∈ D, then∣∣∣∣ − ∂g∂z(z) ∣∣∣∣2 < ∣∣∣∣∂g∂z(z) ∣∣∣∣2 + ∣∣∣∣∂g∂z(z) ∣∣∣∣2 for each z ∈ D. Therefore, L(v)(z,w)(α,β) > 0 , ∀ (z,w) ∈ D × C , ∀ (α,β) ∈ C2\{0}. Consequently, v is strictly psh in D × C. Observe that if k = k1 + k2, where k1,k2 : D → C be 2 analytic functions in the domain D ⊂ C, and u(z,w) = |w − k(z)|2, (z,w) ∈ D × C, then the strict plurisubharmonicity of u is independent of the function k1. On the other hand if we replace the strict inequality < by the large inequality ≤, then the above theorem is false. Remark 2.5. Let k : D → C be an analytic function, D is a domain of C. Put u(z,w) = |w − k(z)|2, v(z,w) = |w − k(z)|2, where (z,w) ∈ D × C. Then u, log(u) and log(v) are not strictly psh functions on any not empty domain of D × C; v is strictly psh in D × C if and only if ∣∣∣∂k∂z∣∣∣ = ∣∣∣∂k∂z∣∣∣ > 0 in D. Example. Let k(z) = exp(z), z ∈ C, and v1(z,w) = |w − exp(z)|2, v2(z,w) = |w − exp(z)|2, for (z,w) ∈ C2; v1 is not strictly psh on any open of C2, but v2 is strictly psh in all C2. Note that log(v2) is not strictly psh on any domain of { (z,w) : |w − exp(z)|2 > 0 } . On the other hand, g1(z) = z and g2(z) = 1 − z (z ∈ C) are analytic functions over C. Set v(z,w) = |w − g1(z)|2 + |w − g2(z)|2, (z,w) ∈ C2. Let (α,β) ∈ C2. The Levi hermitian form of v is L(v)(z,w)(α,β) = αα + ββ + 2 Re [ − αβ ] + αα + ββ + 2 Re [ αβ ] = 2(αα + ββ) > 0 , ∀ (z,w) ∈ C × C , ∀ (α,β) ∈ C2\{0}. strictly plurisubharmonic functions 73 Then v is strictly psh in C2. Observe that in this case if we put u1(z,w) = |w − g1(z)|2, u2(z,w) = |w − g2(z)|2, then u1 and u2 are plurisubharmonic over C2 but not strictly psh functions on any domain of C2. But v = (u1 +u2) is strictly psh in C2. In fact we have the following result. Claim 2.6. Let g1,g2 : D → C be 2 analytic functions, D is a domain of C and v(z,w) = |w − g1(z)|2 + |w − g2(z)|2, where (z,w) ∈ D × C. Then v is strictly psh in D × C if the function Re [ g′1g ′ 2 ] < 0 over D. If D = C, then v is strictly psh in C2 if for example (g′1g ′ 2) is equal a constant c over C and Re(c) < 0. According to the paper Abidi [1], we can prove the following extension. Claim 2.7. Let a,b ∈ C. Put v(z,w) = |(w − z)2 − (a + b)(w − z) + ab|, where (z,w) ∈ C2. Then v is strictly psh on C2 if and only if a = b. In general we can state the following result: For all g : C → C be analytic, if we put u(z,w) = ∣∣(w − g(z))2 − (a + b)(w − g(z)) + ab∣∣ , where (z,w) ∈ C2, then u is strictly psh on C2 if and only if (a = b and∣∣∂g ∂z (z) ∣∣ > 0 for all z ∈ C). Theorem 2.8. Let D be a domain of C and g : D → C be an analytic function. The following statements are equivalent: (a1) |w − g|2 is strictly psh in D × C; (a2) |w − g|2 + |w − g|2 is strictly psh in D × C; (a3) |∂g∂z | > 0 in D; (a4) |w − cg − g|2 is strictly psh in D × C, where c ∈ C\{0}; (a5) |w1 − g|2 + |w2 − g|2 is strictly psh in D × C × C; (a6) for all n ∈ N, ( |w1 − g|2 + · · · + |wn − g|2 + |wn+1 − g|2 ) is strictly psh in D × Cn+1. Proposition 2.9. Let g : D → C be analytic, D is a domain of C. g = h + ik, h = Re(g), k = Im(g). Let a,b ∈ C, (a ̸= 0 or b ̸= 0). Put u(z,w) = |w − g(z)|2, v(z,w) = |w − ah(z)|2 + |w − bk(z)|2, u1(z,w) = |w − h(z)|2, u2(z,w) = |w − k(z)|2, where (z,w) ∈ D × C. We have the equivalents: 74 j. abidi (a) u is strictly psh in D × C; (b) u1 is strictly psh in D × C; (c) u2 is strictly psh in D × C; (d) v is strictly psh in D × C. Observe that in general we can not compare the structure strictly psh of the functions v1 and v2 where v1(z,w) = |w − g(z)|2, v2(z,w) = |w − g(z)|2, g : C → C be analytic and (z,w) ∈ C2. But if we add another function constructed according to the expression of g we have the following extension. Claim 2.10. Let g : Cn → C be analytic g = h + ik, h = Re(g), n ∈ N. Denote by φ(z,w) = |w−g(z)|2, φ1(z,w) = |w−h(z)|2+|w−g(z)|2, φ2(z,w) = |w − h(z)|2 + |w − g(z)|2, φ3(z,w) = |w − g(z)|2, where (z,w) ∈ Cn × C. We have the equivalents: (a) φ1 is strictly psh in Cn × C; (b) φ2 is strictly psh in Cn × C; (c) n = 1 and φ3 is strictly psh in C2. Note that φ is not strictly psh on all not empty domain of C2. At this stage of the development, observe that if f : Cn → R is plurihar- monic (n ≥ 1), and F(z,w) = |w − f(z)|2, where (z,w) ∈ Cn × C, then F is not strictly psh on any not empty domain of Cn × C if and only if (a1) n = 1 and f is constant in C, or (a2) n ≥ 2 and f is an arbitrary prh function over Cn. The function f have real valued is of great importance in this subject. Some fundamental remarks concerning strictly psh functions. At the beginning of this statements we observe the following assertions: Let h : D → C be a function, D is a convex domain of C. If |w − h|2 is psh (resp. convex) in D × C, then ∣∣w − h∣∣2 is psh (resp. convex) in D × C and conversely. But we can obtain |w − h|2 is strictly psh (resp. strictly psh and convex) in D × C and ∣∣w − h∣∣2 is not strictly psh (resp. not strictly psh and convex) on any domain subset of D × C. This is one of the great differences between the classes of functions psh, convex, of the first part and the classes of strictly psh, (strictly psh and convex) functions for the second part. Consequently, if strictly plurisubharmonic functions 75 we replace the large inequality ≤ by the strict inequality < the above result is not true. Now let g1, . . . ,gN : Cn → C be N analytic functions, where n,N ≥ 1. Put u1(z,w) = |w − g1(z)|2 + · · · + |w − gN(z)|2, v1(z,w) = ∣∣w − g1(z)∣∣2 + · · · + ∣∣w − gN(z)∣∣2, (z,w) ∈ Cn × C. If u1 is strictly psh in Cn × C, then{ ∂ ∂z1 (g1, . . . ,gN), . . . , ∂ ∂zn (g1, . . . ,gN) } is linearly independent over CN and n < N (by using the hermitian Levi form of the function u1). If u1 is strictly psh in Cn × C, then v1 is strictly psh in Cn × C. But not conversely. Example. The functions k1(z) = z, k2(z) = z 2 (z ∈ C) are analytic over C. Let v1(z,w) = ∣∣w −k1(z)∣∣2 + ∣∣w −k2(z)∣∣2, where (z,w) ∈ C2; v1 is strictly psh on C2. Put u1(z,w) = |w−z|2+|w−z2|2, where (z,w) ∈ C2. Let α,β ∈ C. The Levi hermitian form of u1 is L(u1)(z,w)(α,β) = |β − α|2 + |β − 2zα|2. If z = 1 2 , then we have L(u1) ( 1 2 ,w ) (α,α) = 0 for each α ∈ C\{0}. Therefore u1 is not strictly psh in C2. Put u2(z,w) = |w1 − g1(z)|2 + · · · + |wN − gN(z)|2, z ∈ Cn, w = (w1, . . . ,wN) ∈ CN; u2 is not strictly psh in any not empty domain of Cn × CN. Now put v2(z,w) = ∣∣w1 − g1(z)∣∣2 + · · · + ∣∣wN − gN(z)∣∣2. If for all fixed z in Cn, the system   ∂g1 ∂z1 (z)α1 + · · · + ∂g1∂zn (z)αn = 0 ... ∂gN ∂z1 (z)α1 + · · · + ∂gN∂zn (z)αn = 0 (α1, . . . ,αn ∈ C) has only the solution (α1, . . . ,αn) = (0, . . . ,0), then v2 is strictly psh in Cn × CN. Therefore u2 and v2 do not have the same structure in the theory of the strictly plurisubharmonic functions. Put u3(z,w) = ∣∣w − φ1(z)∣∣2 + ∣∣w − φ2(z)∣∣2, where (z,w) ∈ C2, φ1,φ2 : C → C are analytic functions, u4(z,w) = ∣∣w − φ1(z)∣∣2 + ∣∣w − φ2(z)∣∣2. Then u3 is strictly psh in C2 if and only if u4 is strictly psh in C2. Question 2.11. An original problem of the theory of functions in several complex variables is now the following. Let f0, . . . ,fk−1 : Cn → C be k 76 j. abidi analytic functions, (n,k ≥ 1). Set u(z,w) = ∣∣wk + fk−1(z)wk−1 + · · · + f1(z)w + f0(z)∣∣ , v(z,w) = ∣∣wk + fk−1(z)wk−1 + · · · + f1(z)w + f0(z)∣∣ , where (z,w) ∈ Cn × C. u is convex in Cn × C if and only if v is convex in Cn × C. Now note that u is psh in Cn × C, but v is not in general (example take v1(z,w) = ∣∣w2 + zw∣∣ is not psh in C2). Find the condition described by the