VICTOR ANANDAM AND KAMALELDIN ABODAYEH Communicated by Toshihiro Nakanishi (Received: May 9, 2013) Abstract. In the classification theory of Riemann surfaces the research for positive or bounded solutions of ∆ u = Pu, where P ≥ 0 is a C 1 –function, has played an important role in establishing the similarities between the solutions of this differential equation and the classical harmonic functions. In the discrete potential theory, the Schrödinger operators are to some extent like the equation △u = Pu. In this note, we develop on an infinite graph, a theory of functions to reflect the properties of the above solutions, without the use of derivatives. This can be used to study discrete Schrödinger and Helmholtz equations in non- locally-finite networks.
1. I NTRODUCTION
In the context of the classification theory of Riemann surfaces, many interesting researches (Ozawa [6]) have been carried around the partial differential equation of elliptic type ∆ u = Pu on a Riemann surface R where P ≥ 0 is a C 1 –function. Myrberg [4],[5] showed that there always exists the Green function of ∆ u = Pu on R (when P is not the zero function); Ozawa [7], [8] studied different classes of positive solutions of ∆ u = Pu on R ; Royden [9] studied the bounded solutions of △u = Pu on R and compared bounded solutions of △u = Pu and △u = Qu when Q ≥ P. Even though many of their results were derived by using partial derivatives on R , yet it is possible to prove them by methods in the Brelot axiomatic potential theory [3] without reference to derivatives of functions. In this note, we illustrate the procedure in the context of discrete potential theory which includes also a discrete version of the Helmholtz equation.
2. Preliminaries
Let T be an infinite tree with {p(x,y)} as a set of nearest transition probabilities. For a real-valued function u on T , define the Laplacian at a vertex x as ∆ u ( x ) = Σ p(x,y)[u(y) − u(x)]. Along with the Laplacian operator, on occasions we consider the Schrödinger operators ∆ u ( x ) = q ( x ) u ( x ) also where q ( x ) ≥ 0 on T and q is not the zero function. These operators are similar to the equation ∆ u = Pu on a Riemann surface. Now writing ∆ q u ( x ) = ∆ u ( x ) − q ( x ) u ( x ), we note that a solution u of ∆ q u = 0 in T is of the form u(x) = ∑ p(x,y) fix a set {p'(x,y)} of weights on the edges of T such that 0 ≤ p'(x,y) ≤ p(x,y) for any pair x, y in T; p'(x,y) > 0 if and only if x and y are joined by an edge; and p(x,y) > p'(x,y) for at least one pair of vertices x,y; and then consider the properties of functions u on T which satisfy the condition u(x) = ∑ p (x,y)u(y), similar to the solutions of the Schrödinger operators. These preliminary remarks are intended to indicate the direction in which the following study of functions on an infinite graph will be carried out.
3. P OTENTIALS ON NON-LOCALLY-FINITE NETWORKS
Let X be an infinite graph with a countably infinite number of vertices and a countably infinite number of edges. If x and y are two vertices joined by an edge [ x, y ], we write x ～ y and say that x and y are neighbours.  We assume that X is connected, that is, given any pair of vertices x,y, there exists a finite path { x = x 0, x 1, . . , x n = y } connecting them. With any pair of vertices x and y is associated a non-negative number c(x,y) such that c(x,y) > 0 if and only if x ～ y and c ( x ) = ∑ c ( x, y ) ≤ 1 for any x ∈ X, with the restriction that c ( x 0 ) < 1 for at least one x 0 E X. Let E be a subset of X. A vertex x ∈ E is said to be an interior vertex of E if all the neighbours of x are also in E . We denote by Ë the set of all interior vertices of E and ∂E = E \ Ë. Note that we do not place the condition that the number of neighbours of any vertex is finite, but we consider in this paper only real-valued functions u such that if u is defined on E , then u satisfies the condition ∑ c(x,y)|u(y)| < ∞ for any x ∈ Ë. Write Au(x) = ∑ c(x,y)u(y). A function u defined on E is s said to be {X,c} –superharmonic or simply c –superharmonic (respectively, c − subharmonic or c − harmonic) on E if u ( x ) ≥ Au ( x ) (respectively u(x) ≤ Au(x) or u(x) = Au(x)) for every x ∈ Ë. If p ≥ 0 is a c − superharmonic function such that any c –subharmonic function majorized by p is non-positive, then p is called a c—potential. Some properties of c –superharmonic functions: We have not assumed the local finiteness of X (X is locally finite if each vertex of X has only a finite number of vertices).  