HARDY AND RELLICH INEQUALITIES WITH ... - UT Math

HARDY AND RELLICH INEQUALITIES WITH REMAINDERS W. D. EVANS AND ROGER T. LEWIS Abstract. In this paper our primary concern is with the establishment of weighted Hardy inequalities in Lp (Ω) and Rellich inequalities in L2 (Ω) depending upon the distance to the boundary of domains Ω ⊂ Rn with a finite diameter D(Ω). Improved constants are presented in most cases.

1. Introduction Recently, considerable attention has been given to extensions of the multi-dimensional Hardy inequality of the form Z Z Z |u(x)|2 2 |∇u(x)| dx ≥ µ(Ω) dx + λ(Ω) |u(x)|2 dx, u ∈ H01 (Ω), 2 δ(x) Ω Ω Ω (1.1) n where Ω is an open connected subset of R and δ(x) := dist(x, ∂Ω). It is known that for µ(Ω) = 14 there are smooth domains for which λ(Ω) ≤ 0, and for λ(Ω) = 0, there are smooth domains for which µ(Ω) < 14 - see M. Marcus, V.J. Mizel, and Y. Pinchover [8] and T. Matskewich and P.E. Sobolevskii [9]. In [2], H. Brezis and M. Marcus showed that for domains of class C 2 inequality (1.1) holds for µ(Ω) =

1 4

and some λ(Ω) ∈ (−∞, ∞)

and when Ω is convex λ(Ω) ≥

1 4D(Ω)2

(1.2)

in which D(Ω) is the diameter of Ω. M. Hoffmann-Ostenhof, T. Hoffmann-Ostenhof, and A. Laptev [6] answered a question posed by H. Brezis and M. Markus in [2] by establishing the improvement to (1.2) that (1.1) holds for a convex domain Date: March 8, 2007. 1991 Mathematics Subject Classification. Primary 47A63; Secondary 46E35, 26D10. Key words and phrases. Rellich inequality, Hardy inequality, Remainder terms. 1

2

W. D. EVANS AND R. T. LEWIS

Ω, with 1 µ(Ω) = , 4

λ(Ω) ≥

K(n)

2 ,

4|Ω| n

and K(n) := n

hs

n−1

n

i2/n

(1.3)

in which sn−1 := |Sn−1 | and |Ω| is the volume of Ω. For a convex domain Ω and µ(Ω) = 1/4, a lower bound for λ(Ω) in (1.1) in terms of |Ω| was also obtained by S. Filippas, V. Maz’ya, and A. Tertikas in [5] as a special case of results on Lp Hardy inequalities. They prove that λ(Ω) ≥ 3Dint (Ω)−2 , where Dint (Ω) = 2 supx∈Ω δ(x), 3 the internal diameter of Ω. Since 3Dint (Ω)−2 ≥ 4n K(n)/|Ω|2/n , their result is an improvement of (1.3) for n = 2, 3, but the estimates don’t compare for n > 3. In this paper we show that (1.1) holds for (1.3) replaced by µ(Ω) =

1 4

and

λ(Ω) ≥

3K(n) 2

2|Ω| n

as well as proving weighted versions of the Hardy inequality in Lp (Ω) for p > 1. In the case p = 2, the following are special cases of our results. If Ω is convex and σ ∈ (0, 1], then Z o σ 2 Z n ¯ ¯2 B(n, 2 − σ) ¡ sn−1 ¢ 2−σ 2 n(1 − σ) n ¯∇u(x)¯ dx ≥ +3 |u(x)|2 dx σ 2−σ 4D(Ω) δ(x) n|Ω| Ω Ω (1.4) for Γ( p+1 ) · Γ( n2 ) 2 B(n, p) := √ . (1.5) π · Γ( n+p ) 2 If σ ∈ [ 2−n , 0] and Ω is convex, then 2 Z Z ¯ ¯2 n(1 − σ)2 σ¯ ¯ δ(x) ∇u(x) dx ≥ B(n, 2 − σ) δ(x)σ−2 |u(x)|2 dx 4 Ω Ω Z CH (n, σ) δ(x)|σ| |u(x)|2 dx. + 2(1−σ) n Ω |Ω| for CH (n, σ) given in (3.4). Similar results for weighted forms of the Hardy inequality in Lp (Ω) are given in section 4. Finally, we show that our one-dimensional inequalities in §2 lead to improved constants for the Rellich inequality obtained by G. Barbatis in [1] for n ≥ 4. 2. One-dimensional inequalities As is the case in [6], our proofs are based on one-dimensional Hardytype inequalities coupled with the use of the mean-distance function introduced by Davies to extend to higher dimensions; see [4]. The basic one-dimensional inequality is as follows:

HARDY AND RELLICH INEQUALITIES

3

Lemma 1. Let u ∈ C01 (0, 2b), ρ(t) := min{t, 2b − t} and let f ∈ C 1 [0, b] be monotonic on [0, b]. Then for p > 1 Z 2b Z 2b |f (ρ(t)) − f (b)|p 0 p 0 p p |f (ρ(t))||u(t)| dt ≤ p |u (t)| dt. (2.1) |f 0 (ρ(t))|p−1 0 0 Proof. First let u := vχ(0,b] , the restriction to (0, b] of some v ∈ C01 (0, 2b). For any constant c ¯b Rb 0 p p¯ − 0 [f (t) − c] |u(t)| dt = −[f (t) − c]|u(t)| ¯ 0 Rb p + 0 [f (t) − c] p2 [|u(t)|2 ] 2 −1 [|u(t)|2 ]0 dt. By choosing c = f (b), we have that Rb Rb − 0 f 0 (t)|u(t)|p dt = p 0 [f (t) − f (b)]|u(t)|p−2 Re[u(t)u0 (t)]dt. (2.2) 1 Similarly, for u = vχ[b,2b) , v ∈ C0 (0, 2b), we have R 2b − b f 0 (2b − s)|u(s)|p ds R 2b = p b [f (2b − s) − f (b)]|u(s)|p−2 Re[u(s)u0 (s)]ds. Therefore, since f is monotonic, for any u ∈ C01 (0, 2b) R 2b 0 |f (ρ(t))||u(t)|p dt 0 R 2b = p 0 |f (ρ(t)) − f (b)||u(t)|p−2 Re[u(t)u0 (t)]dt Rb p−1 (b)| 0 ≤ p 0 |f 0 (ρ(t))| p |u(t)|p−1 |f (ρ(t))−fp−1 |u (t)|dt 0 |f (ρ(t))| p p−1 hR i p hR i p1 b b |f (ρ(t))−f (b)|p 0 p ≤ p 0 |f 0 (ρ(t))||u(t)|p dt |u (t)| dt 0 p−1 0 |f (ρ(t))| on applying H¨older’s inequality. Inequality (2.1) now follows. ¤ The next lemma provides the one-dimensional result needed to improve (1.3), which was proved in [6]. Lemma 2. Let σ ≤ 1 and define µ(t) := 2b−ρ(t). For all u ∈ C01 (0, 2b) Z 2b h ³ 2ρ(t) ´1−σ i2 ³ 1 − σ ´2 Z 2b σ−2 σ 0 2 1+k(σ) ρ(t) |u(t)|2 dt, ρ(t) |u (t)| dt ≥ 2 µ(t) 0 0 (2.3) for ½ £ ¤ 1 −1 −σ σ 1 − 2 , σ < 0, k(σ) := 1, σ ∈ [0, 1]. Proof. On setting f (t) = tσ−1 in (2.1) we get Z 2b Z 2b ¯ £ ρ(t) ¤1−σ ¯p 0 p σ−2 p p p ¯ |u (t)| dt. ρ(t)p+σ−2 ¯1− ρ(t) |u(t)| dt ≤ p |1−σ| b 0 0 (2.4)