functions f0, . . . ,fk−1 such that v is psh in Cn × C. (Observe that we can consider in this study the question of a power series). Remark 2.12. The above proposition is not true if g : D → C is harmonic. For example, if g : C → R, g(z) = x1, z = (x1 + ix2) ∈ C, where x1,x2 ∈ R, then ∣∣w − g∣∣2 is strictly psh in C2. But Im(g) = 0 and |w − 0|2 = |w|2 is not strictly psh on any domain of C2. Theorem 2.13. Let g1, . . . ,gN : C → C, u(z,w) = |w1 − g1(z)|2 + · · · + |wN − gN(z)|2, where (z,w) = (z,w1, . . . ,wN) ∈ C × CN,N ∈ N. u is strictly psh in C × CN if and only if g1, . . . ,gN are harmonic functions in C and∣∣∂g1 ∂z ∣∣2 + · · · + ∣∣∂gN ∂z ∣∣2 > 0 on C. Proof. Assume that u is strictly psh on C×CN. Note that u is a function of class C∞ on C×CN. Let (z,w) = (z,w1, . . . ,wN) ∈ C×CN. Fix w2, . . . ,wN ∈ C. Then the function u(., .,w2, . . . ,wN) is strictly psh on C2. By Abidi [1], g1 is harmonic on C. Consequently, g1, . . . ,gN are harmonic functions on C. Put gj = fj + kj, where fj,kj : C → C be two analytic functions and j ∈ {1, . . . ,N}. Let (α,β) = (α,β1, . . . ,βN) ∈ C × CN\{(0,0)}. The Levi hermitian form of u is now L(u)(z,w)(α,β) = ∣∣∣∣β1 − ∂f1∂z (z)α ∣∣∣∣2 + ∣∣∣∣∂k1∂z (z)α ∣∣∣∣2 + · · · + ∣∣∣∣βN − ∂fN∂z (z)α ∣∣∣∣2 + ∣∣∣∣∂kN∂z (z)α ∣∣∣∣2. Assume that α ̸= 0. Put β1 = ∂f1∂z (z)α,. . . ,βN = ∂fN ∂z (z)α. Then L(u)(z,w)(α,β) = (∣∣∣∣∂k1∂z (z)α ∣∣∣∣2 + · · · + ∣∣∣∣∂kN∂z (z)α ∣∣∣∣2 ) strictly plurisubharmonic functions 77 for each z ∈ C. Thus ∣∣∣∣∂g1∂z (z) ∣∣∣∣2 + · · · + ∣∣∣∣∂gN∂z (z) ∣∣∣∣2 > 0 . The converse is trivial. Observe that the notion u is strictly psh in C × CN on the above theorem is independent of f1, . . . ,fN, where gj = fj + kj, fj,kj : C → C are analytic functions (1 ≤ j ≤ N). Proposition 2.14. For every g : D → C analytic, D is a domain of Cn, (n ≥ 2), u = |g|2 is not strictly psh on any domain D1 ⊂ D. Indeed e|g| 2 , |g|2e|g| 2 , |g|2e|g| 2 ee |g|2 are not strictly psh functions in any domain D2 ⊂ D. For example let v = |g1|2 + · · · + |gn|2, where g1, . . . ,gn : Cn → C are analytic functions. Then v is strictly psh in Cn if and only if the determinant det ( ∂gj ∂zk (z) ) j,k ̸= 0, for all z ∈ Cn. Note that we have the assertion. Let g1, . . . ,gN : D → C be N analytic functions, D is a domain of Cn, n ≥ 2, N ≥ 1. If N < n, then u = |g1|2 + · · · + |gN|2 is not strictly psh on any domain D1 ⊂ D. In fact u is a function of class C∞ in D. The Levi hermitian form of u is L(u)(z)(α) = n∑ j,k=1 ∂2u ∂zj∂zk (z)αjαk = ∣∣∣∣∣ n∑ j=1 ∂g1 ∂zj (z)αj ∣∣∣∣∣ 2 + · · · + ∣∣∣∣∣ n∑ j=1 ∂gN ∂zj (z)αj ∣∣∣∣∣ 2 for each z = (z1, . . . ,zn) ∈ D and α = (α1, . . . ,αn) ∈ Cn. Suppose that u is strictly psh in D. Then for all z ∈ D, for all α1, . . . ,αn ∈ C, L(u)(z)(α1, . . . ,αn) = 0 if and only if  ∂g1 ∂z1 (z)α1 + · · · + ∂g1∂zn (z)αn = 0 ... ∂gN ∂z1 (z)α1 + · · · + ∂gN∂zn (z)αn = 0. Then α1 ( ∂g1 ∂z1 (z), . . . , ∂gN ∂z1 (z) ) + · · · + αn ( ∂g1 ∂zn (z), . . . , ∂gN ∂zn (z) ) = 0 78 j. abidi implies that α1 = · · · = αn = 0. Therefore the subset of vectors{( ∂g1 ∂z1 (z), . . . , ∂gN ∂z1 (z) ) , . . . , ( ∂g1 ∂zn (z), . . . , ∂gN ∂zn (z) )} is a free family of n vectors of CN, and N < n. This is a contradiction. Consequently, u is not strictly psh on any domain D1 ⊂ D. But we have the following result: For all n ∈ N, there exists u1, . . . ,un : C2n → C be n pluriharmonic functions such that v = ( |u1|2 + · · · + |un|2 ) is strictly psh in C2n. Example. Put uj(z) = zj + zn+j, 1 ≤ j ≤ n, where z = (z1, . . . ,z2n) ∈ C2n. uj is in fact prh in C2n; |u1(z)|2 = |z1+zn+1|2 = |z1|2+|zn+1|2+z1zn+1+ z1zn+1. Note that the function K1(z) = z1zn+1 +z1zn+1, K1 is pluriharmonic in C2n and therefore the Levi hermitian form of K1 is equal 0 over C2n × C2n. Then L ( |u1|2)(z)(α1, . . . ,α2n) = |α1|2 + |αn+1|2. Then L(v)(z)(α1, . . . ,α2n) = 2n∑ j=1 |αj|2 > 0 if (α1, . . . ,α2n) ∈ C2n\{0}. Then v is strictly psh in C2n, but n < 2n. In fact for all n ≥ 1, there exists a function u : Cn → R pluriharmonic such that |u|2 is not strictly psh in Cn, u is not constant. Observe that we have if h : C3 → C is pluriharmonic, then |h|2 is not strictly psh in C3. Exactly we have for all h1, . . . ,hs : Cn → C prh, if s < n2 , then (|h1| 2 + · · · + |hs|2) is not strictly psh in Cn. Now if one of the function have real valued, one of the above result is not true. For example, if u : C2 → R is a pluriharmonic function, then u2 is not strictly psh on C2. Theorem 2.15. Let u1, . . . ,un : C2n → R be n pluriharmonic functions, n ∈ N. Set u = u21 + · · · + u 2 n. Then u is not strictly psh on any domain of C2n. Proof. The functions u21, . . . ,u 2 n and u are of class C ∞ in C2n. Denote by L(u)(z)(α1, . . . ,α2n) = 2n∑ j,k=1 ∂2u ∂zj∂zk (z)αjαk for all z = (z1, . . . ,z2n) ∈ C2n and for all α = (α1, . . . ,α2n) ∈ C2n. We have L(u)(z)(α1, . . . ,α2n) = L(u 2 1)(z)(α1, . . . ,α2n) + · · · + L(u 2 n)(z)(α1, . . . ,α2n) strictly plurisubharmonic functions 79 and L(u21)(z)(α1, . . . ,α2n) = 2n∑ j,k=1 ∂2(u21) ∂zj∂zk (z)αjαk = 2 2n∑ j,k=1 ∂u1 ∂zj (z) ∂u1 ∂zk (z)αjαk = 2 ( 2n∑ j=1 ∂u1 ∂zj (z)αj )( 2n∑ k=1 ∂u1 ∂zk (z)αk ) = 2 ∣∣∣∣∣ 2n∑ j=1 ∂u1 ∂zj (z)αj ∣∣∣∣∣ 2 . Consequently, L(u)(z)(α1, . . . ,α2n) = 2 ∣∣∣∣∣ 2n∑ j=1 ∂u1 ∂zj (z)αj ∣∣∣∣∣ 2 + · · · + 2 ∣∣∣∣∣ 2n∑ j=1 ∂un ∂zj (z)αj ∣∣∣∣∣ 2 . Fix z = (z1, . . . ,z2n) ∈ C2n. L(u)(z)(α1, . . . ,α2n) = 0 if and only if 2n∑ j=1 ∂u1 ∂zj (z)αj = 0 , . . . , 2n∑ j=1 ∂un ∂zj (z)αj = 0. Then   ∂u1 ∂z1 (z)α1 + · · · + ∂u1∂z2n (z)α2n = 0 ... ∂un ∂z1 (z)α1 + · · · + ∂un∂z2n (z)α2n = 0. Thus α1 ( ∂u1 ∂z1 (z), . . . , ∂un ∂z1 (z) ) + · · · + α2n ( ∂u1 ∂z2n (z), . . . , ∂un ∂z2n (z) ) = (0, . . . ,0) ∈ Cn, where α1, . . . ,α2n ∈ C. We have 2n vectors of Cn (considered a vector space). Therefore the subset of the above 2n vectors is not a linearly in- dependent family in the C-vector space Cn of dimension n. Then there exists (α1, . . . ,α2n) ∈ C2n\{0} such that L(u)(z)(α1, . . . ,α2n) = 0. Consequently, u is not strictly psh on any not empty domain of C2n. Definition 2.16. (Klimek [12]) Let u : D → R be a psh function, where D is an open of Cn, n ≥ 1. u is maximal psh on D if for all relatively compact open G subset of D and for each upper semi continuous function v on G such that v is psh on G and v ≤ u on ∂G, we have v ≤ u on G. 80 j. abidi Remark 2.17. (a) Let n ∈ N, n ≥ 2. Given u1, . . . ,un−1 : D → R be n − 1 pluriharmonic functions, where D is a domain of Cn. Then u =( u21 + · · · + u 2 n−1 ) is not strictly psh on any domain D1 ⊂ D. (b) Let n ∈ N and D a domain of Cn. Consider h1, . . . ,hn : D → R be n pluriharmonic functions and put u = h21 + · · · + h 2 n. u is psh on D. Then u is strictly psh on D if and only if det ( ∂hj ∂zk (z) ) 1≤j, k≤n ̸= 0 for all z ∈ D. (c) Let g : D → C be analytic, D is a domain of Cn, (n ≥ 2). u = |g|2 is maximal plurisubharmonic (in the sense of Klimek [12] or Sadullaev [20]). But if k : C → C is analytic not constant, then |k|2 = v is not maximal subharmonic because v is not harmonic. This is one of the great differences between the theory of functions of one complex variable and the same theory in several complex variables. In one complex variable, the sum of 2 maximal subharmonic functions is maximal subharmonic. If now g1,g2 : C2 → C be 2 analytic functions such that |g1|2 + |g2|2 = φ is strictly psh in C2, then |g1|2 + |g2|2 is psh but not maximal plurisubharmonic on any open of C2. In this case |g1|2 and |g2|2 are maximal plurisubharmonic functions on C2. But the sum |g1|2 + |g2|2 = φ is not maximal psh on