Optimal power for an elliptic equation related to someCaffarelli-Kohn-Nirenberg inequalities

In this paper we analyze the following elliptic problem related to some Caffarelli-Kohn-Nirenberg inequalities: 
 
\begin{eqnarray} 
-div(|x|^{-2\gamma}\nabla u)-\lambda 
\frac{u}{|x|^{2(\gamma+1)}}=|\nabla u|^p|x|^{-\gamma p}+cf,\; u>0\; 
\mbox{ in }\; \Omega, \qquad u_{|\partial 
\Omega}\equiv0, 
\end{eqnarray} 
 
where $\Omega \subset R^N$ is a domain such that $0\in\Omega$, $N\geq 3$, and $c, 
\lambda, \gamma, p $ are positive constants verifying $0 0$. Our study concerns to existence of solutions to the former problem. More precisely, first we determine a critical thereshold for the power $p$, in the sense that, beyond this value it does not exist any positive supersolution to our problem, not even in a very weak sense. In addition, we show existence of solutions for all the values $p>0$ below this threshold, with the restriction $\gamma>-\frac{N(1-p)+2}{2}$, whenever the righthand side verifies $f(x)\leq |x|^{-2(\gamma+1)}$ if $\gamma>-1$. When $-\frac{N(1-p)+2}{2}<\gamma\leq -1$ it suffices that $f\in L^{2/p}(\Omega)$. The existence of solutions for $0 < p < 1$ and $\gamma\leq -\frac{N(1-p)+2}{2}$ is an open question.

1. Introduction and main results. We devote these notes to study the problem      −div(|x| −2γ ∇u) − λ u |x| 2(γ+1) = |∇u| p |x| −γp + cf, in Ω, where Ω ⊂ R N , N ≥ 3, is a domain that contains the origin. We assume that f is a nonnegative function under certain hypothesis that will be specified later and c, λ, α, γ are positive constants such that 0 < λ ≤ Λ N,γ = N −2(γ+1) and p > 0. Our analysis is motivated in part by the work [3], where the authors consider equations of type − div(|x| −2γ ∇u) = g(u, x), with −∞ < γ < N −2 2 . The function g satisfies certain conditions that vary along the paper, depending on the subject to analyze. In fact, the authors treat here a more general operator of p-laplacian type. Notice that if we take g(∇u, u, x) = λ u |x| 2(γ+1) + |∇u| p |x| −γp + cf our problem is related to (2), it has an additional gradient term that is not considered in [3].
A suitable frame to study existence of solutions to this kind of equations is the following energy setting. The weighed Sobolev Space D 1,2 γ (Ω) is defined as the completion of C ∞ (Ω) with respect to the norm u 2,γ = Ω |u| 2 + |∇u| 2 |x| −2γ dx 1 2 . (3) Analogously, D 1,2 0,γ (Ω) is the completion of C ∞ 0 (Ω) with respect to the norm above. For these spaces there exists a critical exponent for the Sobolev inequality. It is a consequence of the well known Caffarelli-Kohn-Nirenberg inequality, see [13]. , optimal and not achieved.
Indeed different kind of estimates on the remaining terms in the previous inequality have been shown. In this direction, the following result is suitable for the problem in consideration, see [1] and references therein for further literature on this topic. , being 1 < r < 2, and r ≤ β < ∞ if γ ≤ 0, while if γ > 0 then 1 ≤ β < 2 + δ(N, r, γ), for some positive constant δ.
