HARDY S INEQUALITY AND CURVATURE

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1 HARDY S INEQUALITY AND CURVATURE A. BALINSKY, W.D. EVANS, AND R.T. LEWIS Abstract. A Hardy inequality of the form ( ) 1 f(x) dx {1 + a(δ, )(x)} f(x) δ(x) dx, for all f C0 ( \ R()), is considered for (1, ), where is a domain in R n, n, R() is the ridge of, and δ(x) is the distance from x to the boundary. The main emhasis is on determining the deendance of a(δ, ) on the geometric roerties of. A Hardy inequality is also established for any doubly connected domain in R in terms of a uniformization of, that is, any conformal univalent ma of onto an annulus. 1. Introduction This aer is a contribution to the much studied Hardy inequality f(x) f(x) dx c(n,, ) δ(x) dx, f C 0 (), (1.1) where is a domain (an oen connected set) in R n, n, 1 < <, and δ is the distance function δ(x) := dist(x, R n \ ) = inf x y, x. y R n \ In order to ut the roblems we address in context and to summarise our main results, we recall some of the highlights amongst the known results to be found in the literature. For a convex domain in R n, n, the otimal constant in (1.1) is ( ) 1 c(n,, ) = ; (1.) see [1] and []. In all cases equality is only achieved by f = 0. In the case = the inequality was imroved by Brézis and Marcus in [6] to one of the form f(x) dx 1 f(x) dx + λ() f(x) dx, (1.3) 4 δ(x) Date: May 1, Mathematics Subject Classification. Primary 35J85; Secondary 35R45, 49J40. Key words and hrases. Hardy inequality, Distance function, Curvature, ridge, skeleton, uniformization. 1

2 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS where 1 λ() 4diam(). Further imrovements along these lines were made in [15], [10], [3], including ones for (1, ) in [10] and [3], and for Hardy-Sobolev inequalities in [1]. Further ertinent references may be found in these cited aers. For non-convex domains, a shar constant in (1.1) is not known in general, but some shar results were obtained in [13], [8] and [3]. For a lanar simly connected domain, Ancona in [1] roved the celebrated result that c(,, ) (1.4) By assuming certain quantifiable degrees of convexity on a simly connected, lanar domain, Latev and Sobolev in [18] strengthened the Kobe one-quarter theorem used by Ancona in his roof and imroved the lower bound in (1.4). Other results of articular relevance to the resent aer are those in [3] for annular regions. Our objective was to consider inequalities of the form f(x) dx c(n,, ) {1 + a(δ, )(x)} f(x) dx (1.5) δ(x) in which the function a(δ, ) deends on δ and geometric roerties of the boundary of. We are articularly interested in domains which are either convex or have convex comlements. In these cases, we determine a(δ, ) exlicitly in terms of δ and the rincial curvatures of the boundary of. Our analysis makes it necessary to consider the skeleton S() and ridge R() of : these will be defined in. A samle result is the following secial case of our Corollary where is a convex domain with C boundary, = n =, and a condition on the regularity of the ridge of holds (see 3.4): } f(x) f(x) dx 1 4 { 1 + κδ 1 + κδ (x) dx, (1.6) δ(x) for f C0 (), where κ is the curvature of. For = B R, the oen disc of radius R and center the origin, the condition (3.4) holds, and the inversion y = x/ x, with ρ = 1/R, yields R \B ρ f(y) dy 1 4 R \B ρ { 1 y + 1 ( y ρ) } f(y) dy. (1.7) This inequality is given in [3], Remark 1. A significant feature of (1.6) with resect to (1.7) is that the resence of the alien term 1/ y in (1.7) is exlained by the curvature of the boundary. Other results in [3] are recovered from theorems in Section below by taking the convex sets involved therein to be a ball. In Section 4 non-convex domains are considered. Examles are given of Hardy inequalities on a torus and

