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2 2
ai,j uxi uxj dx C ∆u dx , R2 ≤ R2 | | Z i,j=1 Z X ∞ R2 where ai,j = const and u is an arbitrary complex-valued function in C0 ( ), may hold if and only if a11 +a22 = 0. We also find the best constant C in the last inequality. This is a particular case of Theorem 1 proved in Section 1. In Section 2 we establish a new weighted G˚arding type inequality
2 2 −1 2 −1 2u log(e x ) dx Re ∆ u u log x dx (3) R2 |∇ | | | ≤ R2 · | | Z Z ∞ R2 for all u C0 ( 0 ). Estimates of such a kind proved to be useful in the study of boundary∈ behavior\{ of} solutions to elliptic equations (see [M4], [MT], [M6], [M7], [E], [MM]). Before turning to the contents of the next section, we introduce some notation. Rn Rn Rn−1 By + we denote the half-space x = (x1,...xn) , xn > 0 . Also let = Rn ∞ Rn ∞ {Rn ∈ } ∂ +. As usual, C0 ( +) and C0 ( +) stand for the spaces of infinitely differentiable Rn Rn functions with compact support in + and +, respectively. In Section 3 we are concerned with the inequality
2 2 u ∞ Rn xn u dx Λ 2 | | 2 1/2 dx, u C0 ( +). (4) Rn |∇ | ≥ Rn (x + x ) ∈ Z + Z + n−1 n It was obtained in 1972 by one of the authors and proved to be useful in the study of the generic case of degeneration in the oblique derivative problem for second order elliptic differential operators [M2]. −1/2 By substituting u(x)= xn v(x) into (4), one deduces with the same Λ that
2 2 2 1 v dx v dx v dx | | 2 +Λ 2| | 2 1/2 (5) Rn |∇ | ≥ 4 Rn xn Rn xn(x + x ) Z + Z + Z + n−1 n for all v C∞(Rn ) (see [M5], Sect. 2.1.6). ∈ 0 + Another inequality of a similar nature obtained in [M5] is
2 2 1 v γ 2 v dx | | dx + C x v Rn . (6) 2 n Lq( +) Rn |∇ | ≥ 4 Rn xn k k Z + Z + (This is a special case of inequality (2.1.6/3) in [M5].) Without the second term in the right-hand sides of (5) and (6), these inequalities reduce to the classical Hardy inequality with the sharp constant 1/4 (see [Da]). An interesting feature of (5) and (6) is their dilation invariance. Variants, extensions, and refinements of (5) and (6), usually called Hardy’s inequal- ities with remainder term, became the theme of many subsequent studies ([ACR], [Ad], [AGS], [BCT], [BFT1], [BFT2], [BFL], [BM], [BV], [CM], [DD], [DNY], [EL],
2 [FMT1]–[FMT3], [FT], [FTT], [FS], [GGM], [HHL], [TT], [TZ], [Ti1], [Ti2], [VZ], [YZ] et al). In Theorem 3, proved in Section 3, we find a condition on the function q which is necessary and sufficient for the inequality
2 2 2 1 v dx xn v dx v dx | | 2 C q 2 2 1/2 2| | 2 1/2 , (7) Rn |∇ | − 4 Rn xn ≥ Rn (xn−1 + xn) xn (xn−1 + xn) Z + Z + Z + ∞ Rn where v is an arbitrary function in C0 ( +). This condition implies, in particular, that the right-hand side of (5) can be replaced by
v 2dx C | | 2 . Rn xn + 2 Z xn 1 log 2 2 1/2 − (xn−1 + xn) The value Λ = 1/16 in (4) obtained in [M2] is not best possible. Tidblom replaced it by 1/8 in [Ti2]. As a corollary of Theorem 3, we find an expression for the optimal value of Λ.