In the sequel, we will also consider weighted Sobolev spaces with more general powers, namely, D 1,q γ (Ω) with analogous definition. The space W 1,r (Ω; |x| β dx), for some β ∈ R, denotes the completion of C ∞ (Ω) with the norm u W 1,r (Ω;|x| β dx) = Ω |u| r + |∇u| r |x| β dx 1 r . Turning again our attention to equation (2) with g(u, x) = u α |x| 2(γ+1) and α > 1, it is shown that there exists no positive entropy solution . Moreover, the authors prove that the problem exhibits complete blow-up. Conversely, there exists a nontrivial supersolution whenever the second term is λ u α |x| 2(γ+1) , with α ≤ 1 and 0 < λ ≤ Λ N,γ . Inspired in this work, we wish to find out how adding a gradient term in the equation affects the existence of solutions to these problems. As a preceding study we refer to [5], where the problem under consideration is ∆u = λ u |x| 2 + |∇u| p + cf, with 0 < λ ≤ Λ N = N −2 2 2 , c > 0 and f ≥ 0. The authors find the critical exponent, in the sense that, beyond this value there do not exist positive supersolutions, not even very weak supersolutions, concept that we will precise later on. In addition, they show that the nonexistence is due to global blow-up of the approximations to the problem. Our model extends these results corresponding to the case γ = 0. The lack of spatial symmetry of our operator if γ = 0 leads to develop new techniques to achieve some results related with comparison.
Another foregoing work in this direction is [11], where the optimality of the power p is determined to have very weak solutions to ∆u = λ u |x| 2 + u p + cf.
Furthermore, regarding existence and optimal summability of solutions to the elliptic equation div (M (x)∇u) = θ u |x| 2 + f, we refer to [9]. See also [18] for the analysis of regularity of this kind of operators.
We organize this work as follows. In Section 2 we precise what we understand by very weak super and subsolutions to our problem and determine the candidate to be the critical exponent for the nonexistence result. We also give some preliminary results, concerning the integrability of our solutions. Section 3 is devoted to establish a comparison principle, that has interest itself since it is an alternative proof of the comparison result shown in [6]. Moreover, the concern of our proof resides in its application to more general operators. In Section 4 we show that above the critical exponent we have nonexistence of solutions. This nonexistence result is the strongest we can expect, since we prove that it is not possible to have any very weak positive supersolution, namely the class of solutions in the distributional sense. Furthermore, we see that in the range of values for which there is not existence of solutions, the approximating problems exhibit complete blow-up. This is the purpose of Section 5. Finally, in Section 6 we conclude this paper by showing that below this critical exponent, we have existence of positive very weak supersolutions.

2.
Nonexistence results if p ≥ p + (λ). The aim of this work is to characterize the nonexistence of very weak positive solutions to (1) in terms of p. We specify this notion of solutions. Note that it is the more general setting for which the equation has meaning in a distributional sense.
Definition 2.1. We say that u ∈ L 1 loc (Ω) is a very weak supersolution to problem (1) if u ∈ L 1 loc (Ω; |x| −2(γ+1) dx), |∇u| ∈ L p loc (Ω; |x| −γp dx) and it holds that The definition of a subsolution is the analogous reversing the inequalities. If u is a very weak super and subsolution we say that u is a very weak solution.
It is well known that the problem (1) has not positive very weak supersolutions , as a consequence of the optimality of the constant Λ N,γ in the Hardy inequality, see for instance [2]. Therefore, we will assume that 0 < λ ≤ Λ N,γ in the sequel.
To search for the optimal exponent we consider a radial solution to (1) with f ≡ 0, namely u(x) = A|x| −a . It is easy to check that a = (2−p)(γ+1) p−1 and A p−1 a p = −a 2 + (N − 2(γ + 1))a − λ = q(a). The left hand side is positive, then q(a) must be positive. Thus a ∈ (a − , a + ) where a + , a − are the two roots of q given by The fact that a − < a < a + is equivalent to p − (λ) < p < p + (λ), that is However, we observe that if γ < −1 the above exponents will be no longer well defined for all values of λ. This fact is new with respect to the problem treated in [5], (case γ = 0). Indeed, the optimal power should be written as p + (λ, γ), though we omit the dependence on γ to simplify the notation. We will see that if p ≥ 1 = p + (λ) and γ ≤ −1, there do not exist very weak supersolutions to (1). Hence the candidate to be the optimal power is Remark 1. Notice that whenever γ ≥ −1 the function p + (λ) is nonincreasing in λ. If λ → 0, then p + (λ) → 2, whereas p + (λ) = 2(γ+1)+N N if λ = Λ γ,N and γ ≥ −1.
