Table Of ContentWEIGHTED INEQUALITIES AND STEIN-WEISS POTENTIALS
WILLIAM BECKNER
Abstract. Sharp extensions of Pitt’s inequality and bounds for Stein-Weiss fractional integrals are
7
obtained that incorporate gradient forms and vector-valued operators. Such results include Hardy-
0
Rellichinequalities.
0
2
n
Weightedinequalitiesprovidequantitativeinformationtocharacterizeintegrabilityfordifferentialand
a
J integral operators and intrinsically are determined by their dilation character. In the classical context,
4 weighted inequalities for the Fourier transform provide a natural measure of uncertainty. For functions
onRn theissueisthebalancebetweentherelativesizeofafunctionanditsFouriertransformatinfinity.
] An inequality that illustrates this principle at the spectral level is Pitt’s inequality:
P
A
(1) Φ(1/x)f(x)2dx C Φ(y )f(y)2dy
h. ZRn | | | | ≤ ΦZRn | | | |
at whereΦisanincreasingfunction,thefunctionf isintheSchwartzclbass (Rn)andtheFouriertransform
S
m is defined by
[ ( f)(y)=f(y)= e2πixyf(x)dx .
2 F ZRn
v Such inequalities may be fully determined bybdilation invariance, and some cases may be realized with
9 explicit gradient forms as Hardy-Rellich inequalities. In earlier work (see [3]) the effective calculation
0 fortheconstantinPitt’sinequalitywasreducedtoYoung’sinequalityforconvolutiononanon-compact
3
unimodular group. The objective here will be to study more general forms of Pitt’s inequality
7
0
6 (2) Φ(1/x) f 2dx 4π2DΦ Φ(y )y 2 f(y)2dy
0 ZRn | | |∇ | ≤ ZRn | | | | | |
/ usingthestructureofStein-WeisspotentialsandconvolutionestimatesinbconcertwiththeHecke-Bochner
h
t representationfor L2(Rn). The previous work is described by the following three theorems.
a
m Theorem 1 (Pitt’s inequality). For f (Rn) and 0 α<n
∈S ≤
:
v (3) x−α f(x)2dx C y α f(y)2dy
Xi ZRn| | | | ≤ αZRn| | | |
r n α n+αb 2
a C =πα Γ − Γ .
α
4 4
h (cid:16) (cid:17). (cid:16) (cid:17)i
Since the above inequality becomes an identity for α = 0, a differentiation argument provides a loga-
rithmic form that controls the uncertainty principle by using dimensional asymptotics.
Theorem 2 (logarithmic uncertainty). For f (Rn)
∈S
(4) ln x f(x)2dx+ ln y f(y)2dy D f(x)2dx
ZRn | || | ZRn | || | ≥ Z | |
b d
D =ψ(n/4) lnπ , ψ(t)= lnΓ(t) .
− dt
Logarithmic integrals are indeterminate so D may take negative values. The proof of Pitt’s inequality
(3)followsfromasharpestimateforanequivalentintegralrealizationasaStein-Weissfractionalintegral
on Rn.
1
Theorem 3. For f L2(Rn) and 0<α<n
∈
1 1 1
(5) f(x) f(y)dxdy B f(x)2dx
(cid:12)ZRn×Rn |x|α/2 |x−y|n−α|y|α/2 (cid:12)≤ αZRn| |
(cid:12) (cid:12)
(cid:12) α n α n α(cid:12) n+α 2
B =πn/2 Γ Γ − Γ − Γ .
α
2 2 4 4
h (cid:16) (cid:17). (cid:16) (cid:17)ih (cid:16) (cid:17). (cid:16) (cid:17)i
The development of the sharp estimate for the Stein-Weiss integral rests on using symmetrization to
reduce the problem to radial functions, and then as a consequence of dilation invariance the estimate
can be convertedto Young’s inequality for convolution on the multiplicative groupR (or alternatively
+
on R).
ϕ f ϕ f
L2(G) L1(G) L2(G)
k ∗ k ≤k k k k
When ϕ is non-negative, the inequality is sharp with no extremal functions. One objective here will
be to extend this inequality to cases when ψ takes both positive and negative values by using the
Hecke-Bochner formulas.