Hence all the properties of △ – superharmonic functions given in [1], using the Laplacian operator △ defined in Section 2, may not be valid in the present situation. For example, the limit of a sequence of c –superharmonic functions may not be c –superharmonic. The prop- erties of c –superharmonic functions needed in this paper are listed below. They can be proved with slight modifications from the case of a locally finite network. (3.1) If u n is an increasing sequence of non-negative c − superharmonic (respectively, c –harmonic) functions on a subset E of X  and if u = sup u n is finite at each vertex, then u is c – superharmonic (respectively, c – harmonic) on E . For, ∑ c(x,y)un(y), if x ∈ Ë. Write the neighbours of x as y1,y2,... and un(x) ≥ c(x, y i )Un (y i ). Then amn (x) is increasing in m and un (x) ≥ let amn(x) = lim α mn ( x ). Note that for each m fixed, α mn ( x ) is an increasing sequence in n. Consequently,
The analogous property for the limit of non-negative c —harmonic functions is proved similarly. (3.2) Let { u i} be a family of c –superharmonic functions on a subset E of X . Suppose u i ≥ v on E for each i , where v is a function on E such that ∑ c(x,y)|v(y)| < ∞ for each x ∈ Ë. Let u(x) = inf u i (x). Then u(x) is y c—superharmonic on E Since | u(x)| ≤ | u i (x)| + | v(x)| for any x ∈ E , and all u i ∑ c(x,y)u(y) is well-defined for each x on Ë. If x ∈ Ë, for every i,
and hence u(x) = inf u i (x) ≥  c(x,y)u(y). A sequence f n ( x ) in X is said to be locally uniformly convergent to f ( x ) if for any ε > 0 and any x in X , there exists an integer m (depending on ε and x ) such that if n ≥ m, then | f n (y) − f (y)| < ε for all y ∈ V ( x ), where V ( x ) is the neighbourhood set of x consisting of x and al y ~ x. (3.3) Proposition.  Let u n (x) be a sequence of harmonic functions on X con- verging locally uniformly to a finite function u ( x ). Then u is harmonic on X. Proof. For any x, if y ～ x, then
Consequently u(x) = lim un(x) = lim ∑ c(x,y)un(y) = ∑ c(x,y)u(y). Thus, it is proved that u is harmonic on X. (3.4) Domination Principle Let p be a c—potential with c—harmonic support E (that is, p ( x ) is c − harmonic at every very vertex x̸E ). Suppose s > 0 is a c − superharmonic function on X such that s ≥ p on E . Then s ≥ p on X. For, let q = inf(s,p) and write u = p − q. Then u ≥ 0 on X and u = 0 on E . Hence u(x) is c –subharmonic at each vertex x ∈ E ; and if x 0̸∈ E , then p(x) is c—harmonic at x 0 and q(x) is c—superharmonic at x 0 so that u(x) is c − subhharmonic at x 0 . In other words, u ( x ) is c − subharmonic in X , also u ( x ) is majorised by the c − potential p ( x ). Hence u ≤ 0, so that u = 0 on X . Now p = q on X implies that s ≥ p on X. (3.5) Minimum Principle Let u be a c—superharmonic function defined on a finite set E. If u ≥ 0 on ∂E, then u ≥ 0 on E. For, suppose u takes negative values on E . Let −m = inf u(x),m > 0 Then there exists a vertex x 0 ∈ E such that u(x 0) = −m. Since u ≥ 0 on ∂E, x 0 should be in Ë. Let z be a vertex in X \ E . Since X is connected, there is a path { x 0 , x 1 , . . , x n = z } connecting x 0 and z . Let i be the smallest index such that x i ∈ Ë and x i+1 & Ë . In this case since x i ∈ Ë and x i+1 ～ x i we should have x i+1 E ∂E. Now −m = u(x 0) ≥  ∑ c(x 0,y)u(y) ≥ c(x 0)(−m) which gives c(x 0) ≥ 1. But by assumption on c(x,y),c(x) ≤ 1 for all x. Hence c(x 0) = 1, and [ c(x 0,y)[u(y) + m] = 0. Since each term in this sum is non-negative we conclude u(y) = −m if y ～ x 0 . In particular u(x 1 ) = −m. Proceeding simi- larly we obtain u(x 0) = u(x 1) = · · = u(x i) = u(x i+1) = −m; but x i+1 ∈ ∂E so that u(x i+1) ≥ 0. This contradiction shows that u does not take negative values on E, that is u ≥ 0 on E. Uniqueness: From the above Minimum Principle, it follows that if h is a c − harmonic function on a finite set E such that h = 0 on ∂E , then h = 0 on E . Riesz Decomposition Let s > 0 be a c –superharmonic function on X. Take the family ? of all c –subharmonic functions u such that u < s. Then ? is an upper directed family of c –subharmonic functions. Hence as in (3.2), h( x ) = sup u ( x ) is c − subharmonic in X . In fact, h ( x ) is c − harmonic in X . For, consider for an arbitrary vertex z in X.