4


¡ ¢1−σ With u ∈ C01 (0, 2b), let p = 2 and substitute v(t) = [1 − ρ(t) ]u(t) b in (2.4). We claim that this gives Z 2b ³ 1 − σ ´2 Z 2b h ³ ρ(t) ´1−σ i−2 σ 0 2 ρ (t)|v (t)| dt ≥ ρ(t)σ−2 1 − |v(t)|2 dt 2 b 0 0 (2.5) for any real number σ. The substitution gives £ ¡ ρ(t) ¢1−σ ¤ 0 ρ(t)σ/2 v 0 (t) = −(1−σ)bσ−1 ρ(t)−σ/2 ρ0 (t)u(t)+ρ(t)σ/2 1− u (t). b Consequently, £ ¡ ¢1−σ ¤2 0 2 ρ(t)σ |v 0 (t)|2 = (1 − σ)2 b2σ−2 ρ(t)−σ |u(t)|2 + ρ(t)σ 1 − ρ(t) |u (t)| £ ¡ ρ(t) ¢1−σ ¤ 2 0 b σ−1 0 −(1 − σ)b ρ (t) 1 − b [|u| ] which implies that £ ¡ ρ(t) ¢1−σ ¤2 0 2 R 2b R 2b σ 0 2 σ ρ(t) |v (t)| dt = ρ(t) 1 − |u (t)| dt b 0 0R 2b 2 2σ−2 −σ + 0 (1 − σ) b ρ(t) |u(t)|2 dt h £ ¡ ¢1−σ ¤i 2 R 2b +(1 − σ)bσ−1 0 dtd ρ0 (t) 1 − ρ(t) |u| dt b ¢ ¤ £ ¡ R 2b 1−σ 2 = 0 ρ(t)σ 1 − ρ(t) |u0 (t)|2 dt b (2.6) since ρ0 (t) = 1 in (0, b) and −1 in (b, 2b). Therefore, (2.5) follows from (2.4). Since 2b = µ(t) + ρ(t) h i2 £ ¡ ¢1−σ ¤−2 ρ(t)1−σ 1 − ρ(t) = 1 + b b1−σ −ρ(t)1−σ i2 h (2.7) ¡ ρ(t) ¢1−σ ρ(t) kσ ( µ(t) ) = 1 + 21−σ µ(t) for kσ (x) :=

1 , (1 + x)1−σ − (2x)1−σ

x ∈ [0, 1),

σ 6= 1.

For σ < 1, kσ (x) > 0 in (0, 1), kσ (0) = 1 and kσ (x) → ∞ as x → 1− . By examining the derivative of kσ (x) kσ0 (x) =

−(1 − σ)((1 + x)−σ − 21−σ x−σ ) [(1 + x)1−σ − (2x)1−σ ]2

we see that

( lim+ kσ0 (x) =

x→0

−(1 − σ), σ < 0, 1, σ = 0, ∞, 0 < σ < 1.

For σ < 0, kσ (x) is minimized at 1

xσ := 1/(21− σ − 1) < 1. Calculations show that

¤−σ £ 1 =: k(σ). kσ (xσ ) = 1 − 2 σ −1


5

For σ ∈ [0, 1), kσ0 (x) is never zero in (0, 1) indicating that kσ (x) is minimized at x = 0 for σ ∈ [0, 1) and x ∈ [0, 1). The inequality (2.3) now follows. ¤ In order to treat the case in which p 6= 2, we make use of the methods of Tidblom [11] and prove a weighted version of Theorem 1.1 in [11]. Lemma 3. Let u ∈ C01 (0, 2b), p ∈ (1, ∞), and σ ≤ p − 1. Then R 2b 0

σ

0

p

h

ρ(t) |u (t)| dt ≥

p−σ−1 p

ip R

2b 0

{ρ(t)σ−p + (p − 1)bσ−p } |u(t)|p dt.

Proof. We may assume that σ 6= p − 1 since otherwise the conclusion is trivial. According to (2.2) for a monotonic function f and a positive function g, Rb 0

Rb |f 0 (t)||u(t)|p dt ≤ 0 p|f (t) − f (b)||u(t)|p−1 |u0 (t)|dt ¸1−1/p hR ´1/(p−1) i1/p ·R ³ b b |f (t)−f (b)|p p 0 p ≤ p 0 g(t)|u (t)| dt |u(t)| dt . g(t) 0

Consequently, Z

b

pp 0

³R

´p |f 0 (t)||u(t)|p dt g(t)|u0 (t)|p dt ≥ µ ³ ¶p−1 . R b |f (t)−f (b)|p ´1/(p−1) p |u(t)| dt g(t) 0 b 0

Now, as in [11], using a corollary to Young’s inequality, namely Ap /B p−1 ≥ pA − (p − 1)B, with A = lows that pp

Rb 0

Rb 0

|f 0 (t)||u(t)|p dt, B =

R b ³ |f (t)−f (b)|p ´1/p−1 g(t)