Since the form
f(y)2 y αdy =(2π)−α ( ∆)α/4f 2dx
ZRn| | | | ZRn| − |
canbe regardedinthe family ofgradientestimates,Pitt’sinequality hasbeencharacterizedasa Hardy-
b
Rellich inequality in some parts of the recent literature with alternative proofs and extensions (see [1],
[7], [14], and [17]). Some natural generalizations of Pitt’s inequality can now be viewed in the context
of the arguments developed in [3] where proofs of Theorems 1, 2 and 3 are given.
1. Pitt’s inequality with gradient terms
Theorem 4. For f (Rn) and 0<α<n, n>1
∈S
(6) f 2 x−αdx 4π2D f(y)2 y α+2dy
α
ZRn|∇ | | | ≤ ZRn| | | |
n+2k α+2 n+2k+αb+2 2 4kα
D =παmax Γ − Γ 1+
α k 4 4 (n+2k α 2)2
(cid:26)h (cid:16) (cid:17). (cid:16) (cid:17)i (cid:16) − − (cid:17)(cid:27)
By convention the last term in the line above is one when k = 0. This result is interesting for several
aspects:
(a) for the gradient term with radial functions (i.e. k = 0) the constant is reduced from that in
equality (3) since Γ(x+β)/Γ(y+β) is decreasing in β for x < y; this reduction in constant is
also apparent by the Plancherel theorem;
(b) the corresponding Stein-Weiss integral does not have a positive kernel though it does have
symmetry in the angular variables;
(c) the limiting logarithmic uncertainty is sharper
ln x f 2dx+ ln y (4π2 y 2)f(y)2dy E f 2dx
ZRn | ||∇ | ZRn | | | | | | ≥ ZRn|∇ |
3 b
ψ lnπ 1 , n=2
2 − −
E =
(cid:16) (cid:17)
n 1
ψ + lnπ , n 3
4 2 − ≥
(cid:16) (cid:17)
(d) for α=2 and n>4 one obtainsthe Hardy-Rellich inequality
4
(7) f 2 x−2dx ∆f 2dx ;
ZRn|∇ | | | ≤ n2 ZRn| |
and (e) forsomevaluesoftheparametersnandα,thesharpboundisobtainedbyconsideringnon-radial
functions. A sharper version of (7) appears in [14].
2
By using the Fourier transform on Riesz potentials
n α α
x−α =π−n2 +α Γ − Γ x−n+α ,
F | | 2 2 | |
one easily sees that Pitt’s in(cid:2)equali(cid:3)ty in Theohrem(cid:16) 4 is e(cid:17)q.uiv(cid:16)alen(cid:17)tito a Stein-Weiss fractional integral
inequality on Rn.
Theorem 5. For f L2(Rn) and 0<α<n, n>1
∈
1 x y 1 1
(8) f(x) · f(y)dxdy
(cid:12)ZRn×Rn |x|α/2 |x||y| |x−y|n−α |y|α/2 (cid:12)
(cid:12) (cid:12)
(cid:12) πn2−αΓ α Γ n−α Dα f 2dx (cid:12)
≤h (cid:16)2(cid:17). (cid:16) 2 (cid:17)i ZRn| |
with D as in Theorem 4.
α
Proof of Theorems 4 and 5. The main difficulty with showing these estimates is that the Stein-Weiss
kernel is not positive. This obstruction is addressed in two steps.
Step 1. Assume f is radial and set t= x, h(t)= xn/2f(x); then (8) reduces to
| | | |
ds ds n α n α dt
(9) h(t)ψ (s/t)h(s) D Γ Γ 2παΓ − h2
α α
R ×R s t ≤ 2 2 2 R | | t
(cid:12)Z + + (cid:12) h (cid:16) (cid:17) (cid:16) (cid:17). (cid:16) (cid:17)iZ +
(cid:12) (cid:12)
with (cid:12) (cid:12)
1 −(n−α)/2
ψ (t)= ξ t+ 2ξ dξ
α 1 1
ZSn−1 h t − i
where dξ denotes normalized surface measure, ξ is the first component of ξ and D is specified to be
1 α
the best constant in (6). Note that by monotonicity and symmetry ψ (t) is positive, and moreoverany
α
kernel derived from (6) will be positive-definite. Then using Young’s inequality
kψα∗hkL2(R+) ≤kψαkL1(R+)khkL2(R+)
which is sharp, it suffices to calculate the L1 norm of ψ to obtain the constant D
α α
n n α
kψαkL1(R+) =Dα Γ 2 Γ(α/2) 2παΓ −2 .