3 HARDY S INEQUALITY AND CURVATURE 3 on a 1-sheeted hyerboloid, which is unbounded with an unbounded interior radius. In Theorem 7 we establish a Hardy inequality for any doubly connected domain in R in terms of a uniformization of, i.e. any conformal univalent ma of onto an annulus B R \ B ρ in R. This is a rich source of examles of Hardy s inequalities on non-convex domains. For examle, Hardy s inequality is readily derived for the domain {z : ρ < Φ(z) < R }, z = x + iy, where Φ(z) = (z 1)(z + 1). In this case Φ(z) is an aroriate uniformization. The authors are grateful to Ruert Frank and Junfang Li for comments on an earlier version of the aer which led to significant imrovements.. Curvature and distance to the boundary The inequalities to be considered in the next section require the determination of the Lalacian of the distance function in terms of the rincial curvatures of the boundary of the domain. We first recall the following facts which may be found in [9], Section 5.1. The skeleton of a domain is the set S() := {x : card N(x) > 1} where N(x) = {y : y x = δ(x)}, the set of near oints of x on. The function δ is differentiable at x if and only if x / S(). In \ S(), δ is continuous and δ(x) = (y x)/ y x, where N(x) = {y}. (When N(x) = {y}, we sometimes abuse the notation and write y = N(x).) Since δ is Lischitz continuous, and hence differentiable almost everywhere by Rademacher s Theorem, S() is of Lebesgue measure zero. In [9], Corollary 5.1.4, it is shown that if x and y N(x) then N(y + t[x y]) = {y} for all t (0, λ), where λ := su{t (0, ) : y N(y + t[x y])}. The oint (x) := y + λ[x y] is called the ridge oint of x and R() := {(x) : x } is called the ridge of. For further details and roerties of S() and R() we refer to [9], 5.1. In articular, note that R() can be much larger than S() and S() R() S(). We shall be assuming throughout, without further mention, that R() is closed relative to, and so R() = S(); it is roved in [9], Theorem , that this is equivalent to the functions and δ o being continuous on. Note that in [0], Theorem 1.1, it is roved that if has a C,1 boundary (cf. next aragrah) then δ o is Lischitz continuous as a function defined on the boundary. Let be a domain in R n, n, with boundary C : this means that locally, after a rotation of co-ordinates, is the grah

4 4 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS x n = φ(x 1, x,, x n 1 ) of a function φ C. We consider a change of co-ordinates defined for x by Γ : (s 1, s,, s n ) x = (x 1, x,, x n ) x = γ(s 1, s,, s n 1 ) + s n n(s 1, s,, s n 1 ). (.1) Here γ(s 1, s,, s n 1 ), and n(s 1, s,, s n 1 ) is the internal unit normal to at γ(s 1, s,, s n 1 ), i.e. ointing in the direction of x. The co-ordinates (s 1, s,, s n 1 ) are chosen with resect to rincial directions through the (unique) near oint N(x) of x on, such that, with s = (s 1, s,, s n 1 ) and we have γ s i =: v i = (v 1 i, v i,, v n i ), i = 1,,, n 1, v i, v j = δ ij, v i, n = 0, n(s ) = κ s i i (s ) γ(s ) = κ s i i (s )v i (s ), (.) where κ i, i = 1,,, n 1, are the rincial curvatures of at the near oint y of x, and the angular notation denotes scalar roduct. In (.), the signs of the rincial curvatures are determined by the direction of the normal n. If is convex, the rincial curvatures of are non-ositive, while if the domain under consideration is c = R n \, the rincial curvatures are non-negative. We set s n = δ; in (.1), δ is equal to the distance δ(x) of x to. The following result may be found in Gilbarg and Trudinger [14], Lemma 14.17, for oints close to the C boundary of a bounded domain. For our reader s convenience, we give our roof, which is designed for our needs. Note the roof of Lemma. in [19], from which it follows that if has a C -boundary, then δ C on \ R(). We have seen that S() is of zero measure and hence so is R() if S() is closed. We caution the reader that in [14] and [19] comutations are made with resect to the outward unit normal (rather than the inward unit normal as in this aer) causing a different sign for the rincial curvatures κ i, i = 1,..., n 1. Lemma 1. Let be a domain in R n, n, with C boundary, and set δ(x) := dist(x, ). Then δ C ( \ R()), and for g(x) = g(δ(x)), g C (R + ), x g(x) = g δ (x) + n 1 ( κi 1 + δκ i ) g (x), (.3) δ where the κ i are the rincial curvatures of at the near oint N(x) of x. The equation (.3) holds for all x \ R().

