On semilinear elliptic equations with diffuse measures

We consider semilinear equation of the form $-Lu=f(x,u)+\mu$, where $L$ is the operator corresponding to a transient symmetric regular Dirichlet form ${\mathcal E}$, $\mu$ is a diffuse measure with respect to the capacity associated with ${\mathcal E}$, and the lower-order perturbing term $f(x,u)$ satisfies the sign condition in $u$ and some weak integrability condition (no growth condition on $f(x,u)$ as a function of $u$ is imposed). We prove the existence of a solution under mild additional assumptions on ${\mathcal E}$. We also show that the solution is unique if $f$ is nonincreasing in $u$.


Introduction
Let E be a locally compact separable metric space, m be a positive Radon measure on E such that supp [m] = E, and let (E, D(E)) be a regular transient symmetric Dirichlet form on L 2 (E; m). In this paper, we consider semilinear equations of the form − Lu = f (·, u) + µ. (1.1) In (1.1), f : E × R → R is a Carathéodory function satisfying the so-called "sign condition": f (x, 0) = 0, f (x, y)y ≤ 0, x ∈ E, y ∈ R, (1.2) and µ is a diffuse measure on E with respect to the capacity associated with E, i.e. a bounded signed Borel measure on E which charges no set of capacity zero. As for L, we assume that it is the operator corresponding to E, i.e. the unique nonpositive self-adjoint operator on L 2 (E; m) such that where (·, ·) stands for the usual scalar product in L 2 (E; m) (see [15,Section 1.3]). where E := D is a bounded open subset of R d and ∆ is the Laplace operator, and − ∆ α/2 u = f (·, u) + µ in D, u = 0 on R d \ D, (1.4) where ∆ α/2 is the fractional Laplace operator with α ∈ (0, 2).
The study of problems of the form (1.3) with µ ∈ L 1 (D; dx) was initiated by Brezis and Strauss [9] (in fact, in [9] more general second-order elliptic differential operator is considered). In [9] it is proved that if f satisfies the sign condition and where F a (x) = sup |y|≤a |f (x, y)|, x ∈ E (1. 5) with E = D, then there exists a solution to (1.3) for µ belonging to some class which is "arbitrarily smaller" than L 1 (D; dx). If f satisfies stronger monotonicity condition: (f (x, y 1 ) − f (x, y 2 ))(y 1 − y 2 ) ≤ 0, x ∈ E, y 1 , y 2 ∈ R, (1.6) then the solution exists for any µ ∈ L 1 (D; dx) and is unique. Later, Gallouët and Morel [16] proved the existence of a solution to (1.3) for any µ ∈ L 1 (D; dx) and f satisfying (1.2), (1.5). Orsina and Ponce [28] have subsequently generalized and strengthened this result by showing that a solution to (1.3) exists for any diffuse measure µ and any f satisfying (1.2) and an integrability condition weaker than (1.5).
Equations of the form (1.1) in the case where L is a general, possibly nonlocal, operator associated with a transient regular Dirichlet form were considered by Klimsiak and Rozkosz [22,24] in case f satisfies the monotonicity condition, and by Klimsiak [20] in case f satisfies the sign condition (in fact, in [20] systems of equations with right-hand side satisfying a generalized sign condition are considered).
In [20,22,24] the proofs of the existence results rely heavily on probabilistic methods. In particular, we make an extensive use of the theory of backward stochastic differential equations and we use some results from stochastic analysis and probabilistic potential theory. In the present paper we give new, rather short analytical proofs of some of the results of [20,22]. We are motivated by the desire to make them accessible to people working in PDEs that are not familiar with probabilistic methods.