0

|u(t)|p dt, it fol-

0 g(t)|u½ (t)|p dt ¾ ³ ´ Rb p 1/(p−1) |f (t)−f (b)| ≥ 0 p|f 0 (t)| − (p − 1) |u(t)|p dt. g(t)

Choose f (t) = tσ−p+1 and g(t) = (p − σ − 1)−(p−1) tσ . Then ³

|f (t)−f (b)|p g(t)

´1/(p−1)

· = (p − σ − 1)

|tσ−p+1 −bσ−p+1 |

= (p − σ − 1)tσ−p

p

1 ¸ p−1

tσ

h³ 1−

1 ¡ t ¢p−σ−1 ´p i p−1

b

.

6


Consequently, for t ∈ (0, b) ³ ´1/(p−1) (b)|p p|f 0 (t)| − (p − 1) |f (t)−f g(t) ½ 1 ¾ h³ ¡ t ¢p−σ−1 ´p i p−1 σ−p σ−p 1− b = (p − σ − 1) pt − (p − 1)t ½ µ p ¶¾ h ¡ t ¢p−σ−1 i p−1 σ−p = (p − σ − 1)t 1 + (p − 1) 1 − 1 − b n ¡ t ¢p−σ−1 o σ−p ≥ (p − σ − 1)t 1 + (p − 1) b © σ−p ¡ 1 ¢ª ≥ (p − σ − 1) t + (p − 1) bp−σ . and the inequality follows. In the inequality above we have used the fact that p h ³ t ´p−σ−1 i p−1 ³ t ´p−σ−1 1− ≤1− . b b The proof is completed by following the last part of the proof of Lemma 1. ¤ For a certain range of values taken by σ, σ ∈ [−cσ , 1) with cσ > 0, the inequality in L2 (Ω) given by Lemma 2 gives a better bound than Lemma 3 with p = 2. In fact for σ < 1 h ³ 2ρ(t) ´1−σ i2 22−σ k(σ) 22−2σ k(σ)2 ρ(t)σ−2 1 + k(σ) = ρ(t)σ−2 + + µ(t) ρ(t)µ(t)1−σ ρ(t)σ µ(t)2−2σ with 22−σ k(σ) ρ(t)µ(t)1−σ

22−2σ k(σ)2 (2.8) ρ(t)σ µ(t)2−2σ ½ 5 σ−2 b , ¤ σ ∈ [0, 1), 2σ £ ≥ σ |σ| σ−2 2 − σ + b k(σ)ρ(t) k(σ)b , σ < 0. +

Since k(σ) decreases to 0 for σ < 0 as |σ| → ∞ and k(−3) ≈ 0.22, then the left-hand side of (2.8) is greater than bσ−2 for σ ∈ [−3, 1). 3. A Hardy inequality in L2 (Ω) We need the following notation (c.f.[6]). For each x ∈ Ω and ν ∈ Sn−1 , τν (x) := min{s > 0 : x + sν 6∈ Ω}; Dν (x) := τν (x) + τ−ν (x); ρν (x) := min{τν (x), τ−ν (x)}; µν (x) := max{τν (x), τ−ν (x)} = Dν (x) − ρν (x); D(Ω) :=

sup x∈Ω, ν∈Sn−1

Dν (x);

Ωx := {y ∈ Ω : x + t(y − x) ∈ Ω, ∀t ∈ [0, 1]}. Note that D(Ω) is the diameter of Ω and Ωx is the part of Ω which can be “seen” from the point x ∈ Ω. The volume of Ωx is denoted by |Ωx |.


7

n−1 (so that 1 = R Let dω(ν) denote the normalized measure on S dω(ν)) and define Sn−1 Z ρ(x; s) := ρν (x)s dω(ν). (3.1) Sn−1

Hence ρ−1/2 (x; −2) = ρ(x) the ”mean-distance” function introduced by Davies in [4]. For Z Γ( p+1 ) · Γ( n2 ) | cos(e, ν)|p dω(ν) = √ 2 e ∈ Rn , (3.2) B(n, p) := n+p , n−1 π · Γ( 2 ) S it is known that