Then v(z) = ∑ c(z,y)h(y) ≥ h(z) so that v ≥ h on X. It is easy to see then v(x) is c−subharmonic in X and also v ≤ s on X, so that v ∈ S. Hence by the definition of h, v ≤ h on X. We conclude therefore v = h on X. In particular h( x ) is c—harmonic at x = z . Since z is arbitrary in X, h( x ) ≤ s( x ) for all x in X. Clearly, if h 1 (x) ≥ 0 is a c − harmonic function such that h 1 (x) ≤ s(x), then h 1 ( x ) ≤ h ( x ) so we call h ( x ) as the greatest c − harmonic minorant of s(x) and p(x) = s(x) − h(x) as a c − potential. Further the decomposition of s(x) = p(x) + h(x) as sum of a c − potential and a non-negative c − harmonic function is unique. We shall agree to say that a c –superharmonic function p(x) ≥ 0 is a c − potential if and only if f for any non-negative c − subharmonic function u(x) ≤ p(x), we have u(x) = 0 for any x in X . Consequently, if p(x) is a c−potential and u( x ) is any c−subharmonic function such that u( x ) ≤ p( x ) then u ( x ) ≤ 0. For, sup(u, 0) = u + ≤ p and since u + is c − subharmonic, u+ = 0 so that u ≤ 0. (3.6) Let E be a subset of X. Suppose u is c – superharmonic and v is c –subharmonic on E such that u ≥ v. Let & be the family of all c − subharmonic functions v i on E such that v i ≤ u. Then h = sup v i is a v i ∈ S c –harmonic function on E (named the greatest c –harmonic minorant of u ) such that u ≥ h ≥ v. Using the above result, we can obtain the following Dirichlet solution as in [1, Theorem 3.1.7]. (Dirichlet solution) Let F be an arbitrary subset of X, and E C F. Let f be a real-valued function on F \ E . Suppose there exist two functions u and v on F such that at each vertex in E , u is c – superharmonic and v is c − subharmonic; u ≥ f ≥ v on F \ E ; and u ≥ v on F . Then there exists a function h on F such that h = f on F \ E and h is c − harmonic at each vertex in E. In particular, if E is a finite set in X and if f is a real-valued function on ∂E , then there exists a unique c –harmonic function h on E such that h = f on ∂E . For, since E is a finite set, for some M > 0, −M ≤ f(x) ≤ M on ∂E . Note that c(x) ≤ 1 implies that M is c –superarmonic. Hence the Dirichlet solution h with boundary value f exists on E . The uniqueness of h follows from the Minimum Principle. (3.7) Proposition. Let E be a finite set in X . Then any c—harmonic function on E is a linear combination of n c –harmonic functions where n is the cardinality of the set ∂E. Proof.  For y ∈ ∂E , let P y ( x ) denote the Dirichlet solution in E with boundary values δν(x) on ∂E. Let now h be a c—harmonic function on E. Define u(x) = ∑ h(y)P v(x). Then u(x) is c − harmonic on E such that u(y) = h(y) for YEGE every y ∈ ∂E . Hence u = h on E by the Minimum Principle.  Consequently, ■ h is a linear combination of {Py(x)}y∈∂E (3.8) Let f ≥ 0 be a real-valued function on X. Let & be the family of all su- perharmonic functions u on X such that f ≤ u on X. If the family & is non-empty, then Rf = inf u is a c –superharmonic function on X such that Rf is c—harmonic at every vertex x where f is c—subharmonic (in partic- ular, if f ( z ) = 0, then Rf is c –harmonic at the vertex z ). If & contains a c—potential on X, then Rf is a c—potential on X. A consequence: Since 1 is c –superharmonic but not c –harmonic on X (by our assumptions that c(x) = ∑ c(x,y) ≤ 1 for each x ∈ X and that there exists some x 0 in X such that c ( x 0 ) < 1), then there are always c − potentials on X. Take g(x) = δ e (x) where e is a vertex in X and δ e is the Dirac function. Then Rg(x) is c—superharmonic (not c—harmonic) at the ver- tex e and c –harmonic at every vertex x̸= e. That is, A[Rg(x)] = Rg(x) if x̸= e and A[Rg(x)] < Rg(x) if x = e. Choose the constant α such that α(I − A)Rg(x) = δε(x). Then αRg(x) denoted now by G e(x) is the c−Green function on X  with point c – harmonic singularity at e , that is G e ( x ) is a c − potential on X , c − harmonic at every vertex x̸= e and ( I − A ) G e ( x ) = δ e ( x ) for all x E X. Proposition 3.1. Let s > 0 be a c—superharmonic function in X. Then s(x) = 
Z [(1 − A)s(y)] Gy(x) + h(x) where h(x) is a non-negative c−harmonic function