5 HARDY S INEQUALITY AND CURVATURE 5 Proof. From (.1), for i = 1,,..., n, x i s j = γi s j + δ ni s j, j = 1,, n 1, x i δ = ni, (.4) and so, by (.), x s j = (1 + δκ j)v j, j = 1,,, n 1, x = n. (.5) sn Therefore, on recalling that s n = δ = = It follows from (.) that x 1 x 1 s 1 s n.. x n x n s 1 s n s 1 s 1 x 1 x n.. s n s n x 1 x n (1 + δκ 1 )v1 1 (1 + δκ n 1 )vn 1 1 n 1... (1 + δκ 1 )v1 n (1 + δκ n 1 )vn 1 n n n s 1 s 1 x 1 x n.. (.6) s n s n x 1 x n (1 + δκ 1 ) 1 v1 1 (1 + δκ 1 ) 1 v n 1.. (1 + δκ n 1 ) 1 vn 1 1 (1 + δκ n 1 ) 1 vn 1 n n 1 n n Therefore, for j = 1,, n, i = 1,, n 1, = s 1 x 1 s1 x n. s n x 1 sn x n (.7) s i x j = [1 + δκ i] 1 v j i, δ x j = nj (.8) and, emloying the usual summation convention, δ (x j ) = nj s i s i x j = n 1 [1 + δκ i ] 1 v j i n j s i + nj δ nj.

6 6 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS Consequently δ = = = n j=1 { n 1 [1 + δκ i ] 1 v j i } n j nj + nj si δ n 1 [1 + δκ i ] 1 v i, n n + n, si δ n 1 κ i [1 + δκ i ] 1 (.9) by (.). From the Chain Rule, we have g x = g s i j s i x + g j δ and, on using (.8), x g = [ ] g δ s k s k δ x j x j [ g = s k δ nj + g n j δ s k n n 1 = g (δ ) + g δ The lemma is therefore roved. j=1 k=1 k=1 δ x = g j δ ] s k x j δ x j κ k v j k [1 + δκ k] 1 v j k = g (δ ) + g n 1 κ k [1 + δκ k ] 1. δ Remark 1. It follows from (.8) that the terms [κ i /(1 + δκ i )] (y), y = N(x), i = 1,,, n 1, in (.10), are the rincial curvatures of the 1 level surface of δ through x at x. Furthermore, n 1 n 1 [κ i/(1 + δκ i )] (y) is the mean curvature of this level surface at x. Remark. If is convex, then S( c ) = R( c ) =. Remark 3. If is a convex domain with a C -boundary, we have noted that the rincial curvatures of c are non-negative and hence n 1 δ(x) = κ(y) := [κ i /(1 + δκ i )] (y) 0 for all x c (.10) where {y} = N(x).

7 HARDY S INEQUALITY AND CURVATURE 7 We claim that for convex, we also have n 1 δ(x) = κ(y) := [κ i /(1 + δκ i )] (y) 0 for all x \ R(). (.11) To see this, let x 0 be an arbitrary oint in, y 0 = N(x 0 ), and let h(s) be a rincial curve through y 0 on with curvature κ at y 0 : thus h(0) = y 0, h (0) = 1, h (0) = κn(0), where n is the inward normal to at y 0. The function f(s) := h(s) x 0 has a minimum at s = 0 and so at s = 0, we have f (s) = h (s) [h(s) x 0 ] = 0 f (s) = h (s) [h(s) x 0 ] + h (s) 0. Consequently, since [h(0) x 0 ] = δn, we have 1 + δκ 0. This is true for all rincial directions and so, since the rincial curvatures are non-ositive, our claim (.11) is established. Remark 4. If κ i (y) 0, then 1 + δ(x)κ i (y) 1, N(x) = {y}. (.1) Therefore, if is the comlement of a closed convex domain with C boundary, then (.1) holds for all i throughout, since then R() = S() =. Suose is convex and C. Then for all x \ R() and for all i 1 + δ(x)κ i (y) = 1 δ(x) κ i (y) > 0, N(x) = {y}. (.13) This is roved as follows. We saw in Remark 3 that 1 + δκ i 0. Suose for some i and some x with N(x) = {y}, that δ(x) = 1/ κ i (y). If x lies outside R(), then it follows from [9], Corollary 5.1.4, that there exists a oint w \ R() on the ray from y through x such that N(w) = {y} and δ(w) > δ(x). But, this would imly that 1 + δ(w)κ i (y) < 0 which is a contradiction. Remark 5. Suose is convex with C. Then x R()\S() if and only if for N(x) = {y} 1 + δ(x)κ i (y) = 0, for some i. (.14) The ridge in this case has zero Lebesgue measure.