Z

ρ(x; −p) := Sn−1

1 B(n, p) dω(ν) ≥ p ρν (x) δ(x)p

(3.3)

for convex domains Ω – see Exercise 5.7 in [4], [3], and [11]. Note that B(n, 2) = n−1 . This fact can be applied to most of the results below when Ω is convex. For a Hardy inequality in L2 (Ω) with weights we will need to define ³ s ´ 2(1−σ) n n−1 CH (n, σ) := n k(σ)[2|σ| + 22|σ|−1 k(σ)](1 − σ)2 (3.4) n , 0] and n ≥ 2 where as given in Lemma 2 for σ ∈ [ 2−n 2 ½ £ ¤ 1 −1 −σ σ 1 − 2 , σ < 0, k(σ) := 1, σ ∈ [0, 1]. Note that CH (n, 0) = 32 K(n) for K(n) defined in (1.3). Theorem 1. If 2−n ≤ σ ≤ 0, then for any u ∈ C01 (Ω) 2 Z Z ¯ ¯2 n(1 − σ)2 σ¯ ¯ δ(x) ∇u(x) dx ≥ ρ(x; σ − 2)|u(x)|2 dx 4 Ω Ω Z δ(x)|σ| + CH (n, σ) |u(x)|2 dx. (3.5) 2(1−σ) Ω |Ωx | n If 0 < σ ≤ 1, then ¯ R ¯ ¯∇u(x)¯2 dx ≥ Ω n 2σ n(1−σ)2 R 4D(Ω)σ

Ω

ρ(x; σ − 2) + 3

o ¡ sn−1 ¢ 2−σ n n|Ωx |

|u(x)|2 dx.

(3.6)

If Ω is convex, then for any u ∈ C01 (Ω) Z Z 2 ¯ ¯2 n(1 − σ) σ¯ δ(x) ∇u(x)¯ dx ≥ B(n, 2 − σ) δ(x)σ−2 |u(x)|2 dx 4 Ω Ω Z CH (n, σ) + δ(x)|σ| |u(x)|2 dx. 2(1−σ) Ω |Ω| n

8


when σ ∈ [ 2−n , 0] and 2 ¯ R ¯ ¯∇u(x)¯2 dx ≥ Ω n σ 2 R 2 n(1−σ) 4D(Ω)σ

Ω

B(n, 2 − σ)δ(x)σ−2 + 3

o ¡ sn−1 ¢ 2−σ n |u(x)|2 dx. n|Ω|

when σ ∈ (0, 1]. Proof. Let ∂ν u, ν ∈ Sn−1 , denote the derivative of u in the direction of ν, i.e., ∂ν u = ν · (∇u). It follows from Lemma 2 that for σ ∈ (−∞, 1] ¯ R σ ¯ ¯∂ν u¯2 dx ρ (x) ν Ω¡ ¢2 R ¡ £ 2ρν (x) ¤(1−σ) ¢2 (3.7) σ−2 ≥ 1−σ ρ (x) 1 + k(σ) |u(x)|2 dx. ν 2 µν (x) Ω Expanding the integrand in (3.7), we have ¡ £ ν (x) ¤(1−σ) ¢2 ρν (x)σ−2 1 + k(σ) 2ρ µν (x) −σ

(3.8)

−σ

ν (x) ν (x) ρν (x)σ−2 + 22−σ (τνk(σ)ρ + 22(1−σ) k(σ)2 µνρ(x) 2(1−σ) . (x)τ−ν (x))1−σ

= If σ ≤ 0

£ ¡ ν (x) ¢(1−σ) ¤2 ρν (x)σ−2 1 + k(σ) 2ρ µν (x) |σ|

|σ|

k(σ)δ(x) δ(x) 2(1−σ) ≥ ρν (x)σ−2 + 22−σ (τν (x)τ k(σ)2 τν (x)2(1−σ) 1−σ + 2 +τ−ν (x)2(1−σ) −ν (x)) (3.9) since ρν (x)−σ ≥ δ(x)|σ| in this case. As in [6], we note that since R R (τ (x)τ−ν (x))1−σ dω(ν) ≤ Sn−1 (τν (x))2(1−σ) dω(ν) Sn−1 ν £R ¤ 2(1−σ) n n ≤ (τ (x)) dω(ν) ν n−1 £ Sn ¤ 2(1−σ) = sn−1 |Ωx | n

for σ ≥ R

2−n , 2

then

1 dω(ν) Sn−1 (τν (x)τ−ν (x))1−σ

¤−1 (τν (x)τ−ν (x))1−σ dω(ν) £ n ¤− 2(1−σ) n ≥ sn−1 |Ωx | .

