Hardy inequalities for general elliptic operators with improvements

We obtain optimal generalized versions of Hardy inequalities, which as special cases contain Hardy's inequality and Hardy's inequality involving the distance function to the boundary of $ \Omega$. In addition we obtain neccesary and sufficient conditions to add improvements in the form of non negative potentials.

for all u ∈ H 1 0 (Ω). Moreover the constant 1 4 is optimal and not attained. We will refer to this inequality as Hardy's boundary inequality.
Recently Hardy inequalities involving more general distance functions than the distance to the origin or distance to the boundary have been studied (see [BFT]). Suppose Ω is a domain in R n and M a piecewise smooth surface of co-dimension k, k = 1, ..., n. In case k = n we adopt the convention that M is a point, say, the origin. Set d(x) := dist(x, M ). Suppose k = 2 and −∆d 2−k ≥ 0 in Ω\M then for all u ∈ H 1 0 (Ω\M ). We comment that the above inequalities all have L p analogs. In the last few years improved versions of the above inequalities have been obtained, in the sense that non-negative terms are added to the right hand sides of the inequalities; see [BV], [BM], [BFT], [BMS], [FT], [FHT], [VS]. One common type of improvement for the above Hardy inequalities are the so called potentials; we call 0 ≤ V (x), defined in Ω, a potential for (4) provided Most of the results in this direction are explicit examples of potentials V where, in the best results, V is an infinite series involving complicated inductively defined functions. Very recently Ghoussoub and Moradifam [GM] gave the following necessary and sufficient conditions for a radial function V (x) = v(|x|) to be a potential in the case of Hardy's inequality (4) on a radial domain Ω: V is a potential if and only if there exists a positive function y(r) which solves y ′′ + y ′ r + vy = 0 in (0, sup x∈Ω |x|).
In another direction people have considered Hardy inequalities for operators more general than the Laplacian. One case of this is the results obtained by Adimurthi and A. Sekar [AS]: Suppose Ω is a smooth domain in R n which contains the origin, A(x) = ((a i,j (x))) denotes a symmetric, uniformly positive definite matrix with suitably smooth coefficients and for ξ ∈ R n we define |ξ| 2 A := |ξ| 2 A(x) := A(x)ξ · ξ. Now suppose E is a solution of L A,p (E) := −div |∇E| p−2 A A∇E = δ 0 in Ω with E = 0 on ∂Ω where δ 0 is the Dirac mass at 0. Then for all u ∈ W 1,p 0 (Ω) Improvements of this inequality were also obtained and they posed the following question: Is p−1 p p optimal? We show this is the case, even for a more general inequality.
After completion of this work we noticed that various people had taken a similar approach to generalized Hardy inequalities, see [DL], [KMO], [LW].