8 8 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS 3. Inequalities inside and outside domains We first establish the following general inequality Theorem 1. Let R n, n, be a domain having a ridge R() and a sufficiently smooth boundary for Green s formula to hold. Let δ(x) = dist(x, R n \ ). Then for all f C0 ( \ R()) and (1, ), ( ) 1 δ f dx { 1 δ δ 1 } f Proof. For any vector field V we have the identity [ ] (divv ) f dx = Re (V f) f fdx for all f C 0 ( \ R()). Choose Then, for any ε > 0, ( divv f dx ε V = δ/δ 1. dx. (3.1) δ (3.) ) 1/ ( ) δ f f 1 1/ dx δ dx δ f dx + ( 1)ε /( 1) f δ dx which gives, since divv = ( 1)δ δ 1 δ for x \ R(), [ δ f dx ε ( 1) ( 1)ε /( 1) δ δ ] f δ dx. The roof of (3.1) is comleted on choosing ε = [/( 1)] ( 1). Corollary 1. If C, then for all f C0 ( \ R()) ( ) { 1 δ f dx 1 δ κ } f dx (3.3) 1 δ where κ := n 1 κ i 1+δκ i. Proof. The roof follows from Lemma 1 and Theorem 1. In many cases, we are able to rove an inequality for all f C 0 (). Theorem. Assume the hyothesis of Theorem 1. Let R() be the intersection of a decreasing family of oen neighborhoods {S ɛ : ɛ > 0} with smooth boundaries, and let η ɛ (x) denote the unit inward normal at x S ɛ. If ( δ η ɛ )(x) 0, x S ɛ (3.4) for all ɛ sufficiently small and 1 [δ δ](x), x \ R(), (3.5)

9 HARDY S INEQUALITY AND CURVATURE 9 then (3.1) holds for all f C 0 (). Proof. We roceed as in (3.), but now with f C 0 (), and account for the contribution of the boundary of S ɛ. On using (3.4) we have for all f C 0 () [ ] (divv) f dx Re (V f) f fdx. (3.6) \S ɛ \S ɛ On roceeding as in the roof of Theorem 1 we obtain δ f dx \S ɛ δ f dx ( ) [ ] 1 1 δ δ f χ 1 δ \Sɛ dx. The roof concludes on using (3.5) and the monotone convergence theorem. Corollary. Suose that the hyothesis of Theorem is satisfied and that is convex with a C boundary. Then for all f C0 () ( ) { 1 δ f dx 1 + δ κ } f dx. (3.7) 1 δ Proof. Since is convex, it follows from Lemma 1 that δ = κ 0 for x \ R(); see Remark 3. Therefore, (3.5) must hold and the result then follows from Theorem. Corollary 3. Let be a ball B R := {x R n : x < R}. Then for > 1, ( ) ( ) 1 f f dx B R BR 1 1 δ dx (n 1) f BR dx x δ 1 (3.8) for all f C0 (B R ). Proof. In this case we have that R(B R ) = S(B R ) = {0} and δ = R x. We now have that S ɛ = B ɛ and on S ɛ, δ = x = η x ɛ. Therefore (3.4) holds imlying that (3.8) is valid since κ = (n 1)/ x. In [11], Theorem 3.1, it is roved that for R n convex, u dx 1 u 4 δ dx c αd (α+) int δ α u dx, (3.9) for any α >, where { c α = α (α + 3), if α 1 α (α + ), if α (, 1) and D int := su{δ(x) : x }. A comarison of the right-hand side of (3.9), when = B R, with that in the case = of (3.8), is now