We give some preliminaries concerning to the regularity and the admissible weights for the solutions to the operator in divergence form. The proofs of these results are analogous to the Lemmas 2.2 and 2.3 in [5] and we omit them.
in a distributional sense. Then there exists a positive constant c and a small ball , with a − given by (5).
with g ∈ L 1 loc (Ω) nonnegative and β ≤ Λ N,γ . If equation (7) has a very weak supersolution, then |x| −a − g ∈ L 1 loc (Ω), with a − defined in (5). Remark 2. If we consider equation (1) with λ < Λ N,γ we can split β = λ + (β − λ) for some λ < β ≤ Λ N,γ . Then, applying this result we get Our nonexistence proof is strongly based on the contradiction of the Hardy-Sobolev inequality in Theorem 1.2, if we admit the existence of a very weak supersolution. This inequality holds only for functions vanishing on the boundary. However, in the case p = p + (λ) and λ = Λ N,γ we will need to use an analogous argument for functions that do not vanish on the boundary. To overcome this fact we introduce a version of the well known Picone's inequality for measures (see also [21] or its version for the p-laplacian [7]). Lemma 2.4. Let φ be a positive function verifying that −div(|x| −qβ |∇φ| q−2 ∇φ) is a positive Radon measure if 1 < q < N and −∞ < β < N −q q . Then, for all v ∈ D 1,q β (Ω) it holds that For the proof of this lemma we refer to [3,4]. Now we show with the norm · β,q defined in (3), being β and q as in Lemma 2.4.

3.
A comparison principle. The following comparison principle will be the key to prove both, existence of solutions for the corresponding values of the parameter p and, in case of nonexistence of solutions, to show the global blow-up for the approximating problems. We note that this is the analogous comparison result due to N. Alaa and M. Pierre (see [6]), for our operator. However, such an extension is not straight forward, since in [6] the proof is based on the isoperimetric inequality. To skip this difficulty we follow the ideas in [19], which are strongly based in the Schauder fixed point. We have devoted this section to see this proof in detail.
Throughout this section, we define the exponent which is related to the integrability of the gradient of the solutions to the following problem. More precisely If g ∈ L 1 loc (Ω) then ∇w ∈ Lq(Ω; |x| −2γ dx), for anyq < q, with q given in (10). For the proof of this Lemma we refer to Lemma 2.6 in [3]. We also recall the extension of the Sobolev inequality for this setting, see Theorem 1.1, with critical Sobolev exponent We have all the preliminaries to show the main result of this section. Proposition 1. Let w ∈ L 1 loc (Ω) be such that ∇w ∈ Lq(Ω; |x| −2γ dx), for anyq < q with q given in (10), satisfying where h : Then w ≤ 0.
We begin with this existence result, for which we establish the following notation. We denote by H * the dual space of D 1,2 0,γ (Ω) with the norm Lemma 3.2. There exists a unique solution u ∈ D 1,2 0,γ (Ω), u ≥ 0 in Ω to the problem where h : R N → R N is such that |h| ∈ L q (Ω; |x| 2γ q q dx) with q given in (10) and f ∈ H * . In particular we have Proof. Let us define the operator such that v verifies −div(|x| −2γ ∇v) = c h, ∇u + f. The continuity of T easily follows. We show that T is compact. Note that the term c h, ∇u ∈ H * , since for any ϕ ∈ D 1,2 0,γ (Ω) it holds that We have applied the Sobolev inequality in Lemma 1.1 for the exponent (11). The fact that also f ∈ H * yields that the sequence v n = T (u n ) is uniformly bounded in D 1,2 0,γ (Ω). By Rellich Theorem we get the existence of some v ∈ D 1,2 0,γ (Ω) such that for a subsequence, still labeled with n, it holds that v n → v strongly in L 2 (Ω; |x| −2γ dx) and weakly in D 1,2 0,γ (Ω).
To show the strong convergence in D 1,2 0,γ (Ω), we take v n − v as test function in (14) and get due to (16) and f ∈ H * . With the remaining term we argue as follows: being T m (s) = max{−m, min(s, m)} and Ω m,n = Ω ∩ {|v n − v| > m}. Observe that
Finally we show the last estimate of the proposition. Set v 1 = v ||f || H * . Then, repeating the same computation as in the previous step, we reach that , and hence (15) follows.