Let D e (E) denote the extended Dirichlet space of (E, D(E)). In the present paper we provide a proof of the existence of a solution u in the sense of duality (or, equivalently, renormalized solution; see Section 3) to (1.1) for f satisfying (1.2) and (1.5) under the following additional assumption on E: if {u n } ⊂ D e (E) and sup n≥1 E(u n , u n ) < ∞, then, up to a subsequence, {u n } converges m-a.e. (1.7) We also show that if u is a solution to (1.1), then T k (u) = ((−k) ∨ u) ∧ k ∈ D e (E) for every k > 0 and where µ T V is the total variation norm of µ. Furthermore, if (1.6) is satisfied, then the solution u is unique. Condition (1.7) holds true in many interesting situations. For instance, it holds if the embedding V 1 ֒→ L 2 (E; m) is compact, (1.8) where V 1 denotes the space D(E) equipped with the norm determined by the form E 1 (·, ·) := E(·, ·) + (·, ·). Another condition, which is often satisfied in practice and implies (1.7), is the so called absolute continuity condition saying that R α (x, ·) ≪ m for any α > 0 and x ∈ E, (1.9) where R α (x, ·) is the resolvent kernel associated with E. For symmetric forms considered in this paper, condition (1.9) is equivalent to the condition P t (x, ·) ≪ m for any t > 0 and x ∈ E, (1.10) where P t (x, ·) is the transition kernel associated with E.
The main idea of our proofs resembles the idea used in case of problem (1.3) (see the proof of Theorem B.4 in Brezis, Marcus and Ponce [8] and also Ponce [31,Chapter 19]). Let V denote the extended space D e (E) equipped with the norm determined by E. We first prove the existence of a solution to (1.1) is the set of all diffuse measures on E and V ′ is the dual of V . This step can be viewed as some modification of the result of Brezis and Browder [7] on absorption problems (1.3) with µ ∈ H −1 (D). To get the existence for general µ ∈ M 0,b , we approximate it by a suitably chosen sequence {µ n } ⊂ M 0,b ∩ V ′ and show that solutions u n of (1.1) corresponding to the measures µ n converge to a solution of (1.1). In this second step we use some a priori estimates for u n in V and condition (1.7). In [25] it is proved that any µ ∈ M 0,b admits decomposition of the form µ = g + ν with g ∈ L 1 (E; m) and ν ∈ M 0,b ∩ V ′ (this generalizes the corresponding result proved by Boccardo, Gallouët and Orsina [5] for the form associated with ∆). Therefore, similarly to [8], in the second step of the proof it is enough to approximate by {µ n } the measure µ = g · m. This, however, does not simplify the reasoning, so in the present paper we give a direct approximation of µ ∈ M 0,b (without recourse to [25]).
In the present paper we confine ourselves to single equation with operator corresponding to symmetric regular Dirichlet forms. For results (proved with the help of probabilistic methods) for quasi-regular, possibly nonsymmetric forms, we refer the reader to [24], and for results for systems of equations to [20]. Also note that equations with f = 0 but ∆ replaced by the Schrödinger operator are treated in [29] and [31,Chapter 22].
In the paper we deal exclusively with equations with diffuse measures. The theory of semilinear equations with general bounded measures is much more subtle. In this case (1.3) with f satisfying (1.6) need not have a solution (see [2,3,8]). Results on (1.3) with general bounded measure µ and f satisfying the monotonicity condition are found in [3,8,13], and for equations with f satisfying the sign condition (1.2) in [31,Chapter 21]. The Dirichlet problem for linear equations with nonlocal operators and bounded measure µ is studied in [19,26,30]. In Klimsiak [21] general equations of the form (1.1) with general bounded measure µ and f satisfying (1.6) are considered. The question whether one can extend the existence results of [31] to some nonlocal operators or extend some existence results of [21] to f satisfying (1. 2) remains open.

Preliminaries
In this paper, E is a locally compact separable metric space and m is a Radon measure such that supp[m] = E, i.e. m is a nonnegative measure on the σ-field of Borel subsets of E which is finite on compact sets and strictly positive on nonempty open sets.
In what follows (E, D(E)) is a symmetric regular Dirichlet form on L 2 (E; m). We denote by (·, ·) the usual inner product in L 2 (E; m). As usual, for λ ≥ 0 we set In the whole paper we assume that (E, D(E)) is transient. Recall that this means that there exists a bounded strictly m-a.e. positive g ∈ L 1 (E; m) such that (2.1) For an equivalent formulation, see [15,Section 1.5]. The extended Dirichlet space associated with (E, D(E)) (see [15,Section 1.5] for the definition) will be denoted by (E, D e (E)). Note that D e (E) with the inner product E is a Hilbert space (see [15,Theorem 1.5.3]). In the sequel this space will be denoted by V . We denote by V ′ the dual space of V . The duality pairing between V ′ and V will be denoted by ·, · . In the paper we define 0-order quasi notions with respect to E (capacity Cap (0) , exceptional sets, nests, quasi-continuity) as in [  We say that A ⊂ E is exceptional if Cap (0) (A) = 0, and we say that a property of points in E holds quasi-everywhere (q.e. in abbreviation) if it holds outside some exceptional subset of E.
