On the structure of Hardy Sobolev Maz ya inequalities

Transcription

1 J. Eur. Math. Soc., c European Mathematical Society 2009 Stathis Filippas Achilles Tertikas Jesper Tidblom On the structure of Hardy Sobolev Maz ya inequalities Received October, 2007 and in revised form February 6, 2008 Abstract. We establish new improvements of the optimal Hardy inequality in the half-space. We first add all possible linear combinations of Hardy type terms, thus revealing the structure of this type of inequalities and obtaining best constants. We then add the critical Sobolev term and obtain necessary and sufficient conditions for the validity of Hardy Sobolev Maz ya type inequalities.. Introduction One version of the Hardy inequality states that for convex domains R n the following estimate holds: u 2 4 d(x) 2, u C 0 (), where d(x) = dist(x, ) and the constant /4 is best possible. This result has been improved and generalized in many different ways: see for example [], [2], [4] [9], [2], [3]. One pioneering result due to Brezis and Marcus [4] is the following improved Hardy inequality: u 2 4 d(x) 2 C 2(), u C0 (), (.) valid for any convex domain R n. This estimate has recently been extended in [7]: u 2 4 ( ) 2/q d(x) 2 C q() u q, u C0 (). (.2) S. Filippas: Department of Applied Mathematics, University of Crete, 7409 Heraklion, Greece, and Institute of Applied and Computational Mathematics, FORTH, 70 Heraklion, Greece; filippas@tem.uoc.gr A. Tertikas: Department of Mathematics, University of Crete, 7409 Heraklion, Greece, and Institute of Applied and Computational Mathematics, FORTH, 70 Heraklion, Greece; tertikas@math.uoc.gr J. Tidblom: The Erwin Schrödinger Institute (ESI), Boltzmanngasse 9, A-090 Vienna, Austria; Jesper.Tidblom@esi.ac.at

2 66 Stathis Filippas et al. Moreover, it is shown in [7] that there exist constants c and c 2 only depending on q and the dimension n of such that the best constant C q () satisfies c D n 2 2n/q C q () c 2 D n 2 2n/q, where D = sup x d(x) < and 2 q < 2n/(n 2). We note that the critical Sobolev exponent q = 2 := 2n/(n 2) is not included in the above theorem. For results in the critical case we refer to [8]. Let S n = πn(n 2)(Ɣ(n/2)/ Ɣ(n)) 2/n, n 3, denote the best constant in the Sobolev inequality ( ) 2/2 u 2 S n, u C0 (). The first inequality that combines both the critical Sobolev exponent term and the Hardy term, the latter with best constant, is due to Maz ya [0], and is the following Hardy Sobolev Maz ya inequality: u 2 ( ) 2/2 C n, u C0 4 (Rn ), (.3) where = {(x,..., x n ) : x > 0} denotes the upper half-space, C n is a positive constant and 2 = 2n/(n 2), n 3. Recently, it was shown in [3] that in the 3- dimensional case, the best constant C 3 coincides with the best Sobolev constant S 3! On the other hand, when n 4 one has C n < S n (see []). We next mention an improvement of Hardy s inequality that involves two distance functions: u 2 R n 4 R n x 2 C(τ) x 2 τ (x 2, u C x2 2 )τ/2 0 (Rn ), where 0 < τ. This is a special case of a more general inequality proved in [3]. In this work we study improvements of Hardy s inequality that involve various distance functions. Working in the upper half-space, we obtain Hardy type inequalities that involve constant multiples of the inverse square of the distance to linear submanifolds of different codimensions of the boundary. Actually, we are able to give a complete description of the structure of this kind of improved Hardy inequalities. In particular, we have a lot of freedom in choosing these constants and we will show that all our configurations of constants are, in a natural sense, optimal. More precisely, our first result reads: Theorem A (Improved Hardy inequality). (i) Let α,..., α n be arbitrary real numbers and β = α 2 /4, β m = α 2 m (α m /2) 2, m = 2,..., n. Then for any u C0 (Rn ), u 2 ( β β 2 x2 2 β n x 2 x2 n ) u 2.