Notice that the solutions to (12) do not necessarily belong to the energy space D 1,2 0,γ (Ω), thus we must look for a test function within L ∞ (Ω). This is the reason for the next result.
Proof. We use again Schauder fixed point theorem and define the operator The continuity and compactness follow straight forward as in Lemma 3.2. Finally, To see that there exists a constant M > 0 such that The existence of such a solution is guaranteed by Lemma 3.2 if we take f = |φ| 2 * γ −2 φ|x| −2γ . Indeed, multiplying by any nontrivial ϕ ∈ D 1,2 0,γ (Ω), we get In particular, Multiplying by v and φ equations (20) and (21) respectively, and subtracting we get where we have also used (15). Hence, To get a bound for the gradient, we multiply (20) by φ. Integrating by parts, we obtain Taking into account (13) and (22) this implies that φ D 1,2 0,γ (Ω) ≤ C. By the fixed point Schauder theorem, we have shown that there exists φ ∈ D 1,2 0,γ (Ω) a unique solution to (19).
The positivity of φ follows from a comparison principle for problem (19), analogous to Lemma 3.3 and the fact that zero is a subsolution to (19).
More precisely, if we denote as g = w 1 − w 2 , the difference of a sub and a supersolution to (19), respectively, it verifies where we have applied also Kato's inequality. We multiply again by g k = (g + − k) + to obtain (17) which shows that g + ≤ 0. Then φ ≥ 0, since 0 is a subsolution to (19).
To conclude the proof we verify that φ ∈ L ∞ (Ω). To this end, we argue as in [3] and consider in (19) the following test function, v = sign(φ)(|φ| − k) + . We find that where We observe as well that µ(A(k)) → 0 as k → ∞, otherwise we would contradict the fact that φ ∈ D 1,2 0,γ (Ω). Our purpose is to show that indeed, there exists k 0 such that µ(A(k)) = 0 for any k ≥ k 0 . We treat first the integral on the right hand side, where we have used the Sobolev inequality and the identity ∇v = ∇φ in A(k). We deal now with the second term on the left. We take the following Hölder exponents: (10), to see that Substituting this estimate and (24) into (23), and applying once more Sobolev inequality, it follows Taking into account that A(m) ⊂ A(k) if 0 < k < m, it is easy to deduce that Plugging this inequality into (25) yields Then, the classical Stampacchia Theorem implies that there exists k 0 such that µ(A(k)) = 0 for any k ≥ k 0 as we wished to prove.
Proof of Proposition 1. We take as test function φ ∈ L ∞ (Ω), φ ≥ 0 verifying (19), to find that Integrating by parts in the first of the integrals and canceling terms we arrive to Thus w ≤ 0 and the proposition holds. 4. Nonexistence results if p ≥ p + (λ). We state the main result of this section. (6), then the equation (1) has not any positive very weak supersolution.
In case f ≡ 0, then the unique nonnegative very weak supersolution is u ≡ 0.
Proof. We separate into cases the proof of this result. In all of them we argue by contradiction. We assume that there exists a solution and finally we contradict Hardy inequality. Non existence result for γ > −1 and p + (λ) > 1.
Case p = p + (λ) and λ < Λ N,γ : As before, we assume that there exists a very weak supersolution to problem (1). Again by Lemma 2.2, there exists a positive constant, c 0 such that Furthermore, since λ < Λ N,γ , applying Lemma 2.3 we get the estimates given in (8). Without loss of generality, we fix some η ≤ e −1 .
Therefore, we deduce that a nonnegative bounded function in B η (0).
We consider u 1 = c 1 u for some c 1 > 0 such that c 0 c 1 ≥ log 1 η β . This ensures that u 1 ≥ w in ∂B η (0). Let us denote by v = ω − u 1 . Then v ≤ 0 on ∂B η (0) and we claim that v ≤ 0 in B η (0), namely On the other hand, u 1 satisfies Therefore, in a similar way we obtain an estimate analogous to (26), But then which contradicts the optimality of the constant Λ N,γ in Hardy inequality.