any not empty open of C2. But we have the following result. Let g1, . . . ,gN : D → C be N analytic functions (N ≥ 1). Then if N < n, u = |g1|2 + · · · + |gN|2 is maximal plurisubharmonic on D. Proposition 2.18. There exists a function u : C2 → R, u real analytic on C2, u is maximal plurisubharmonic on C2, but eu is plurisubharmonic on C2 and not maximal plurisubharmonic on any not empty domain of C2. Moreover, for all v : D → R prh, (D is a domain of Cn, n ≥ 2) the function ev is maximal plurisubharmonic on D. Proof. Let u(z1,z2) = x 2 1 + x2, where z1 = (x1 + ix3), z2 = (x2 + ix4) ∈ C (x1,x2,x3,x4 ∈ R). u is plurisubharmonic in C2 and real analytic. We have the determinant det ( ∂2u ∂zj∂zk (z) ) j,k = 0 for each z ∈ C2. By Klimek [12], u is maximal plurisubharmonic in C2. Now det ( ∂2(eu) ∂zj∂zk (z) ) j,k = 1 8 ̸= 0 for every z ∈ C2. By Klimek [12, Proposition 3.1.6], eu is not maximal psh on any domain of C2. strictly plurisubharmonic functions 81 Lemma 2.19. Let u : D → R be plurisubharmonic, where D is a domain of Cn, n ≥ 1. If eu is maximal psh on D, then u is maximal psh on D. Proof. Let G be a relatively compact open subset of D and v : G → [−∞,+∞[ be an upper semi continuous function such that v is psh on G and v ≤ u on ∂G. Then ev ≤ eu on ∂G and consequently, ev ≤ eu on G. It follows that v ≤ u on G. Remark 2.20. For all n ≥ 1, for all domain D of Cn, there exists u : D → R be C∞ psh such that eu is strictly psh on D but u is not strictly psh on any domain D1 ⊂ D. In general we have the following lemma. Lemma 2.21. Let A,B two hermitian matrix of type (n,n) with coeffi- cients in C. Suppose that A and B are positive semi-definite. (a) If A is positive definite then A + B is positive definite on Cn. (b) If the determinant det(A) ̸= 0 then A is positive definite on Cn. (c) If A+B is positive definite, we can not conclude that A or B is positive definite on Cn if n ≥ 2. Example. Let D be a domain of C2. Let F = {(g1,g2) / g1,g2 : D → C be analytic functions such that (|g1|2 + |g2|2) is strictly psh in C2}. Let (g1,g2) ∈ F. Fix z = (z1,z2) ∈ D. Put A = ( ∂2|g1|2 ∂zj∂zk (z) ) j, k , B = ( ∂2|g2|2 ∂zj∂zk (z) ) j, k . A and B are hermitian matrix positive semi definite on C2. Then A+B is an hermitian matrix positive definite, but A and B are not positive definite over C2. Now we can prove the following result. Theorem 2.22. Let u : D → R be a function of class C2, D is a domain of Cn, n ≥ 1. Suppose that u is psh on D. Then eu is maximal psh on D if and only if ee u is maximal psh on D. Therefore if eu is maximal psh on D, then Fs(u) is maximal psh on D, for each s ∈ N, where Fs = exp ◦ exp ◦ · · · ◦ exp (s times). 82 j. abidi This theorem have good and several applications in problems and exercises. Proof. If eu is maximal psh on D. Since eu is psh and of class C2 in D, then det ( ∂2(eu) ∂zj∂zk (z) ) j,k = 0 for all z ∈ D. Fix z ∈ D. Thus the matrix A = ( ∂2(u) ∂zj∂zk (z) + ∂u ∂zj (z) ∂u ∂zk (z) ) j,k is not an injection. Hence there exists α = (α1, . . . ,αn) ∈ Cn\{0} such that Aα = 0. If < .,. > is the hermitian habitual product on Cn, then < α,Aα >= 0, n∑ j,k=1 ∂2(u) ∂zj∂zk (z)αjαk + n∑ j,k=1 ∂u ∂zj (z) ∂u ∂zk (z)αjαk = 0 . As a consequence n∑ j,k=1 ∂2(u) ∂zj∂zk (z)αjαk + ∣∣∣∣∣ n∑ j=1 ∂u ∂zj (z)αj ∣∣∣∣∣ 2 = 0 . Since u is psh and of class C2 in D, thus n∑ j,k=1 ∂2(u) ∂zj∂zk (z)αjαk ≥ 0. Now since ∣∣∣∣∣ n∑ j=1 ∂u ∂zj (z)αj ∣∣∣∣∣ 2 ≥ 0, it follows that n∑ j,k=1 ∂2(u) ∂zj∂zk (z)αjαk = 0 and∣∣∣∣∣ n∑ j=1 ∂u ∂zj (z)αj ∣∣∣∣∣ 2 = n∑ j,k=1 ∂u ∂zj (z) ∂u ∂zk (z)αjαk = 0, and thus ( 1 + eu(z) ) n∑ j,k=1 ∂u ∂zj (z) ∂u ∂zk (z)αjαk = 0 . Consequently, n∑ j,k=1 ∂2(u) ∂zj∂zk (z)αjαk + ( 1 + eu(z) ) n∑ j,k=1 ∂u ∂zj (z) ∂u ∂zk (z)αjαk = n∑ j,k=1 ( ∂2(u) ∂zj∂zk (z) + ( 1 + eu(z) ) ∂u ∂zj (z) ∂u ∂zk (z) ) αjαk = 0 . strictly plurisubharmonic functions 83 Now the matrix B = ( ∂2(u) ∂zj∂zk (z) + (1 + eu(z)) ∂u ∂zj (z) ∂u ∂zk (z) ) j, k is an hermitian matrix positive semi definite because ee u is psh on D. If det(B) ̸= 0, then B is positive definite on Cn. But there exists α ∈ Cn\{0} such that < α,Bα >= 0. Then B is not definite positive in Cn. Consequently, det(B) = 0 and we have ee u is maximal psh on D. The converse is trivial. Example. Let h : D → R be prh, where D is a domain of Cn, n ≥ 2. We denote by Fs = exp ◦ exp ◦ · · · ◦ exp (s times), for s ∈ N (and F0 is the identity operator). Then Fs(h) is maximal plurisubharmonic in D. Now let s,t ∈ N. Thus the function Fs(h) − Ft(h) is maximal plurisubharmonic in D in the case s ≥ t. (We prove that det ( ∂2(Fs(h)−Ft(h)) ∂zj∂zk ) 1≤j,k≤n = 0. By Klimek [12, Corollary 3.1.8] we conclude the required property). The following two theorems have several applications in the theory of func- tions. Theorem 2.23. Let f : D → R be a function, D is a domain of Cn, n ≥ 1. Put u(z,w) = |w − f(z)|2, where (z,w) ∈ D × C. The following two conditions are equivalent: (a) u is strictly psh in D × C; (b) n = 1, f is harmonic in D and ∂f ∂z (z) ̸= 0 for each z ∈ D. Proof. (a) ⇒ (b) Since u is strictly psh in D × C, then u is psh in D × C and consequently, f is pluriharmonic in D. Therefore u is a function of class C∞ in D. Suppose that n ≥ 2. Let z0 = (z01, . . . ,z 0 n) ∈ D. Consider now R > 0 such that P(z0,R) = D(z01,R) × D(z 0 2,R) × · · · × D(z 0 n,R) ⊂ D. We consider the function f(., .,z03, . . . ,z 0 n) defined and prh in D(z 0 1,R) × D(z 0 2,R) = A. Let f1 = f(., .,z 0 3, . . . ,z 0 n) and u1(z1,z2,w) = u(z1,z2,z 0 3, . . . ,z 0 n,w) = ∣∣w − f(z1,z2,z03, . . . ,z0n)∣∣2 = |w − f1(z1,z2)|2 , 84 j. abidi where (z1,z2,w) ∈ D(z01,R) × D(z 0 2,R) × C. Note that f1 is prh in A and u1 is strictly psh in A × C, u1(z1,z2,w) = |w|2 + |f1(z1,z2)|2 − wf1(z1,z2) − wf1(z1,z2) . Fix w0 = 0 ∈ C. The Levi hermitian form of u1(., .,0) is L(u1)(z1,z2,0)(α1,α2) = ∂2u1 ∂z1∂z1 (z1,z2,0)α1α1 + ∂2u1 ∂z2∂z2 (z1,z2,0)α2α2 + 2 Re [ ∂2u1 ∂z1∂z2 (z1,z2,0)α1α2 ] > 0 , for all (z1,z2) ∈ A and for all (α1,α2) ∈ C2\{(0,0)}. Moreover 2 ∣∣∣∣∂f1∂z1 ∣∣∣∣2α1α1 + 2 ∣∣∣∣∂f1∂z2 ∣∣∣∣2α2α2 + 2 Re [ 2 ∂f1 ∂z1 ∂f1 ∂z2 α1α2 ] > 0 , on A for all (α1,α2) ∈ C2\{0}. Then ∣∣∣2∂f1∂z1 ∂f1∂z2 ∣∣∣2 < 4∣∣∣∂f1∂z1 ∣∣∣2∣∣∣∂f1∂z2 ∣∣∣2 over A. But we have ∣∣∣∂f1∂z1 ∂f1∂z2 ∣∣∣ = ∣∣∣∂f1∂z1 ∣∣∣∣∣∣∂f1∂z2 ∣∣∣ = ∣∣∣∂f1∂z1 ∣∣∣∣∣∣∂f1∂z2 ∣∣∣ < ∣∣∣∂f1∂z1 ∣∣∣∣∣∣∂f1∂z2 ∣∣∣ in A (because f1 has real valued). A contradiction. Then n = 1. Now the Levi hermitian form of u is L(u)(z,w)(α,β) = ∂2u ∂z∂z (z,w)αα + ∂2u ∂w∂w (z,w)ββ + 2 Re [ ∂2u ∂z∂w (z,w)αβ ] = 2 ∣∣∣∣∂f∂z (z) ∣∣∣∣2αα + ββ + 2 Re [ − ∂f ∂z (z)αβ ] > 0 , for all (z,w) ∈ D × C and for all (α,β) ∈ C2\{0}. Then∣∣∣∣∂f∂z (z) ∣∣∣∣2 < 2 ∣∣∣∣∂f∂z (z) ∣∣∣∣2 for each z ∈ D. Thus ∂f ∂z (z) ̸= 0 for every z ∈ D. Consequently, n = 1, f is harmonic in D and { z ∈ D : ∂f ∂z (z) = 0 } = ∅. (b) ⇒ (a) The Levi hermitian form of u is L(u)(z,w)(α,β) = 2 ∣∣∣∣∂f∂z (z) ∣∣∣∣2αα + ββ + 2 Re [ − ∂f ∂z (z)αβ ] > 0 strictly plurisubharmonic functions 85 for each (z,w) ∈ D × C and (α,β) ∈ C2\{0}. We have L(u)(z,w)(α,β) > 0 ∀ (z,w) ∈ D × C , ∀ (α,β) ∈ C2\{0} if and only if ∣∣∣∣∂f∂z (z) ∣∣∣∣2 < 2 ∣∣∣∣∂f∂z (z) ∣∣∣∣2. But this is equivalent to ∂f ∂z (z) ̸= 0 for all z ∈ D. Now the case where the function is complex valued, we prove the following extension. Theorem 2.24. Let g : D → C be a function, D is a domain of Cn, n ≥ 1. Put v(z,w) = |w−g(z)|2, where (z,w) ∈ D×C. The following two conditions are equivalent: (a) v is strictly psh in D × C; (b) n = 1, g is harmonic in D and { z ∈ D : ∂g ∂z (z) = 0 } = ∅. Proof. (a) ⇒ (b) Since v is strictly plurisubharmonic in