10 10 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS made to seek further evidence of the significance of the curvature in these inequalities. Set α = + ε. Then the terms to be comared from (3.8) and (3.9), resectively, are I 1 = (n 1)/δ(x) x and I = c α D (α+) int δ α (x), with δ(x) = R x. It is readily shown that { n 1 ε, if ε 1, 4δ x I 1 I (n )R (n 1) x, if 0 < ε < 1. 4δ x A similar comarison can be made in the L case using Theorem 3. of [11] with = q and α >. Also see [1]. Examle 1. The infinite cylinder. Let = B 1 (0) R, where B 1 (0) is the unit ball, center the origin, in R. Clearly, is convex and R() is the z-axis. The distance function is δ = 1 x + y, δ = (x, y, 0)(1 δ) 1, η = (x, y, 0)/(1 δ) on S ɛ, and δ = 1, where S 1 δ ɛ := {x = (x, y, z) : x +y < ɛ}. Therefore, (3.7) holds for this cylinder with κ = δ = 1 1 δ. Theorem 3. Under the conditions of Theorem 1, we have for all f C0 ( \ R()), f(x) dx 1 { (n ) (1 + δ δ ) + 4 x δ + (n ) x δ } f(x) dx. x δ (3.10) In articular, if is a convex domain with a C boundary, the conditions of the theorem are met and δ = n 1 κ i/(1 + δκ i ) in \ R(). Proof. Let V (x) = δ(x) δ(x) + (n ) x x. Then divv(x) = δ(x) δ(x) (n ) + 0 δ(x) x and 1 4 V (x) = 1 (n ) + (n ) x δ δ x x δ. For any ε > 0, and f C0 ( \ R()) [ ] (divv) f dx = Re (V f)fdx ) 1/ ( ( f dx ε f dx + ε V f dx V f dx. ) 1/

11 HARDY S INEQUALITY AND CURVATURE 11 The result follows on choosing ε =. When = B R, we have on substituting in Theorem 3, δ(x) = R x, κ i (x) = 1/R, i = 1,,, n 1, and so δ(x) = (n 1)/ x from Lemma 1 (or by direct calculation), { (n ) x + 1 δ(x) + x δ(x) } f(x) dx f(x) dx 1 B R 4 B R (3.11) which is given in Corollary in [3]. Note that (3.11) is valid for all f C0 (B R ) see Corollary 3. The alication of Lemma 1 to Theorem 1 also yields the following Hardy inequality in the comlement of a closed convex domain. Recall that in this case R(R n \ ) =. Theorem 4. Let R n, n, be convex with a C boundary. Then for all f C0 (R n \ ), ( ) { 1 f(x) dx 1 κδ } f(x) dx, (3.1) 1 δ(x) R n \ where κ = n 1 κ i 1+δκ i 0. R n \ Note that if = B ρ, the integrand on the right-hand side of (3.1) is non-negative if and only if (n 1)ρ x n 3. The following is another form of Hardy inequality, reminiscent of that derived in [5], Theorem 3.1. Theorem 5. Let be a convex domain in R n with a C boundary. Then for all f C0 (R n \ ), δ δ f dx 1 [1 + κδ] f dx, R n \ where κ = n 1 κ i 1+δκ i 0. R n \ Proof. The roof follows the lines of that of Theorem 3.1 in [5]. From (3.), ( ) 1/ ( ) ( 1)/ (divv) f dx V f dx f dx R n \ R n \ R n \ ε V f dx + ( 1)ε /( 1) f dx. R n \ R n \

12 1 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS On choosing V = δ, we have ɛ δ δ f dx + ( 1)ɛ 1 R n \ = by Lemma 1. Hence, as in (3.6) of [5], where R n \ R n \ R n \ [ δ ] f dx f dx [1 + κδ]dx (3.13) R n \ δ δ f dx K(ɛ) R n \ f dx +ɛ R n \ κδ f dx K(ɛ) = ɛ ( 1)ɛ 1 has a maximum value of (/) at ɛ = (/) ( 1)/. The roof is comleted by making the substitution for this value of ɛ. When =, it is readily shown that the substitution u = δf in Theorem 5 yields (3.1). Examle. If = B ρ in Theorem 5, then R n \B ρ ( x ρ) f dx 1 for all f C 0 (R n \ B ρ ). R n \B ρ [ 1 + (n 1) x ρ x ] f dx We have the following analogue of Theorem 1 in [3] for an annulus bounded by convex domains. Theorem 6. Let 1,, be convex domains in R n, n, with C boundaries and 1. For x := \ 1 denote the distances of x to 1, by δ 1, δ, resectively. Then for all f C0 ( \ R()) { (n 1)(n 3) \ 1 f(x) dx 1 4 \ 1 x + 1 δ δ δ 1 δ 1 δ δ δ 1 δ δ 1 δ + (n 1) x δ 1 x δ 1 + (n 1) x δ x δ } f(x) dx. (3.14)