Outline and approach
Our approach will be similar to the one taken by Adimurthi and A. Sekar but we mostly concentrate on the quadratic case (p = 2) and for this we define L A (E) := −div(A∇E).
We now motivate our main inequality. Suppose E is a smooth positive function defined in Ω. Let u ∈ C ∞ c (Ω) and set v := E −1 2 u. Then a calculation shows that in Ω and after integrating this over Ω we obtain If we further assume that L A (E) ≥ 0 in Ω then From this we see that the optimal constant C(E) It is possible to show that for all non-zero u ∈ H 1 0 (Ω) we have where v is defined as above. Using this and (7) one sees that if C(E) = 1 4 then C(E) is not attained and hence if C(E) is attained then C(E) > 1 4 . This shows that needs to be singular if we want C(E) = 1 4 . In fact one can show that H 1 0 (Ω) is compactly embedded in L 2 (Ω, ∈ L p (Ω) for some p > n 2 and so one could then apply standard compactness arguments to show that C(E) is attained. We are only interested in the case where C(E) = 1 4 and hence we need to ensure is singular and this can be done in two obvious ways. This naturally leads one to consider the following two classes of functions E (weights).
Definition 1.1. Suppose 0 < E in Ω and L A (E) is a nonnegative nonzero finite measure in Ω denoted by µ. 1) If in addition E ∈ H 1 0 (Ω) then we call E a boundary weight on Ω. 2) If in addition E ∈ C ∞ (Ω\K) where K ⊂ Ω denotes the support of µ, E = ∞ on K and dim box (K) < n − 2 (see below) then we call E an interior weight on Ω.
Given a compact subset K of R n we define the box-counting dimension (entropy dimension) of K by dim box (K) := n − lim rց0 log(H n (K r )) log(r) provided this limit exists and where K r := {x ∈ Ω : dist(x, K) < r} and H α is the αdimensional Hausdorff measure.
Remark 1.1. It is possible to show that C 0,1 c (Ω\K) is dense in W 1,p 0 (Ω) provided K is compact and dim box (K) < n − p (use appropriate Lipschitz cut off functions).
From here on µ will denote the measure L A (E) and in the case where E is an interior weight on Ω, K will denote the support of µ.
We now list the main results. We show that if E is either an interior or a boundary weight in Ω then we have the following inequality: with optimal constant which is not attained.
In the case that E is a boundary weight on Ω we obtain Moreover 1 2 is optimal (once one fixes 1 4 ) and is not attained.
Using the methods developed in [GM] we obtain necessary and sufficient conditions on 0 ≤ V (x) to be a potential for (9) in the case where E is an interior weight. We show that the following are equivalent: 1) For all u ∈ H 1 0 (Ω) 2) There exists some 0 < θ ∈ C 2 (Ω\K) such that If we further assume that E = γ ≥ 0 (constant) on ∂Ω and if we are only interested in potentials of the form In practice this ode classification is more useful because of the shear abundance of ode results in the literature. We obtain weighted versions of (9) (respectively (10)) in the case that E is an interior weight (respectively boundary weight) on Ω which can be viewed as generalized versions of the Cafferelli-Kohn-Nirenberg inequality. To be more precise we obtain: Suppose E is an interior weight on Ω and t = 1 2 . Then for all u ∈ C 0,1 c (Ω\K). Moreover the constant is optimal and not attained in the naturally induced function space. Suppose E is a boundary weight on Ω and 0 = t < 1 2 . Then (13) holds for all u ∈ C ∞ c (Ω) and is not attained in the naturally induced function space. Similarly for all u ∈ C ∞ c (Ω). Moreover the constant on the right is optimal and not attained in the natural function space. In addition we show that the class of potentials for (13) is given by {E 2t V : V is a potential for (9)}.
We also examine generalized Hardy inequalities which are valid for functions u ∈ H 1 (Ω). Suppose E a positive function with L A (E) + E a nonnegative nonzero finite measure denoted by µ, E = ∞ on the K (as before K denotes the support of µ) and where we assume that E satisfies a Neumann boundary condition. Then Moreover these constants are optimal (in the sense that if one is fixed then the other is optimal). Improvements of (15) are also obtained. Assuming the same conditions on E we show that for 0 ≤ V we have if and only if there exists some 0 < θ ∈ C ∞ (Ω\K) such that with A∇θ · ν = 0 on ∂Ω. Weighted version of (15) are established. Assuming the same conditions on E we show that for t = 1 2 we have for all u ∈ C ∞ c (Ω\K). Moreover the constants are optimal and not obtained in the naturally induced function space.
We establish optimal Hardy inequalities which are valid on exterior and annular domains. Suppose Ω is a exterior domain in R n , E > 0 in R n , lim |x|→∞ E = 0, −∆E = µ in R n where µ is a nonzero nonnegative finite measure with compact support K. In addition we assume that dist(K, Ω) > 0 and ∂ ν E ≥ 0 on ∂Ω. We define D 1 (Ω ∪ ∂Ω) to be the completion of C ∞ c (Ω ∪ ∂Ω) with respect to the norm ∇u L 2 (Ω) . Then Moreover the constant is optimal and not attained.
Now suppose Ω = Ω 2 \Ω 1 where Ω 1 ⊂⊂ Ω 2 are both connected and Ω is connected. Suppose 0 < E in Ω 2 and −∆E = µ in Ω 2 where µ is a nonnegative nonzero finite measure compactly supported in Ω 1 . In addition we assume that E = 0 on ∂Ω 2 and ∂ ν E ≤ 0 on ∂Ω 1 . Then (17) is optimal and not attained over Optimal non-quadratic Hardy inequalities are also obtained in both the interior and boundary cases.

Examples
We now look at various examples of Hardy inequalities (and applications of) which can be obtained after making suitable choices of weights E and matrices A. In most of the examples we will take A to be the identity matrix.
Since Ω is convex one can show δ is concave and hence −∆δ ≥ 0 in Ω. Putting E into (10) gives an improved version of (5).

Hardy's boundary inequality in the unit ball:
Let B denote the unit ball in R n and set E(x) := 1 − |x|. Putting E into (10) gives

Intermediate case:
Set E(x) := d(x) 2−k where d and k are as in (6). Since −∆E ≥ 0 we obtain (6) after subbing E into (9). 6. Hardy's boundary inequality in the half space: Let R n + denote the half space and set E(x) := dist(x, R n + ) = x n . Then putting E into (9) gives Maz'ja (see [M]) obtained the following improvement One might ask whether we can take a more symmetrical potential in the improvement, say something like V (x) = f (x n ) where f is strictly positive. Using our ode classification of potentials we will see that this is not possible.
7. Hardy's inequality valid for u ∈ H 1 (Ω): Let B denote the unit ball in R 3 and set E(x) := |x| −1 e |x| . Then a computation shows that and where ∂ ν E = 0 on ∂B. Here δ 0 is the Dirac mass at 0. Putting E into (15) we see that Also the constants are optimal (in the sense mentioned in (15)) and are not attained.

Hardy's inequality in a annular domain:
Assume that 0 ∈ Ω 1 ⊂⊂ B R ⊂ R 2 where Ω 1 is connected and B R is the open ball centered at 0 with radius R. In addition we assume that x · ν(x) ≥ 0 on ∂Ω 1 where ν is the outward pointing normal. Define Ω := B R \Ω 1 , which we assume is connected, and set E(x) := − log(R −1 |x|). Then by the above mentioned results on annular domains one has for all u ∈ H 1 0 (Ω ∪ Ω 1 ). Moreover the constant is optimal and not attained. 10. Suppose E > 0 in Ω, let f : (0, ∞) → (0, ∞) and setẼ := f (E). PuttingẼ into (7) for E gives An important example will be when f (E) := E t where 0 < t < 1; in fact we will use E(x) := δ(x) t (δ(x) := dist(x, ∂Ω)) to show that if one drops the requirement that µ is a finite measure (and just assumes µ a locally finite measure) (9) need not be optimal.
11. Eigenvalue bound: Let Ω be a bounded subset of R n and E > 0, L A (E) ≥ 0 in Ω with |∇E| 2 A = 1 a.e. in Ω. Let λ A (Ω) denote the first eigenvalue of L A in H 1 0 (Ω). Then λ A (Ω) E 2 L ∞ ≥ π 2 4 . To show this one puts f (z) := sin 2 ( πz 2 E L ∞ ) into the above result and drops the term involving the measure.
12. Suppose E is an interior weight on Ω with E = 1 on ∂Ω. Then by using the above result with f (E) := (log(E)) 1 2 one obtains the inequality .