D × C, then v is plurisubharmonic in D × C. Consequently, g is pluriharmonic in D. Let z0 = (z01, . . . ,z 0 n) ∈ D, R > 0 such that D(z01,R) × · · · × D(z 0 n,R) = A ⊂ D. Put g = g1 +g2 in the convex domain A, where g1,g2 : A → C be two analytic functions. Now we use the following fundamental decomposition v(z,w) = |w − g1(z) − g2(z)|2 = |w − g1(z)|2 + |g2(z)|2 − (w − g1(z))g2(z) − (w − g1(z))g2(z) for each (z,w) ∈ A × C. Suppose that n ≥ 2. Case 1: n = 2. We have v1(z,w) = |w − g1(z)|2, v2(z,w) = |g2(z)|2, v3(z,w) = −(w − g1(z))g2(z) − (w − g1(z))g2(z), where (z,w) ∈ A × C; v1,v2 and v3 are C ∞ functions in the domain A × C, and v3 is pluriharmonic in A × C. Then the Levi hermitian form of v3 is L(v3)(z1,z2,w)(α1,α2,β) = 0 86 j. abidi for all (z1,z2,w) ∈ A × C and for all (α1,α2,β) ∈ C3. The Levi hermitian form of v2 is L(v2)(z1,z2,w)(α1,α2,β) = ∣∣∣∣∂g2∂z1 (z) ∣∣∣∣2α1α1 + ∣∣∣∣∂g2∂z2 (z) ∣∣∣∣2α2α2 + 2 Re [ ∂g2 ∂z1 (z) ∂g2 ∂z2 (z)α1α2 ] = ∣∣∣∣∂g2∂z1 (z)α1 + ∂g2∂z2 (z)α2 ∣∣∣∣2 for each (z,w) = (z1,z2,w) ∈ A× C and (α1,α2,β) ∈ C3. The Levi hermitian form of v1 is L(v1)(z1,z2,w)(α1,α2,β) = ∣∣∣∣∂g1∂z1 (z) ∣∣∣∣2α1α1 + ∣∣∣∣∂g1∂z2 (z) ∣∣∣∣2α2α2 + ββ + 2 Re [ ∂g1 ∂z1 (z) ∂g1 ∂z2 (z)α1α2 ] + 2 Re [ − ∂g1 ∂z1 (z)α1β ] + 2 Re [ − ∂g1 ∂z2 (z)α2β ] = ∣∣∣∣∂g1∂z1 (z)α1 + ∂g1∂z2 (z)α2 ∣∣∣∣2 + |β|2 + 2 Re [ − ( ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ) β ] = ∣∣∣∣β − [ ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ]∣∣∣∣2, where (z,w) = (z1,z2,w) ∈ A × C. Now we have L(v)(z,w)(α1,α2,β) = ∣∣∣∣β− [ ∂g1 ∂z1 (z)α1+ ∂g1 ∂z2 (z)α2 ]∣∣∣∣2+ ∣∣∣∣∂g2∂z1 (z)α1+∂g2∂z2 (z)α2 ∣∣∣∣2 where (z,w) = (z1,z2,w) ∈ A × C and (α1,α2,β) ∈ C3. Let z ∈ A. Choose (α1,α2) ∈ C2 \{(0,0)} such that ∂g2∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 = 0. Now let β = ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2. We have (α1,α2,β) ∈ C3\{(0,0,0)} and L(v)(z,w)(α1,α2,β) = 0. This proves for example that v is not strictly psh on any open of D × C. A contradiction. strictly plurisubharmonic functions 87 Case 2: n ≥ 3. We deduce by in fact the formula L(v)(z,w)(α1, . . . ,αn,β) = ∣∣∣∣∣β − n∑ j=1 ∂g1 ∂zj (z)αj ∣∣∣∣∣ 2 + ∣∣∣∣∣ n∑ j=1 ∂g2 ∂zj (z)αj ∣∣∣∣∣ 2 , where (z,w) = (z1, . . . ,zn,w) ∈ A × C and (α1, . . . ,αn,β) ∈ Cn+1. Let z ∈ A. Now it is possible to choose (α1, . . . ,αn) ∈ Cn\{0} such that n∑ j=1 ∂g2 ∂zj (z)αj = 0 (because n ≥ 3) . Let β = n∑ j=1 ∂g1 ∂zj (z)αj. We have (α1, . . . ,αn,β) ∈ Cn+1\{0} and L(v)(z,w)(α1, . . . ,αn,β) = 0 . Therefore in fact v is not strictly psh on any domain of A×C. A contradiction. Consequently, n = 1. By the above theorem, g is harmonic in D and∣∣∂g ∂z ∣∣ > 0 in D. (b) ⇒ (a) By the above theorem we deduce this assertion in fact. Theorem 2.25. Let g1, . . . ,gN : C2 → C, v(z,w) = |w1 − g1(z)|2 + · · · + |wN − gN(z)|2, where z ∈ C2, w = (w1, . . . ,wN) ∈ CN and N ∈ N. The following conditions are equivalent: (a1) v is strictly psh in C2 × CN; (a2) gj is pluriharmonic in C2, gj = fj + kj, where fj,kj : C2 → C are analytic functions, for all 1 ≤ j ≤ N (N ≥ 2). The functions k1, . . . ,kN satisfies an algebraic condition, that is for each z ∈ C2, the set {s1, . . . ,sN} is a generating family of the C-vector space C2, s1 = ( ∂k1 ∂z1 (z), ∂k1 ∂z2 (z) ) , . . . ,sN = ( ∂kN ∂z1 (z), ∂kN ∂z2 (z) ) ; (a3) gj is pluriharmonic in C2, gj = fj + kj, where fj,kj : C2 → C are analytic functions for all j ∈ {1, . . . ,N}, N ≥ 2 and for all z ∈ C2, there exist R > 0 and there exists s,t ∈ {1, . . . ,N} (s ̸= t) such that v1 is strictly psh in B(z,R)×C2, where v1(z,t) = |ws −gs(z)|2 +|wt −gt(z)|2, for (z,w) ∈ B(z,R) × C2 and w = (ws,wt); 88 j. abidi (a4) gj is pluriharmonic on C2, gj = fj + kj, where fj,kj : C2 → C are analytic functions for all 1 ≤ j ≤ N. k1, . . . ,kN satisfies {( ∂k1 ∂z1 (z), . . . , ∂kN ∂z1 (z) ) , ( ∂k1 ∂z2 (z), . . . , ∂kN ∂z2 (z) )} is a free family in the C-vector space CN, for all fixed z ∈ C2. Proof. (a1) ⇒ (a2) Firstly we prove that g1, . . . ,gN are continuous func- tions over C2. Let z0 ∈ C2. Put ζ1 = g1(z0), . . . ,ζN = gN(z0) ∈ C; v(z0,ζ1, . . . ,ζN) = 0. Let ϵ > 0. Since v is upper semi-continuous in the point (z0,ζ1, . . . ,ζN) then there exists δ > 0 such that ∥z − z0∥ + |w1 − ζ1| + · · · + |wN − ζN| < δ implies that |w1 − g1(z)|2 + · · · + |wN − gN(z)|2 ≤ ϵ2. Let j ∈ {1, . . . ,N}. If we put w1 = ζ1, . . . ,wj−1 = ζj−1,wj+1 = ζj+1, . . . ,wN = ζN, then we have ∥z − z0∥ + |wj − ζj| < δ implies that |wj − gj(z)|2 < ϵ2. Let wj = ζj = gj(z 0). Then ∥z−z0∥ < δ implies that |gj(z)−gj(z0)| < ϵ. Then gj is continuous in the point z0 ∈ C2. Consequently, g1, . . . ,gN are continuous functions on C2. We have v is strictly psh in C2 × CN, therefore v is psh in C2 × CN. Therefore the function of two variables v(., .,0, . . . ,0) is psh in C2 × C, where v(z,w1,0, . . . ,0) = v1(z,w1) = |w1 − g1(z)|2 + |g2(z)|2 + · · · + |gN(z)|2. Let φ : C2 → R+, φ is of class C∞ and have a compact support in C2. Let ∆ = 4 ( ∂2 ∂z1∂z1 + ∂2 ∂z2∂z2 ) the Laplace operator on C2. Then we have∫ |w1 − g1(z)|2∆φ(z) dm4(z) + N∑ j=2 ∫ |gj(z)|2∆φ(z) dm4(z) ≥ 0 for each w1 ∈ C. Let w1 ∈ R. Then we have −w1 ∫ [g1(z) + g1(z)]∆φ(z) dm4(z) + N∑ j=1 ∫ |gj(z)|2∆φ(z) dm4(z) ≥ 0 for all w1 ∈ R. If ∫ [g1(z) + g1(z)]∆φ(z) dm4(z) > 0, then we obtain a contra- diction by letting w1 to +∞. If ∫ [g1(z) + g1(z)]∆φ(z) dm4(z) < 0, then we have a contradiction by letting w1 go to (−∞). Consequently,∫ [g1(z) + g1(z)]∆φ(z) dm4(z) = 0. strictly plurisubharmonic functions 89 Since g1 + g1 is a continuous function in C2, then g1 + g1 is harmonic in C2. Let w1 ∈ iR. Then w1 = −w1. In this case we prove that (g1 − g1) is harmonic in C2. Now since g1 = 12 [ (g1 + g1) + (g1 − g1) ] , then g1 is harmonic in C2. Let T1 : C2 → C2 be a C-linear bijective transformation. Consider now T(z,w1) = (T1(z),w1), where z ∈ C2 and w1 ∈ C. Note that T : C3 → C3 is a C-linear bijective transformation. v1 is psh in C2 × C, then v1 ◦ T is psh in C2 × C, v1 ◦ T(z,w1) = |w1 − g1 ◦ T1(z)|2 + |g2 ◦ T1(z)|2 + · · · + |gN ◦ T1(z)|2 , (z,w1) ∈ C2 × C. By the above development we have g1 ◦ T1 is harmonic in C2. Consequently, g1 is a pluriharmonic function on C2. Therefore g1, . . . ,gN are pluriharmonic functions on C2; gj = fj + kj,fj,kj : C2 → C are analytic functions , 1 ≤ j ≤ N. Consider now a1(z,w) = |w1 −g1(z)|2, where (z,w1) ∈ C2 ×C; a1(z,w1) = |w1−f1(z)−k1(z)|2. We consider now the following decomposition a1(z,w1) = |w1 −f1(z)|2 +|k1(z)|2 −k1(z)(w1 −f1(z))−k1(z)(w1 − f1(z)). a1 is a function of class C∞ in C2 × C. Let H1(z,w1) = k1(z)(w1 − f1(z)) + k1(z)(w1 − f1(z)), where (z,w1) ∈ C2 × C; H1 is pluriharmonic in C2 × C. Therefore the Levi hermitian form of H1 is L(H1)(z,w1)(α1,α2,β1) = 0, for all (z,w1) ∈ C2 × C and for all (α1,α2,β1) ∈ C3. Then the Levi hermitian form of a1 is L(a1)(z,w1)(α1,α2,β1) = L(b1)(z,w1)(α1,α2,β1) + L(c1)(z,w1)(α1,α2,β1) , where b1(z,w1) = |w1 − f1(z)|2, c1(z,w1) = |k1(z)|2 and (z,w1) ∈ C2 × C. b1 and c1 are in particular functions of class C ∞ in C2 × C. We have L(b1)(z,w1)(α1,α2,β1) = ∂2b1 ∂z1∂z1 (z,w1)α1α1 + ∂2b1 ∂z2∂z2 (z,w1)α2α2 + 2 Re [ ∂2b1 ∂z1∂z2 (z,w1)α1α2 ] + ∂2b1 ∂w1∂w1 (z,w1)β1β1 + 2 Re [ ∂2b1 ∂z1∂w1 (z,w1)α1β1 + ∂2b1 ∂z2∂w1 (z,w1)α2β1 ] = ∣∣∣∣∂f1∂z1 (z) ∣∣∣∣2α1α1 + ∣∣∣∣∂f1∂z2 (z) ∣∣∣∣2α2α2 + 2 Re [ ∂f1 ∂z1 (z) ∂f1 ∂z2 (z)α1α2 ] + β1β1 + 2 Re [ − ∂f1 ∂z1 (z)α1β1 − ∂f1 ∂z2 (z)α2β1 ] 90 j. abidi = ∣∣∣∣∂f1∂z1 (z)α1 + ∂f1∂z2 (z)α2 ∣∣∣∣2 + ∣∣β1∣∣2 − 2 Re [( ∂f1 ∂z1 (z)α1 + ∂f1 ∂z2 (z)α2 ) β1 ] = ∣∣∣∣β1 − [ ∂f1 ∂z1 (z)α1 + ∂f1 ∂z2 (z)α2 ]∣∣∣∣2. Since