h (cid:16) (cid:17) . (cid:16) (cid:17)i
To compute this integral, observe that
∞ 1 −(n−α)/2 dt
kψαkL1(R+) =Z0 hZSn−1ξ1ht+ t −2ξ1i dξi t
2πn/2 −1 x y 1 1
= · dy
hΓ(n/2)i ZRn |x||y| |x−y|n−α |y|(n+α)/2
for x = 1 and n > 1. The second integral will be calculated for the set of values n 2 > α > 0. But
| | −
noticethatthefirstintegralisananalyticfunctionoftheparametersn 2andβ =n αforsomerange
≥ −
of values. Hence any computation for some parameter domain will determine by analytic continuation
the value of kψαkL1(R+) for the desired parameter interval 0 < α < n with n > 1. Then (noting that
x =1)
| |
2x y
· x y −(n−α) y −(n+α)/2dy =
ZRn |x||y| | − | | |
x y −(n−α) y −(n+α)/2 −1dy+ x y −(n−α) y −(n+α)/2 +1dy
ZRn| − | | | Z | − | | |
x y −(n−α−2) y −(n+α)/2−1dy =I +I I
1 2 3
−ZRn| − | | | −
3
These integrals are computed by using the formula for the convolution of two Riesz potentials
Γ(n−β)Γ(n−δ)Γ(β+δ−n)
(10) x−β x−δ =πn/2 2 2 2 x−(β+δ−n)
| | ∗| | " Γ(β)Γ(δ)Γ(2n−β−δ) #| |
2 2 2
with 0<β <n, 0<δ <n and n<β+δ <2n. Then
Γ(α)Γ(n−α−2)Γ(n−α+2)
I =πn/2 2 4 4
1 Γ(n−α)Γ(n+α+2)Γ(n+α−2)
(cid:20) 2 4 4 (cid:21)
Γ(α)Γ(n−α+2)Γ(n−α−2)
I =πn/2 2 4 4
2 Γ(n−α)Γ(n+α−2)Γ(n+α+2)
(cid:20) 2 4 4 (cid:21)
Γ(α+2)Γ(n−α−2)Γ(n−α−2)
I =πn/2 2 4 4
3 Γ(n−α−2)Γ(n+α+2)Γ(n+α+2)
(cid:20) 2 4 4 (cid:21)
and
Γ(α) Γ(n−α−2) 2 n α 2 n+α 2 α n α 2
I +I I =πn/2 2 4 2 − − − − −
1 2− 3 Γ(n−α) Γ(n+α+2) 4 4 − 2 2
2 (cid:20) 4 (cid:21) (cid:20) (cid:16) (cid:17)(cid:16) (cid:17) (cid:16) (cid:17)(cid:21)
Γ(α) Γ(n−α−2) 2 n α 2 2 Γ(α) Γ(n−α+2) 2
=πn/2 2 4 2 − − =2πn/2 2 4 .
Γ(n−α) Γ(n+α+2) 4 Γ(n−α) Γ(n+α+2)
2 (cid:20) 4 (cid:21) (cid:20) (cid:16) (cid:17) (cid:21) 2 (cid:20) 4 (cid:21)
This demonstrates that
Γ(n)Γ(α) Γ(n−α+2) 2
kψαkL1(R+) = 2Γ2(n−α2) Γ(n+α4+2)
2 (cid:20) 4 (cid:21)
and that for radial functions in (6)
Γ(n−α+2) 2
D =πα 4 .