Trace theorem:
Let Ω denote a domain in R n where n ≥ 3 and such that B ⊂⊂ Ω (here B is the unit ball). Define A computation shows that −∆E = cσ where c > 0 and where σ is the surface measure associated with ∂B. Putting this E into (10) and dropping a couple of terms gives 17. Regularity: Suppose E ∈ L ∞ loc (Ω) is a positive solution to L A (E) = µ in Ω where µ is locally finite measure. Then using (9) we see that E ∈ H 1 loc (Ω). 18. Baouendi-Grushin operator: Here we mention that various operators can be put into the form we are interested in. Suppose Ω is an open subset of R N = R n × R k and ξ ∈ Ω is written ξ = (x, y) using the above decomposition of R N . For γ > 0 one defines the vector field ∇ γ := (∇ x , |x| γ ∇ y ) and the Baouendi-Grushin operator where I n , I k are the identity matrices of size n and k. Then |∇ γ E| 2 = |∇E| 2 A and −div(A∇E) = L A (E).

Main Results
Throughout this article we shall assume that Ω is a bounded connected domain in R n (unless otherwise mentioned) with smooth boundary and A(x) = ((a i,j (x))) is a n × n symmetric, uniformly positive definite matrix with a i,j ∈ C ∞ (Ω) and for ξ ∈ R n we define |ξ| 2 If E is an interior weight or a boundary weight on Ω we have, by the strong maximum principle (see [V]), E bounded away from zero on compact subsets of Ω.
The following theorem gives the main inequalities. In addition we consider a slight generalization of the case where E is a boundary weight on Ω.