c1(z,w1) = |k1(z)|2, then L(c1)(z,w1)(α1,α2,β1) = ∂2c1 ∂z1∂z1 (z,w1)α1α1 + ∂2c1 ∂z2∂z2 (z,w1)α2α2 + 2 Re [ ∂2c1 ∂z1∂z2 (z,w1)α1α2 ] = ∣∣∣∣∂k1∂z1 (z) ∣∣∣∣2α1α1 + ∣∣∣∣∂k1∂z2 (z) ∣∣∣∣2α2α2 + 2 Re [ ∂k1 ∂z1 (z) ∂k1 ∂z2 (z)α1α2 ] = ∣∣∣∣∂k1∂z1 (z)α1 + ∂k1∂z2 (z)α2 ∣∣∣∣2. Consequently, L(a1)(z,w1)(α1,α2,β1) = ∣∣∣β1 − [∂f1∂z1 (z)α1 + ∂f1∂z2 (z)α2]∣∣∣2 +∣∣∣∂k1∂z1 (z)α1 + ∂k1∂z2 (z)α2∣∣∣2 for each (z,w1) ∈ C2 × C and (α1,α2,β1) ∈ C3. Since v is a function of class C∞ in C2 × CN, then we have for each (z,w1, . . . ,wN) ∈ C2 × CN, z = (z1,z2) ∈ C2 and all (α1,α2,β1, . . . ,βN) ∈ CN+2, the Levi hermitian form of v is L(v)(z,w1, . . . ,wN)(α1,α2,β1, . . . ,βN) = ∣∣∣∣β1 − [ ∂f1 ∂z1 (z)α1 + ∂f1 ∂z2 (z)α2 ]∣∣∣∣2 + ∣∣∣∣∂k1∂z1 (z)α1 + ∂k1∂z2 (z)α2 ∣∣∣∣2 + · · · + ∣∣∣∣βN − [ ∂fN ∂z1 (z)α1 + ∂fN ∂z2 (z)α2 ]∣∣∣∣2 + ∣∣∣∣∂kN∂z1 (z)α1 + ∂kN∂z2 (z)α2 ∣∣∣∣2. Fix z ∈ C2. If L(v)(z,w1, . . . ,wN)(α1,α2,β1, . . . ,βN) = 0, then  ∂k1 ∂z1 (z)α1 + ∂k1 ∂z2 (z)α2 = 0 ... ∂kN ∂z1 (z)α1 + ∂kN ∂z2 (z)α2 = 0. Therefore if α1,α2 ∈ C such that α1 ( ∂k1 ∂z1 (z), . . . , ∂kN ∂z1 (z) ) + α2 ( ∂k1 ∂z2 (z), . . . , ∂kN ∂z2 (z) ) = (0, . . . ,0) ∈ CN, strictly plurisubharmonic functions 91 then α1 = α2 = 0. Thus N ≥ 2 and there exists s,t ∈ {1, . . . ,N} (s ̸= t) such that {( ∂ks ∂z1 (z), ∂ks ∂z2 (z) ) , ( ∂kt ∂z1 (z), ∂kt ∂z2 (z) )} is a basis of the C-vector space C2. Then {( ∂k1 ∂z1 (z), ∂k1 ∂z2 (z) ) , . . . , ( ∂kN ∂z1 (z), ∂kN ∂z2 (z) )} is a generating family of the C-vector space C2. Observe that locally (s,t) is independent of z ∈ C2, but not globally if N ≥ 3. (a2) ⇒ (a1) Let z ∈ C2. Since {( ∂k1 ∂z1 (z), ∂k1 ∂z2 (z) ) , . . . , ( ∂kN ∂z1 (z), ∂kN ∂z2 (z) )} is a generating family of the C-vector space C2, then N ≥ 2 and we can exhibit a family of 2 vectors which is a basis of C2. Without loss of generality we suppose that {( ∂k1 ∂z1 (z), ∂k1 ∂z2 (z) ) , ( ∂k2 ∂z1 (z), ∂k2 ∂z2 (z) )} is a basis of C2. Therefore the matrix (λµν)1≤µ, ν≤2 have a determinant det(λµν)1≤µ, ν≤2 = φ(z) ̸= 0, where λµν = ∂kµ ∂zν (z). Since the function φ is analytic in C2, then |φ| > 0 on a neighborhood B(z,r) of the point z (r > 0). Then for all ξ ∈ B(z,r) and (α1,α2) ∈ C2, we have   ∂k1 ∂z1 (ξ)α1 + ∂k1 ∂z2 (ξ)α2 = 0 ∂k2 ∂z1 (ξ)α1 + ∂k2 ∂z2 (ξ)α2 = 0 if and only if α1 = α2 = 0. Thus if (α1,α2,β1,β2) ∈ C4, ξ ∈ B(z,r),∣∣∣β1 − [∂f1∂z1 (ξ)α1 + ∂f1∂z2 (ξ)α2]∣∣∣2 + ∣∣∣∂k1∂z1 (ξ)α1 + ∂k1∂z2 (ξ)α2∣∣∣2 + ∣∣∣β2 − [∂f2∂z1 (ξ)α1 + ∂f2 ∂z2 (ξ)α2 ]∣∣∣2 + ∣∣∣∂k2∂z1 (ξ)α1 + ∂k2∂z2 (ξ)α2∣∣∣2 = 0, then  ∂k1 ∂z1 (ξ)α1 + ∂k1 ∂z2 (ξ)α2 = 0 ∂k2 ∂z1 (ξ)α1 + ∂k2 ∂z2 (ξ)α2 = 0. It follows that α1 = α2 = 0. Thus β1 = β2 = 0. Consequently, φ1(ξ,w1,w2) = |w1 − g1(ξ)|2 + |w2 − g2(ξ)|2 is strictly psh in B(z,r) × C × C. In fact we can prove that φ1 is strictly psh in (C2\A) × C2, where A is an analytic subset of C2. Now the above proof implies that the assertions (a1), (a3) and (a4) are equivalent. Corollary 2.26. Let g1,g2 : C2 → C be two analytic functions. Put u(z,w1,w2) = |w1 − g1(z)|2 + |w2 − g2(z)|2, where (z,w1,w2) ∈ C2 × C × C. Let A ⊂ C2, A closed and bounded in C2. Suppose that u is strictly psh in C2 × (C2\A). Then u is strictly psh in C2 × C2. 92 j. abidi Proof. Note that u is a function of class C∞ on C2 × C2. Assume that u is not strictly psh at the point (z0,w0) ∈ C2 × C2. Then there exists( (α1,α2),(β1,β2) ) ∈ C2 × C2\{(0,0)} such that the Levi hermitian form of u verify L(u)(z0,w0) ( (α1,α2),(β1,β2) ) = ∣∣∣∣∣β1 − 2∑ j=1 ∂g1 ∂zj (z0)αj ∣∣∣∣∣ 2 + ∣∣∣∣∣β2 − 2∑ j=1 ∂g2 ∂zj (z0)αj ∣∣∣∣∣ 2 = 0 . Let b0 ∈ C2\A. Since u is strictly psh on C2 × (C2\A), then u is strictly psh at the point (z0,b0). But we have L(u)(z0,b0) ( (α1,α2),(β1,β2) ) = ∣∣∣∣∣β1 − 2∑ j=1 ∂g1 ∂zj (z0)αj ∣∣∣∣∣ 2 + ∣∣∣∣∣β2 − 2∑ j=1 ∂g2 ∂zj (z0)αj ∣∣∣∣∣ 2 = 0 . and ( (α1,α2),(β1,β2) ) ∈ C2 × C2\{(0,0)}. A contradiction. Consequently, u is strictly psh on C2 × C2. Corollary 2.27. Let g1,g2 : C2 → C be two analytic functions. Set u(z,w) = |w1 −g1(z)|2 + |w2 −g2(z)|2, v(z,w) = |w1 −g1(z)|2 + |w2 −g2(z)|2, φ(z,ζ) = |ζ − g1(z)|2 + |ζ − g2(z)|2, where z ∈ C2, w = (w1,w2) ∈ C2 and ζ ∈ C. Then u and v are not strictly plurisubharmonic functions in C2 × C2. We have, φ is strictly psh in C2×C if and only if |g1|2+|g2|2 (or |g1+g2|2) is strictly psh on C2. Proof. We have the fundamental decomposition (complex structure) u(z,w) = |w1 − g1(z)|2 + |w2|2 + |g2(z)|2 − w2g2(z) − w2g2(z), for any (z,w) = (z,w1,w2) ∈ C2 × C × C where z = (z1,z2) ∈ C2. Put u1(z,w) = w2g2(z)+w2g2(z); u1 is a pluriharmonic function in C2×C2. Therefore the Levi hermitian form of this function is equal to 0 over C4. Let u2(z,w) = |w1 − g1(z)|2; u2 is a function of class C∞ in C2 × C, u2(z,w) = |w1|2 +|g1(z)|2 −w1g1(z)−w1g1(z). Then the Levi hermitian form of u2 is now L(u2)(z,w)(α1,α2,β1,β2) = ∣∣∣∣β1 − [ ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ]∣∣∣∣2 strictly plurisubharmonic functions 93 for each z = (z1,z2) ∈ C2, w = (w1,w2) ∈ C2 and (α1,α2,β1,β2) ∈ C4. Let u3(z,w) = |w2|2 + |g2(z)|2; u3 is a function of class C∞ in C4. The Levi hermitian form of u3 is L(u3)(z,w)(α1,α2,β1,β2) = ∣∣β2∣∣2 + ∣∣∣∣∂g2∂z1 (z)α1 + ∂g2∂z2 (z)α2 ∣∣∣∣2. The function u is of class C∞ in C2 × C2. We have L(u)(z,w)(α1,α2,β1,β2) = −L(u1)(z,w)(α1,α2,β1,β2) + L(u2)(z,w)(α1,α2,β1,β2) + L(u3)(z,w)(α1,α2,β1,β2) = ∣∣∣∣β1 − [ ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ]∣∣∣∣2 + ∣∣β2∣∣2 + ∣∣∣∣∂g2∂z1 (z)α1 + ∂g2∂z2 (z)α2 ∣∣∣∣2. Case 1: |g1|2 + |g2|2 (or equivalently |g1 + g2|2) is not strictly psh on C2. Note that |g1|2 and |g2|2 are functions of class C∞ in C2. The Levi hermitian form (in C2) of |g1|2 is L ( |g1|2 ) (z)(δ1,δ2) = 2∑ j,k=1 ∂2 ( |g1|2 ) ∂zj∂zk δjδk = 2∑ j,k=1 ∂g1 ∂zj (z) ∂g1 ∂zk (z)δjδk = ( 2∑ j=1 ∂g1 ∂zj (z)δj )( 2∑ k=1 ∂g1 ∂zk (z)δk ) = ∣∣∣∣∣ 2∑ j=1 ∂g1 ∂zj (z)δj ∣∣∣∣∣ 2 , where z = (z1,z2) ∈ C2 and (δ1,δ2) ∈ C2. Therefore L ( |g1|2+|g2|2 ) (z1,z2)(α1,α2) = ∣∣∣∣∂g1∂z1 (z)α1 + ∂g1∂z2 (z)α2 ∣∣∣∣2 + ∣∣∣∣∂g2∂z1 (z)α1 + ∂g2∂z2 (z)α2 ∣∣∣∣2 for each (α1,α2) ∈ C2. Now fix z = (z1,z2) ∈ C2. Since |g1|2 + |g2|2 is not strictly psh in C2, then there exists (α1,α2) ∈ C2\{0} such that  ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 = 0 ∂g2 ∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 = 0. 94 j. abidi Fix w = (w1,w2) ∈ C2. Take now β1 = β2 = 0 ∈ C. Then we have L(u)(z,w)(α1,α2,β1,β2) = 0 but (α1,α2,β1,β2) ∈ C4\{0}. Consequently, u is not strictly psh in C2 × C2. Case 2: |g1|2 + |g2|2 (or equivalently |g1 + g2|2) is strictly psh in C2. L(u)(z,w)(α1,α2,β1,β2) = 0 if and only if β1 = ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2,β2 = 0 and ∂g2 ∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 = 0. Fix (α1,α2) ∈ C2\{0} such that ∂g2∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 = 0. Define β1 ∈ C by β1 = ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 (β2 = 0). Then (α1,α2,β1,β2) ∈ C4\{0} and we have L(u)(z,w)(α1,α2,β1,β2) = 0. Consequently, u is not strictly psh on any domain D ⊂ C4. Concerning the function v, we have v is defined on C4 and of class C∞. The Levi hermitian form of v is L(v)(z,w)(α1,α2,β1,β2) = ∣∣∣∣β1 − [ ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ]∣∣∣∣2 + ∣∣∣∣β2 − [ ∂g2 ∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 ]∣∣∣∣2, where z = (z1,z2) ∈ C2, w = (w1,w2) ∈ C2 and (α1,α2,β1,β2) ∈ C4. Fix (z,w) ∈ C2 × C2. Let (α1,α2) ∈ C2\{0} such that ∂g1∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 = 0. Put β1 = 0, β2 = [ ∂g2 ∂z1 (z)α1+ ∂g2 ∂z2 (z)α2 ] . Then (α1,α2,β1,β2) ∈ C4\{0} and L(v)(z,w)(α1,α2,β1,β2) = 0. Consequently, v is not strictly