α Γ(n+α+2)
(cid:20) 4 (cid:21)
Step 2. The Hecke-Bochner representation for L2(Rn) is used to reduce the study of inequality (6)
f 2 x−αdx 4π2D f(y)2 y α+2dy
α
ZRn|∇ | | | ≤ ZRn| | | |
to estimates for radial functions. For f (Rn) b
∈S
∞
f(x)= f (x)P (x)
k k
| |
k=0
X
where P is a harmonic polynomial of degree k,
k
P (x)= xkY (ξ) , ξ = x , Y (ξ)2dξ = ωn−1+2k
k k k
| | |x| ZSn−1| | ωn−1
Y is a spherical harmonic of degree k, ω = surface area of the unit sphere Sm, and dξ is normalized
k m
surface measure on Sn−1. Then
∞
f 2dx= f (x)2dx .
k
ZRn| | k=0| | | |
X
Let denote the Fourier transform on Rn. Bochner’s relation for spherical harmonics is
n
F
(11) f (x)P (x) =ik (f (x))P .
n k k n+2k k k
F | | F | |
and the integral on the right-hand(cid:0)side of (6) b(cid:1)ecomes
∞
f(y)2 y α+2dy = f (y )2 y α+2dy .
k
ZRn| | | | k=0ZRn+2k| | | | | |
X
b b
4
For the integral on the left-hand side in (6)
∞
f 2 x−αdx= [f (x)P (x)]2 x−αdx
k k
ZRn|∇ | | | k=0ZRn|∇ | | | | |
X
∞ ∞
= f (x)2 x−αdx+ kα f (x)2 x−α−2dx .
k k
k=0ZRn+|∇2k | | | | | k=1 ZRn+|2k | | | | |
X X
Using inequality (6) for radial functions from Step 1,
Γ(n+2k−α+2) 2
f (x)2 x−αdx 4π2+α 4 f (y)2 y α+2dy
ZRn+|∇2k k | | | | | ≤ "Γ(n+2k4+α+2)# ZRn+|2kk | | |
and using inequality (3) b
Γ(n+2k−α−2) 2
f (x)2 x−α−2dx π2+α 4 f (y)2 y α+2dy
ZRn+|2kk | | | | | ≤ "Γ(n+2k4+α+2)# ZRn+|2kk | | |
one obtains b
f 2 x−αdx
ZRn|∇ | | |
∞ Γ(n+2k−α+2) 2 kα Γ(n+2k−α−2) 2
4π2+α 4 + 4 f (y)2 y α+2dy
≤ k=0"(cid:20)Γ(n+2k4+α+2)(cid:21) 4 (cid:20)Γ(n+2k4+α+2)(cid:21) #ZRn+|2kk | | |
X
Γ(n+2k−α+2) 2 4kα b
4π2+αmax 4 1+ f(y)2 y α+2dy
≤ k ((cid:20)Γ(n+2k4+α+2)(cid:21) (cid:20) (n+2k−α−2)2(cid:21))ZRn| | | |
which demonstrates inequality (6), Theorem 4 and the equivalent Theorem 5. b
Corollary 1. For α=2
144
, if n=3 (k =1 term)
25
(12) Dπ22 = 34 , if n=4 (k =1 term)
16
, if n>4 (k =0 term)
This result recovers a “classicalHardy-Relnli2ch inequality” for n>4 (see remarks in [14])
4
f 2 x−2dx ∆f 2dx .
ZRn|∇ | | | ≤ n2 ZRn| |
Notice that for n=8 this inequality is entirely elementary since integrating by parts
1
f 2 x−2dx= f(∆f)x−2dx 8 f 2 x−4dx ∆f 2dx .
|∇ | | | − | | − | | | | ≤ 16 | |
Z Z Z Z
In comparing terms to evaluate constants explicitly, the following fact is useful: for 0 < x < y, the
ratio Γ(x+β)/Γ(y+β) is decreasing for β >0. Set F(β)=lnΓ(x+β) lnΓ(y+β); then
−
∞
F′(β)=ψ(x+β) ψ(y+β)= (y x) (x+β+k)−1(y+β+k)−1 <0 .
− − −
k=0
X
For n=2, the expression
Γ(n+2k−α+2) 2 4kα
4 1+
"Γ(n+2k4+α+2)# (cid:18) (n+2k−α−2)2(cid:19)
is decreasing for k 1 since both terms in the product are decreasing so the value of D is found by
α
≥
comparing the terms for k =0 and k =1.