Theorem 2.1. (i) Suppose E is either an interior or a boundary weight on Ω.
Then for all u ∈ H 1 0 (Ω). Moreover 1 4 is optimal and not attained.
Remark 2.1. One can consider more general functions E. Most of the results (including the above one) concerning interior weights on Ω can be generalized to the case where L A (E) = µ + h, here µ is again a nonnegative nonzero finite measure and h is a suitably smooth non-negative function.
Proof. (i) Since E is smooth away from K and noting the supports of both u and v the integration by parts used in obtaining (7) is valid.
(ii) Now suppose E in a boundary weight. Extend E to all of R n by setting E = 0 outside of Ω and let in Ω and uF ε ⇀ uµ in H −1 (Ω). Using these results along with Fatou's lemma allows us to pass to the limit.
Remark 2.2. When we prove our various Hardy inequalities, which all stem from (7) we will generally drop the term To show the given inequality does not attain we will generally just not drop this term. This term is positive for non-zero u provided u is not a multiple of √ E. Since √ E / ∈ H 1 0 (Ω) this will not be an issue. In theorem 2.2 this will be a concern.
As usual we will need an ample supply of test functions for best constant calculations. The next lemma provides this. When E is an interior weight we let g denote a solution to L A (g) = 0 in Ω with g = E on ∂Ω.
Then v t,τ ∈ H 1 0 (Ω) for 0 < t < 1 2 and τ > 1 2 . Moreover for each 0 < t < 1 2 we have J t (τ ) → ∞ as τ ց 1 2 . Proof. We prove the results up to some unjustified integration by parts; which can be justified by regularizing the measure, integrating by parts and passing to limits. (i), (ii) Fix 0 < t < 1 2 and then note that |∇u t | 2 ≤ CE 2t−2 |∇E| 2 A + Cg 2t−2 |∇g| 2 A where C is some uniform constant. The term involving g is harmless. Now multiply L A (E) = µ by E 2t−1 and integrate over Ω to obtain We also see that lim tր 1 2 I(t) = ∞. (iii) Take 0 < t < 1 2 , τ > 1 2 and v t,τ defined as above. One easily sees that v t,τ is continuous near ∂Ω and vanishes on ∂Ω. So to show v t,τ ∈ H 1 0 (Ω) it is sufficient to show These functions are only singular near K and ∂Ω. Now set W τ := E 2t−2 |∇E| 2 log 2τ −2 (γ −1 E) and so w 2 = W τ and w 1 = W τ +1 . Now suppose t ′ ∈ (t, 1 2 ) and so where K ε is a small neighborhood of K. Now note that w 2 is better behaved than w 1 near K and so we also have w 2 ∈ L 1 (K ε ).
Proof of theorem 2.1: (i) Using lemma 2.1 and, in the case where E is a interior weight on Ω, the fact that C 0,1 c (Ω\K) is dense in H 1 0 (Ω) we obtain (19). We now show the constant is optimal. Suppose E is an interior weight on Ω and define where the constants C k possibly depend on ε. From this we see that lim tր 1 2 Q t,ε = 1 4 after recalling Q t,ε ≥ 1 4 . Now fix ε > 0 and let u ∈ C ∞ c (Ω) be non-zero. Then a simple computation shows which, when combined with the above facts, gives the desired best constant result. To see 1 4 is not attained use (21). Now suppose E is a boundary weight on Ω, ε > 0 and t > 1 2 . Define f ε (z) := z 2t−1 − ε 2t−1 for z > ε and 0 otherwise. Using f ε (E) ∈ H 1 0 (Ω) as a test function in the pde associated with E one obtains, after sending ε ց 0, which shows that E t ∈ H 1 0 (Ω) for 1 2 < t ≤ 1. To see 1 4 is optimal in (19) use E t (as t ց 1 2 ) as a minimizing sequence.
(ii) Suppose E is a boundary weight on Ω. Let 1 2 < t < 1 and so E t ∈ H 1 0 (Ω). Using (23) we have which shows that 1 2 is optimal. (iii) Suppose E is as in the hypothesis. The only issue is whether 1 4 is optimal. Without loss of generality assume that 0 ∈ ∂Ω and B(0, 2R) ∩ ∂Ω ⊂ Γ. Suppose 0 < r < R and define where Ω(r) := B(0, r) ∩ Ω. Define u t := E t φ which can be shown to be an element of H 1 0 (Ω) for t > 1 2 . One uses u t as t ց 1 2 as a minimizing sequence along with arguments similar to the above to show 1 4 is optimal. 2 The following example shows that if we just assume that 0 < E ∈ H 1 0 (Ω) with L A (E) a locally finite measure then (19) need not be optimal.
Example 2.1. Take Ω a bounded convex domain in R n and set δ(x) := dist(x, ∂Ω). Fix 1 2 < t < 1 and set and so putting E into (19) gives . This shows that (19) was not optimal. This apparent failure of theorem 2.1 is due to the fact µ not a finite measure; use the co-area formula to show δ t−2 / ∈ L 1 (Ω).
Fix t ≥ 1 and put E 2 := E t . Then we have and so we see that ( t 2 , t 2 − t 2 4 ) ∈ C for all t ≥ 1. From this we see that the curve α = β − β 2 for β ≥ 1 2 is contained in C. It is straightforward to see the remaining portion of ∂C ′ is contained in C. To see the inequality does not attain when (β, α) ∈ ∂C\Γ use the fact that (9) does not attain in H 1 0 and the fact that µ ≥ 0. To see the inequality does attain on the remaining portion of ∂C note that (25) attains at u := E t 2 ∈ H 1 0 (Ω) for t > 1.
We now give a result relating to the first eigenvalue of L A on subdomains of Ω. Suppose (E, λ A (Ω)) is the first eigenpair (with E > 0) of L A on H 1 0 (Ω) and for B ⊂ Ω we let λ A (B) denote the first eigenvalue of L A on H 1 0 (B). Corollary 2.1. Let E be as above. For B ⊂ Ω we set Proof. Let B ⊂ Ω and let u ∈ C ∞ c (B) with B u 2 = 1. Using (25) gives for 0 < t < 2. If t > 2 then we get the same expression but with the infimum replaced with supremum. Now take the infimum over u and in case (i) set t := 1 + λA(Ω) α(B) < 2 and in case (ii) set t := 1 + λA(Ω) α(B) > 2 to see the result.

Weighted versions
We now examine weighted versions of the above inequalities which, as mentioned earlier, can be seen as analogs of Cafferelli-Kohn-Nirenberg inequalities. We now introduce the spaces we work in.
Definition 2.1. For t ∈ R we define u 2 t := Ω E 2t |∇u| 2 A dx. Suppose E is an interior weight on Ω. We define X t to be the completion of C 0,1 c (Ω\K) with respect to · t . In the case that E is a boundary weight on Ω we define X t to be the completion of C 0,1 c (Ω) with respect to the same norm.
Remark 2.3. One should note that if E is an interior weight on Ω and t > 1 2 then X t does not contain C ∞ c (Ω). To see this use (26) to see that if C ∞ c (Ω) ⊂ X t then E t ∈ H 1 loc (Ω) which we know to be false. For t < 1 2 we do have C ∞ c (Ω) ⊂ X t . Theorem 2.3. Suppose t = 1 2 and E an interior weight on Ω. Then for all u ∈ X t . Moreover the constant is optimal and not attained.
Proof. Let t = 0, 1 2 , u ∈ C 0,1 c (Ω\K) and define w : and re-group to obtain (26). We now show the constant is optimal. Let v m ∈ C 0,1 c (Ω\K) be such that and since D m → 1 4 we see that (t − 1 2 ) 2 is optimal. For the case γ := min ∂Ω E > 0 we can show the constant is not obtained by using later results on improvements. If γ = 0 we then sub w into (7) instead of (19) and hold onto the extra term to see the optimal constant is not attained.
Theorem 2.4. (i) Suppose 0 = t < 1 2 and E is a boundary weight on Ω. Then for all u ∈ X t . Moreover the constant is optimal and not attained.
(ii) Suppose 0 = t < 1 2 and E is a boundary weight on Ω. Then for all u ∈ X t . Moreover the constant on the right is optimal and not attained.
Proof. We first prove (28) for u ∈ C 0,1 c (Ω) which then gives us (27) for the same class of u's. Suppose 0 = t < 1 2 and E is a boundary weight on Ω. We now use the notation introduced in the proof of lemma 2.1; namely E ε is the standard mollification of E and F ε := L A (E ε ). Recall that for any u ∈ C 0,1 c (Ω) we have uF ε → uµ in H −1 (Ω) and that we have for all v ∈ H 1 0 (Ω). Now let u ∈ C 0,1 c (Ω) and set v := E t ε u ∈ C 0,1 c (Ω). Putting v into the above gives and using similar ideas from the proof of lemma 2.1 one can show that So using these results, sending ε ց 0 in (29) and after an application of Fatou's lemma we arrive at (28) for u ∈ C 0,1 c (Ω). Now we show the constants are optimal. Recalling the proof of theorem 2.1 there exists v m ∈ C ∞ c (Ω) such that Define u m := E −t v m which one easily sees is an element of X t . Then Using E ε , F ε as defined above one can show, using similar methods, that So from this we see that we see that (27) is optimal. Similarly one sees using (30) that Ψ m = F m − t and hence (28) is optimal.
To show the constants are not obtained we as usual hold on to the extra term that we dropped in the above calculations.
Since Ω E −1 |∇E| 2 A dx = ∞ one can show this extra term is positive for u ∈ X t \{0}. (iii) Now take t > 1 2 and E a boundary weight on Ω. For ε, τ > 0 but small define Then u ε,τ ∈ X t . Now use the sequence u m where u m := u εm,τm to see desired result where ε m := m −m and τ m := m −1 .