Let the measure µb be defined by dx µ (K)= (8) b x b ZK | | for any compact set K in Rn. In Section 4 we obtain the best constant in the inequality
1/p p dx u Lτ,q(µb) C u(x) a , k k ≤ Rn |∇ | x Z | | where the left-hand side is the quasi-norm in the Lorentz space (µ ), i.e. Lτ,q b ∞ q/τ 1/q u = µ x : u(x) t d(tq ) . k kLτ,q(µb) b{ | |≥ } Z0 As a particular case of this result we obtain the best constant in the Hardy-Sobolev inequality 1/q 1/p q dx p dx u(x) b u(x) a , (9) Rn | | x ≤C Rn |∇ | x Z | | Z | | first proved by Il’in in 1961 in [Il] (Th. 1.4) without discussion of the value of . Our result is a direct consequence of the capacitary integral inequality from [M7] combinedC with an isocapacitary inequality. For particular cases the best constant was found in [CW] (p = 2), in [M3], Sect. 2 (p = 1, a = 0), in [GMGT] (p = 2, n =C 3, a = 0), in [L] (p = 2, n 3, a = 0), and in [Na] (1
3 1 Estimate for a quadratic form of the gradient
n Theorem 1 Let n 2 and let A = ai,j i,j=1 be an arbitrary matrix with constant complex entries. The≥ inequality k k
n+2 2 A u, u Cn dx C ( ∆) 4 u dx , (10) Rn h ∇ ∇ i ≤ Rn − Z Z ∞ Rn where C is a positive constant, holds for all complex-valued u C0 ( ) if and only if the trace of A is equal to zero. The best value of C is given∈ by
(4π)−n/2 C = max ai,j ωiωj , (11) Γ n +1 ω∈Sn−1 2 1≤i,j≤n X where Sn−1 is the (n 1)-dimensional unit sphere in Rn. − (The notation ( ∆)s in (10) stands for an integer or noninteger power of ∆.) − − Proof. By we denote the unitary Fourier transform in Rn defined by F h(ξ)=(2π)−n/2 h(x) e−ix·ξ dx. (12) F Rn Z We set h = ( ∆)(n+2)/4u and write (10) in the form − 2 2 ξ ξ dξ h(ξ) A , n C h(x) dx . (13) Rn |F | ξ ξ Cn ξ ≤ Rn | | Z D | | | |E | | Z
The singular integral in the left-hand side exists in the sense of the Cauchy principal value, since −1 n−1 Aω,ω Cn dsω = n S Tr A =0, n−1 h i | | ZS where Tr A is the trace of A (see, for example, [MP], Ch. 9, Sect. 1 or [SW], Theorem 4.7). Let ξ ξ k(ξ)= ξ −n A , . | | ξ ξ Cn D | | | | E The left-hand side in (13) equals
−1 k(ξ) h (ξ) (x) h(x) dx = (2π)−n/2 ( −1k) h (x) h(x) dx Rn F F Rn F ∗ Z Z with meaning the convolution. Thus, inequality (13) becomes ∗ 2 ( −1k) h (x) h(x) dx (2π)n/2 C h(x) dx . (14) Rn F ∗ ≤ Rn | | Z Z We note that for ξ Rn, ∈ n n k(ξ)= ξ −n−2 a ξ2 n−1 ξ 2 + a ξ ξ . (15) | | jj j − | | ij i j j=1 i,j=1 X Xi=6 j Hence, for n> 2 n 1 ∂2 k(ξ)= a ξ 2−n, (16) n (n 2) ij ∂ξ ∂ξ | | i,j=1 i j − X
4 and for n =2 2 1 ∂2 k(ξ)= a log ξ −1. (17) 2 ij ∂ξ ∂ξ | | i,j=1 i j X Applying −1 to the identity F ∂2 ξ 2−n ∂2 ∆ξ n−|1| = δ(ξ), − ∂ξi∂ξj S (n 2) ∂ξi∂ξj | | − where n> 2 and δ is the Dirac function, we obtain from (16)
Sn−1 n x x −1k (x)= −| | a i j . (18) F n (2π)n/2 ij x 2 i,j=1 X | | Here Sn−1 stands for the (n 1)-dimensional measure of Sn−1: | | − n n−1 2π 2 S = n . (19) | | Γ( 2 ) Hence 2−n/2 n x x −1k (x)= − a i j . (20) F Γ(1 + n ) ij x 2 2 i,j=1 X | | Similarly, we deduce from (17) that (20) holds for n = 2 as well. Now, (10) with C given by (11) follows from (20) inserted into (14). Next, we show the sharpness of C given by (11). Let θ denote a point on Sn−1 such that −1k (θ) = max −1k(ξ) . (21) F ξ∈Rn\{0} |F | In order to obtain the required lower estimate for C, it suffices to set x h(x)= η( x ) δ , | | θ x | | ∞ n−1 where η C0 [0, ), η 0, and δθ is the Dirac measure on S concentrated at θ, into the∈ inequality∞ (14).≥ (The legitimacy of this choice of h can be easily checked by approximation.) Then estimate (14) becomes ∞ ∞ ρ r −1k − θ η(r) rn−1 η(ρ) ρn−1drdρ 0 0 F ρ r Z Z | − | ∞ 2 (2π)n/2 C η(ρ) ρn−1dρ . (22) ≤ 0 Z In view of (18) and (20),
−1k ( θ)= −1k (θ) F ± F which together with (21) enables one to write (22) in the form