5
Corollary 2. For n=2
Γ(3 α) 2 4+α2
(13) D =πα 2 − 4 .
α Γ(3 + α) (2 α)2
(cid:20) 2 4 (cid:21) (cid:18) − (cid:19)
Proof. Set β =α/4 with 0 β < 1 since α<n=2. Then to see that the k =1 term is largerthan the
≤ 2
k =0 term, consider the log of the ratio of these two terms and it suffices to show that
1 3 1 1
F(β)=lnΓ β +lnΓ(1+β) lnΓ(1 β) lnΓ +β + ln +β2
2 − − − − 2 2 4
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
is positive for β >0 which will follow by showing that F(β) is increasing with f(0)=0.
1 3 4β
′
F (β)= 4 β +ψ(1+β)+ψ(1 β) ψ +β +
− 2 − − − 2 1+4β2
(cid:16) (cid:17) (cid:16) (cid:17)
∞
1 1 −1 3 −1
= +2β β+k (1+β+k)−1 (1 β+k)−1 +β+k
2 2 − − − 2
(cid:16) (cid:17)kX=0(cid:20)(cid:16) (cid:17) (cid:16) (cid:17) (cid:21)
4β
+ >0
1+4β2
′
F (β)>0 and F(0)=0 ensure that F(β) is positive.
Corollary 3. For n 3
≥
Γ(n+2k−α+2) 2 4kα
(14) D =πα max 4 1+ .
α k=0,1((cid:20)Γ(n+2k4+α+2)(cid:21) (cid:18) (n+2k−α−2)2(cid:19))
Proof. The objective here is to show that only the k = 0 and k = 1 terms in Theorem 4 need to be
compared. That is, the only functions that are necessary to consider in Theorem 4 are those contained
in the span of functions with spherical harmonics up to degree one. Set β =2k and show that
Γ(n+β−α+2) 2βα 1/2
G(β)=ln 4 1+
((cid:20)Γ(n+β+4α+2)(cid:21)(cid:18) (n+β−α−2)2(cid:19) )
is decreasing for β 2 with 0<α<n; then
≥
1 n+β α+2 n+β+α+2
′
G(β)= ψ − ψ
4 4 − 4
(cid:20) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21)
+α 1 2β (n+β α 2)2+2βα −1
− n+β α 2 − −
(cid:20) − − (cid:21)
(cid:2) (cid:3)
1 ∞ n+β+2 2 α2 −1
=α +k
"− 8k=0(cid:20)(cid:18) 4 (cid:19) − 16(cid:21)
X
+ (n−β−α−2) (n+β α 2)2+2βα −1 .
(n+β α 2) − − #
− −
(cid:2) (cid:3)
This derivative is clearly negative for n = 3,4 and β 2, n > α > 0 since n β α 2 < 0. Now
≥ − − −
consider n 5 with 0<α<n, and use a Riemann sum to approximate the first term from above:
≥
1 ∞ n+β+2 2 α2 −1 1 ∞ n+β+2 −2
+k > +k
8k=0"(cid:18) 4 (cid:19) − 16# 8k=0(cid:18) 4 (cid:19)
X X
∞ −2
1 n+β+2 1
> +x dx= (n+β+2)−1 .
8 4 2
Z0 (cid:18) (cid:19)
6
Then
G′(β)<α 1(n+β+2)−1+ n−β−α−2 (n+β 2)2+α2 2α(n 2) −1 .
−2 n+β α 2 − − −
(cid:20) − − (cid:21)
(cid:2) (cid:3)
The right-hand expression is negative if
n β α 2 (n+β 2)2+α2 2α(n 2)
− − − < − − − ;
n+β α 2 2(n+β+2)
− −
this is clearly the case if n β α 2<0 so set δ =n β α 2 and consider the expression
− − − − − −
δ (2β+δ+α)2+α2 2α(δ+β+α)
− + −
δ+2β 2(δ+2β+α+4)
or
H(δ)=(δ+2β) (2β+δ+α)2+α2 2α(δ+β+α) 2δ(δ+2β+α+4)
− −
=(δ+2β)((cid:2)δ+2β+α)2 (δ+2β)(α2+2αδ+(cid:3)2αβ) 2δ(δ+2β+α+4)
− −
=(δ+2β) δ2+4β2+2βα+(4β 2δ 2δ(α+4)>0
− −
′
for β 2 and the positivity(cid:2) is clear for the case δ > 0(cid:3). H(δ) > 0 implies that G(β) < 0, and this
≥
completes the proof of Corollary 3.