yex in X. Proof. We have already remarked that s is the unique sum of a c – potential p and a non-negative c —harmonic function h in X . Hence it remains to show that p(x) has the series expansion given above. Let { E n } be an increasing sequence of f nite sets such that X = U E n . Let p n (x) =  ∑  (1 − A)p(y)G y (x). Then p n is a c − potential on X . Let s n (x) = p(x) − p n (x) on X. Since (1 − A)G y (x) = δ y (x), we see that for y ∈ E n , (1 − A)p n (y) = (1 − A)p(y) so that (1 − A)sn(y) = 0 if y ∈ En; if x̸∈ En, (1 − A)pn(x) = 0 while (1 − A)p(x) ≤ 0 so that (1 − A)s n (x) ≤ 0. Consequently, (1 − A)s n (x) ≤ 0 for all x ∈ X . That is s n ( x ) is c − superharmonic and p n + s n = p ≥ 0 so that − s n ≤ p n on X . This implies that −s n ≤ 0 since −s n is c−subharmonic and p n is a c−potential. Consequently, p = p n + s n ≥ p n for any n. Allow n → ∞ to obtain the inequality p(x) ≥  ∑ (1 − A)p(y)Gy(x). Denoting the right side infinite sum as q(x), we remark that q(x) is c – superharmonic on X since it is the limit of an increasing sequence of c − potentials. Since p ( x ) is a c − potential and q ( x ) ≤ p ( x ), we conclude that q ( x ) also is a c − potential. Let v ( x ) = p ( x ) − q ( x ). Then (1 − A)v(x) = (1 − A)p(x) − (1 − A)q(x) = 0, so that v(x) is c−harmonic on X. Thus p(x) = q(x) + v(x). Introduce now the uniqueness of decomposition of a non-negative c—superhhamronoic function as the sum of a c – potential and a c – harmonic function to conclude that v = 0. Hence p(x) =  ∑  [(1 − A)p(y)] Gy(x). Finally, we have the expression s(x) = ∑ [(1 − A)s(y)] Gy(x) for all x ∈ X, since (1 − A)s(y) = (1 − A)p(y) for any yEX y in X. ■
4. N ETWORK CLASSIFICATION
Since 1 is c—superharmonic on X , by the representation given above, 1 is the sum of a c − potential p and a non-negative c − harmonic function h on X . It is possible that h is 0 on X . There are many differences in the study of c − potentials on X , depending on whether h = 0 or h > 0, that is whether the c –superharmonic function 1 is a c – potential on X or not. Accordingly we introduce a definition as follows: Definition 4.1. X is said to be parahyperbolic if and only if the constant 1 is a c—potential on X. Otherwise X is said to be bounded hyperbolic. Note: The term parahyperbolic is chosen to indicate that though X has positive c – potentials, X manifests many characteristics of a parabolic Riemannian man- ifold.  The term bounded hyperbolic indicates that there are non-zero bounded c —harmonic functions in X. For a real-valued function f ≥ 0 on X, we say that  lim f ( x ) = 0 if for any given ϵ > 0 and any finite set A, f(x) < ε for some x & A. lim f(x) = α > 0 would mean that f(x) ≥ = for all x outside a finite set. Proposition 4.2. X  is bounded hyperbolic if and only if f or any c—potential p > 0 on X,  lim p(x) = 0. x → 00 Proof. Suppose  lim p(x) = α > 0 for some c−potential p on X. Then p(x) ≥元outside a finite set A . This implies that p(x) ≥ β on X for some β > 0. Since p is a c—potential and β is c—superharmonic on X, then β (and hence 1) is a c − potential on X . That is, X is parahyperbolic. Hence if X is bounded hyperbolic, then  lim p(x) = 0 for any c – potential > 0 in X. On the other hand, since 1 is a c—potential in a parahyperbolic network, the condition that  lim p ( x ) = 0 for any c − potential p > 0 on X would mean that X is bounded hyperbolic. ■ Remark 4.3. For some vertex e in X, suppose the c—Green function G e ( x ) with pole e satisfies the condition  lim G e ( x ) > 0. Then X is parahyperbolic and for x → 00 any c − superharmonic function s > 0 on X, lim s(x) > 0. For by the Domination s(e) Principle, s(x) ≥ G e ( c) for any x ∈ X . Again, by the Domination Principle, if  lim G e ( x ) = 0, then  lim p ( x ) = 0 for any c − potential p with finite c − harmonic support in X . For, if E is the finite c −harmonic support of p , then p ( x ) ≤ αG e ( x ) on E for some α > 0, consequently p(x) ≤ αG e (x) on X. In particular, if for some e in X,  lim Ge(x) = 0 (respectively,  lim Ge(x) > 0 ) then for any vertex a, lim G α ( x ) = 0 (respectively,  lim G α ( x ) > 0.) The following Proposition 4.4, Lemma 4.7 and Theorem 4.8 are proved in the non-locally finite case as was done in the locally finite case [1, Section 4.3]. Proposition 4.4. Let X be a parahyperbolic network. Let E be an arbitrary subset of X. Suppose s is a lower bounded c—superharmonic function on E such that s ≥ 0 on ∂E. Then s ≥ 0 on E. Proof. Let u = inf(s,0) on E , extended by 0 outside E . Then u is a c − superharmonic function on X ; if s ≥ − m on E for some m > 0, then u ≥ − m on X . Then − u is c − subharmonic and − u ≤ m on X . Since X is parahyperbolic, m is a c − potential on X and hence − u ≤ 0 on X. This implies that s ≥ 0 on E . ■ Remark 4.5. The above Minimum Principle is valid in a bounded hyperbolic net- work if E is a finite subset of X , but not necessarily if E is an infinite set. For an example to show that the Minimum Principle is not valid when E is an infinite set in a bounded hyperbolic network, consider a fixed vertex e with a finite number of neighbours and let V ( e ) be the set of e and al its neighbours. Take E = X \ { e }. Then V(e) \ {e} is the boundary ∂E of E. Now, since X is bounded hyperbolic, there exists a c − harmonic function h, 0 < h < 1, on X. Let u = h − R H ′ ( e ), where R V ( x ) = inf s ( x ) for x ∈ X , S being the family of all positive c –superharmonic functions on X such that s ( x ) ≥ h ( x ) on V ( ε ). Note that R V ( e ) ≤ h on X, R V ( e ) = h on V ( e ), R V ( e ) is c − harmonic at each vertex in X \ V ( e ) and R V ( e ) is a c − potential since V ( e ) is a finite set.  