13 HARDY S INEQUALITY AND CURVATURE 13 Proof. The starting oint is again (divv) f(x) dx ε f dx + ε \ 1 \ 1 \ 1 V f dx. Guided by the roof of Corollary 1 in [3], the theorem follows on setting and ε =. V = (n 1) x x If 1 = B ρ, = B R, R > ρ, we have δ 1 δ 1 δ δ δ 1 = n 1, δ = n 1 x x by Lemma 1, and δ 1 = δ = x/ x. On substituting in (3.14), we derive Corollary 1 in [3], namely, f(x) dx 1 { (n 1)(n 3) } f(x) dx, B R \B ρ 4 B R \B ρ x δ1 δ δ 1 δ (3.15) where δ 1 (x) = x ρ, δ (x) = R x. 4. Non-convex domains 4.1. Torus. We show that Theorem can be alied to give a Hardytye inequality on a torus. Corollary 4. Let R 3 be the interior of a ring torus with minor radius r and major radius R > r. Then δ < 0 in \ R() and ( ) 1 δ f f dx δ dx ( ) 1 ( ) (r δ) 1 f dx x 1 + x δ 1 (4.1) for all f C 0 (), where x has co-ordinates (x 1, x, x 3 ), and the last integrand is ositive. Proof. The domain under consideration is the doughnut-shaed domain generated by rotating a disc of radius r about a co-lanar axis at a distance R from the center of the disc. The fact that δ 0 was roved by D.H. Armitage and Ü. Kuran []. We give a different roof here which meets our uroses using a curvature argument. The ridge of the torus is R() = {x : ρ(x) = 0},

14 14 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS where ρ(x) is the distance from the oint x in to the center of the cross-section and δ(x) = r ρ(x). Moreover, in the notation of Theorem, S ɛ = {x : ρ(x) < ɛ}, and oints on the surface of S ɛ are on the level surface ρ(x) = ɛ, so that the unit inward normal to S ε is η ε = ρ(x)/ ρ(x) = δ. Therefore δ η > 0. For x \ R(), let y = N(x) = (y 1, y, y 3 ) have the arametric co-ordinates y 1 = (R + r cos s ) cos s 1 y = (R + r cos s ) sin s 1 y 3 = r sin s, where s 1, s ( π, π]. The rincial curvatures at y are κ 1 = 1 r, κ cos s = R + r cos s, e.g., see Kreyszig [17],.135, and so, by Lemma 1, ( ) κi R + (r δ) cos s δ(x) = (y) = 1 + δκ i (r δ)(r + (r δ) cos s ) x = 1 + x + (r δ) cos s (r δ) x 1 + x < 0 since R + r cos s = x 1 + x + δ(x) cos s and R > r. The inequality (4.1) follows from Theorem sheeted Hyerboloid. Next, we aly Theorem 1 to the 1- sheeted hyerboloid = {(x 1, x, x 3 ) R 3 : x 1 + x < 1 + x 3}. (4.) This is non-convex and unbounded with infinite volume and infinite interior diameter D int (). To calculate the rincial curvatures, we choose the following arametric co-ordinates for y : y 1 (s, t) = y (s, t) = s + 1 cos t, s + 1 sin t, y 3 (s, t) = s, for t [0, π) and s (, ). A calculation then gives (see [17],. 13) 1 κ 1 = [s + 1], κ 1 3/ = s + 1,