Theorem 6 (Hardy-Rellich trace inequality). For f (Rn), n 2
∈S ≥
D
(15) f 2 x−1dx 1 ( ∆)3/4f 2dx
ZRn|∇ | | | ≤ 2π ZRn| − |
5 Γ(5) 2
4 , n=2 (k =1 term)
2 Γ(7)
(cid:20) 4 (cid:21)
(16) D2π1 =π4 , n=3 (k =1 term)
1 Γ(n+1) 2
4 , n 4 (k =0 term).
Proof. Using Corollary 3, determine2th(cid:20)eΓ(mna+4x3i)m(cid:21)um of t≥he two terms k = 0,1 in Theorem 4. From
Corollary 2 for n = 2 and by explicit calculation for n = 3, one observes that the k = 1 term is larger.
For higher dimensions, consider the log of the ratio of the k = 1 term to the k = 0 term: set w = n/4
and define for w 1
≥
3 4 1 2 5 2 4
Λ(w)=ln Γ w+ Γ w+ Γ w+ 1+ .
4 4 4 (4w 1)2
(cid:26)(cid:20) (cid:16) (cid:17) . (cid:16) (cid:17) (cid:16) (cid:17) (cid:21)(cid:18) − (cid:19)(cid:21)
Observe that by Stirling’s formula, Λ(w) 0 as w and
→ →∞
117 3 4 1 4
Λ(1)=ln Γ Γ 2.796 .
25 4 4 ≃−
(cid:20) (cid:16) (cid:17) . (cid:16) (cid:17) (cid:21)
Since Λ(1) is negative, the k =0 term is largest for n=4.
3 1 8(4w 1) 32w
′
Λ(w)=4 ψ w+ ψ w+ + −
4 − 4 (4w 1)2+4 − 16w2 1
(cid:20) (cid:16) (cid:17) (cid:16) (cid:17)(cid:21) − −
∞ 1 2 1 −1 8(4w 1) 32w
=2 k+w+ + −
2 − 16 (4w 1)2+4 − (16w2 1)
Xk=0(cid:20)(cid:16) (cid:17) (cid:21) − −
4 8(4w 1) 32w
> + − >0
2w+1 (4w 1)2+4 − (16w2 1)
− −
7
for w 1 since
≥
4 8(4w 1) 32w 4 8 8
+ − > + >0
2w+1 (4w 1)2+4 − 16w2 1 2w+1 4w+1 − 4w 1
− − −
for this range of values. Hence Λ(w) is increasing for w 1 or n 4 and since the limit at infinity is
≥ ≥
zero, Λ(w) must be negative for all w 1 and the k = 0 term is largest for n 4. This completes the
≥ ≥
argument for Theorem 6.
Theorem 7.
(A) For n 2 α<n
− ≤
2
n α n+α 4α
D =πα Γ − +1 Γ +1 1+ .
α 4 4 (n α)2
(cid:20) (cid:16) (cid:17). (cid:16) (cid:17)(cid:21) (cid:18) − (cid:19)
(B) For n 3 and α sufficiently near 0
≥
n α 1 n+α 1 2
D =πα Γ − + Γ + .
α
4 2 4 2
(cid:20) (cid:16) (cid:17). (cid:16) (cid:17)(cid:21)
Moreover, for n 4 this value holds when 0<α n 3.
≥ ≤ −
(C) For large n and fixed α, D (4π)α.
α ≃ n
Proof. Thecasen=2iscontainedinCorollary2. Thefirststeptoprovepart(A)willbetosetα=n 2
−
and consider the log of the ratio of the k =1 term to the k =0 term. Then for
Γ(3)Γ(n) 2
Λ=ln 2 2 (n 1) ,
((cid:20)Γ(n+21)Γ(1)(cid:21) − )
treat n=w as a continuous variable and calculate Λ′(w).