Then u is a bounded c − harmonic function on E , such that u = 0 on ∂E . If the Minimum Principle is valid on X , u should be 0 on E . Since u = 0 on V ( ε ), then u is identically 0 on X , which would imply that h = R ν ( ε ) on X . This is not possible, since the left side is a positive c –harmonic function on X and the right side is a positive c—potential on X. Remark 4.6. (Uniqueness of the Dirichlet solution in an infinite subset): From our earlier reference to the Dirichlet solution, it is clear that if E is an arbitrary set in X and if f is a bounded function on ∂E , then there exists a bounded c—harmonic function H on E such that H = f on ∂E . It follows from Proposition 4.4 that this function H is uniquely determined if X is parahyperbolic. On the other hand, if X is bounded hyperbolic, then this Dirichlet solution may not be unique if E is an infinite subset. For, as in the example given in Remark 2, take E = X \ {e}. Then ∂E = V(e) \ {e}. If f = 0 on ∂E, then the bounded Dirichlet solution on E can be H = 0 or H = h − R (e) Lemma 4.7. Let c(x) =  ∑ c(x,y) for each x in X. If Gy(x) is the c−Green function with vertex c–harmonic singularity at y, then there exists a non-negative c – harmonic function h on X  such that  ∑ [1 − c(z)]G z (x) + h(x) = 1 for all x in X. Proof. If s is the constant function 1, then for any z ∈ X,
Now, writing 1 as the unique sum of a c —potential and a non-negative c —harmonic function h on X, we have for all x in X,
Theorem 4.8. X is parahyperbolic if and only if  [1 − c(z)] Gz(x) = 1 for all x E X. Proof. By definition, X is parahyperbolic if and only if 1 is a c –potential on X . Now from the above lemma, 1 is a c − potential if and only if h = 0, that is if and ■ only if  ∑ [1 − c(z)] G z ( x ) = 1. Theorem 4.9. Let {X,c(x,y)} be an infinite network as above. Let {c'(x,y)} be another system of conductance on X such that 0 ≤ c (x,y) ≤ c (x,y) for every pair x,y in X  and c'(x,y) > 0 if and only if x ～ y. If {X,c} is parahyperbolic, then {X,c'} is also parahyperbolic. Proof. First note that if u ≥ 0 is { X, c } − superharmonic at a vertex x, then u is {X,c'}=superharmonic at x. For
Now 1 is a {X,c}-potential by hypothesis. Then 1 is {X,c'}-superharmonic on X. Let v be any {X,c'} —subharmonic function such that 0 ≤ v ≤ 1 on X. Note that v is {X,c} − subharmonic also; since 1 is {X,c} − potential, v = 0. This means that 1 is a {X,c'}—potential. Hence {X,c'} is parahyperbolic. Corollary 4.10. If there is a non-zero bounded {X, c'}—harmonic function on X, then there is at least one bounded positive {X,c}—harmonic function on X. Proof. By hypothesis, there is a non-zero { X, c ′} —harmonic function b on X such that | b | ≤ 1 on X. Since | b | is { X, c ′} − subharmonic function on X, the constant 1 cannot be a {X, c'} − potential. That is, X is {X, c'} − bounded hyperbolic. Then, by the above theorem, X is {X,c} – bounded hyperbolic. Consequently, there ■ exists a {X,c}−harmonic function on X, 0 < h < 1.
5. NETWORKS WITHOUT POSITIVE SUPERHARMONIC FUNCTIONS
In the introduction, we mentioned about the equation ∆ u = Pu in the context of the classification of Riemann surfaces R . The condition that P is non-negative, non- zero ensures that the positive constants are P –superharmonic (not P –harmonic) functions on R. Consequently, potential-theoretic methods can be used here. How- ever for the existence of positive P –superharmonic functions on R , the condition Δξ(x) is sufficient, where ξ > 0 is a C 2 —function on R . This condition that P(x) ≥ permits P ( x ) to possibly take non-positive values also. In the context of an infinite network X , with conductance t ( x, y ) ≥ 0 such that t(x) = ∑ t(x,y) < ∞ for all x , this situation is described as follows: Take ∆ u(x) =
graph with symmetric conductance.) Recall that u is said to be q − superharmonic at x if q(x)u(x) = [t(x) + P(x)]u(x) ≥ Au(x). However, it is possible that q(x) = t(x) + P(x) > 0 for every x ∈ X, yet there is no positive q(x)—superharmonic on X. Consider the following Example: Let X = {. . . , x − 2 , x − 1 , x 0 , x 1 , x 2 , . . } be a linear tree with conductance t(x i , x i+1) = t(x i+1, x i ) = = for all i . Let P(x) = − 2 for all x in X. Then q(x) = 1
うfor all x . Suppose s(x) ≥ 0 is a q − superharmonic on X . Then we t(x) + P(x) = should have for all i ,