psh on any open of C4. Now we have the decomposition φ(z,ζ) = |ζ − g1(z)|2 + |ζ − g2(z)|2 = |ζ − g1(z)|2 + |ζ|2 + |g2(z)|2 − ζg2(z) − ζg2(z) , for every ζ ∈ C and z = (z1,z2) ∈ C2; φ is a function of class C∞ in C3. Put φ1(z,ζ) = ζg2(z) + ζg2(z). Then φ1 is pluriharmonic in C3 and consequently, the Levi hermitian form of this function is 0. Let φ2(z,ζ) = |ζ − g1(z)|2; φ2 is a function of class C∞ in C3 and L(φ2)(z,ζ)(α1,α2,β) = ∣∣∣∣β − [ ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ]∣∣∣∣2 for each (z,ζ) ∈ C2 × C and (α1,α2,β) ∈ C3. Let φ3(z,ζ) = |ζ|2 + |g2(z)|2; φ3 is a function of class C∞ in C3 and L(φ3)(z,ζ)(α1,α2,β) = ∣∣β∣∣2 + ∣∣∣∣∂g2∂z1 (z)α1 + ∂g2∂z2 (z)α2 ∣∣∣∣2 strictly plurisubharmonic functions 95 for every (z,ζ) ∈ C2 × C and (α1,α2,β) ∈ C3. It follows that L(φ)(z,ζ)(α1,α2,β) = L(φ2)(z,ζ)(α1,α2,β) + L(φ3)(z,ζ)(α1,α2,β) = ∣∣∣∣β − [ ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ]∣∣∣∣2 + ∣∣β∣∣2 + ∣∣∣∣∂g2∂z1 (z)α1 + ∂g2∂z2 (z)α2 ∣∣∣∣2. Therefore L(φ)(z,ζ)(α1,α2,β) = 0 if and only if β = 0, ∂g2 ∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 = 0 and ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 = 0. Observe now that φ is strictly psh in C3 if and only if |g1|2 + |g2|2 is strictly psh in C2. Corollary 2.28. Let g1,g2 : C2 → C be two pluriharmonic functions. Put g1 = f1 + k1, where g2 = f2 + k2, f1,f2,k1,k2 : C2 → C be four analytic functions. Let u(z,w1,w2) = |w1 − g1(z)|2 + |w2 − g2(z)|2 , v(z,w1,w2) = ∣∣w1 − k1(z)∣∣2 + ∣∣w2 − k2(z)∣∣2, where (z,w1,w2) ∈ C2 × C × C. The following conditions are equivalent (a) u is strictly psh in C4; (b) v is strictly psh in C4. That is the strict plurisubharmonicity of u is independent of the choice of the analytic functions f1 and f2. Corollary 2.29. Let gj,kj : D → C be analytic functions, where 1 ≤ j ≤ N and D is a domain of Cn, n,N ≥ 1. Put u = N∑ j=1 ∣∣gj + kj∣∣2 and v = N∑ j=1 ∣∣gj∣∣2 + N∑ j=1 ∣∣kj∣∣2. Then u is strictly psh in D if and only if v is strictly psh in D. Corollary 2.30. Let g1, . . . ,gN : D → C be N analytic functions, where N ≥ 1 and D is a domain of Cn (n ≥ 1). Set u(z,w) = N∑ j=1 |w−gj(z)|2, where (z,w) ∈ D × C. If N ≤ n, then u is not strictly psh on any domain of D × C. 96 j. abidi Proof. Fix z = (z1, . . . ,zn) ∈ D and w ∈ C. Let uj(z,w) = |w − gj(z)|2, 1 ≤ j ≤ N. Then uj is a function of class C∞ in D×C. If now (α1, . . . ,αn) ∈ Cn and β ∈ C, we have the Levi hermitian form of uj is L(uj)(z,w)(α1, . . . ,αn,β) = ∣∣∣∣∣β − n∑ s=1 ∂gj ∂zs (z)αs ∣∣∣∣∣ 2 . Therefore, the Levi form of u is L(u)(z,w)(α1, . . . ,αn,β) = N∑ j=1 L(uj)(z,w)(α1, . . . ,αn,β) = N∑ j=1 ∣∣∣∣∣β − n∑ s=1 ∂gj ∂zs (z)αs ∣∣∣∣∣ 2 . Let v = [|g1|2 + · · · + |gN|2]; v is a function of class C∞ on D. Case 1: v is strictly psh on D. We have L(v)(z)(α1, . . . ,αn) = 0 imply that (α1, . . . ,αn) = (0, . . . ,0). The Levi form of v is L(v)(z)(α1, . . . ,αn) = ∣∣∣∣∣ n∑ s=1 ∂g1 ∂zs (z)αs ∣∣∣∣∣ 2 + · · · + ∣∣∣∣∣ n∑ s=1 ∂gN ∂zs (z)αs ∣∣∣∣∣ 2 . Since L(v)(z)(α1, . . . ,αn) = 0 then (α1, . . . ,αn) = 0. Thus the system of equations in (α1, . . . ,αn) ∈ Cn satisfies  ∂g1 ∂z1 (z)α1 + · · · + ∂g1∂zn (z)αn = 0 ... ∂gN ∂z1 (z)α1 + · · · + ∂gN∂zn (z)αn = 0 if and only if (α1, . . . ,αn) = (0, . . . ,0). Since N ≤ n, then N = n. Thus the matrix (λjk)1≤j,k≤n is invertible; where λjk = ∂gj ∂zk (z). Now we have L(u)(z,w)(α1, . . . ,αn,β) = 0 if and only if  ∂g1 ∂z1 (z)α1 + · · · + ∂g1∂zn (z)αn = β ... ∂gn ∂z1 (z)α1 + · · · + ∂gn∂zn (z)αn = β. strictly plurisubharmonic functions 97 Fix β ∈ C\{0}; the above system has a unique solution (α1, . . . ,αn) ∈ Cn\{0}. Consequently, (α1, . . . ,αn,β) ∈ Cn+1\{0} and L(u)(z,w)(α1, . . . ,αn,β) = 0. Case 2: v is not strictly psh on D. Then there exists (α1, . . . ,αn) ∈ Cn\{0} such that L(v)(z)(α1, . . . ,αn) = 0. Take β = 0 ∈ C. Then (α1, . . . ,αn,β) ∈ Cn+1\{0} and L(u)(z,w)(α1, . . . ,αn,β) = 0 . Consequently, u is not strictly psh in D × C. Example. Let (z1,z2) ∈ C2 and w ∈ C. Put g1(z) = z1, g2(z) = z2, g3(z) = z1 + z2; g1,g2,g3 are analytic functions in C2. Put u(z,w) = 3∑ j=1 |w − gj(z)|2. Then u is a function of class C∞ and strictly psh in C2 × C. If (w1,w2,w3) ∈ C3, we put v(z1,z2,w1,w2,w3) = 3∑ j=1 Aj|wj − gj(z)|2, where (A1,A2,A3 ∈ R+\{0}). Then v is not strictly psh on any domain of C2 × C3. In fact v is a function of class C∞ in C2 × C3 and the Levi form of v is L(v)(z,w1,w2,w3)(α1,α2,β1,β2,β3) = A1 ∣∣∣∣β1 − [ ∂g1 ∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 ]∣∣∣∣2 + A2 ∣∣∣∣β2 − [ ∂g2 ∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 ]∣∣∣∣2 + A3 ∣∣∣∣β3 − [ ∂g3 ∂z1 (z)α1 + ∂g3 ∂z2 (z)α2 ]∣∣∣∣2, for every (α1,α2,β1,β2,β3) ∈ C5. Let (α1,α2) ∈ C2\{0} such that ∂g1∂z1 (z)α1 + ∂g1 ∂z2 (z)α2 = 0. Put β1 = 0, β2 = [ ∂g2 ∂z1 (z)α1 + ∂g2 ∂z2 (z)α2 ] , β3 = [ ∂g3 ∂z1 (z)α1 + ∂g3 ∂z2 (z)α2 ] . Then (α1,α2,β1,β2,β3) ∈ C5\{0} and L(v)(z,w1,w2,w3)(α1,α2,β1,β2,β3) = 0 . Therefore v is not strictly psh on any not empty open of C2 × C3. Observe that here in fact we have for all k1, . . . ,kN : Cn → C analytic functions, where n,N ∈ N, if N ≥ n, then v1 is not strictly psh in any domain of Cn × CN, where v1(z,w1, . . . ,wN) = N∑ j=1 Bj|wj − kj(z)|2 ( B1, . . . ,BN ∈ R+\{0} ) , z ∈ Cn and (w1, . . . ,wN) ∈ CN. 98 j. abidi A fundamental application concerning analytic functions and the complex structure is now the following extension. Theorem 2.31. Let g1, . . . ,gN : D → C be N analytic functions, D is a domain of Cn, (n ≥ 1) and (N ≥ 1). Put u(z,w) = |w − g1(z)|2 + · · · + |w − gN(z)|2, v(z,w) = |w − g1(z)|2 + · · · + |w − gN(z)|2, v1(z,w) = |w − h1(z)|2 + · · · + |w − hN(z)|2, where (z,w) ∈ D × C, and hj = Re(gj), for 1 ≤ j ≤ N. (a) Suppose that u is strictly psh in D × C. Then v and v1 are strictly psh in D × C and N ≥ n + 1, but the converse is false. (b) In fact, v is strictly psh in D×C if and only if v1 is strictly psh in D×C. For the proof of this theorem, we use Lemma 2.2. Example. Let g1(z) = z, g2(z) = z 2, where z ∈ C. u1(z,w) = |w − z|2 + |w − z2|2, u2(z,w) = |w − z|2 + |w − z2|2, for (z,w) ∈ C2; u1 is not strictly psh on any domain of the form D ( 1 2 ,r ) × C (for every r > 0); u2 is strictly psh on C2. On the other hand, the minimal number N of analytic functions k1, . . . ,kN : Cn → C (n ≥ 1) such that if u1(z,w) = ∣∣w − k1(z)∣∣2 + · · · + ∣∣w − kN(z)∣∣2 is strictly psh on Cn × C is in fact N = n. But for all φ1, . . . ,φN : Cn → C be N analytic functions, u2(z,w) = |w − φ1(z)|2 + · · · + |w − φN(z)|2 satisfies u2 is not strictly psh on Cn × C if N ≤ n. Now there are a great differences between the class of functions defined analogues to u1 and the class of functions defined similar of u2. Now we are in position to prove the following result. Theorem 2.32. Let g1, . . . ,gn : D → C, D is a domain of Cn, n ∈ N. Set v(z,w) = |w1 − g1(z)|2 + · · · + |wn − gn(z)|2, where (z,w) ∈ D × Cn, w = (w1, . . . ,wn). The following conditions are equivalent: (a) v is strictly psh in D × Cn; strictly plurisubharmonic functions 99 (b) g1, . . . ,gn are prh functions in D and for all z = (z1, . . . ,zn) ∈ D (fixed), the system   ∂g1 ∂z1 (z)α1 + · · · + ∂g1∂zn (z)αn = 0 ... ∂gn ∂z1 (z)α1 + · · · + ∂gn∂zn (z)αn = 0 has only the solution (α1, . . . ,αn) = (0, . . . ,0). That is strictly plurisub- harmonic functions and partial differential equations have a rigid rela- tion to discover here for example. Question 2.33. Let g1 : C2 → C be a prh function. Find a condition satisfied by g1 such that there exists g2 : C2 → C prh and satisfying u is strictly psh on C2 × C2, where u(z,w) = A1|w1 − g1(z)|2 + A2|w2 − g2(z)|2, z ∈ C2, w = (w1,w2) ∈ C2 and A1,A2 ∈ R+\{0}. In general this problem have no solution and an affirmative answer is given by the following result. Proposition 2.34. Let g : C2 → C, g(z1,z2) = k1(z1)k2(z2), where (z1,z2) ∈ C2, k1,k2 : C → C be two analytic not constant functions, k1(0) = k2(0) = 0. For all A1,A2 ∈ R+\{0}, there does not exists a function k : C2 → C be analytic such that v = A1|g|2 + A2|k|2 is strictly psh on C2. Proof. Let k : C2 → C be a analytic function. Put v = A1|g|2 + A2|k|2. v, |g|2 and |k|2 are functions smooth of class C∞ in C2. The Levi hermitian form of |g|2 is L(|g|2)(z1,z2)(α1,α2) = ∣∣∣∣ ∂g∂z1 (z1,z2)α1 + ∂g∂z2 (z1,z2)α2 ∣∣∣∣2 for each z = (z1,z2) and (α1,α2) ∈ C2. Therefore, the Levi hermitian form of v is L(v)(z1,z2)(α1,α2) = A1 ∣∣∣∣k′1(z1)k2(z2)α1 + k1(z1)k′2(z2)α2 ∣∣∣∣2 + A2 ∣∣∣∣ ∂k∂z1 (z)α1 + ∂k∂z2 (z)α2 ∣∣∣∣2. Take z1 = z2 = 0. L(v)(0,0)(α1,α2) = A2 ∣∣∣ ∂k∂z1 (0)α1 + ∂k∂z2 (0)α2∣∣∣2. Now take (α1,α2) ∈ C2\{0} such that ∂k∂z1 (0)α1 + ∂k ∂z2 (0)α2 = 0. Then L(v)(0,0)(α1,α2) = 0. Consequently, v is not strictly psh on C2. 100 j. abidi It follows that, for all k : C2 → C be analytic, for all A1,A2 ∈ R+\{0}, if u1(z,w) = A1|w1 − g(z)|2 + A2|w2 − k(z)|2, where z = (z1,z2) ∈ C2,w = (w1,w2) ∈ C2. Then u1 is not strictly psh on C2 × C2. Consequently, the above question globally has a negative answer. But locally we have a positive answer. Because in fact, by using all the notation of the question 2.33, we have if g2 exists, then |g1|2 + |g2|2 is strictly psh on C2. By the above proposition, there exists a function g : C2 → C be analytic such that A1|g|2 + A2|k|2 is not strictly psh on C2, for any k : C2 → C be analytic, for every A1,A2 ∈ R+\{0}. Now locally, if z0 = (z01,z 0 2) ∈ C 2 we can write g1 = f1 + k1, g2 = f2 + k2, where f1,f2,k1,k2 : C2 → C be 4 analytic functions. In fact we can prove that, the functions f1 and f2 do not have any role on the subject of the strict plurisubharmonicity of u; u is a function of class C∞ in C2 × C2. The Levi hermitian form of u is L(u) ( z0,w1,w2 ) (α1,α2,β1,β2) = A1 ∣∣∣∣β1 − [ ∂f1 ∂z1 ( z0 ) α1 + ∂f1 ∂z2 ( z0 ) α2 ]∣∣∣∣2 + A1 ∣∣∣∣∂k1∂z1 (z0)α1 + ∂k1∂z2 (z0)α2 ∣∣∣∣2 + A2 ∣∣∣∣β2 − [ ∂f2 ∂z1 ( z0 ) α1 + ∂f2 ∂z2 ( z0 ) α2 ]∣∣∣∣2 + A2 ∣∣∣∣∂k2∂z1 (z0)α1 + ∂k2∂z2 (z0)α2 ∣∣∣∣2, where (w1,w2) ∈ C2, (α1,α2,β1,β2) ∈ C4. If u is strictly psh on a neighbor- hood G of (z0,w0),w0 ∈ C2, then L(u)(z,w)(α1,α2,β1,β2) = 0 implies that (α1,α2,β1,β2) = 0, for every (z,w) ∈ G = G1 × G2, G1 and G2 are convex domains of C2, where z0 ∈ G1, w0 ∈ G2. But L(u)(z,w)(α1,α2,β1,β2) = 0 has only the solution (α1,α2,β1,β2) = 0 (for every (z,w) ∈ G), if and only if the system   ∂k1 ∂z1 (z)α1 + ∂k1 ∂z2 (z)α2 = 0 ∂k2 ∂z1 (z)α1 + ∂k2 ∂z2 (z)α2 = 0 (where z is fixed on G1 and (α1,α2) is the variable in C2) has only the solution (α1,α2) = (0,0). Observe that this condition is independent of w0 ∈ C2. Therefore if( ∂k1 ∂z1 ( z0 ) , ∂k1 ∂z2 ( z0 )) ̸= (0,0), there exists a ball B(z0,r) ⊂ C2 (r > 0) such that ( ∂k1 ∂z1 (z), ∂k1 ∂z2 (z) ) ̸= (0,0) for each z ∈ B(z0,r). Suppose for example that ∂k1 ∂z1 (z) ̸= 0, for every z ∈ B(z0, t), where 0 < t < r. Let k2(z1,z2) = z2, where (z1,z2) ∈ C2; k2 is analytic on C2. Put g2 = k2; g2 is pluriharmonic on strictly plurisubharmonic functions 101 C2. We have ∂k2 ∂z1 (z) = 0, ∂k2 ∂z2 (z) = 1. The above system has only the solution (α1,α2) = (0,0). Then u is strictly psh on B(z 0, t) × C2. Proposition 2.35. Let g1,g2 : C2 → C. Put u(z,w1,w2) = A1|w1 − g1(z)|2 + A2|w2 − g2(z)|2, where z ∈ C2, (w1,w2) ∈ C2, A1,A2 ∈ R+\{0}; u1(z,w1) = |w1 − g1(z)|2 + |g2(z)|2, u2(z,w2) = |w2 − g2(z)|2 + |g1(z)|2. The following conditions are equivalent: (a) u is strictly psh on C2 × C2; (b) u1 and u2 are strictly psh functions on C2 × C; (c) g1 and g2 are prh functions over C2, g1 = f1 + k1, g2 = f2 + k2 (f1,k1,f2,k2 : C2 → C be analytic) and the antiholomorphic parts of g1 and g2 satisfies |k1|2 + |k2|2 is stictly psh on C2. Moreover observe that if the holomorphic parts of g1 and g2 satisfies |f1|2+|f2|2 is strictly psh on C2 (therefore here |g1|2 + |g2|2 is strictly psh on C2) but we can not conclude that u is strictly psh on C2. Example. Let g1 : C2 → C be a prh function and let N ∈ N, N ≥ 2. Prove that there exists g2, . . . ,gN+1 : C2 → C be N prh functions such that if u(z,w1,w2, . . . ,wN+1) = N+1∑ j=1 |wj − gj(z)|2, where z ∈ C2, then u is strictly psh on C2 ×CN+1. In fact, the answer is very simple, if we consider the family of prh functions g2(z) = z1, g3(z) = z2, g4(z) = · · · = gN+1(z) = 0, where z = (z1,z2) ∈ C2. We have in this case |w2 − g2|2 + |w3 − g3|2 is strictly psh in C2 × C2. Then u is strictly psh in C2 × CN+1. 3. Convex and strictly plurisubharmonic functions We consider in this section a classical family of psh functions, that is the class of convex and strictly psh functions. Theorem 3.1. Let g1,g2 : C → C be two analytic functions. Assume that u(z,w,w1,w2) = A|w1 − g1(w − z)|2 + B|w2 − g2(w − z)|2, v(z,w1,w2) = A|w1 − g1(z)|2 + B|w2 − g2(z)|2, where (z,w,w1,w2) ∈ C4, A,B ∈ R+\{0}. The following statements are equivalent: 102 j. abidi (a) u is psh on C4; (b) g1 and g2 are analytic affine functions; (c) v is convex on C3. Proof. (a) ⇒ (b) Fix w1,w2 ∈ C. Put u1(z,w) = |w1 − g1(w − z)|2 + |w2 − g2(w − z)|2, where (z,w) ∈ C2. Since u1 is of class C∞ and psh on C2, then the Levi hermitian form of u1 is L(u1)(z,w)(α,β) = ( |g′1(z)| 2 + |g′2(z)| 2 ) αα + ( |g′1(z)| 2 + |g′2(z)| 2 ) ββ + 2 Re ([ (w1 − g1(z))g′′1(z) + (w2 − g2(z))g ′′ 2(z) ] αβ ) ≥ 0 , for all (α,β) ∈ C2. Thus∣∣(w1 − g1(z))g′′1(z) + (w2 − g2(z))g′′2(z)∣∣ ≤ |g′1(z)|2 + |g′2(z)|2, for all z ∈ C and all (w1,w2) ∈ C2. Now fix z ∈ C. If g′′1(z) ̸= 0. Fix w2 = g2(z) ∈ C. Then |(w1 − g1(z))g ′′ 1(z)| ≤ |g ′ 1(z)| 2+|g′2(z)| 2, for any w1 ∈ C. It follows that C is bounded. A contradiction. Consequently, g′′1 = 0, g ′′ 2 = 0 over C. Therefore g1 and g2 are analytic affine functions over C. Comparison theorems. We prove in this context that there exists an infinite number F1 of C ∞ functions defined on C2, such that for each F ∈ F1, the function F satisfy F has a fixed type, F is convex and strictly psh on C2, but F is not strictly convex on C2. Denote by < .,. > the habitual hermitian product over Cn in all of this section. Let f : Cn → C be a analytic function. Set u(z,w) = |w−f(z)|2, v(z,w) = |w−f(z)|2, u1(z,w) = A1|w−(< z,a > +b)|2 +A2|w−(< z,a >+b)|2, where (z,w) ∈ Cn × C, a ∈ Cn, b ∈ C, A1,A2 ∈ R + \{0}. We study now the structure of the functions u, v and u1. We have the following 3 assertions: (a) u is psh in Cn × C, but u is not strictly psh on any domain of Cn × C. (b) v is strictly psh on Cn × C if and only if n = 1 and |f ′| > 0 over C. But v is not strictly convex in all not empty convex domain of Cn × C for every n ≥ 1 and for any f : Cn → C be analytic. (c) u1 is not strictly convex in all not empty euclidean open ball subset of Cn × C, for A1,A2 ∈ R+\{0} and (a,b) ∈ Cn × C. strictly plurisubharmonic functions 103 But if we consider u2(z,w) = |w − f(z)|2 + |w − f(z)|2 + |w − g(z)|2, where g : C → C be analytic, n = 1, (z,w) ∈ C2, we have the following result. Proposition 3.2. u2 is strictly convex in C2 if and only if f and g are analytic affine functions, f(z) = a1z + b1, g(z) = a2z + b2, for z ∈ C, where a1,a2,b1,b2 ∈ C such that ((a1,a2 ∈ C\{0} and a2a1 ̸= 1) or (a1 = 0, a2 ̸= 0) or (a1 ̸= 0, a2 = 0)). Proof. Suppose that u2 is strictly convex in