≥

£R

Sn−1

For the third term in inequality (3.9) R R 2(1−σ) 2(1−σ) (τ (x) + τ (x) )dω(ν) = 2 τ (x)2(1−σ) dω(ν) ν −ν n−1 S Sn−1 ν implying that for σ ≥ R Sn−1

2−n 2

(τν (x)2(1−σ) + τ−ν (x)2(1−σ) )−1 dω(ν) ≥

£ n ¤− 1 |Ω | x 2 sn−1

Consequently, for 2−n ≤ σ ≤ 0 we have that 2 £ ¡ ν (x) ¢(1−σ) ¤2 R ρ (x)σ−2 1 + k(σ) 2ρ dω(ν) µν (x) Sn−1 ν £ 2(1−σ) ¤ ≥ ρ(x; σ − 2) + CH (n, σ)δ(x)|σ| / n|Ωx | n .

2(1−σ) n

.

(3.10)


9

Upon combining this fact with (3.7) we have ³ ´2 R © |σ| ª 1−σ H (n,σ)δ(x) ρ(x; σ − 2) + Cn|Ω |u(x)|2 dx 2(1−σ)/n 2 Ω | x ¯ ¯2 R R ≤ Ω Sn−1 ρν (x)σ ¯∂ν u(x)¯ dω(ν)dx ¯ ¯2 R R = Ω δ(x)σ Sn−1 | cos(ν, ∇u(x))|2 dω(ν)¯∇u(x)¯ dx for σ ≤ 0. Since

Z | cos(ν, α)|2 dω(ν) = Sn−1 n−1

1 n

(3.11)

(3.12)

for any fixed α ∈ S (see Tidblom [11], p.2270), inequality (3.5) follows. For 0 < σ ≤ 1, we consider first the third term on the right-hand side of (3.8). We have R ρ (x)σ µν (x)2(1−σ) dω(ν) Sn−1R ν ≤ Sn−1 2−σ (τν (x) + τ−ν (x))σ (τν (x) + τ−ν (x))2(1−σ) dω(ν) = 2−σ kτν (x) + τ−ν (x)k2−σ L2−σ (Sn−1 ) £ ¤2−σ −σ ≤ 2 kτνR(x)kL2−σ (Sn−1 ) + kτ−ν (x)kL2−σ (Sn−1 ) = 22(1−σ) Sn−1 τν (x)2−σ dω(ν) £R ¤ 2−σ ≤ 22(1−σ) Sn−1 (τν (x))n dω(ν) n £ n ¤ 2−σ = 22(1−σ) sn−1 |Ωx | n for n ≥ 2 by the Minkowski and H¨older inequalities. Therefore, the term ¢ 2−σ ¡ R ρν (x)−σ dω(ν) 2(σ−1) sn−1 n ≥ 2 . n−1 2(1−σ) n|Ωx | S µν (x) Similarly, in the second term of (3.8) R ρν (x)µν (x)1−σ dω(ν) Sn−1 R ≤ 21 Sn−1 (τν (x) + τ−ν (x))(τν (x) + τ−ν (x))1−σ dω(ν) £ n ¤ 2−σ ≤ 21−σ sn−1 |Ωx | n as before implying that R Sn−1

dω(ν) ρν (x)µν (x)1−σ

≥ 2σ−1

¡ sn−1 ¢ 2−σ n n|Ωx |

.

For 0 < σ < 1 we now have that £ ¡ ν (x) ¢(1−σ) ¤2 R ρ (x)σ−2 1 + k(σ) 2ρ dω(ν) µν (x) Sn−1 ν ¡ sn−1 ¢ 2−σ ≥ ρ(x; σ − 2) + 3 n|Ωx | n since k(σ) = 1 in this case. Consequently, ¯ ¯2 R R ρ (x)σ | cos(ν, ∇u(x))|2 dω(ν)¯∇u(x)¯ dx Ω ³Sn−1´ ν 2R £ ¡ sn−1 ¢ 2−σ ¤ n ρ(x; σ − 2) + 3 n|Ω ≥ 1−σ |u(x)|2 dx. 2 | Ω x