This is possible only if s = 0. Definition 5.1. Let q(x) be a real-valued function defined on a network X. Let us write Aqu ( x ) = Au ( x ) − q ( x ) u ( x ) for x in X . A real-valued function u is said to be A q −superharmonic on a subset E , if q(x)u(x) ≥ Au(x) for each x ∈ Ë. Ag −harmonic and Ag −subharmonic functions are defined accordingly. Proposition 5.2. If q(x 0 ) ≤ 0 for some x 0 in X , then 0 is the only non-negative Ag –superharmonic function on X. Proof. Suppose s ≥ 0 is A g − superharmonic on X. Then, As( x 0 ) ≤ q(x 0 ) s(x 0 ) ≤ 0. This implies that s(y) = 0 for every y ～ x 0 . We know that if a non-negative A g —superharmonic function s on X takes the value 0 at a vertex in X , then s is the zero function.  For if s(a) = 0 then 0 = q(a)s(a) ≥  ∑ t(a,y)s(y) ≥ 0 which implies that s(y) = 0 if y ～ a . This leads to the conclusion s = 0, since X is connected. Corollary 5.3. Let q(x 0) ≤ 0 for some x 0 in X. If u is A q –superharmonic and v is Aq – subharmonic on X such that u ≥ v, then u = v is Aq – harmonic on X. Proof. For s = u − v ≥ 0 is A q −superharmonic on X , so that s = 0. Hence u = v is A g −harmonic on X. In this section, we assume that 0 is the only non-negative q-superharmonic function on X and develop a potential theory on X. Proposition 5.4. Let f be a real-valued function on X such that [t(x,y)|f(y)| < ∞ for each x ∈ X. Suppose the family S of A g –superharmonic functions s ma- jorizing f  is non-empty.  Then Rf ( x ) = inf s ( x ) is Aq − superharmonic on X  and Ag –harmonic at each vertex a where f is Ag –subharmonic and q(a) < 0. Proof. First note that S is lower directed. For, if s 1 , s 2 ∈ S then set s = inf(s 1 , s 2 ). At a vertex z, suppose s(z) = s 1 (z). Then
Hence s ∈ S so that S is lower directed.  Consequently, since X has only a countable number of vertices, there exists a decreasing sequence s n in ? such that Rf(x) = inf s n ( x ) = lim s n ( x ), for x ∈ X . Then for any vertex z in X ,
Let now f(x) be Ag –subharmonic at a vertex a . Then,
Since q(a) < 0, we have then f(a) ≥ Rf(a) and hence f(a) = Rf(a). This shows that q(a)Rf(a) = ∑ t(a,y)Rf(y), that is Rf(x) is Aq —harmonic at the vertex ■ a. Remark 5.5. It is possible that q(x) > 0 for all x, yet there is no positive q − superharmonic function on X . In that case, we remove the condition q( a ) < 0 and prove the above proposition 5.4 by using the Poisson modification (similar to the one used above while obtaining the Riesz decomposition ). In the sequel, to prove certain results, our method requires that there are no closed paths in the network X. Hence we are going to assume that X is an infinite tree T , as before possibly non-locally finite and { t ( x, y )} as conductance. We make the assumption that every non-terminal vertex in T has at least two non- terminal vertices as neighbours and that q(y) ≠ 0 if y is a terminal vertex in T . Apart from the trees without terminal vertices like homogeneous trees and binary trees, some other examples satisfying the first assumption are: i) a linear tree [ x 0 , x 1 , x 2 , . . . ] where x i ～ x i+1 for i ≥ 0; ii) a tree consisting of non-terminal vertices {. . . , x_2, x_1, x 0, x 1, x 2, . . . }, x i ~ x i+1 together with terminal vertices {ai} and {bi}, −∞ < i < ∞, where ai and bi have xi as neighbours. With this assumption on T , we can prove the following lemma as indicated in [1, p.112]: (Recall that for a subset E in T , V ( E ) stands for the set consisting of E and the neighbours of each vertex in E .) Lemma 5.6. Let E be an arbitrary connected subset of T and F = V ( E ). Suppose u is a real-valued function defined on F. Then there exists a function v on T such that v = u on F and Aqv(x) = 0 for each x ∈ T \ F. Proof. The proof is similar to that of [1, Theorem 5.1.2].  Let z ∈ ∂F. Then z has only one neighbour x 0 in F . Note that z is not a terminal vertex. Since by the assumption on T , the non-terminal vertex z should have at least two non- terminal vertices as neighbours, z should have at least one non-terminal vertex as a neighbour outside F. Let A = { Y 1 , Y 2 , . . . } be the neighbours of z outside F . Let A 1 denote the set of all terminal vertices in A and A 2 = A \ A 1 . Then A 2 ≠ 0 . t(y, z) -u(z); and if Define v(x) = u(x) if x ∈ F; if y ∈ A 1 , then take v(y) = y ∈ A 2 then take v (y) = λ , a constant where the constant λ is chosen so that Agv(z) = 0. That is
This procedure can be used with respect to each one of the vertices on ∂F to get an extension of u from F to a function v on the set V(F) such that Aqv(z) = 0 at every z ∈ ∂F. Thus, v is a function defined on V(F) such that v = u on F and A gv (z) = 0 for each z ∈ ∂F. Then v is similarly extended to V [V(F)]. Since T is connected, eventually v is defined at any vertex x in T such that A q v(x) = 0 at each x ∈ T \ F and v = u on F . Theorem 5.7. For any e in T , there exists a q(x) – superharmonic function ge(x) on T  such that (− Aq)ge(x) = q(x)ge(x) − Age(x) = δe(x) for x ∈ T. Proof. Let the neighbours of e be { z 1 , . . , z i , z 1 , . . , z i } of which { z 1 , . . , z i } are non-terminal vertices and { z i , . . . , z j } are terminal vertices. Note that i or j or both can be infinite. For a terminal vertex z', let u(z') = u( z i ) = λ where λ and u( e ) are chosen so that u( e )