For a measure µ on E and a function u : E → R, we use the notation whenever the integral is well defined. For a signed Borel measure µ, we denote by µ + and µ − its positive and negative parts, and by |µ| the total variation measure, i.e. If f ∈ V ′ , then by Riesz's theorem there is a unique element Gf ∈ D e (E) such that In particular, if µ ∈ M 0,b ∩ V ′ , then the function Gµ is well defined and belongs to V . Let B + (E) (resp. B b (E)) denote the set of all positive (resp. bounded) real Borel functions on E, and let (G α ) α>0 denote the strongly continuous resolvent on L 2 (E; m) associated with (E, D(E)). Recall that αG α is Markovian for each α > 0, i.e. 0 ≤ αG α f ≤ 1 m-a.e. whenever f ∈ L 2 (E; m) and 0 ≤ f ≤ 1 m-a.e. Since G α is positivity preserving, we can extend it to any positive f ∈ B + (E) by where {f n } ⊂ L 2 (E; m) is a nondecreasing sequence of positive functions converging m-a.e. to f . It is clear that G α f does not depend on the choice of the sequence {f n }.
where µ n = 1 Fn · µ and {F n } is a generalized nest such that µ n ∈ M 0,b ∩ V ′ , n ≥ 1. Note that Gµ is defined uniquely up to m-equivalence.
In the paper, for a function u on E, we denote byũ its quasi-continuous m-version (whenever it exists). We will freely use, without explicit mention, the following fact: if u 1 ≤ u 2 m-a.e. and u 1 , u 2 have quasi-continuous m-versions, thenũ 1 ≤ũ 2 q.e. (see [15,Lemma 2.1.4]).
Let β > 0. In the proof of the lemma below we will need the symmetric form E (β) defined by Since βG β is a symmetric linear operator on L 2 (E; m), by [15, Lemma 1.4.1] there exists a unique nonnegative symmetric Radon measure σ on the product space E × E such that for any Borel functions u, v ∈ L 1 (E; m), with respect to m. Then 0 ≤ s β ≤ 1 m-a.e., and by a direct computation one can check that for a Borel u ∈ L 2 (E; m) one can rewrite E (β) (u, u) in the form , and let ψ : R → R be an increasing function such that ψ(0) = 0 and ψ is Lipschitz continuous with Lipschitz constant 1. Then Proof. By [15, Theorem 1.5.3(δ)], ψ(u) ∈ D e (E). Furthermore, for any β > 0 we have and From this we conclude that for any β > 0, Letting β → ∞ and using [15, Theorem 1.5.2(ii)] we obtain the desired inequality.
For k ≥ 0 and u : E → R, we write Since ψ(y) = ((−k)∨y)∧y, y ∈ R, satisfies the assumptions of Lemma 2.1, if u ∈ D e (E), then T k u ∈ D e (E) and for every k ≥ 0, Then Gµ has a quasi-continuous m-version.
Proof. Let {H n } be a generalized nest such that µ n = 1 Hn µ ∈ M 0,b ∩ V ′ , n ≥ 1, and It is clear that u n ր u m-a.e. By the 0-order version of [15, (2.1.10)], for all ε, δ > 0 we have Sinceũ n+1 −ũ n ∈ D e (E), it follows from the above inequality and (2.10) that Taking ε = 2 −n and letting δ ց 0 we get (2.12) Proof. By Lemma 2.2, Gµ + , Gµ − are finite m-a.e., so Gµ is well defined m-a.e. Let {F n } be a generalized nest for µ such that µ n = 1 Fn µ ∈ V ′ , n ≥ 1. Set u n = Gµ n , u = Gµ. Then u n ∈ D e (E) and by (2.2) and (2.3), In particular, {T k (u n )} n is weakly relatively compact in V . Taking a subsequence if necessary, we can assume that T k (u n ) → v weakly in V as n → ∞. By the Banach-Saks theorem, there is a subsequence (n l ) such that the Cesàro mean {v N : . On the other hand, u n → u m-a.e., so v N → T k (u) m-a.e. Consequently, T k (u n ) → T k (u) weakly in V as n → ∞. Therefore letting n → ∞ in (2.13) yields (2.12).
