Inequalities for Jacobi polynomials

0.2 MB·English

by  Uffe Haagerup

INEQUALITIES FOR JACOBI POLYNOMIALS
UFFEHAAGERUPANDHENRIKSCHLICHTKRULL
2 Abstract. ABernsteintypeinequalityisobtainedfortheJacobipolynomials
1 Pn(α,β)(x), which is uniform for all degrees n ≥ 0, all real α,β ≥ 0, and all
0 values x ∈ [−1,1]. It provides uniform bounds on a complete set of matrix
2 coefficients for theirreduciblerepresentations of SU(2) withadecay of d−1/4
in the dimension d of the representation. Moreover it complements previous
n
resultsofKrasikovonaconjectureofErd´elyi,MagnusandNevai.
a
J

0
3
1. Introduction
]
T For α,β R, α,β > 1, and n a non-negative integer we denote by P(α,β) the
n
R ∈ −
Jacobi polynomial with the standard normalization. Recall that in terms of the
h. Gauss hypergeometric function,
t
a Γ(n+α+1) 1 z
P(α,β)(x)= F ( n,n+α+β+1;α+1; − ).
m n Γ(α+1)Γ(n+1) 2 1 − 2
[
Recall also that for a fixed pair (α,β) these functions are orthogonal polynomials

2 on [ 1,1] for the weight function
v −
5 w(α,β)(x)=(1 x)α(1+x)β
−
9
4 with the explicit values
0
1 2α+β+1 Γ(n+α+1)Γ(n+β+1)
1. Pn(α,β)(x)2w(α,β)(x)dx= 2n+α+β+1 Γ(n+1)Γ(n+α+β+1)
0 Z−1
2 (see [12], eq. (4.3.3)).
1
For x [ 1,1] and α,β 0 let
: ∈ − ≥
v
Xi g(α,β)(x)= Γ(n+1)Γ(n+α+β+1) 1/2 1−x α/2 1+x β/2P(α,β)(x),
n Γ(n+α+1)Γ(n+β+1) 2 2 n
r (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
a then these functions are orthogonal on [ 1,1] for the constant weight. Moreover
−
1 1 1
(1) g(α,β)(x)2dx= .
2 n 2n+α+β+1
Z−1
(α,β)
Insuitablecoordinatesthefunctionsg witharbitrarynon-negativeintegersα,β
n
andn comprise a naturalandcomplete set ofmatrix coefficients for the irreducible
representations of SU(2) (see Section 2 below). The value 2n+α+β+1 in (1) is
exactly the dimension of the corresponding irreducible representation.
We shall prove the following uniform upper bound
Theorem 1.1. There exists a constant C >0 such that
|(1−x2)14 gn(α,β)(x)|≤C(2n+α+β+1)−14
for all x [ 1,1], all α,β 0 and all non-negative integers n.
∈ − ≥
Date:January30,2012.
2000 Mathematics Subject Classification. 22E46,33C45.
1
2 HAAGERUPANDSCHLICHTKRULL
We have not made a serious effort to find the best value of C, but at least our
proof shows that C <12.
With standard normalization the inequality in Theorem 1.1 amounts to the fol-
lowing uniform bound for the Jacobi polynomials
(sinθ)α+12(cosθ)β+12|Pn(α,β)(cos2θ)|
(2) C Γ(n+α+1)Γ(n+β+1) 1/2
1
(2n+α+β+1)−4
≤ √2 Γ(n+1)Γ(n+α+β+1)
(cid:18) (cid:19)
for 0 θ π/2. The decay rate of 1/4 in Theorem 1.1 is optimal when α and β
≤ ≤
tend to infinity, see Remark 4.4. However,if the pair (α,β) is fixed, then P(α,β)(x)
n
is O(n 1/2) for each x = 1, cf. [12], Thm. 7.32.2. In particular, in Legendre’s
−
6 ±
case α= β =0 where P(α,β)(x) specializes to the Legendre polynomial P (x), the
n n
Bernstein inequality (refined by Antonov and Kholshevnikov)
(3) (1 x2)1/4 P (x) (4/π)1/2(2n+1) 1/2, x [ 1,1],
n −
− | |≤ ∈ −
is known to be sharp, see [12], Thm. 7.3.3, and [10]. We refer to [2] for a further
discussion of the sharpest constant in (2), with a subset of the current parameter
range.