Thus u is defined on F = V(e), Aqu(zk) = 0 for k = 1, . . . ,j and (− Aq)u(e) > 0. Now use the above lemma 5.6 to construct v on T such that v = u on F and Aqv(x) = 0 for each x ∈ T \ F. This means that v is defined on T such that ν( x ) for x E T (−Aq)v(e) > 0 and (−Aq)v(x) = 0 if x̸= e. Define ge(x) = ( — Ag)v(e) and remark that (− Aq)ge(x) = δe(x) for all x in T. Remark 5.8. The only non-negative A g − harmonic function in T is 0. However there are many non-zero A q −harmonic functions in T . This can be seen as follows: In the proof of the above Theorem 5.7, choose λ and the non-zero value u( e ) so that
which is possible since   t(e,zk) > 0. Then we construct the function v on T k=1 such that v = u on F = V(e) and Aqv(x) = 0 if x ∈ T \ F. Consequently, v is a A g —harmonic function on T . Another remark, as a consequence of Lemma 5.6, concerns the analytic property of harmonic functions on T : we have termed a class of functions on a network X as harmonic functions if these functions have the sheaf property, the property of local solvability of the Dirichlet problem and the Harnack property. Thus we have con- sidered Laplace harmonic functions, Schrödinger harmonic functions, c – harmonic functions etc. on X . We shall say that a class of S of harmonic functions on X has the analytic property if for any h ∈ S , h = 0 in a neighbourhood V ( ε ) of a vertex e implies that h = 0 on X. Proposition 5.9. Let E be a connected subset of T with more than one vertex and F = V ( E ). Suppose some e in E has at least two non-terminal vertices in F \ E as its neighbours.  Then the A g —harmonic functions in T do not possess the analytic property. Proof. Let a,b ∈ F \ E be two non-terminal vertices that are neighbours of e in E. Note that a,b ∈ ∂F. Let e̸z ∈ E. Since E is connected neither a nor b is a neighbour of z . Define u on F such that u ( a )̸= 0, u ( b ) = − t(e, b) u(a) and u( x ) = 0 for all other x in F . Note that Agu(y) = 0 for all y ∈ F . Then, by Lemma 5.6, there exists a function v on T such that v = u on F and A q v ( x ) = 0 for all x ∈ T \ F . Consequently, v is A q − harmonic on T such that v = 0 on FつE . In particular, v is a non-zero A g − harmonic function on T and v = 0 on V(z). Hence the Ag —harmonic functions in this case do not possess the analytic property.日
6. A DISCRETE VERSION OF THE HELMHOLTZ EQUATION
As an illustration of what was discussed above in a connected infinite network X (which may not be locally finite), let us consider now a discrete Helmholtz equation of the type ∆ u(x) + k(x)u(x) = 0 in X, where k(x) ≥ 0 and ∆ u(x) = E t(x,y)[u(y) – u(x)]. Let us write N
and say that a real-valued function u defined on a subset E of X such that A|u|(x) < ∞ for each x ∈ Ë is a Helmholtz superharmonic ( HH −superharmonic ) function on E if and only if [t(x) − k(x)]u(x) ≥ Au(x) for each x ∈ Ë. That is ∆ ku(x) ≤ 0 if x ∈ Ë . Define similarly HH −harmonic functions and HH −subharmonic functions on E. A HH –superharmonic function s ≥ 0 on E is said to be a HH –potential if for any HH –subharmonic v on E such that v ≤ s on E , we have v ≤ 0. If there is a HH−potential p > 0 on X, we say that X is HH−hyperbolic; otherwise X is called a HH –parabolic network. In the above nomenclature we leave out the prefix HH if k = 0. Unless mentioned otherwise, k stands for a function k( x ) ≥ 0. such that k(x 0 ) > 0 for at least one x 0 in X. In the Euclidean space R or R 2 , let u ( x ) be a non-negative C 2 − function and k ( x ) be continuous such that ∆ k u ( x ) = ∆ u ( x ) + k ( x ) u ( x ) ≤ 0. Then u ( x ) is a classical △−superharmonic function, so that u ( x ) is a constant which should necessarily be 0. A discrete version of this result is the following: Proposition 6.1. Let u ≥ 0 be a HH – superharmonic function on a parabolic network X. Then u = 0. Proof. By assumption Δ k u(x) = Δ u(x) + k(x)u(x) ≤ 0 on X. This implies that △u( x ) ≤ 0 on X . That is u( x ) is a superharmonic function on X . Since u ≥ 0 and X is parabolic, we should have u = c, a constant. Necessarily u = 0. ■ Proposition 6.2. Suppose t(x 0) ≤ k(x 0) for some x 0 in X. Then 0 is the only non-negative HH – superharmonic function on X. Proof. Let s ≥ 0 be a HH –superharmonic function on X. Then, 0 ≥ [t(x 0 ) – k( x 0 )] s( x 0 ) ≥ ∑ t( x 0 , y ) s( y ). This means that s( y ) = 0 if y ～ x 0 . Then, for such a vertex y ～ xo,