Proof. We can and do assume that µ n ≤ 2 −2n , n ≥ 1. Then by Lemma 2.4, which proves the lemma.

14)
where Gη is a quasi-continuous m-version of Gη.

Existence and uniqueness of solutions
Throughout this section, we assume that µ ∈ M 0,b and f : E×R → R is a Carathéodory function, i.e. f (·, y) is measurable on E for each fixed y ∈ R, and f (x, ·) is continuous on R for each fixed x ∈ E.
Following [22] we adopt the following definition. Recall that an increasing sequence {F n } of closed subsets of E is called a generalized nest if Cap (0) (K \ F n ) → 0 for any compact K ⊂ E. Proposition 3.3. Let u be a measurable function such that f (·, u) ∈ L 1 (E; m). Then the following assertions are equivalent: is a solution to (1.1).
is the unique solution of the problem is the unique solution of (3.2).
Note that from Lemma 3.6(i) with a = 0 the following absorption estimate follows: Corollary 3.7. If f satisfies (1.6), then there exists at most one solution to (1.1).
Proposition 3.8. If u is a solution of (1.1) with f satisfying (1.2), then for every k ≥ 0, T k (u) ∈ D e (E) and Proof. By Proposition 3.3(ii) and Lemma 2.3, T k (u) ∈ D e (E) and which when combined with (3.5) yields (3.6). Proof. We divide the proof into two steps.
Step 1. We first assume that µ + µ − ∈ M 0,b ∩ V ′ . For a positive g ∈ L 2 (E; m) ∩ V ′ set f n = ng 1+ng T n (f ), n ∈ N. Under the hypothesis (1.5) the operator A n : V → V ′ defined as A n (u) = −Lu − f n (u) is pseudomonotone (see, e.g., [32, Section 2.1] for the definition). Indeed, it is clear that A n maps bounded sets of V into bounded sets of V ′ . Next, suppose that u k → u weakly in V . Then for any v ∈ V , Furthermore, by (1.7), we can assume that u k → u m-a.e. Consequently, we can assume that f n (·, u k ) → f n (·, u) in V ′ . Therefore, for any v ∈ V , Accordingly, A n is pseudomonotone. Since by (1.2), for u ∈ V we have A n u, u = E(u, u) − (f n (·, u), u) ≥ E(u, u), the operator A n is also coercive. Therefore A n is surjective by standard result in the theory of pseudomonotone mappings (see, e.g., [32,Theorem 2.6]). Thus, there exists a weak solution u n ∈ D e (E) of the equation i.e. for any v ∈ D e (E), Taking u n as a test function in (3.8) we get By (1.2) and (3.9), By (1.7) and (3.10) there is u ∈ D e (E) and a subsequence (still denoted by n) such that u n → u m-a.e. and weakly in V . Then, by the definition of f n , f n (·, u n ) → f (·, u) m-a.e. By (3.10), for any Borel subset B of E and a > 0 we have From the above inequality and (1.5) we conclude that the sequence {f n (·, u n )} is equiintegrable and tight. Hence f n (·, u n ) → f (·, u) in L 1 (E; m) by Vitali's convergence theorem (see, e.g., [14,Theorem 2.24]). Therefore letting n → ∞ in (3.8) we see that for any bounded v ∈ D e (E). Let η ∈ L 1 (E; m) be such that G|η| ∞ < ∞, and let {F n } be a generalized nest such that η n = 1 Fn η ∈ V ′ . Then Gη n is bounded and Gη n ∈ V . Therefore taking v = Gη n as a test function in (3.11) we get By Lemma 2.5, Gη n → Gη as n → ∞. Therefore letting n → ∞ in the above equation we obtain (3.1). Thus u is a solution of (1.1).