15 HARDY S INEQUALITY AND CURVATURE 15 and if y = N(x), x \ R(), then by Lemma 1, κ i δ(x) = κ := = δκ i w 3 δ + 1 w + δ, (4.3) where w = s + 1 is the distance of y from the origin, and the ridge is R() = {(x 1, x, x 3 ) : x 1 = x = 0, x 3 (, )}. Therefore δ(x) changes sign in. To find y = N(x), we first determine the vector normal to at y, namely i j k y s y t = s cos t s sin t 1 s +1 s +1 s + 1 sin t s + 1 cos t 0 = [ s + 1 cos t]i + [ s + 1 sin t]j + sk. The inward unit normal vector at y is therefore n = {[ s + 1 cos t]i + [ s + 1 sin t]j + sk}/ s + 1. The distance from y to the ridge oint (x) of x (see Section ) is given by s + 1/ cos θ, where cos θ = (z n)/ z, and z = [ s + 1 cos t]i + [ s + 1 sin t]j. Hence s + 1/ cos θ = s + 1 = w. Consequently, the near oint of x is the oint on the boundary of which is equidistant from the ridge oint (x) of x and the origin. We therefore have from Theorem 1. Corollary 5. Let R 3 be the 1-sheeted hyerboloid (4.). Then, for all f C0 ( \ R()), ( ) δ ( ) 1 f 1 f dx 1 f dx κ dx, (4.4) δ δ 1 where κ is given in (4.3), with w = y = δ((x)), y = N(x) and (x) the ridge oint of x Doubly connected domains. A domain R C is doubly connected if its boundary is a disjoint union of simle curves. If it has a smooth boundary then it can be maed conformally onto an annulus ρ,r = B R \ B ρ = {z C : ρ < z < R}, for some ρ, R; see [4], Theorem 1.. Lemma. Let 1 C and B ρ B R C, 0 < ρ < R, where B r is the disc of radius r centered at the origin. Let F : \ 1 B R \ B ρ

16 16 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS be analytic and univalent. Then for z = x 1 +ix, x = (x 1, x ) \ 1, F(z) := F { } (z) F (z) + F (z) 1 F (z) ρ + 1 (4.5) R F (z) is invariant under scaling, rotation, and inversion. Hence, F does not deend on the choice of the maing F, but only on the geometry of \ 1. Proof. The fact that F is invariant under scaling and rotations is straightforward. To see that it is also invariant under inversions suose that F (z) = 1/G(z). Then, under inversion F(z) becomes { } G (z) + G (z) 1 G(z) G(z) 4 1 = G (z) G(z) + G (z) G(z) { = G (z) G(z) + G (z) G(z) { = G (z) G(z) + G (z) { 1 + G(z) ρ 1 R 1 1 G(z) } ρ + R ρ G(z) G(z) R } (ρ R) G(z) (ρ G(z) )( G(z) R) } ρ G(z) G(z) R imlying that F is invariant under inversions. The rest of the lemma follows from [16], In alying the last Lemma we regard 1, as domains in R with z = x + iy and x = (x, y). Theorem 7. For := \ 1 R, u(x) dx 1 4 F(x) u(x) dx. Proof. From Corollary 1 of [3] it follows that for all u H0(B 1 R \ B ρ ), u(y) dy 1 [ ( 1 1 B R \B ρ 4 y + δ ρ (y) + 1 ) ] u(y) dy, δ R (y) B R \B ρ where δ ρ (y) := y ρ and δ R (y) := R y. Let F : ρ,r be analytic and univalent, and set y = F (x), with y = (y 1, y ), x = (x 1, x ). Then, with F denoting the comlex derivative, ( ) dy = det (y1, y ) dx = F (x) dx, (x 1, x ) [ ] t (y1, y ) x u = y u, (x 1, x ) imlying that x u = y u F (x). The theorem follows from Lemma.