It is of interest also to express our inequality in terms of the orthonormal poly-
nomials defined by
1/2
(2n+α+β+1)Γ(n+1)Γ(n+α+β+1)
Pˆ(α,β)(x)= P(α,β)(x)
n 2α+β+1Γ(n+α+1)Γ(n+β+1) n
(cid:18) (cid:19)
for which
1
Pˆ(α,β)(x)2w(α,β)(x)dx=1.
n
Z−1
Here our estimate reads
C
(1−x2)41 wα,β(x)|Pˆn(α,β)(x)|≤ √2(2n+α+β+1)14.
q
ThefollowinggeneralizationofBernstein’sinequality(3)wasconjecturedbyErd´elyi,
Magnus and Nevai, [3],
(4) (1−x2)41 wα,β(x)|Pˆn(α,β)(x)|≤C′(α+β+2)1/4
q
for all α,β 1 and all integers n 0, with a uniform constant C > 0. A
≥ −2 ≥ ′
stronger form of the conjecture was recently established by Krasikov, [7], but only
in the parameter range α,β 1+√2, n 6. Our estimate is valid for a more
≥ 4 ≥
general range, but it involves 2n+α+β rather than α+β. Note however that by
combiningourresultswiththoseof[7],onecanremoveKrasikov’srestrictionn 6
≥
in the parameter range for the validity of (4).
The estimate (2) implies a similar estimate for the ultrasperical (Gegenbauer)
polynomialsC(λ)(x),asthesearedirectlyrelatedtotheJacobipolynomialsP(α,β)(x)
n n
with α = β = λ 1. Previous to [7] this case had been considered in [8], and as
− 2
above (2) allows the removal of a restriction on the degree.
The proof of Theorem 1.1 is based on an expression for P(α,β)(x) as a contour
n
integral,forwhichwecanestimatetheintegrandbyelementaryanalysis. Theproof
is simpler when α and β are integers. In this case, which is treated in Section 3,
the contour is just a circle. The general case is the discussed in Section 5.
JACOBI POLYNOMIALS 3
2. Motivation from representation theory
It is well known that the irreducible representations of SU(2) can be expressed
by Jacobi polynomials. In the physics literature it is customary to denote the cor-
responding matrix representations as Wigner’s d-matrices. We recall a few details
(see [14], 38, [13], Ch. 3, or [6]). The irreducible representations π of SU(2) are
l
§
parametrized by the non-negative integers or half-integers l = 0,1,1,..., where
2
2l+1 is the corresponding dimension. The standard representation space for π is
l
the space ofpolynomialsin twocomplex variablesz ,z ,homogeneousofdegree
l 1 2
P
2l, on which the representation is given by
a b
[π f](z ,z )=f(az +cz ,bz +dz ).
l c d 1 2 1 2 1 2
(cid:18) (cid:19)
Let
eiφ 0 cosθ sinθ
kφ = 0 e iφ and tθ = sinθ −cosθ
−
(cid:18) (cid:19) (cid:18) (cid:19)
for φ,θ R, then every element A SU(2) allows a decomposition of the form
∈ ∈
A = k t k . The monomials zjzk with j +k = 2l form a basis for , and it is
φ θ −ψ 1 2 Pl
convenient to use the notation
hlp(z1,z2)=z1l−pz2l+p
where p= l, l+1,...,l. Notice that these are weight vectors
− −
π (k )hl =e i2pφhl, (p= 1,...,l).
l φ p − p −
Choosing the inner product on so that π is unitary, the functions hl form an
Pl l p
orthogonalbasis. We denotebyhˆl thecorrespondingnormalizedbasisvectors. For
p
A SU(2) the matrix elements
∈
ml (A)= π (A)hˆl,hˆl
pq h l q pi
withp,q= l,...,l,formtheso-calledWigner’sd-matrix. OurresultfortheJacobi
−
polynomials implies the following.
Theorem 2.1. Let C be the constant from Theorem 1.1. Then
(5) sin2θ 1/2 ml (k t k ) C(2l+1) 1/4
| | | pq φ θ −ψ |≤ −
for allφ,θ,ψ R,alll =0,1,1,... andallp,q= l,...,l. Moreover, theexponent
∈ 2 −
1/4 on the right hand side is best possible.