Proof of the claim. We can take β small enough such that c The regularity of ω and the fact that u 1 verifies (8), assure that v ∈ W 1,p+(λ) (B η (0)) and It is not difficult to check that v satisfies Notice that m has not the required regularity to apply the comparison principle stated in Proposition 1. We proceed then using Kato's inequality: Moreover, by (29) it holds that Since N > (γ +1)p+a − , applying Caffarelli-Kohn-Nirenberg inequalities (see [13]), it gives Let 2θ = β 1 2 p + (λ). For β small |x| −2θ is an admissible weight for Caffarelli-Kohn-Nirenberg inequalities. Then, equation (30) becomes into We consider now some nonnegative function ϕ verifying the following problem The idea is to take ϕ as test function in (33). Integrating by parts we would get that as we wanted to prove. An easy computation shows that We recall that λ < Λ N,γ . Thus taking β small we can assure that a is a real number.
Since ϕ lacks of the regularity to be used as test function in (33), we take instead the approximation ϕ n ∈ C ∞ (B η (0)): It is not difficult to see that we just need the following two conditions to perform the integration by parts using ϕ n (x), To fulfill the first condition we observe that, as a function of θ, a verifies a(0) = a − and a θ (θ) > 0 for θ > 0. Therefore, a(θ) > a − for θ > 0, small. We claim that for some σ > 2(γ + 1) + a − . This shows (34) 1 for θ sufficiently small. We proceed now with the proof of the claim. Note that .

MAYTE PÉREZ-LLANOS
The first integral is bounded by (32) and to show that the second integral is also bounded we define It is not difficult to check that since λ < Λ N and (35) is proved.

But taking into account Lemma 2.2 this implies that
which is a contradiction.
Case p > p + (λ) = 1: The proof can be performed exactly as in the case γ > −1.
On the other hand, a simple calculation shows that, for any b > 0 the function If we denote by g = w n − u and apply Kato's inequality, we obtain that in B η1 (0). We show that g + = 0 arguing as in (33). We consider

which is admissible as test function if
by (39) and the fact that γ ≤ −1.
As before, we can deduce that g + = 0 and so u ≥ w n . But then, (39) gives that for any n . In this section we will show that if p ≥ p + (λ), problem (40) arises complete blow-up as n → ∞, as consequence of the nonexistence of very weak supersolutions to (1). To achieve this aim, we study the following approximating problems    −div(|x| −2γ ∇u n ) = λu n a n (x) + |∇u n | p b n (x) + cf, in Ω, being a n (x) = 1 and f ≥ 0, f = 0. For existence of solutions to those problems see [3]. The main result of this section is enclosed in the following theorem.
Before performing the proof we need some preliminaries. The first one consists of a property satisfied by solutions to the divergence operator, see [3,10].
Lemma 5.2. Let u be the unique positive solution to problem: where g is a positive function such that g|x| 2γ ∈ L ∞ (Ω). Then, for any ball B r ⊂ Ω such that B 4r ⊂ Ω, there exists a positive constant c = c(r, N, γ) such that B2r g(y)dist(y, ∂Ω) dy, for every x ∈ Ω.
Remark 3. Note that, by this result for any compact set K ⊂⊂ Ω, there exists a constant c = c(K, N, γ) such that for every x ∈ Ω, since there exists a finite number of balls B ri with i = 1, · · · , , such that K ⊂ ∪ i B 2ri and B 4ri ⊂ Ω.
Finally, thanks to the comparison principle shown in Lemma 1, we get the following result, whose proof is similar to Lemma 3.4 in [5].
With this comparison result we can argue as in [8] with an iteration procedure, to show the existence of the minimal solution v j ∈ D 1,2 By Lemma 5.3 we can prove that v j ≤ v j+1 and v j ≤ u n for every j. We define We deduce that ω n ≤ ω n+1 , since a n , b n are nondecreasing sequences.
Proof of Theorem 5.1. We perform this proof arguing by contradiction. We assume that there exists x 0 ∈ Ω such that u n (x 0 ) ≤ C for all n. We get a contradiction after several steps.