3 Hardy Sobolev Maz ya inequalities 67 (ii) Suppose that for some real numbers β,..., β n the inequality ( u 2 β β ) 2 β n R n x 2 x2 2 x 2 u 2 x2 n holds for any u C 0 (Rn ). Then there exist nonpositive constants α,..., α n such that β = α 2 /4, β m = α 2 m (α m /2) 2, m = 2,..., n. We next investigate the possibility of adding Sobolev type remainder terms. It turns out that almost every choice of the constants in Theorem A allows one to add a positive Sobolev term as well. The details are in our second main theorem. Theorem B (Improved Hardy Sobolev Maz ya inequality). Let α,..., α n be arbitrary nonpositive real numbers and β = α 2 /4, β m = α 2 m (α m /2) 2, m = 2,..., n. If α n < 0 then there exists a positive constant C such that for any u C 0 (Rn ), u 2 ( β ( C β ) 2 β n x 2 x2 2 x 2 u 2 x2 n ) 2/2 If α n = 0 then there is no positive constant C such that (.4) holds.. (.4) It is interesting to note that the Sobolev term vanishes precisely when the constant β n, in front of the Hardy-type term containing the point singularity, is chosen optimal. It is a bit curious that the size of the other constants, β,..., β n, does not matter at all for this question. Only the relative size of β n compared to the other constants matters. Our results depend heavily on the Gagliardo Nirenberg Sobolev inequality and also on an interesting relation between the existence of an L Hardy inequality and the possibility of adding a Sobolev type remainder term to the corresponding L 2 inequality. The precise result reads: Theorem C. Let R n, n 3, be a smooth domain. Assume that φ C 2 (), φ > 0, and the following weighted L inequality holds: φ 2(n ) n 2 v C φ n 2 n φ v, v C0 (). (.5) Then there exists c > 0 such that ( u 2 φ φ u 2 c ) 2/2, u C0 (). (.6)

4 68 Stathis Filippas et al. The regularity assumptions on φ can be weakened, but for our purposes it is enough to restrict ourselves to φ C 2 (). We note that under the sole assumption φ > 0 and φ C 2 () the inequality u 2 φ φ u 2, u C0 (), (.7) is always true; see Lemma 2.. It is the validity of (.5) that makes possible the addition of the Sobolev term in (.7). An easy example where both (.5) and (.6) fail is the case where φ is taken to be the first Dirichlet eigenfunction of the Laplacian of, for bounded. Our methods are not restricted to the case =. In the last section of the paper we give an example of how to apply the method to get some results for the quarter-space. Moreover, as one can easily check, our results remain valid even for complex-valued functions. The paper is organized as follows. In Section 2 we give the proof of Theorem A. In Section 3 we prove Theorems B and C. Finally, in the last section we obtain some results for the quarter-space. 2. Improved Hardy inequalities in the half-space The half-space has some nice features that are not present for an arbitrary convex domain. The fact that the boundary has zero curvature is very useful when one tries to prove certain sorts of inequalities, as we shall see below. We start with a general auxiliary lemma. Lemma 2.. (i) Let F C (). Then u 2 = (div F F 2 ) u Fu 2, u C0 (). (2.) (ii) Let φ > 0, φ C 2 () and u = φv. Then u 2 φ = φ u2 φ 2 v 2, u C0 (). (2.2) Proof. By expanding the square we have u Fu 2 = u 2 F 2 u 2 F u 2. Identity (2.) now follows by integrating by parts the last term. To prove (2.2) we apply (2.) to F = φ/φ. Elementary calculations now yield the result.