1 w w+1
′
Λ(w)= +ψ ψ
w 1 2 − 2
w+1 − w (cid:16) (cid:17) ∞ (cid:16)e−wt (cid:17)
ψ ψ =2 dt .
2 − 2 1+e−t
(cid:16) (cid:17) (cid:16) (cid:17) Z0
ThisformulafollowsfromtheGaussintegralrepresentationforψ(seeWhittakerandWatson,page247).
Set δ =w 1 and write for δ >1
−
1 ∞ e−δt 2 ∞ ete−δt
′
Λ = 1 2δ dt = dt>0 .
δ − 1+et δ (1+et)2
(cid:20) Z0 (cid:21) Z0
Since Λ′ >0,Λ asacontinuousfunctionofw increasesfrom toln(π/2). Λ(2)=0thenimplies that
−∞
Λ(n)>0 for n 3 and verifies the claim in part (A) for α=n 2.
≥ −
The estimates obtained here for ψ using both Riemann sums and the Gauss integralare expressedin
the following lemma.
Lemma. For w >1
1 2
+ , 1<w 3
2 w+1 w w w(w+1) ≤
(17) <ψ ψ <
2w+1 (cid:16) 2 (cid:17)− (cid:16)2(cid:17) 1 , 3 w .
w 1 ≤
−
To complete the proof of part (A), set α=n 2+2δ with 0<δ <1 and consider
−
Γ[n+δ]Γ[3−δ] 2 n+δ2 1
Λ(δ)=ln 2 2 − .
"Γ[n+δ+1]Γ[1 δ]# (1 δ)2
2 − 2 (cid:16) − (cid:17)
8
Note by using the lemma above
n+δ n+δ+1 δ 3 δ 2δ 2
′
Λ(δ)=ψ ψ +ψ 1 ψ + +
2 − 2 − 2 − 2 − 2 n+δ2 1 δ 1
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) − −
2δ 1 1
> + >0
n+δ2 1 1 δ − n+δ 1
− − −
so Λ(δ) is increasing for 0 < δ < 1 and since Λ(0) > 0 Λ(δ) is positive and the result in part (A) is
verified.
The purpose of parts (A) and (B) is to demonstrate that for dimension at least four or larger there
are definite ranges of the parameter α where either the k = 0 or k = 1 terms give the precise constant
for Theorem4. For n=2 the k =1 termis largestfor all α. For n=3, the k=1 term is largestexcept
for a small neighborhoodof α=0. For n 4 one can identify a definite interval in the parameterrange
≥
for α where the transition between the two terms for determining the maximum value for D occurs.
α
Consider the log of the ratio of the two terms
Γ(n−α +1) Γ(n+α + 1) 2 4α
Λ=ln 4 4 2 1+
((cid:20)Γ(n+4α +1) Γ(n−4α + 12)(cid:21) (cid:16) (n−α)2(cid:17))
which for n=3 becomes
Γ(7−α)Γ(5+α) 2 α2 2α+9
Λ=ln 4 4 − .
((cid:20)Γ(7+4α)Γ(5−4α)(cid:21) (cid:16) (3−α)2 (cid:17))
For small values of α, Λ is still positive which means that the k = 1 term is largest, e.g. for α = 0.2,
Λ 0.0021145,but for α=0,1, Λ 0.00103461. Note
≃ ≃−
′
Λ(0)= ψ(7/4)+ψ(5/4)+4/9 0.0304815
− ≃−
which requires since Λ(0)=0 that near zero, Λ(α)<0 and the k=0 term is largest.
To show the case n 4 for part (B), set n=α+2δ with δ 3/2; then
≥ ≥
2
Γ(δ +1)Γ(α+δ + 1) α
Λ=ln 2 2 2 1+
"Γ(δ + 1)Γ(α+δ +1)# δ2
2 2 2 (cid:16) (cid:17)
which can now be viewed as a function of α for 0<α n 2δ and δ 3/2. For this range of values of
≤ − ≥
δ
α+δ+1 α+δ 1
′
Λ(α)=ψ ψ +1 +
2 − 2 α+δ2
(cid:18) (cid:19) (cid:18) (cid:19)
2 1
< + <0 .