Step 1. First of all, we prove that |∇T k (u n )| is uniformly bounded (for k fixed) in L 2 loc (Ω; |x| −2γ dx). In fact, if we take T k (u n )φ as test function in (40), being φ ≥ 0 such that φ ∈ C ∞ 0 (K), with K some compact set K ⊂⊂ Ω, integrating by parts we get Here we used Lemma 5.2 at x 0 ∈ supp(φ) = K and the assumption u(x 0 ) ≤ C. Note that the second integral can be estimated Since the last integral is bounded, together with (43) it gives that |∇T k (u n )| is bounded in L 2 loc (Ω; |x| −2γ dx), as we wanted to show.
Step 2. T k (v j ) → T k (ω n ) strongly in D 1,2 loc,γ (Ω). In particular, this ensures that ∇v j → ∇ω n a. e. in Ω. The following identity Thanks to the first step we can conclude that T k (v j ) is uniformly bounded in D 1,2 0,γ (Ω) and T k (v j ) T k (ω n ) weakly in D 1,2 0,γ (Ω). Using the semicontinuity of the norm Moreover, taking into account again that −div(|x| −2γ ∇T k (v j )) ≥ 0, we get that

It yields lim sup
j→∞ Ω and the strong convergence is shown.
Step 3. T k (ω n ) is uniformly bounded (with respect to n) in (D 1,2 γ ) loc (Ω). We take T k (v j )φ as test function in (42), for some nonnegative function φ as in the first step. Let us denote g k (x) = λv j a n (x) + Using Lemma 5.2 for problem (42) and the fact that v j (x 0 ) ≤ u n (x 0 ) ≤ C we deduce that g k is uniformly bounded (with respect to n and j) in L 1 loc (Ω). Proceeding as in (43), we can prove then that |∇T k (v j )| is uniformly bounded (with respect to j and n) in L 2 loc (Ω; |x| −2γ dx). By the strong convergence we conclude that |∇T k (ω n )| is uniformly bounded (with respect to n) in (D 1,2 γ ) loc (Ω).
Step 4. |∇ω n | is uniformly bounded in Lq loc (Ω; |x| −2γ dx), withq < q given in (10). In the previous step we deduced the boundedness in L 1 loc (Ω) of the second member of equation (42). Then, by Lemma 3.1 we obtain that |∇v j | is uniformly bounded in Lq loc (Ω; |x| −2γ dx), so it is |∇w n |, by the a.e. convergence of the gradients.
Thanks to the a.e. convergence of the gradients, we can apply Fatou's lemma to obtain (45) But then, we can take some nonnegative test function, φ ∈ C ∞ 0 (Ω), in (42) and using once more Fatou's Lemma together with the convergence of the gradients obtained in Step 4, to end with the proof of (44).
Step 6. We pass to the limit in (44) and find a contradiction. By the monotonicity of ω n we can conclude that a n (x)ω n ω |x| 2γ in L 1 loc (Ω).
By (45) we have as well K b n (x)|∇ω n | p dist(x, ∂Ω) dx < C uniformly with respect to n for any K ⊂⊂ Ω.
To finish the proof of the previous result we show the following lemma.
But on the other hand, the previous limit is nonnegative by (46), thus it goes to zero concluding the proof.
Proof. We first note that in this case γ > −1. We perform the proof in several steps.
Step 5: Let us consider again the approximation problems in Ω, being a n (x) and b n (x) defined in (41). Note that u n ∈ D 1,2 0,γ (Ω) ∩ L ∞ (Ω). Applying then the comparison principle in Section 3, we have that u n ≤ u n+1 ≤ω for all n. Let us denote byū = lim n→∞ u n ≤ω.
We have shown that the last three integrals in (56) go to zero. Hence summing up and substituting (54) and (55) into (53), we deduce that Taking into account that φ n ≥ 1 2 + φ n , this proves the strong convergence of the gradients. In particular, ∇u n → ∇ū a.e. in Ω.
Step 8:ū is a supersolution to (1). If we take ψ n = (1 + G k (u n )) s − 1 as test function in (52), noting that u n ≤ ω, it is easy to see that By the regularity of ω, the first integral on the righthand side goes to zero as k → ∞. Furthermore, Moreover, since ∇u n → ∇ū as n → ∞, Vitali's Theorem implies that ∇u n → ∇ū, strongly in L p (Ω; |x| −γp dx).