5 Hardy Sobolev Maz ya inequalities 69 We especially want to study inequalities of the type u 2 ( β β 2 x2 2 β n x 2 x x2 n ), u C 0 (Rn ), where β = (β,..., β n ) is a vector of nonnegative constants. The case when β = /4 is especially interesting since it corresponds to the term in the standard Hardy inequality. So every legitimate choice of β with β = /4 corresponds to an improved Hardy inequality. Let us introduce some notation. Let X k := (x,..., x k, 0,..., 0) so that X k 2 = x2 k. We now give the proof of the first part of Theorem A: Proof of Theorem A(i). Let γ,..., γ n be arbitrary real numbers and set An easy calculation shows that With this choice of F, we get and F 2 = We then get where n m= γm 2 n X m 2 2 φ := X γ... X n γ n, m= j= F = div F = m γ m γ j n γ m m= n m= X m X m 2. γ m m 2 X m 2 F := φ φ. X m X m 2 X j X j 2 = n φ φ = div F F 2 = n m= m= γm 2 n X m 2 2 m m= j= γ m γ j X j 2. β m X m 2, (2.3) m β = γ (γ ), β m = γ m (2 m γ m 2 γ j ), m = 2,..., n. We next set j= γ = α /2, γ m = α m α m /2, m = 2,..., n. With this choice of γ s the β s are given as in the statement of the theorem.

6 70 Stathis Filippas et al. As a consequence of Lemma 2. we have u 2 (div F F 2 )u 2. (2.4) The result then follows from (2.3) and (2.4). Remark. It is easy to check that for any choice of n real numbers α,..., α n, we can find n nonpositive real numbers α,..., α n that give the same constants β,..., β n. Consequently, without loss of generality, we may assume that α,..., α n are nonpositive. In the above theorem we have a lot of freedom. We can choose the γ s in many different ways, each choice giving a different inequality. We may, for instance, first maximize β and then β 2 and so on. More generally, we might try to make the first m β m s equal to zero and then maximize the β m s in increasing order. In fact, we have the following corollary: Corollary 2.2. Let k =,..., n. Then ( k u 2 2 R n 4 x 2 x2 4 k x 2 x2 k ) 4 x 2 u 2, u C x2 0 (Rn ). n Proof. In the case k = we choose α = = α n = 0. In this case all β k s are equal to /4. In the general case k > we choose α m = m/2 when m =,..., k, and α m = 0 when m = k,..., n. We next prove the second part of Theorem A: Proof of Theorem A(ii). We will first prove that β /4, therefore β = α 2 /4 for suitable α 0. Then, for this β, we will prove that β 2 (α /2) 2, and therefore β 2 = α2 2 (α /2) 2 for suitable α 2 0, and so on. Step. Let us first prove the estimate for β. To this end we set Q [u] := R n u 2 n i=2 β i u 2 R n x 2 We clearly have β inf u C 0 ( ) Q [u]. We will show that whence β /4. u 2 (x 2 x2 i ). (2.5) inf Q [u] /4, (2.6) u C0 (Rn )

7 Hardy Sobolev Maz ya inequalities 7 At this point we introduce a family of cutoff functions for later use. For j =,..., n and k j > 0 we set and Note that 0, t < /k j 2, φ j (t) = ln k j t, /kj 2 ln k t < /k j, j, t /k j, h kj (x) := φ j (r j ) where r j := X j = ( x2 j )/2. h kj (x) 2 = ln 2 k j rj 2, /kj 2 r j /k j, 0, otherwise. We also denote by φ(x) a radially symmetric C 0 (Rn ) function such that φ = for x < /2 and φ = 0 for x >. To prove (2.6) we consider the family of functions We will show that as k, R n u k 2 n u 2 i=2 β i R n k (x 2 x2 i ) u 2 R n k x 2 u k (x) = x /2 h k (x)φ(x). (2.7) = R n u k 2 u 2 R n k x 2 o(). (2.8) To see this, let us first examine the behavior of the denominator. For k large we easily compute u 2 k = /2 x h2 k φ 2 > C x > C ln k. /k (2.9) On the other hand, the terms n u 2 i=2 β i R n k are easily seen to be bounded (x 2 x2 i ) as k, by the Lebesgue dominated convergence theorem. From this and (2.9) we deduce (2.8). We now estimate the gradient term in (2.8): u k 2 = 4 x h2 k φ 2 x h k 2 φ 2 x h 2 k φ 2 mixed terms. (2.0)