Proof. Explicitly the matrix elements are given as follows (see [14], [13], [6]). For
p,q = l,...,l such that q p,
− | |≤
ml (k t k )=e i2pφei2qψg(α,β)(cos2θ),
pq φ θ ψ − n
−
where
α=p q, β =p+q, n=l p.
− −
For other values of p and q there are similar expressions, and in all cases one has
ml (k t k ) = g(α,β)(cos2θ)
| pq φ θ −ψ | | n |
where α= p q , β = p+q and n=l max p, q . Moreover
| − | | | − {| | | |}
dimπ =2l+1=2n+α+β+1.
l
Thus(5)followsdirectlyfromTheorem1.1. Forthe laststatementofTheorem2.1,
see Remark 4.4. (cid:3)
4 HAAGERUPANDSCHLICHTKRULL
Remark 2.2. For l integral π descends to a representation of SO(3), and the
l
matrix elements ml with q = 0 descend to spherical harmonic functions on S2
p0 ≃
SO(3)/SO(2). Withthecommon normalization from quantummechanics thespher-
ical harmonics Ym with l m l satisfy
l − ≤ ≤
(2l+1)1/2
Ym(θ,φ)= g(α,α)(cosθ)eimφ,
l ± (4π)1/2 l α
−
where α= m. From Theorem 1.1 we obtain the uniform estimate
| |
C
sinθ 1/2 Ym(θ,φ) (2l+1)1/4
| | | l |≤ (4π)1/2
for all θ,φ and all integers l,m with m l.
| |≤
TheJacobipolynomialsarealsorelatedtotheharmonicanalysisonthecomplex
spheres with respect to the action of the unitary group. The spherical functions
for the pair (U(q),U(q 1)) are functions on the unit sphere in Cq, and in suitable
−
coordinatestheycanbeexpressedbymeansofJacobifunctionsP(α,β)withα=q 2
n
−
(see [11], [5]). The direct motivation for the present paper was an application of
this observationforq =2 to astudy ofSp(2,R). In[4] the firstauthorandde Laat
apply the uniform estimates of the present paper for the case α = 0, to show that
Sp(2,R) does not have the approximation property. Earlier, Bernstein’s inequality
(3) had been used in [9] with a similar purpose for the group SL(3,R).
3. Integral parameters
The proof is based on the following integral expression, which is obtained by
applyingCauchy’sformulatoRodrigues’formulaforP(α,β)(x)(see[12],eq.(4.3.1)),
n
(6) (1 x)α(1+x)βP(α,β)(x)=( 1)nI(α,β)(x)
− n −2 n
for x ( 1,1), where
∈ −
1 (1 z)n+α(1+z)n+β dz
(7) I(α,β)(x)= − .
n 2πi (z x)n z x
Zγ(x) − −
Here γ(x) is any closed contour encircling x in the positive direction. We assume
in this section that α and β are integers 0. Without this assumption one would
≥
have to request also that γ(x) does not enclose the points z = 1. We shall take
±
γ(x)=C(x,r), thecirclecenteredatxandwitharadiusr >0tobespecifiedlater.
The case n=0 will be treated separately in Lemma 4.3 below. Here we assume
n 1 and let a=α/n and b=β/n, then
≥
1 (1 z)a+1(1+z)b+1 n dz
I(α,β)(x)= −
n 2πi z x z x
ZC(x,r)(cid:18) − (cid:19) −
1 (1 x s)a+1(1+x+s)b+1 n ds
= − − .
2πi s s
ZC(0,r)(cid:18) (cid:19)
In order to select a suitable radius r we look for the stationary points of the
expression inside the parentheses, as a function of s. We let
ψ(s)=(a+1)log(1 x s)+(b+1)log(1+x+s) logs
− − −
for s C, and analyze the derivative
∈
a+1 b+1 1
ψ (s)= + ,
′
s+x 1 s+x+1 − s
−
which is independent of the branch cut used for the complex logarithm. Now
As2+B(x)s+C(x)
ψ (s)=
′
(s+x 1)(x+s+1)s
−
JACOBI POLYNOMIALS 5
where
A=a+b+1, B(x)=(a+b)x+a b, C(x)=1 x2.