Proof. We note that the existence in the case p ≤ 1 follows almost straightforward if λ ≤ Λ N,γ . In this case we can use u n as test function to get Ω |x| −2γ |∇u n | 2 dx = Ω λu 2 n a n (x)dx + Ω |∇u n | p |x| −pγ u n dx + Ω cf u n dx.
Using Hölder inequality, it follows that We absorb again the first term on the righthand side and apply once more Hölder inequality to the next terms, to find Observe that since f ≤ |x| −2(γ+1) and N ≥ 2(γ + 1), the last integral is bounded. Since p(λ) > 1, then γ > −1 and the second integral on the left hand side is also bounded. Finally we just notice that (see [14]), so we can absorb this term in the left hand side to obtain a uniform bound for u n in D 1,2 0,γ (Ω). We conclude passing to the limit as in previous cases.
Proof. Proceeding as before we verify (57). We use the same Hölder exponents to deal with the second integral on the lefthand side, while for the last term we take To ensure that the second integral on the right hand side is also bounded we have to assume that γ > − N (1−p)+2

2
. Arranging terms as in the previous case we show that u n is uniformly bounded in D 1,2 0,γ (Ω). We obtain the desired solution passing to the limit in (52). Proof. We restrict ourselves to the case Ω = B r (0), since the result for a general domain follows as in Step 4 in the previous result. We consider the approximating problems given in (52), for λ = Λ N,γ and Ω = B r (0). Namely, To ensure the existence of a minimal solution to these problems we just point out that λ 1 (a n ) < Λ N,γ , where λ 1 (a n ) denotes the first eigenvalue for the divergence operator with weight a n (x), and then we can argue as before. Now we construct a supersolution to (1) in B r (0). For certain R > r > 0, consider the radial function It is tedious but simple to show that Since p < N +2(γ+1) N we can find a suitable constant such that cω is a supersolution to (1) in B r (0), with λ = Λ N,γ and ω = 0 on ∂B r (0). Moreover, ω ∈ D 1,q 0,γ (B r (0)), for every q < 2. Applying the comparison principle in Section 3, we deduce that u n ≤ u n+1 ≤ ω. Thus u n converges pointwise to u ≤ ω, u ∈ L q (B r (0); |x| −2γ ), for all q < 2 * γ . Let us denote by H γ (B r (0)) the completion of C ∞ 0 (B r (0)) with respect to the norm H γ (B r (0)) is a Hilbert space and verifies D 1,2 0,γ (Ω) ⊂ H γ (B r (0)) ⊂ D 1,q 0,γ (B r (0)), for all q < 2. Let us show that {u n } is uniformly bounded in H γ (B r (0)). Indeed, using u n as test function in (58) and taking into account that u n ≤ ω, we get that We observe that ω ∈ L 1 (Ω; |x| −2(γ+1) dx), hence the last integral is bounded if f (x) ≤ c|x| −2(γ+1) . If γ ≤ −1 it suffices taking f ∈ L 2/p (Ω) to ensure the boundedness of this term. To estimate the second integral, we apply Hölder, Young and the improved Hardy-Sobolev inequalities (see [1]), to deduce that It is easy to see that the last integral is bounded since p < N +2(γ+1) N . Therefore, taking ε small we have shown that u n Hγ (Br(0)) ≤ C and thus u n u in H γ (B r (0)). Furthermore, Hγ (Br(0)) ≤ u n 2 Hγ (Br(0)) .
This estimate together with (60) give us that u n Hγ (Br(0)) ≤ u Hγ (Br(0)) . Then, (59) implies the strong convergence u n → u in H γ (B r (0)). Now, it is easy to check that u is supersolution to (1) in B r (0), passing to the limit in (58) thanks to the strong convergence, and applying Fatou's Lemma.
Open problem: Note that the question about existence of solutions in the range of parameters 0 < p < 1 and −∞ < γ ≤ − N (1−p)+2 2 remains open.