Hence s( z ) = 0 if z ～ y . This leads to the conclusion s = 0, since X is connected. ■ Proposition 6.3. There exists a positive HH—superharmonic function on X if Δξ(x) for x and only if there exists some function ξ > 0 on X  such that k ( x ) ≤ ξ(x) in X. Proof. Suppose p > 0 is a HH –superharmonic function on X . Then [t(x) – Δp(x) k(x)]p(x) ≥ Ap(x) for each x in X. That is −k(x) ≥ . Conversely, p(x) Δξ(x) for some function ξ > 0. Then ξ( x ) is a positive suppose k(x) ≤ E(x) HH—superharmonic function on X. Corollary 6.4. If there exists a positive HH – superharmonic function on X , then X is hyperbolic. Δξ(x) Proof. For there exists a function ξ > 0 such that — ≥ k(x). Since k(xo) > 0 ξ(x) for some x 0 , Δξ(x) ≤ 0 and Δξ(x 0 ) < 0. That is, ξ(x) is superharmonic but not harmonic on X. Hence X is hyperbolic. ■ 34 Example of a hyperbolic network that is not HH –hyperbolic: Let X be a homo- geneous tree, each vertex having q+1 neighbours, q ≥ 2, and the conductance being t(x,y) =  — q + 1 X the HH –harmonic functions defined by ∆ k where k(x) = 1 for all x ∈ X. Fix a vertex e and let | x | denote the distance between e and x . Define g ( e ) = 1 q 
q − 1 and g(x) = q m− 1 (q − 1) if | x | = m ≥ 1. Then ∆ g(x) = − δ e (x) so that X is
q − 1 hyperbolic, where ∆ u(x) = ∑ ≠ [u(y) − u(x)] . Take now k = 1 and write Δ k u ( x ) = Δ u ( x ) + k u ( x ) so that Δ k u ( x ) = g+1 ∑ u(y). Suppose now that there exists a HH –superharmonic function s > 0 on X , so that ∆ k s ( x ) ≤ 0 for each x in X . But ∆ k s ( x ) ≤ 0 for each x means that s = 0, a contradiction. Hence X is not HH–hyperbolic. Remark 6.5. Analogous to the example above, we can consider in the line segment X = (−π,π) the operators Δy = y ′′ and Δ 1y = y ′′ + y . Then X is Δ−hyperbolic but 0 is the only non-negative Δ 1 − superharmonic function on X . The following remarks are easy to verify: i) Any non-negative HH –superharmonic function on E is superharmonic on E . ii) Any non-negative subharmonic function on E is HH –subharmonic on E . iii) Any HH—potential on E is a potential on E . In particular, if X is H H —hyperbolic then X is hyperbolic also. iv) In a HH –hyperbolic network X , for any vertex e in X , there exists a unique HH –potential G e ( x ) called the HH –Green potential on X , such that (-Ak)Ge(x) = δe(x). Construction: Let S be the family of all positive HH –superharmonic func- tions s on X such that s(x) ≥ δ e (x). Let u (x) = inf s(x). Then u (x) is HH H –superharmonic on X , since & is a lower directed family of positive HH H –superharmonic functions. Actually, u is a HH –potential on X, since there are HH –potentials in the family S. Note that ∆ k u ( x ) = 0 if x̸= e since u(x) δε(x) = 0 if x̸= e. Take Gε(x) = (-Ak)u(e) Proposition 6.6. Let X be a HH-hyperbolic network, e ∈ X. Then the Green potential g e (x) with harmonic support at e exists on X and g e (x) ≤ G e (x) where Ge (x) is the HH – Green potential on X  with HH – harmonic support at e. Proof. Since X is HH –hyperbolic, as remarked in iii) above, X is hyperbolic also. Hence the Green potential g e ( x ) exists. Now
Hence, g e (x) − G e (x) = v (x) is subharmonic on X. Since v (x) ≤ g e (x), we conclude □ that v(x) ≤ 0 so that ge(x) ≤ Ge(x). Remark 6.7. Actually, by using the Domination Principle, we can show that g e ( x ) ≤ ge(e) Ge(e) Ge(x) ≤ Ge(x). Acknowledgement: We thank the referee for the valuable comments.
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[1] V. Anandam, Harmonic functions and potentials on finite or infinite networks, Lecture Notes, UMI 12, Springer, 2011. [2] E. Bendito, A.Carmona and A. M. Encinas, Potential theory for Schrödinger operators on finite networks,  Rev.Mat.Iberoamericana, 21(2005), 771-818. [3] M. Brelot, Axiomatique des fonctions harmoniques, Les Presses de l ′ Université de Montréal, 1966. [4] L. Myrberg, Über die Integration der Differentialgleichung ∆ u = c(P)u auf offenen Rie- mannschen Flächen, Ann. Acad. Sci. Fenn., 2(1954) 142-152. [5] L. Myrberg, Über die Existenz der Greenshen Funktion der Gleichung ∆ u = c(P)u auf Riemannschen Flächen, Ann. Acad. Sci. Fenn., AI170 (1954). [6] M. Ozawa, Classification of Riemann surfaces, Kōdai Math. Sem. Rep., 4(1952), 63-76. [7] M. Ozawa, Some classes of positive solutions of ∆ u = Pu on Riemann surfaces I , Kōdai Math. Sem. Rep., 6(1954), 121-126. [8] M. Ozawa, Some classes of positive solutions of ∆ u = Pu on Riemann surfaces II, Kōdai Math. Sem. Rep., 7(1955), 15-20. [9] H. L. Royden, The equation ∆ u = Pu and the classification of open Riemann surfaces, Ann. Acad. Sci. Fenn., AI271 (1959). V ICTOR A NANDAM : T HE I NSTITUTE OF M ATHEMATICAL S CIENCES , C.I.T. C AMPUS , T ARA - MANI, CHENNAI 600113, INDIA E-mail address: vanandam@imsc.res.in K AMALELDIN A BODAYEH : D EPARTMENT OF M ATHEMATICAL S CIENCES , P RINCE S ULTAN U niversity P.O. Box 66833, Riyadh 11586, S audi A rabia E-mail address: kamal@psu.edu.sa