Step 2. We now show how to dispense with the assumption that µ + , µ − ∈ M 0,b ∩ V ′ . By the 0-order version of [ n . By Step 1, there exists a solution u n ∈ D e (E) of the equation − Lu n = f (·, u n ) + µ n . (3.12) In particular, for any η ∈ L 1 (E; m) such that G|η| is bounded. By Lemma 3.6, for any Borel subset B of E and a > 0 we have (3.14) By Proposition 3.8, If k > a, then {|u n | > a} = {|T k (u n )| > a}, so for any k > a, Hence Let ε > 0. As |µ| ∈ M 0,b , by Lemma 3.9 there exists δ > 0 such that From this and (3.14) it follows that if m(B) ≤ γ, then B |f (x, u n (x))| m(dx) ≤ ε. Furthermore, by (1.5) and σ-finitness of m, there exists a Borel set E 0 ⊂ E such that m(E 0 ) < ∞ and E\E 0 F a (x) m(dx) < ε/2. Therefore taking B = E \ E 0 in (3.14) we get E\E 0 |f (x, u n (x))| m(dx) ≤ ε. This shows that the sequence {f (·, u n )} is equiintegrable and tight. On the other hand, by (1.7) and (3.15), for each k > 0 the sequence {T k (u n )} n is, up to a subsequence, convergent m-a.e., so using the diagonal argument, one can find a subsequence, still denoted by (n), such that {u n } converges m-a.e. to some u. Hence f (·, u n ) → f (·, u) m-a.e. Consequently, by Vitali's theorem, f (·, u n ) → f (·, u) in L 1 (E; m). Let k > 0, and let g be a strictly positive function such that Gg ∞ < ∞ and g ∈ L 1 (E; m). Taking η = g1 {un≥k} and η = −g1 {un≤−k} as test functions in (3.13) we obtain We already know that the sequence {f (·, u n )} is equi-integrable and tight. Furthermore, the functions G(g1 {|un|≥k} ) is bounded q.e., and by Lemma 2.5, G(g1 {|un|≥k} ) ց 0 q.e. as k → ∞. Therefore from (3.17) it follows that the sequence {u n } is equi-integrable with respect to the finite measure ν = g · m. Since u n is a solution of (3.12), for any η ∈ L 1 (E; m) such that G|η| ∞ < ∞ and any k ≥ 0 we have where g k = kg 1+kg . By what has already been proved, letting n → ∞ in (3.18) yields with F = ∞ n=1 F n . Since µ(E \ F ) = 0 by (2.6), letting k → ∞ and using Lemma 2.5 shows that u is a solution to (1.1).
By the above, {f (·, u n )} is convergent in L 1 (E; m). The rest of the proof runs as the proof of Theorem 3.10 (see the reasoning following the statement that {f (·, u n )} is equi-integrable).
If (1.8) is satisfied, then the following Poincaré-type inequality holds true: there exists c > 0 such that (see [18,Corollary 2.5]). Hence, under (1.8), D e (E) = D(E) and the norms determined by E and E 1 are equivalent. It follows in particular that under the assumptions of Theorem 3.10 the solutions of (1.3) belong to D(E).
In general, solutions of (1.3) are not even locally integrable (see [22,Example 5.7]). Below we shall see that a simple condition guaranteeing their integrability is This condition is sometimes expressed by saying that E is Green-bounded (see, e.g., [6,10]; note that in the case where problem (1.3) (resp. (1.4)) is considered, G is the Green function for ∆ (resp. ∆ α/2 ) on D. The Green-bounded domain need not be bounded. For instance, if L = ∆, then the infinite strip {(x, y) ∈ R 2 : |x| < a} (a > 0) in R 2 is Green-bounded (see [10, p. 39]).
Proposition 3.13. Assume that E is Green-bounded. If u is a solution to (1.1), then u ∈ L 1 (E; m).
Proof. To see this it is enough to consider an increasing sequence of compact sets {F n } such that ∞ n=1 F n = E, take η n = 1 Fn sign(u) as test functions in (3.1), and use (3.5) and Fatou's lemma.

Applications
In this section we provide some examples of local and nonlocal symmetric transient regular Dirichlet forms satisfying condition (1.7). Before proceeding, we make some general comments on conditions (1.8) and (1.9).