More general weighted inequalities
We now investigate the possibility of inequalities of the form Theorem 2.5. Suppose E is an interior weight on Ω with γ := min ∂Ω E and 0 < f ∈ C ∞ (γ, ∞). Then for all u ∈ C 0,1 c (Ω\K). In addition this is optimal (in the sense that the optimal constant is 1) if lim inf z→∞ f ′′ (z) > 0 or if lim z→∞ Proof. Let u ∈ C 0,1 c (Ω\K) and define w := f (E)u ∈ C 0,1 c (Ω\K). Putting w into (19), integrating by parts and re-grouping gives (31). Let v m ∈ C 0,1 c (Ω\K) be such that Without loss of generality we can assume the supports of v m concentrate on K. Define u m := vm f (E) ∈ C 0,1 c (Ω\K). Then a computation shows that Now suppose lim inf z→∞ f ′′ (z) > 0. Then using the monotonicity of x → α+x β+x , where α and β are positive constants, shows Q m → 1. Now suppose lim z→∞ z 2 f ′′ (z) f (z) = 0. Using this and the fact that the v m 's support concentrates on K one easily sees that Using this one sees that Q m → 1.

Improvements
We now investigate the possibility of improving (19) in the sense of potentials. The method we employ was first used by Ghoussoub and Moradifam (see [GM]). We now define precisely what we mean by a potential. Suppose E is an interior weight on Ω and 0 ≤ V ∈ C ∞ (Ω\K) (recall K is the support of µ). We say V is a potential for E provided for all u ∈ H 1 0 (Ω). We analogously define a potential V for the case that E is a boundary weight on Ω except we restrict our attention to 0 ≤ V ∈ C ∞ (Ω). The next theorem gives necessary and sufficient conditions for V to be a potential of E in terms of solvability of a singular linear equation. For the necessary direction we will need to assume some conditions on Ω. (B1) Suppose E is an interior weight on Ω. We assume that that there exists a sequence (Ω m ) m of non-empty subdomains of Ω which are connected, have a smooth boundary, Ω m ⊂⊂ Ω\K, Ω m ⊂⊂ Ω m+1 and Ω\K = ∪ m Ω m . (B2) Suppose E is a boundary weight on Ω. We assume that there exists a sequence (Ω m ) m of non-empty subdomains of Ω which are connected, have a smooth boundary, Ω m ⊂⊂ Ω m+1 and Ω = ∪ m Ω m .
(i) Suppose there exists some 0 < φ ∈ C 2 (Ω\K) such that Then V is a potential for E. After the change of variables θ := E 1 2 φ one sees that it is sufficient to find a 0 < θ ∈ C 2 (Ω\K) such that (ii) Suppose V is a potential for E and Ω satisfies (B1). Then there exists some 0 < θ ∈ C ∞ (Ω\K) which satisfies (34).
It is important to note that the above theorem can be used (in theory) for best constant calculations; without the need for constructing appropriate minimizing sequences. To see this suppose 0 ≤ V is a potential for the interior weight E and let C(V ) > 0 denote the associated best constant, ie Then one sees that After theorem 2.9, which is analogous result to the above theorem but phrased in terms of solvability of a linear ode, this remark on best constants will be of more importance because of the shear magnitude of results concerning solvability of ode's.
(ii) Now suppose V ∈ C ∞ (Ω\K) is a potential for E and (Ω m ) m is the sequence of domains from assumption (B1). Define the elliptic operator P by Using a standard constrained minimization argument along with the strong maximum principle there exists some 0 < θ m ∈ H 1 0 (Ω m ) such that where 0 ≤ λ m , ie. (θ m , λ m ) is the first eigenpair of P in H 1 0 (Ω m ). Since H 1 0 (Ω m ) ⊂ H 1 0 (Ω m+1 ) we see that λ m is decreasing and hence there exists some 0 ≤ λ such that λ m ց λ. Let x 0 ∈ ∩ m Ω m and suitably scale θ m such that θ m (x 0 ) = 1 for all m. Now fix k and let m > k + 1. Then and we now apply Harnacks inequality to the operator P − λ m to see there exists some C k such that So we see that (θ m ) is bounded in L ∞ loc (Ω\K). Now applying elliptic regularity theory and a bootstrap argument one sees that (θ m ) m>k+1 is bounded in C 1,α (Ω k ) for α < 1 and after applying a diagonal argument one sees that there exists some non-zero 0 ≤ θ ∈ C 1,α (Ω\K) such that θ m → θ in C 1,α (Ω k ) for all k. Using this convergence one can pass to the limit in (38) to see that P (θ) = λθ in Ω\K and after applying the strong maximum principle on Ω m one sees that θ > 0 in Ω\K. Now applying regularity theory one sees that θ ∈ C ∞ (Ω\K).
2 Proof of theorem 2.7. (i) The proof is essentially unchanged from the proof of theorem 2.6. (ii) Again the proof is the same as in theorem 2.6 except now the measure µ does not drop out. 2 The next theorem gives some explicit examples of potentials.