max −1k(ξ) (2π)n/2 C. ξ∈Rn\{0} |F |≤ By (18) and (20) this can be written as
n−1 S n/2 | | max aij ωiωj (2π) C. n (2π)n/2 ω∈Sn−1 ≤ 1≤i,j≤n X
5 The result follows from (19).
Remark 1 Let P and Q be functions, positively homogeneous of degrees 2m and m + n/2 respectively, m > n/2. We assume that the restrictions of P , Q, and P Q −2 to Sn−1 belong to L−1(Sn−1), By the same argument as in Theorem 1, one concludes| | that the condition P (ω) 2 dsω = 0 (23) n−1 Q(ω) ZS | | is necessary and sufficient for the inequality
2 P (D)u u dx C Q(D) u dx (24) Rn · ≤ Rn Z Z ∞ Rn to hold for all u C 0 ( ). Moreover, using the classical formula for the Fourier transform of a positively∈ homogeneous function of degree n (see Theorem 4.11 in [SW]1), one obtains that the best value of C in (24) is given− by
iπ P (θ) sup sgn(θ ω) + log θ ω 2 dsθ . (25) ω∈Sn−1 Sn−1 2 · | · | Q(θ) Z | |
In particular, if P (ω)/ Q(ω) 2 is a spherical harmonic, the best value of C in (24) is equal to | | (4π)−n/2Γ(m) P (ω) n max | | , (26) Γ + m ω∈Sn−1 Q(ω) 2 2 | | which coincides with (11) for m = 1, P(ξ)= Aξ ξ, and Q(ξ)= ξ 1+n/2. · | | 2 Weighted G˚arding inequality for the biharmonic operator
We start with an auxiliary Hardy type inequality.
Lemma 1 Let u C∞(R2). Then the sharp inequality ∈ 0
dx 2 Re x1ux1 + x2ux2 ∆u 2 ∆u dx (27) R2 x ≤ R2 | | Z | | Z holds.
Proof. Let (r, ϕ) denote polar coordinates in R2 and let
∞ ikϕ u(r, ϕ)= uk(r)e . k=X−∞ Then (27) is equivalent to the sequence of inequalities
∞ 2 ∞ 2 2 ′′ 1 ′ k ′ ′′ 1 ′ k Re v + v 2 v v dr v + v 2 v r dr, k =0, 1, 2,... (28) 0 r − r ≤ 0 r − r Z Z 1 Note that the definition of the Fourier transform in [SW] contains exp(−2πix · ξ) unlike (12).
6 ∞ −1 where v is an arbitrary function on C0 ([0, )). Putting t = log r and w(t) = v(e−t), we write (28) in the form ∞
2 Re w′′ k2w w′ e2t dt w′′ k2w e2t dt R1 − ≤ R1 − Z Z which is equivalent to the inequality
2 Re g′′ 2g′ + (1 k2)g g′ g dt g′′ 2g′ + (1 k2)g dt, (29) R1 − − − ≤ R1 − − Z Z where g = etw. Making use of the Fourier transform in t, we see that (29) holds if and only if for all λ R1 and k =0, 1, 2 ... ∈ 2 Re λ2 +1 k2 2iλ 1 iλ λ2 1+ k2 +4λ2, − − − − ≤ −