By [15,Lemma 1.3.4], α(w k n , w k n − αG α w k n ) ≤ E(w k n , w k n ) for every α > 0. Moreover, By the above estimates, αG α w k n − w k n 2 L 2 (E;m) ≤ α −1 E(w k n , w k n ) for α > 0, which when combined with (4.1) shows that there is a constant c(k, v) depending only on k and v such that for every η ∈ L 1 (E; m). In particular, for every x ∈ E, Since |αG α w k n | ≤ kαG α |v|, it follows from (4.3) that the sequence{αG α w k n ′ } converges in L 2 (E; m) for any fixed α > 0, k > 0. This and (4.2) imply that there exists a subsequence (n ′′ ) ⊂ (n ′ ) such that {w k n ′′ } converges in L 2 (E; m). Using the diagonal procedure one can find a further subsequence (n ′′′ ) ⊂ (n ′′ ) such that {u n ′′′ } converges m-a.e. on E.
, be a nonempty bounded open set, and let a ij : D → R be locally integrable functions such that a ij (x) = a ji (x) for x ∈ D, i, j = 1, . . . , d, and for some λ > 0, (i) (Dirichlet boundary conditions) The form defined by with D(E) = H 1 0 (D) is a regular symmetric Dirichlet form on L 2 (D) (see, e.g., [15,Section 3.1]). The generator L of (E, D(E)) is of the form The form (E, D(E)) is transient by Poincaré's inequality, and (1.8) is satisfied by Rellich's theorem. Also note that by classical results (see [1]), assumption (1.10) (and hence (1.9)) is satisfied as well. Therefore Theorem 3.10 applies to the Dirichlet problem Note that we impose no regularity assumption on the boundary ∂D of D. Note also that by Poincaré's inequality, D e (E) = H 1 0 (D). Consequently, T k (u) ∈ H 1 0 (D) for every k > 0. Furthermore, since D is Green-bounded (see, e.g., [10, Theorem 1.17]), u ∈ L 1 (D; dx) by Proposition 3.13.
(ii) (Neumann boundary conditions) Assume additionally that ∂D is Lipschitz. Consider the form E defined by (4.4), but with domain H 1 (D). Then (E, H 1 (D)) is a regular symmetric Dirichlet form on L 2 (D; dx) withD = D ∪ ∂D (see [15,Example 4.5.3]), and clearly so is (E λ , H 1 (D)) with λ ≥ 0. Moreover, if λ > 0, then (E λ , H 1 (D)) is transient because D is Green-bounded (see Lemma 3.12). The generator L λ of (E λ , H 1 (D)) is equal to L − λ, where L is the generator of (E, H 1 (D)). By Rellich's theorem, H 1 (D) ֒→ L 2 (D; dx) is compact, so the results of the paper apply to equation (1.1) with L replaced by the operator L − λ defined above. A solution u to such equation can be viewed as a solution to the Neumann problem −Lu = −λu + f (·, u) + µ in D, ∂u ∂(a · n) = 0 on ∂D, where n denotes the unit outward normal to ∂D. (i) Let ν t , t > 0, be a probability measure on R d such thatν t (x) = e −tψ(x) , x ∈ R d . Then the semigroup (P t ) t>0 associated with E has the form P t f (x) = R d f (x+y) ν t (dy) for f ∈ L 2 (R d ; dx) ∩ B b (E) (see [15,Example 1.4.1]). It follows in particular that if ν t are absolutely continuous with respect to the Lebesgue measure, then (1.10) is satisfied. For instance, this is the case when ψ(ξ) = |ξ| α , ξ ∈ R d , with α ∈ (0, 2] (see [15,Example 1.4.1]). For such ψ, the operator corresponding to E is the fractional Laplacian ∆ α/2 . If α < d, then (4.7) is satisfied, so the form E is transient. Therefore  [15,Theorem 4.4.4]. Let L denote the generator of (E, D(E)) and L D denote the generator of (E D , D(E D )). By virtue of (4.8), the solution u of (1.1) with E = D and operator L D can be interpreted as a solution of the Dirichlet problem − Lu = f (·, u) + µ in D, u = 0 on R d \ D. Therefore, if (4.7) and (4.10) are satisfied, then by Theorem 3.10 there exists a solution u to (4.9). Since (1.8) implies (3.19), D e (E D ) = D(E D ). Consequently, T k (u) ∈ H ψ,1 0 (D) for k > 0.