8 72 Stathis Filippas et al. The first integral of the right hand side behaves exactly as the denominator (cf. (2.9)), that is, it goes to infinity like O(ln k ). The last integral is easily seen to be bounded as k. For the middle integral we have x h k 2 φ 2 C ln 2 x k /k 2 x C. /k ln k As a consequence of these estimates, we easily see that the mixed terms in (2.0) are of the order o(ln k ) as k. Hence, as k, u k 2 = x 4 h2 k φ 2 o(ln k ). (2.) From (2.8), (2.9) and (2.) we conclude that as k, Q [u k ] = /4 o(), hence inf u C 0 ( ) Q [u] /4 and consequently β /4. Therefore for a suitable nonnegative constant α we have β = α 2 /4. We also set γ := α /2. (2.2) Step 2. We will next show that β 2 (α /2) 2. To this end, setting R n u 2 (/4 α 2) Q 2 [u] := we will prove that We now consider the family of functions u 2 n x 2 i=3 β i u 2 X 2 2 inf u C 0 (Rn ) Q 2[u] (α /2) 2. u 2 X i 2, (2.3) u k,k 2 (x) := x γ X 2 α /2 h k (x)h k2 (x)φ(x) =: x γ v k,k 2 (x). (2.4) An easy calculation shows that Q 2 [u k,k 2 ] = x 2γ v k,k 2 2 n i=3 β i R n x 2γ X i 2 vk 2,k 2. (2.5) R n x 2γ X 2 2 vk 2,k 2 We next use the precise form of v k,k 2 (x). Concerning the denominator of Q 2 [u k,k 2 ] we have x 2γ X 2 2 vk 2,k 2 = x 2α R n (x 2 x2 2 )α 3/2 h 2 k h 2 k 2 φ 2.

9 Hardy Sobolev Maz ya inequalities 73 Sending k to infinity, using the structure of the cutoff functions and then introducing polar coordinates we get x 2γ R n X 2 2 v,k 2 2 = x 2α R n (x 2 x2 2 )α 3/2 h 2 k 2 φ 2 C C /k 2 < x2 2 </2 x 2α ( x2 2 )α 3/2 2 π /2 0 /k 2 r (sin θ) 2α dr dθ C ln k 2. (2.6) The terms in the numerator that are multiplied by the β i s stay bounded as k or k 2 go to infinity (cf. the estimates related to (2.29) in Step 3). Next, x 2γ v k,k 2 2 = (α /2) 2 x 2γ X 2 2α 3 h 2 k h 2 k 2 φ 2 x 2γ X 2 2α (h k h k2 ) 2 φ 2 x 2γ X 2 2α h 2 k h 2 k 2 φ 2 mixed terms. (2.7) The first integral on the right hand side above is the same as the denominator of Q 2, and therefore is finite as k and increases like ln k 2 as k 2 (cf. (2.6)). The last integral is bounded, no matter how big the k and k 2 are. Concerning the middle term we have M[v k,k 2 ] := x 2γ R n X 2 2α (h k h k2 ) 2 φ 2 = x 2γ R n X 2 2α h k 2 h 2 k 2 φ 2 x 2γ X 2 2α h 2 k h k2 2 φ 2 mixed terms Since we easily get and therefore, since α 0, =: I I 2 mixed terms. (2.8) X 2 2α h 2 k 2 = r 2α 2 φ 2 (r 2 ) C k2, 0 < r 2 <, I C /k (ln k ) 2 x 2α /k 2, I C ln k, k. (2.9)