Proof. (i) Let E be an interior weight on Ω, γ := min ∂Ω E > 0 and suppose 0 < f ∈ C 2 ((γ, ∞)). Put φ := f (E) into (33) to obtain the result. Now take f (E) := log(γ −1 E) to obtain (39) for all u ∈ C 0,1 c (Ω\K) and extend to all of H 1 0 (Ω) by density and by Fatou's lemma. We now show 1 4 is optimal. Fix 0 < t < 1 2 and for τ > 1 2 define u τ := E t log τ (γ −1 E). By lemma 2.2 u τ ∈ H 1 0 (Ω). A computation shows that where J t (τ ) is defined in lemma 2.2. Sending τ ց 1 2 and using results from lemma 2.2 we see 1 4 is optimal. (ii) Suppose E ∈ L ∞ (Ω) is a boundary weight on Ω. Here we use the notation from the proof of lemma 2.1; E ε := η ε * E, F ε := L A (E ε ). Let 0 < f ∈ C 2 ((0, E L ∞ ]). Then starting at (22) for E ε and decomposing v as usual one arrives at for all u ∈ C ∞ c (Ω) after using methods similar to the proof of (i). Now take f (z) := − log( z e E L ∞ ) and let u ∈ C ∞ c (Ω). Then one has where Using methods similar to ones used in the proof of lemma 2.1 one easily sees that lim εց0 I ε ≥ 0. Using this and standard results on convolutions and Fatou's lemma we obtain the desired inequality for u ∈ C ∞ c (Ω) and we then extend to all of H 1 0 (Ω).
We now obtain a more useful (than (34)) necessary and sufficient condition for V to be a potential for E; at least in the case where E is an interior weight on Ω and E = γ ≥ 0 on ∂Ω. As in theorem 2.6 we assume some geometrical properties of Ω.
(i) ⇒ (ii). The proof will be similar to theorem 2.6 (ii). Let γ < t m ր ∞ and define Ω m := {x ∈ Ω : γ + 1 tm < E(x) < t m }. By hypothesis we can take Ω m to be connected and non-empty for each m. Now define one sees that λ m is decreasing and from (43) one sees that λ m ≥ 1 and hence there exists some λ ≥ 1 such that λ m ց λ. By suitably scaling φ m as before and after an application of Harnacks inequality we can assume that φ m → φ in C 1,α loc (Ω\K) where φ ≥ 0 is nonzero and constant on level sets of E. Passing to the limit shows that and a strong maximum principle argument shows that φ > 0 in Ω\K. Since φ constant on level sets of E we have φ = h(E) for some 0 < h in (γ, ∞) and since φ smooth on Ω\K we see that h is smooth on (γ, ∞). Writing the equation for φ in terms of h gives and using Hopfs lemma we can cancel the gradients.
Using the vast knowledge of ode's one can use the above theorem to obtain various results concerning potentials of the form V (x) = |∇E| 2 A f (E). We don't exploit this fact other than to look at one result.
Corollary 2.2. Suppose E is an interior potential on Ω and E = 0 on ∂Ω. Then there no 0 < f ∈ C(0, ∞) such that Proof. Suppose there is such a function f . Using the proof of theorem 2.9 one sees that there is some 0 < h ∈ C 2 (0, ∞) such that in (0, ∞) and y(t) > 0. But oscillation theory from ordinary differential equations shows this is impossible.
Other than some regularity issues this ode approach extends immediately to the case where E is a boundary weight in Ω. Using this corollary (but in the boundary case) one can show the result mentioned in the examples section regarding improvements of Hardy's boundary inequality in the half space; the regularity is not an issue in this example since δ(x) := dist(x, ∂R n + ) = x n is smooth. We now present a result obtained by Avkhadiev and Wirths (see [AW]). Given a domain Ω in R n we say it has finite inradius if δ(x) := dist(x, ∂Ω) is bounded in Ω. We let λ 0 (Lambs constant) denote the first positive zero of J 0 (t) − 2tJ 1 (t) where J n is the Bessel function of order n. Numerically one sees that λ 0 = 0.940.... Now for their result.
Theorem 2.10. (Avkhadiev, Wirths) Suppose Ω is a convex domain in R n with finite inradius. Then This extends a result of H. Brezis and M. Marcus (see [BM]) which said that if Ω is a convex subset of R n then where diam(Ω) denotes the diameter of Ω. Note that there are unbounded convex domains with infinite diameter but finite inradius; for example take a cylinder. We establish a generalized version of this result. Suppose µ is a nonnegative nonzero locally finite measure in Ω (possibly unbounded) and 0 < E ∈ L ∞ (Ω) is a solution to a.e. in Ω E = 0 on ∂Ω.
We then have the following theorem.
Theorem 2.11. Suppose E is as above. Then for all u ∈ C ∞ c (Ω). Proof. Let E be as above. Now extend E to all of R n by setting E = 0 on R n \Ω, let E ε denote the ε mollification of E and F ε := L A (E ε ). Returning to the proof of theorem 2.8 (ii) we have for u ∈ C ∞ c (Ω) and 0 < f ∈ C 2 ((0, E L ∞ ]). Now set λ := Ω) and subbing this f into the above gives after noting that E ε L ∞ ≤ E L ∞ and where I ε := Ω l(E ε )u 2 F ε dx. It is possible to show that l ∈ C ∞ ((0, E L ∞ ]). A standard argument shows that l(E ε )u → l(E)u in H 1 0 (Ω) and uF ε dx ⇀ uµ in H −1 (Ω) and hence one can conclude that lim inf εց0 I ε ≥ 0. Passing to the limit (as ε ց 0) in the remaining integrals gives the desired result.
We now look at improvements of the weighted generalized Hardy inequalities. The next theorem allows us to transfer our knowledge of improvements from the non-weighted case to the weighted case, at least in the case that E is an interior weight.
Theorem 2.12. (Weighted interior improvements) Suppose E is an interior weight on Ω and 0 ≤ V ∈ C ∞ (Ω\K). Then the following are equivalent: (ii) For all t = 1 2 and u ∈ X t Using similar arguments one can obtain a version of theorem 2.12 for the case when E is a boundary weight on Ω; we omit the details since the results is not as clean.