−α+δ+2 α+δ2
Since Λ(0)=0, the desired value of Λ for α=n 2δ must be negative which will imply that the k =0
−
term is largest for 0<α n 3 and n 4. This completes the proof of part (B).
≤ − ≥
Part (C) follows directly as an application of Stirling’s formula
Γ(z+a) √2π e−zzz+a−21 as z
≃ →∞
since for fixed α and large n, the value D will use the k =0 term
α
4π α
D .
α
≃ n
(cid:18) (cid:19)
Note that for α=2, this becomes an exact relation for n 5: D =(4π/n)2.
2
≥
9
2. Logarithmic uncertainty
The basic relation for Pitt’s inequality with gradient terms
f 2 x−αdx 4π2D f(y)2 y α+2dy
α
ZRn|∇ | | | ≤ ZRn| | | |
becomes an equality at α = 0 (D0 = 1) so it can be differenbtiated at this value of α. Corollary 2 and
Theorem 7 express the value of D needed for this calculation.
α
Theorem 8. For f (Rn) and n 2
∈S ≥
(18) ln x f 2dx+4π2 ln y y 2 f(y)2dy E f 2dx
ZRn | ||∇ | ZRn | || | | | ≥ ZRn|∇ |
3 b
ψ lnπ 1 , n=2
2 − −
E =
(cid:16) (cid:17)
n 1
ψ + lnπ , n 3 .
4 2 − ≥
(cid:16) (cid:17)
The increase in the constant here over the corresponding value in Theorem 2 reflects the comparison
of the integrals
ln x f 2dx ln x −1(2π y f(y)) 2dx .
ZRn | ||∇ | ≥ZRn | | F | |
(cid:12) (cid:12)
(cid:12) b (cid:12)
3. Iterated Stein-Weiss potentials
The Stein-Weiss potentials discussed here act at the spectral level, that is, in terms of L2 estimates.
In the context of using these potentials to define a linear operator on L2(Rn), it is natural to examine
iterated applications. For example, the Stein-Weiss potential with 0<α<n
f(x)x−α/2 x y −(n−α) y −α/2f(y)dxdy
ZRn×Rn| | | − | | |
corresponds to the linear operator
g x−α/2(x−(n−α2) g)
−→| | | | ∗
and inequality (5) can be rephrased
Γ(α)Γ(n−α)
(19) |x|−α/2(|x|−(n−α2)∗g) L2(Rn) ≤πn/2 Γ(n 4 α)Γ(4n+α) kgkL2(Rn)
(cid:13) (cid:13) (cid:20) 2 − 4 4 (cid:21)
(cid:13) (cid:13)
or as a weighted S(cid:13)obolev inequality (cid:13)
Γ(n−α)
(20) khkL2(Rn ≤2−α/2 Γ(n+4α) k(−∆)α/4(|x|α/2h)kL2(Rn)
(cid:20) 4 (cid:21)
which further implies by using the T∗T = T 2 argument that
k k k k
Γ(n−α) 2
(21) khkL2(Rn ≤2−α Γ(n+4α) |x|α/2(−∆)α/2(|x|α/2h) L2(Rn) 1
(cid:20) 4 (cid:21)
(cid:13) (cid:13)
These latter inequalities extend to include su(cid:13)ccessive applications of p(cid:13)owers of x and ( ∆)1/2 and
| | −
correspondtoiteratedStein-Weisspotentialssubjecttogrowthconstraintsonthesizeofthepowersand
thedilationconstraintthatthesumofthepowersof x mustequalthesumofthepowersof( ∆)1/2. At
| | −
the second iterationlevel, this algorithmleads to a result that includes the Maz’ya-Eilertseninequality:
(22) h L2(Rn) C ( ∆)ρ/4 xσ/2( ∆)β/4(xα/2h) L2(Rn)
k k ≤ k − | | − | | k
1Thisinequalitycorrespondstothecaseµ=λforinequality(7)in[7].
10