17 HARDY S INEQUALITY AND CURVATURE 17 Examle 3. Let Φ(z) = (z 1)(z + 1) and = {z : ρ < Φ(z) < R } for 0 < ρ < R. The function F (z) = Φ(z) is analytic and univalent in and A calculation gives F(z) = z z 1 + F : ρ,r. z (R ρ) z 1 ( z 1 ρ) (R z 1). Finally, we refer the reader to further develoments along these lines in [19]. References [1] A. Ancona. On strong barriers and an inequality of Hardy for domains in R n, J. London Math. Soc.()34, 74-90, [] D.H. Armitage and Ü. Kuran. The convexity of a domain and the suerharmonicity of the signed distance function. Proc. A.M.S. 93(4), Aril [3] F.G. Avkhadiev and A. Latev. Hardy inequalities for non-convex domains. The Erwin Schrödinger International Institute for Mathematical Physics. Prerint ESI 185, October 009. [4] A.A. Balinsky. Hardy-tye inequalities for Aharonov-Bohm magnetic otentials with multile singularities. Math. res. Lett. 10, , 003. [5] A.A. Balinsky and W.D. Evans. Some recent results on Hardy-tye inequalities. Alied Mathematics & Information Sciences 4(), 010. [6] H. Brezis and M. Marcus. Hardy s inequalities revisited. Ann. Scuola Norm. Pisa 5, 17 37, [7] E.B. Davies. Sectral Theory and Differential Oerators, Cambridge Studies in Advanced Mathematics, Vol. 4, Cambridge University Press, Cambridge, [8] E.B. Davies. The Hardy constant. Quart. J. Math. Oxford () 46, , [9] D.E. Edmunds and W.D. Evans. Hardy Oerators, Function Saces, and Embeddings, Sringer Monograhs in Mathematics, Sringer, Berlin, Heidelberg, New York, 004. [10] W.D. Evans and R.T. Lewis. Hardy and Rellich inequalities with remainders. J. Math. Inequal., 1, , 007. [11] S. Filias, V. Maz ya, and A. Tertikas. On a question of Brezis and Marcus. Calc. Var., 5(4), , 006. [1] S. Filias, V. Maz ya, and A. Tertikas. Critical Hardy-Sobolev inequalities. J. Math. Pures Al., 87, 37 56, 007. [13] S. Filias, L. Moschini, and A. Tertikas. Shar two-sided heat kernel estimates for critical Schrödinger oerators on bounded domains. Commun. Math. Phys., 73, 37-81, 007.

18 18 A. BALINSKY, W. D. EVANS, AND R. T. LEWIS [14] D. Gilbarg and N.S. Trudinger. Ellitic Partial Differential Equations of Second Order, Rerint of the 1998 edition, Sringer-Verlag, Berlin, Heidelberg, New York, 001. [15] M. Hoffmann-Ostenhof, T. Hoffman-Ostenhof, and A. Latev. A geometrical version of Hardy s inequality. J. Funct. Anal. 189, , 00. [16] S. Krantz. Comlex Analysis: the Geometric Viewoint, Carus Mathematical Monograhs, 3, Math. Assoc. of America, Washington, DC, [17] E. Kreyszig. Differential Geometry, Dover Publications, Inc., New York, [18] A. Latev and A. V. Sobolev. Hardy inequalities for simly connected domains. [19] R.T. Lewis, J. Li, and Y. Li. A geometric characterization of a shar Hardy inequality. In rearation. [0] Y. Li and L. Nirenberg. The distance function to the boundary, Finsler geometry, and the singular set of viscosity solutions of some Hamilton-Jacobi equations. Comm. Pure and Alied Math. 58(1), , 005. [1] M. Marcus, V.J. Mizel, and Y. Pinchover. On the best constant for Hardy s inequality in R n. Trans. A.M.S., 350, , [] T. Matskewich and P.E. Sobolevskii. The best ossible constant in a generalized Hardy s inequality for convex domains in R n. Nonlinear Analysis TMA, 8, , [3] J. Tidblom. A geometrical version of Hardy s inequality for W 1, 0. Proc. A.M.S. 13(8), 65 71, 004. [4] G.-C. Wen. Conformal Maings and Boundary- value Problems, Transl, Math. Monograhs, 166, American Math.Soc., Providence, RI, 199. Cardiff School of Mathematics, Cardiff University, Cardiff, CF4 4AG, Wales, UK address: BalinskyA@cardiff.ac.uk address: EvansWD@cardiff.ac.uk Deartment of Mathematics, University of Alabama at Birmingham, Birmingham, AL , USA address: rtlewis@uab.edu