Hardy inequalities valid for u ∈ H 1 (Ω)
Let K be a compact subset of Ω with dim box (K) < n − 2. Standard arguments show that C 0,1 c (Ω\K) is dense in H 1 (Ω).
Definition 2.2. We say E is a Neumann interior weight on Ω provided: there exists some compact K ⊂ Ω, dim box (K) < n − 2, E ∈ C ∞ (Ω\K), inf Ω E > 0, L A (E) + E is a nonnegative nonzero measure µ whose support is K, E = ∞ on K and A∇E · ν = 0 on ∂Ω where ν(x) denotes the outward normal vector at x ∈ ∂Ω.
Theorem 2.13. Suppose E is a Neumann interior weight on Ω. Then holds for all u ∈ H 1 (Ω). Moreover 1 4 and 1 2 are optimal in the sense that if one fixes 1 4 then you can do no better than 1 2 and vice-versa. Also the inequality is not attained. One can again view the best constants in a different manner which is analogous to theorem 2.2; we omit the details.
Proof. Let E be a Neumann interior weight on Ω. (i) Let u ∈ C 0,1 c (Ω\K) and define v := E −1 2 u. Then A 4E 2 u 2 + v∇v · A∇E, and integrating this over Ω gives (ii) Using (i) and the fact that C 0,1 c (Ω\K) is dense in H 1 (Ω) one obtains (49) for all u ∈ H 1 (Ω). We now show the constants are optimal. We first show that E t ∈ H 1 (Ω) for 0 < t < 1 2 . As in the proof of lemma 2.2 the following calculations are only formal but they can be justified as hinted at there; by first regularizing the measure, obtaining approximate solutions and passing to the limit. Fix 0 < t < 1 2 and multiply L A (E) + E = µ by E 2t−1 and integrate over Ω using integration by parts and the fact that E = ∞ on K along with the boundary conditions of E to see that which shows that E t ∈ H 1 (Ω) for 0 < t < 1 2 . To show the constants are optimal we will use as a minimizing sequence E t as t ր 1 2 . A computation shows and we see that 1 4 is optimal. One similarly shows 1 2 is optimal. To show the inequality does not attain we, as usual, just hold on to the extra term that we dropped in the above calculations. This term is positive for non-zero u ∈ H 1 (Ω) provided E 1 2 / ∈ H 1 (Ω) which is the case after one considers (51).
We now examine weighted versions of (49). Suppose E is a Neumann interior weight on Ω and as usual we let K denote the support of µ. For t = 1 2 and u ∈ C 0,1 c (Ω\K) we define and we let Y t denote the completion of C 0,1 c (Ω\K) with respect to this norm. We then have the following theorem.
Theorem 2.14. Suppose E is a Neumann interior weight on Ω and t = 1 2 . Then for all u ∈ Y t . Moreover the constants are optimal and not attained.
Note in particular that for t > 1 2 one only has a gradient term on the left hand side and so we can conclude that C ∞ (Ω) is not contained in Y t for t > 1 2 .
Proof. Suppose E is a Neumann interior weight on Ω, t = 1 2 and let u ∈ C 0,1 c (Ω\K). Putting E t u into (48) gives To show the constants are optimal one takes the same approach as in theorem 2.3. We now show the optimal constants are not obtained. Suppose we have equality for some nonzero u ∈ Y t . Then it is easily seen that √ E ∈ H 1 (Ω) which we know is not the case.
We now examine improvements of (49).
Note that one can go from (52) to (53) by using the change of variables θ = φE 1 2 in the case that A∇φ · ν = 0 on ∂Ω.
Proof. The proof is similar to the proof of theorem 2.6.
Remark 2.5. One can obtain an analogous version of theorem 2.12 for the case where E is an interior weight on Ω satisfying a Neumann boundary condition.