10 74 Stathis Filippas et al. Also, since h 2 k, we similarly get (for any k ) I 2 From (2.8) (2.20) we infer that as k 2, C /k2 (ln k 2 ) 2 r /k2 2 2 dr 2 C, k 2. (2.20) ln k 2 M[v,k2 ] = o(). Returning to (2.7) we see that as k 2, x 2γ v,k2 2 = (α /2) 2 We then have as k 2, x 2γ X 2 2 v 2,k 2 o(ln k 2 ). (2.2) Q 2 [u,k2 ] = (α /2) 2 o(); (2.22) consequently, β 2 (α /2) 2, and therefore β 2 = α 2 2 (α /2) 2 for suitable α 2 0. We also set γ 2 = α 2 α /2. Step 3. The general case. At the (q )th step, q n, we have already established that β = α 2 /4, β m = α 2 m (α m /2) 2, m = 2,..., q, for suitable nonpositive constants a i. Also, we have defined γ = α /2, γ m = α m α m /2, m = 2,..., q. Our goal for the rest of the proof is to show that β q (α q /2) 2. To this end we consider the quotient R Q q [u] := n u 2 n u q =i= β 2 i R n X i 2. (2.23) u 2 R n X q 2 The test function is now given by u k,k q (x) := x γ X 2 γ 2... X q γ q X q α q /2 h k (x)h kq (x)φ(x) =: x γ X 2 γ 2... X q γ q v kq (x). (2.24) A straightforward calculation shows that Q q [u k,k q ] q R = n j= v k,k q 2 n i=q β i q q j= X q 2 v 2 k,k q j= X i 2 vk 2,k q. (2.25)

11 Hardy Sobolev Maz ya inequalities 75 Let us first look at the denominator, D q [u k,k q ] := q X q 2αq 3 h k (x)h kq (x)φ(x). j= Sending k, we find that h k and therefore D q [u,kq ] = q X q 2αq 3 h kq (x)φ(x). j= To see that this is finite we note that with B R := {x Rn : x < R, x 0}, D q [u,kq ] C B {/k2 q r q /k q } j= {/kq 2 r q /k q } j= q X q 2αq 3 q X q 2αq 3... q. (2.26) To estimate this, we introduce polar coordinates (x,..., x q ) (r q, θ,..., θ q ), x = r q sin θ q sin θ q 2... sin θ 2 sin θ x 2 = r q sin θ q sin θ q 2... sin θ 2 cos θ x 3 = r q sin θ q sin θ q 2... cos θ 2. x q = r q cos θ q, where 0 θ < 2π and 0 θ m < π for m = 2,..., q. The surface measure on the unit sphere S q then becomes C(sin θ q ) q 2 (sin θ q 2 ) q 3... sin θ 2 dθ... dθ q. Also, r q = X q and for m q, We then have r m = X m = ( x2 m )/2 = r q sin θ q sin θ q 2... sin θ m. {/kq 2 r q /k q } j= q X q 2αq 3... q = C {/kq 2 r q /k q } (sin θ j ) 2α j dθ... dθ q dr q C ln k q. (2.27) q rq j=

12 76 Stathis Filippas et al. On the other hand, since D q [u,kq ] B /2 {/k q r q /2} j= by practically the same argument we find that as k q, For i = q,..., n, we consider the terms q X i 2 vk 2,k q = j= j= q X q 2αq 3, D q [u,kq ] C ln k q. (2.28) q X q 2αq X i 2 h 2 k h 2 k q φ(x) 2 j= q X q 2αq X q 2 h 2 k h 2 k q φ(x) 2. (2.29) If we let first k and then k q, the above integral converges to I q := q X q 2αq X q 2 φ(x) 2. j= To see this is finite we introduce polar coordinates (x,..., x q ) (r q, θ..., θ q ) and use elementary estimates to get I q C B sin θ q q (sin θ j ) 2α j dθ... dθ q dr q <. j= We next consider the gradient term q v k,k q 2 j= = (α q /2) 2 j= q X q 2αq 3 h 2 k h 2 k q φ 2 j= q X q 2αq (h k h kq ) 2 φ 2 q X q 2αq h 2 k h 2 k q φ 2 mixed terms. (2.30) j=