C2. Recall that if φ : Cm → R be a function of class C2 (m ≥ 1), then φ is strictly convex in Cm if and only if ∣∣∣∣∣ m∑ j,k=1 ∂2φ ∂zj∂zk (z)αjαk ∣∣∣∣∣ < m∑ j,k=1 ∂2φ ∂zj∂zk (z)αjαk for each z ∈ Cm and all (α1, . . . ,αm) ∈ Cm\{0}. We have u2(z,w) = ww + f(z)f(z) − wf(z) − wf(z) + ww + f(z)f(z) − wf(z) − wf(z) + ww + g(z)g(z) − wg(z) − wg(z) , where (z,w) ∈ C2. Let (α,β) ∈ C2; u2 is strictly convex in C2, then u2 is convex in all C2. Now since u2 is of class C∞ in C2, then we have∣∣[f ′′f − wf ′′ + f ′′f − wf ′′ + g′′g − wg′′]α2 − 2αβf ′∣∣ ≤ |β − f ′α|2 + |β|2 + |f ′α|2 + |β − g′α|2 is valid over C for each (α,β) ∈ C2 and w ∈ C. If w ∈ R, then |w(2f ′′(z) + g′′(z)) + φ(z)| ≤ φ1(z), where φ : C → C and φ1 : C → R+ be two functions. The condition 2f ′′+g′′ ̸= 0, imply that R is bounded, which is a contradiction. Thus 2f ′′ + g′′ = 0 over C. Now put w = it, where t ∈ R. Therefore for each t ∈ R, |tg′′ + θ| ≤ θ1, where θ : C → C and θ1 : C → R+ be two functions. Then g′′ = 0 in C. It follows that f ′′ = 0. Consequently, f and g are analytic affine functions over C; f(z) = a1z + b1, g(z) = a2z + b2 for z ∈ C, where a1,a2,b1,b2 ∈ C. Case 1: a1 = 0. In this situation u2(z,w) = |w − b1|2 + |w − b1|2 + |w − g2(z)|2; u2 is a smooth function over C2. Let (z,w) ∈ C2 and (α,β) ∈ C2\{0}. We have ∣∣∣∣∂2u2∂z2 (z,w)α2 + ∂ 2u2 ∂w2 (z,w)β2 + 2 ∂2u2 ∂z∂w (z,w)αβ ∣∣∣∣ = 0 , 104 j. abidi ∂2u2 ∂z∂z (z,w)|α2| + ∂2u2 ∂w∂w (z,w)|β|2 + 2 Re [ ∂2u2 ∂z∂w (z,w)αβ ] = |β − a2α|2 + 2|β|2, and then 0 < |β − a2α|2 + 2|β|2 for each (α,β) ∈ C2\{0}. If β = 0, then α ̸= 0 and 0 < |a2α|2. It follows that a2 ̸= 0. In this case we have 2|β|2 + |β − a2α|2 > 0 for each (α,β) ∈ C2\{0}. Case 2: a2 = 0. In this situation u2(z,w) = |w − f(z)|2 + |w − f(z)|2 + |w − b2|2. Let (z,w) ∈ C2 and (α,β) ∈ C2\{0}; u2 is a function of class C∞ in C2. We have∣∣∣∣∂2u2∂z2 (z,w)α2 + ∂ 2u2 ∂w2 (z,w)β2 + 2 ∂2u2 ∂z∂w (z,w)αβ ∣∣∣∣ = | − 2αβf ′(z)| , ∂2u2 ∂z∂z (z,w)|α2| + ∂2u2 ∂w∂w (z,w)|β|2 + 2 Re [ ∂2u2 ∂z∂w (z,w)αβ ] = |β − a1α|2 + |β|2 + |a1α|2 + |β|2 > |2αβa1| . Assume that a1 = 0. We take β = 0 and α = 1. We obtain 0 > 0, which is a contradiction. It follows that a1 ̸= 0. In this case we have |2αβa1| ≤ |β|2 + |a1α|2. But also we have|β − a1α|2 + |β|2 > 0 for each (α,β) ∈ C2\{0}. Thus |β − a1α|2 + 2|β|2 + |a1α|2 > 2|αβa1| for every (α,β) ∈ C2\{0}. Case 3: a1 ̸= 0 and a2 ̸= 0. By the above development it follows that |2αβa1| < |β − a1α|2 + |β|2 + |a1α|2 + |β − a2α|2, for all (α,β) ∈ C2\{0}. Thus t1(α,β) = |β − a1α|2 + |β − a2α|2 + (|β| − |a1α|)2 > 0 . strictly plurisubharmonic functions 105 Assume that αa1 = βδ, where δ ∈ ∂D(0,1). Then |αa1| = |β|. In this case, |1 − δ|2 + ∣∣1 − a2 a1 δ ∣∣2 > 0. Consequently, t1(α,β) > 0 for each (α,β) ∈ C2\{0} if and only if |1−δ|2 + ∣∣1− a2 a1 δ ∣∣2 > 0 for every δ ∈ ∂D(0,1). Thus t1(α,β) > 0 for each (α,β) ∈ C2\{0} if and only if a2 a1 ̸= 1. Finally, we resume the above development by the following result. Let φ1,φ2 : C → C be 2 analytic functions. Set φ(z,w) = A1|w − φ1(z)|2 + A2|w − φ1(z)|2 + A3|w − φ2(z)|2, where (z,w) ∈ C2, A1,A2,A3 ∈ R+\{0}. Then φ is strictly convex in C2 if and only if φ1(z) = a1z + b1, φ2(z) = a2z + b2, for z ∈ C and a1,a2,b1,b2 ∈ C with a1 ̸= a2 (the stictly convexity of φ is independent of b1 and b2). Note that, for all B1,B2 ∈ R+\{0}, ψ is not strictly convex in C2, where ψ(z,w) = B1|w−φ1(z)|2 +B2|w−φ1(z)|2, (z,w) ∈ C2. But there exists several possible cases (of the analytic function φ1 defined over C) such that ψ is strictly psh over all C2. Question 3.3. Prove that there exists an analytic function g : C → C such that for all A0,A1,A2,A3 ∈ R+\{0}, the function u = A0|g|2 + A1|g′|2 + A2|g′′|2 + A3|g′′′|2 is not convex over C. We can in fact generalize this question for every fixed order m of the derivative of g denoted ∂mg ∂zm or over analytic functions defined on Cn, where n ≥ 2. Remark 3.4. Let g1, . . . ,gN : Cn → C be N analytic functions, where n,N ∈ N. Assume that |g1|2 + · · · + |gN|2 = u is convex and strictly psh in Cn. We can not conclude that u is strictly convex in Cn. But we have the next statement. Theorem 3.5. Let g1, . . . ,gN : Cn → C be N analytic functions and n,N ∈ N. Put u(z,w) = |g1(w1 − z1)|2 + · · · + |gN(wN − zN)|2, v(w1, . . . ,wN) = |g1(w1)|2 + · · · + |gN(wN)|2, where (zj,wj) ∈ Cn × Cn, 1 ≤ j ≤ N and (z,w) = (z1, . . . ,zN,w1, . . . ,wN). The following conditions are equivalent: 106 j. abidi (a) u is strictly psh in (Cn × Cn)N; (b) n = 1 and |g1|2, . . . , |gN|2 are strictly convex functions over C; (c) n = 1 and v is strictly convex in CN. Proof. Recall that by Abidi [2], we have for every function K : Cn → C be analytic, if we put φ(z,w) = |K(w − z)|2, where (z,w) ∈ Cn × Cn. Then φ is psh on Cn × Cn if and only if (K(z) = (< z,a > +b)m for each z ∈ Cn, where a ∈ Cn, b ∈ C and m ∈ N ∪ {0}) or (K(z) = e(+µ), for every z ∈ Cn, where λ ∈ Cn and µ ∈ C). Note that φ is psh on Cn × Cn if and only if |K|2 is convex on Cn. (a) ⇒ (b) Let u1(z,w) = |g1(w1 − z1)|2, . . . ,uN(z,w) = |gN(w − z)|2, where (z,w) = ((z1,w1), . . . ,(zN,wN)) ∈ (Cn × Cn)N. u is strictly psh on (Cn ×Cn)N if and only if u1, . . . ,uN are strictly psh on Cn ×Cn. For example by Abidi [2], u1 is strictly psh on Cn × Cn if and only if n = 1 and g1 is an affine bijective function over C. Therefore |g1|2 is strictly convex on C. It follows that |g1|2, . . . , |gN|2 are strictly convex functions over C. The remainder of the proof of this theorem follows from the above devel- opment. Claim 3.6. Let k1, . . . ,kn : D → C be n analytic functions, D is a domain of Cn, n ≥ 1. The system  α1 ∂k1 ∂z1 (z) + · · · + αn ∂k1∂zn (z) = 0 ... α1 ∂kn ∂z1 (z) + · · · + αn ∂kn∂zn (z) = 0 has only the solution (α1, . . . ,αn) = 0 ∈ Cn (for all z fixed in D), if and only if u is strictly psh in D × Cn, where u(z,w) = A1 ∣∣w1 − k1(z)∣∣2 + · · · + An∣∣wn − kn(z)∣∣2, for z = (z1, . . . ,zn) ∈ D, w = (w1, . . . ,wn) ∈ Cn and A1, . . . ,An ∈ R+\{0}. Now fix f1, . . . ,fn : D → C be n arbitrary analytic functions. The above system has only the solution (α1, . . . ,αn) = (0, . . . ,0) for all z ∈ D if and only if v is strictly psh in D × Cn, where v(z,w) = A1 ∣∣w1 − f1(z) − k1(z)∣∣2 + · · · + An∣∣wn − fn(z) − kn(z)∣∣2, for z = (z1, . . . ,zn) ∈ D, w = (w1, . . . ,wn) ∈ Cn and A1, . . . ,An ∈ R+\{0}. strictly plurisubharmonic functions 107 That is we have a rigid relation between strictly plurisubharmonic func- tions and holomorphic or antiholomorphic partial differential equations in Cn, n ≥ 1. Observe that we have a good relation between the algebraic method for the resolution of a system of holomorphic partial differential equations and the study of the strictly plurisubharmonic of a only one function in complex analysis and conversely. In the case of a power of analytic equations, we have the following result. Theorem 3.7. Let g1,g2, ,k1,k2 : C2 → C be four analytic functions, and let m1,s1,m2,s2 ∈ N. The system  ( α1 ∂k1 ∂z1 (z) + α2 ∂k1 ∂z2 (z) )2m1 + ( α1 ∂g1 ∂z1 (z) + α2 ∂g1 ∂z2 (z) )2s1 = 0( α1 ∂k2 ∂z1 (z) + α2 ∂k2 ∂z2 (z) )2m2 + ( α1 ∂g2 ∂z1 (z) + α2 ∂g2 ∂z2 (z) )2s2 = 0 has only the solution (α1,α2) = (0,0) for each z = (z1,z2) ∈ C2, if and only if u is strictly psh on C2 × C, where u(z,w) = ∣∣w − k1(z)∣∣2 + ∣∣w − k2(z)∣∣2 + ∣∣w − g1(z)∣∣2 + ∣∣w − g2(z)∣∣2 for each (z,w) ∈ C2 × C. Proof. Define v by v(z,w) = 4|w|2 +|k1(z)|2 +|k2(z)|2 +|g1(z)|2 +|g2(z)|2, where (z,w) ∈ C2 ×C; u and v are functions of class C∞ on C2 ×C. 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