10


According to (3.12) it follows that Z D(Ω)σ ρν (x)σ | cos(ν, ∇u(x))|2 dω(ν) ≤ σ . 2 n Sn−1 Therefore, (3.6) holds. The inequalities in the statement of the theorem for the case of a convex domain Ω follow from (3.3) and the fact that |Ωx | = |Ω| for all x ∈ Ω. ¤ 4. An Lp (Ω) inequality With the guidance of Tidblom’s analysis for the Hardy inequality in [11], Lp versions of the weighted Hardy theorem in the last section can be proved by similar techniques. When σ = 0, the next theorem reduces to Theorem 2.1 of [11]. Theorem 2. Let u ∈ C01 (Ω) and p ∈ (1, ∞). If σ ≤ 0, then for B(n, p) defined in (3.2) R δ(x)σ |∇u(x)|p dx ≥ Ω n h o i p−σ (4.1) n [|p−σ−1|/p]p R sn−1 ρ(x; σ − p) + (p − 1) |u(x)|p dx B(n,p) n|Ωx | Ω and if σ ∈ [0, p − 1], then R |∇u(x)|p dx ≥ Ω n h o i p−σ n 2σ [|p−σ−1|/p]p R sn−1 ρ(x; σ − p) + (p − 1) |u(x)|p dx. B(n,p) D(Ω)σ Ω n|Ωx | (4.2) If Ω is convex, ρ(x, σ − p) can be replaced in (4.1) and (4.2) by the term B(n, p − σ)/δ(x)p−σ (in view of (3.3)) and |Ωx | by |Ω|. Proof. From Lemma 3 we have that for σ ≤ p − 1, any ν ∈ Sn−1 , and u ∈ C01 (Ω) R ρ (x)σ |∂ν u(x)|p dx ≥ Ω ν£ ¤ © p−σ ª (4.3) |p−σ−1| p R ρν (x)σ−p + (p−1)2 |u(x)|p dx. p Dν (x)p−σ Ω If σ ≤ 0 we bound ρν (x)σ for any ν ∈ Sn−1 by δ(x)σ in the first integral above. If σ > 0, we bound it by D(Ω)σ /2σ . As in [11] we may use the fact that Z |∂ν u(x)|p dω(ν) = B(n, p)|∇u(x)|p . (4.4) Sn−1

After bounding ρν (x)σ as described above, integrate in (4.3) over Sn−1 with respect to dω(ν). In order to evaluate the integral of (2/Dν (x))p−σ , we proceed as in [11]. Since σ ≤ p − 1, then f (t) = tσ−p is convex for t > 0 and we have that Z ³Z ´σ−p ³ n|Ω | ´− p−σ ³ 2 ´p−σ Dν (x) n x dω(ν) ≥ dω(ν) ≥ 2 sn−1 Sn−1 Sn−1 Dν (x) (4.5)


11

by Jensen’s inequality and Lemma 2.1 of [11]. The conclusion follows. ¤ 5. Rellich’s inequality The methods described above with Proposition 1 below can be used to prove a weighted Rellich inequality which, for n ≥ 4 and without weights, improves the constant given in a Rellich inequality proved recently by Barbatis ([1], Theorem 1.2). A comparison is made below. The methods used by Barbatis depends upon the identity (5.2) first proved by M.P. Owen ([10], see the proof of Theorem 2.3). In order to incorporate weights, our proof requires the point-wise identity (5.1) which does not follow from the proof of Owen. Proposition 1. Let Ω be a domain in Rn . Then, for all u ∈ C 2 (Rn ) Z n ¯ 2 ¯i h X 1 ¯ ∂ u(x) ¯2 2 2 2 |∂ν u(x)| dω(ν) = |∆u(x)| + 2 ¯ ¯ , (5.1) n(n + 2) ∂xi ∂xj Sn−1 i,j=1 and for all u ∈ C02 (Ω) Z Z |∂ν2 u(x)|2 dω(ν)dx = Ω

Sn−1

3 n(n + 2)

Z |∆u(x)|2 dx.

(5.2)

Ω

Proof. For ν = (ν1 , . . . , νn ) we have P ∂ν2 u = (ν · ∇)2 u = Pn`,m=1 ν` νm u`m P n 2 = `=1 ν` u`` + 2

ν` νm u`m

1≤`