13 Hardy Sobolev Maz ya inequalities 77 The first term of the right hand side is the same as the denominator. Using polar coordinates and arguments similar to the ones used in estimating the gradient term in (2.7), all other terms of (2.30) are bounded as k and k q. In particular, we end up with q v,kq 2 = (α q /2) 2 D q [u,kq ] o(ln k q ), k q. j= Putting things together we obtain Q q [u,kq ] = (α q /2) 2 o(), k q, from which it follows that β q (α q /2) 2. This completes the proof of the theorem. The previous analysis can also lead to the following result: Theorem 2.3. Let α,..., α k, k n, be nonpositive constants and β = α 2 /4, β m = α 2 m (α m /2) 2, m = 2,..., k. Suppose that there exists a constant β k such that the inequality u 2 ( β holds for any u C0 (Rn ). Then β 2 x2 2 β k x 2 x2 k ) u 2 (2.3) β k (α k /2) 2. (2.32) Moreover, inf u C 0 (Rn ) R n u 2 β x 2 β k x2 k x 2 x2 k = (α k /2) 2. (2.33) Proof. The proof of the first part, that is, of estimate (2.32), is contained in the proof of Theorem A(ii). To establish (2.33), we first use (2.32) to deduce than the infimum in (2.33) is less than or equal to (α k /2) 2. To obtain the reverse inequality we use Theorem A(i) with a kl = (l )/2, l =,..., n k. For this choice we have β k2 = = β n = 0. The following is an interesting consequence of the previous theorem.

14 78 Stathis Filippas et al. Corollary 2.4. For k n, inf u C 0 (Rn ) u 2 x 2 x2 k = k2 4, (2.34) and for k m < n, inf u C 0 (Rn ) R n u 2 k2 4 x 2 x2 k 4 x 2 x2 k x 2 x2 m 4 x 2 x2 m = 4. (2.35) Proof. To establish (2.34) we use (2.33) with α l = l/2, l =,..., k. To establish (2.35) we again use (2.33) with α l = l/2, l =,..., k, and α l = 0, k l m. With this choice we have β = = β k = 0, β k = k 2 /4 and β l = /4, l = k,..., m. 3. Hardy Sobolev Maz ya inequalities We begin by proving Theorem C. Proof of Theorem C. Our starting point is the Gagliardo Nirenberg Sobolev inequality C n ( ) n f n n n f, f C 0 (). (3.) Let f = φ α w, where α = 2(n ) n 2. This leads to C n ( ) n φ n αn w n n αφ α φ w φ α n w, w C 0 (). We now estimate the first term in the integral according to inequality (.5) and let w = v θ. Then we get ( ) n C φ αn/n v θn/n n The choice φ α v θ v ( ) /2 ( /2 φ 2α 2 v 2θ 2 φ 2 v ) 2. θ = α = 2(n ) n 2

15 Hardy Sobolev Maz ya inequalities 79 gives us the inequality Let u = φv. By Lemma 2. we have u 2 = ( ) n 2 C φ n 2 2n v n 2 2n n φ 2 v 2. (3.2) φ φ u2 φ 2 v 2. We conclude the proof by combining this result with (3.2). Condition (.5) might seem to be unnatural and not easily checked. However, it will be very natural and is easily verified for our choices of φ. To produce Hardy inequalities in the half-space with remainder terms also including the Sobolev term, we will need a weighted version of the Sobolev inequality. Theorem 3.. Let σ,..., σ k be real numbers for some k with k n. Set c l := σ σ l l for l k. Assume that c l = 0 whenever σ l = 0. Then there exists a positive constant C such that for any w C 0 (Rn ), and ( ( ) n ) n x σ X 2 σ 2... X k σ k w C x σ R n X 2 σ 2... X k σ n n k w, (3.3) σ (n 2) (n ) x X 2 σ2(n 2) (n )... X k σ k (n 2) (n ) w 2 ( σ (n 2) ) 2(n ) C (x R n X 2 σ 2(n ) 2(n 2)... X k σ k (n 2) n 2 2(n ) w ) n 2 2n n. (3.4) Proof. For = we let u = xσ v in the Sobolev inequality (3.) to get nσ ( ) n n C n x v n n ( σ x σ R n v x σ v ) n, v C 0 ( ). Using the inequality div F v with the vector field F = (x σ, 0,..., 0) one obtains σ x σ v F v (3.5) x σ v