H 1 (Ω) inequalities for exterior and annular domains
In this section we obtain optimal Hardy inequalities which are valid on exterior and annular domains. Moreover these inequalities will be valid for functions u which are nonzero on various portions of the boundary. For simplicity we only consider the case where A(x) is the identity matrix and hence L A = −∆; the results immediately generalize to the case where A(x) is not the identity matrix. We first examine the exterior domain case.

Condition (Ext.):
We suppose that E > 0 in R n , −∆E is a nonnegative nonzero finite measure (which we denote by µ) with compact support K and we let Ω denote a connected exterior domain in R n with dist(K, Ω) > 0. In addition we assume that the compliment of Ω denoted by Ω c is connected, lim |x|→∞ E = 0 and ∂ ν E ≥ 0 on ∂Ω.
We will work in the following function space. Let D 1 (Ω ∪ ∂Ω) denote the completion of C ∞ c (Ω ∪ ∂Ω) with respect to the norm ∇u L 2 (Ω) . Note we don't require u to be zero on the boundary of ∂Ω. We then have the following theorem.
Moreover the constant is optimal and not attained.
Using a Newtonian potential argument one can show that as R → ∞ the surface integral over the ball B R goes to zero. So using this one sees that and so Ω |∇E t | 2 dx < ∞. With this along with a standard cut-off function argument one sees that E t ∈ D 1 (Ω ∪ ∂Ω). Now one uses E t as t ց 1 2 as a minimizing sequence to show that 1 4 is optimal. We now show the constant is not attained. Now assume that x 0 ∈ ∂Ω is such that E(x 0 ) = min ∂Ω E. Then by Hopf's lemma ∂ ν E(x 0 ) > 0 and so using this along with continuity and (57) one sees that E 1 2 / ∈ D 1 (Ω ∪ ∂Ω). Now to finish the proof it will be sufficient to show that Ω E|∇v| 2 dx > 0 for all nonzero u ∈ D 1 (Ω ∪ ∂Ω). The only nonzero u's for which this integral is zero are multiples of E 1 2 which are not in D 1 (Ω ∪ ∂Ω).

Example 2.2.
Take Ω a exterior domain in R n where n ≥ 3, 0 / ∈ Ω, and such that ν(x) · x ≤ 0 on ∂Ω where ν(x) is the outward pointing normal. Define E(x) := |x| 2−n and use theorem 2.16 to see that Ω |∇u| 2 dx ≥ n − 2 2 2 Ω u 2 |x| 2 dx, for all u ∈ D 1 (Ω ∪ ∂Ω). Moreover the constant is optimal and not attained. In fact using (ii) from the same theorem shows we can add the following nonnegative term to the right hand side of (58): We now examine the annular domain case.

Condition (Annul.):
We assume that Ω 1 ⊂⊂ Ω 2 are two bounded connected domains in R n with smooth boundaries and Ω := Ω 2 \Ω 1 is connected. In addition we assume that E > 0 in Ω 2 with −∆E = µ in Ω 2 where µ is a nonnegative nonzero finite measure supported on K ⊂ Ω 1 . We also assume that ∂ ν E ≤ 0 on ∂Ω 1 . We then have the following theorem.
Theorem 2.17. Suppose Ω, K, E are as in condition (Annul.). Then (i) For all u ∈ H 1 (Ω) with u = 0 on ∂Ω 2 we have Moreover the constant is optimal and not attained if we assume that E = 0 on ∂Ω 2 .
Remark 2.6. These inequalities have analogous weighted versions and using the methods developed earlier one easily obtains results concerning improvements. We leave this for the reader to develop.

The non-quadratic case
For 1 < p ≤ n we define L A,p (E) := −div(|∇E| p−2 A A∇E). As mentioned earlier Adimurthi and Sekar [AS] obtained generalized Hardy inequalities of the form where u ∈ W 1,p 0 (Ω). There approach (as their title suggests) was to look at functions E which solve where 0 ∈ Ω and where δ 0 is again the Dirac mass at 0. They posed the question (see [AS]) as to whether ( p−1 p ) p is optimal in (61)? The next theorem shows this is the case (at least for 1 < p < n); infact we show the result for a more general case.

Interior case
Suppose µ is a nonnegative nonzero finite measure supported on K ⊂ Ω, dim box (K) < n − p (and hence C 0,1 c (Ω\K) is dense in W 1,p 0 (Ω)) and 0 < E is a solution of L A,p (E) = µ in Ω.
By regularity theory (see [D], [T]) there is some 0 < σ < 1 such that E ∈ C 1,σ (Ω\K) and by the maximum principle (see [V]) E > 0 in Ω\K. Now if we assume that µ = δ 0 , as was the case in the question posed in [AS], then one can show E(0) = ∞.
Theorem 2.18. Suppose E is as above but we don't assume that E = ∞ on K.
(i) Then for all u ∈ W 1,p 0 (Ω). (ii) Suppose E = ∞ on K and E = γ on ∂Ω where γ is a non-negative constant. Then the constant in (63) is optimal.