16 80 Stathis Filippas et al. and hence C n nσ ( ) n n x v n n x σ R n v n, v C 0 ( ). Now let v = X 2 σ 2w = ( x2 2 )σ 2/2 w in the above inequality. This gives C n ( n x nσ (x 2 x2 2 ) nσ 2 2(n ) w n n ) n x σ (x2 x2 2 )σ2/2 w σ 2 x σ R n (x2 x2 2 )σ2/2 /2 n w. Letting F = x σ (x2 x2 2 )σ2/2 /2 X 2 in (3.5), we get σ σ 2 x σ (x2 x2 2 )σ2/2 /2 w Combining the previous two estimates we conclude that nσ ( n c x R n (x 2 x2 2 ) nσ 2 2(n ) w n n x σ (x2 x2 2 )σ 2/2 w. (3.6) x σ (x2 x2 2 )σ 2/2 w ) n n. Note that, in case σ 2 = 0, we have the desired result immediately and we do not have to check whether the constant σ σ 2 is zero or not. We may repeat this procedure iteratively. In the l-th step we need the analogue of (3.6) which is c l x σ R n (x2 x2 2 )σ2/2... (x 2 x2 l )(σl )/2 w x σ (x2 x2 2 )σ2/2... (x 2 x2 l )σl/2 w for some positive constant c l. This follows from (3.5) with For this choice we get F = x σ (x2 x2 2 )σ 2/2... ( x2 l )(σ l )/2 X l. c l = σ σ l (l ). So our procedure works nicely in case c l = 0 for those l such that σ l = 0. This proves (3.3). To show (3.4) we apply (3.3) to the function w = v θ. Trivial estimates give C nσ n x (x 2 x2 2 ) nσ 2 2(n )... (x 2 x2 k ) nσ k 2(n ) v n nθ ( ) n θ x σ (x2 x2 2 )σ2/2... (x 2 x2 k )σk/2 v θ n v.

17 Hardy Sobolev Maz ya inequalities 8 We will now apply Hölder s inequality to the right hand side. We want to do it in such a way that one of the factors becomes identical to the left hand side raised to some power. Therefore we need to choose θ so that nθ 2(n ) = 2θ 2 θ = n n 2. Hölder s inequality then immediately gives the result. We are now ready to give the proof of Theorem B: Proof of Theorem B. For φ > 0 and u = φv, Lemma 2. gives us the inequality u 2 φ φ u 2 φ 2 v 2. (3.7) We will choose where φ(x) = (x σ (x2 x2 2 )σ 2/2... ( x2 n )σ n/2 ) n 2 2(n ) = X γ... X n γ n, (3.8) γ = α /2, γ m = α m α m /2, m = 2,..., n, and 2(n ) σ m = n 2 γ m, m =,..., n. We now apply (3.4) of Theorem 3. to obtain provided that ( ) n 2 φ 2 v 2 C φv 2n n n 2, c l := σ σ l l = 0 whenever σ l = 0, (3.9) for l n. Combining this with (3.7) we get On the other hand, by Theorem A(i), ( ) n 2 u 2 φ R n φ u 2 C n n n 2. φ φ = β x 2 β 2 x 2 x2 2 β n x 2, x2 n and the desired inequality follows. It remains to check condition (3.9). After some elementary calculations we see that 2(n ) c l = n 2 α l n l 2(n ), l =,..., n.

18 82 Stathis Filippas et al. Since α l 0 we clearly have c l = 0 for l =,..., n. Moreover, c n = 0 when α n < 0. This completes the proof of (.4). In the rest of the proof we will show that (.4) fails in case α n = 0. To this end we will establish that inf u C 0 (Rn ) R n u 2 u β 2 R n β x 2 n where β n = (α n /2) 2. Let inf v C 0 (Rn ) ( 2n u n 2 ) n 2 n u(x) = x γ X 2 γ 2... X n γ n v(x). x 2 x2 n = 0, (3.0) A straightforward calculation, quite similar to the one leading to (2.5), shows that the infimum in (3.0) is the same as n j= X n 2 v 2 We now choose the test functions n j= v 2 β n ( R n ( n j= X j γ j ) n 2 2n v n 2 2n ) n 2 n. (3.) v k,ε = X n γ nε h k (x)φ(x), ε > 0, (3.2) where h k (x) and φ(x) are as in the first step of the proof of Theorem A(ii). For this choice, after straightforward calculations, quite similar to the ones used in the proof of Theorem A(ii), we obtain the following estimate for the numerator N in (3.): N[v,ε ] = ((α n /2 ε) 2 (α n /2) 2 ) n X n 2γn2ε φ(x) 2 O ε (), = Cε = Cε j= 0 r 2ε n j= r ε dr O ε (). (sin θ j ) 2α j φ(r) 2 dθ... dθ n dr O ε () In the above calculations we have taken the limit k and we have used polar coordinates (x,..., x n ) (θ,..., θ n, r). We then conclude that N[v,ε ] < C as ε 0. (3.3) Similar calculations for the denominator D in (3.) reveal that ( D[v,ε ] = C (sin θ j ) n j n 2 2nα j n 2 φ n 2 2n dθ... dθ n dr n 2εn r n 2 j= ( /2 ) n 2 2εn n C r n 2 dr = Cε n 2 n. 0 ) n 2 n

19 Hardy Sobolev Maz ya inequalities 83 We then have N[v,ε ] 0 as ε 0, D[v,ε ] and therefore the infimum in (3.) or (3.0) is equal to zero. This completes the proof of the theorem. Here is a consequence of Theorem B. Corollary 3.2. Let k < n. For any β n < /4, there exists a positive constant C such that for all u C 0 (Rn ), ( k u 2 2 R n 4 x 2 x2 4 k x 2 x2 k ) ( β n 4 x 2 x2 n x 2 C x2 n ) 2/2. If β n = /4 the previous inequality fails. In case k = n for any β n < n 2 /4, there exists a positive constant C such that for all u C 0 (Rn ), u 2 β n The above inequality fails for β n = n 2 /4. x2 n ( ) 2/2 C. Proof. In Theorem B we make the following choices: In the case k = we choose α = = α n = 0. In this case β k = /4, k =,..., n. The condition α n < 0 is equivalent to β n < /4. In the case < k n we choose α m = m/2 when m =,..., k, and α m = 0 when m = k,..., n. Finally, in case k = n we choose α m = m/2 for m =,..., n. 4. Further generalizations The techniques used in the previous sections can be generalized to other situations as well. For example, consider the subset of R n where x,..., x k > 0. We denote this domain by R n k. Then we can easily prove the Hardy-Sobolev inequality Theorem 4.. There exists a positive constant C such that for any u C0 (Rn k ) u 2 4 R n k R n k ( x 2 k ) ( ) 2/2 C. R n k

20 84 Stathis Filippas et al. Proof. Let φ = x... x k. For u = φw we calculate u 2 = x... x k w ( x... x k,..., ) 2 w R n R k n 2 x x k k = x... x k w 2 ( x... x k 4 x 2 ) xk 2 w 2 R n k 2 R n k R n k x... x k ( x,..., x k ) w 2. By partial integration, we see that the last term is zero. If the second term is expressed in terms of u, we see that it is equal to the Hardy term ( ) 4 xk 2. R n k By Theorem C, the first term may be estimated from below by the Sobolev term provided that we can prove the following L Hardy inequality: ( C (x... x k ) n n 2 ) /2 xk 2 v (x... x k ) n n 2 v. R n k To do this we work as in the previous section, using the inequality div F v F v, R n k with the proper choice of vector field, which turns out to be ( F = (x... x k ) τ x 2 ) β ( xk 2,..., ), x x k where τ = n n 2 R n k and β = 2. We immediately see that F = φ 2τ = (x... x k ) n n 2. Also, ) β R n k div F = (x... x k ) τ ( x 2 k ) β ( τ(x... x k ) τ x 2 xk 2 ( 2β x 4 ) ( xk 4 (x... x k ) τ x 2 ) β xk 2. Since τ > 0 and the last term is positive, we get the result.

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