− −
The numerator is a second order polynomial in s with the discriminant
∆(x)=B(x)2 4AC(x)
−
=(a+b+2)2x2+2(a2 b2)x+(a b)2 4(a+b+1),
− − −
which coincides with the polynomial ∆ defined in [1]. The polynomial ∆(x) has
two real roots
x+ b2 a2 4 (a+1)(b+1)(a+b+1)
= − ±
x (a+b+2)2
−(cid:27) p
for which 1 x <x+ 1. For x <x<x+ we have ∆(x)<0, and thus there
− −
− ≤ ≤
are two conjugate solutions s = s ,s to the equation As2 +B(x)s+C(x) = 0.
1 2
They are
B(x) i ∆(x)
s ,s = − ± − .
1 2
2A
p
Note that
C(x) 1 x2
s 2 = s 2 =s s = = − .
1 2 1 2
| | | | A a+b+1
Hence, if we choose the radius
1 x2
(8) r = − ,
a+b+1
r
then our contour C(0,r) will pass through the stationary points of ψ. We define r
by (8) for all x ( 1,1) (also when ∆(x) 0).
∈ − ≥
We now find
|In(α,β)(x)|≤ 21π 2π (1−x−reiθ)1+a(1+x+reiθ)1+br−1 n dθ,
Z0
(cid:12) (cid:12)
and write (cid:12) (cid:12)
(1 x reiθ)1+a(1+x+reiθ)1+br 1 =ef(cosθ)
−
| − − |
where
a+1
f(t)= ln r2+(1 x)2 2r(1 x)t
2 − − −
(9)
b+1(cid:0) (cid:1)
+ ln r2+(1+x)2+2r(1+x)t ln(r)
2 −
fort [ 1,1]. Noticethatweall(cid:0)owthepossiblevaluef(t)=(cid:1) intheendpoints
∈ − −∞
t= 1. Let
±
r2+(1 x)2 r2+(1+x)2
(10) t = − , t =
2 1
2r(1 x) − 2r(1+x)
−
then t 1 and 1 t . It follows that
1 2
≤− ≤
a+1 b+1
(11) f(t)= ln(t t)+ ln(t t )+K
2 1
2 − 2 −
where
a+1 b+1 a+b a+b+2
(12) K = ln(1 x)+ ln(1+x)+ lnr+ ln2
2 − 2 2 2
is independent of t. With (11) we can extend the domain of definition for f to
[t ,t ] [ 1,1]. For later reference we note that from (10) and (8) it follows that
1 2
⊃ −
(a+b+2) (a+b)x (a+b+2) (a+b)x
(13) t = − − , t = − ,
1 2
2√a+b+1√1 x2 2√a+b+1√1 x2
− −
6 HAAGERUPANDSCHLICHTKRULL
and
a+b+2
(14) t t = .
2 1
− √a+b+1√1 x2
−
We have
1 2π
I(α,β)(x) enf(cosθ)dθ.
| n |≤ 2π
Z0
From (11) we find
a+1 b+1 (a+b+2)(t t)
0
(15) f′(t)= + = − ,
−2(t t) 2(t t ) 2(t t)(t t )
2 1 2 1
− − − −
where t is the convex combination
0
(a+1)t +(b+1)t a+b (a+b)x
1 2
(16) t = = − − (t ,t ).
0 1 2
a+b+2 2√a+b+1√1 x2 ∈
−
Moreover
a+1 b+1
f (t)= <0.
′′
−2(t t)2 − 2(t t )2
2 1
− −
Hencethefunctionf(t)isconcaveandhasaglobalmaximumatt . Wethusobtain
0
the initial estimate
1 π
(17) I(α,β)(x) enf(cosθ)dθ enf(t0).
| n |≤ π ≤
Z0
Since
(a+1)(t t ) (b+1)(t t )
2 1 2 1
(18) t t = − , t t = −
2 0 0 1
− a+b+2 − a+b+2
we find
a+1 (a+1)(t t ) b+1 (b+1)(t t )
2 1 2 1
f(t )= ln − + ln − +K,
0
2 a+b+2 2 a+b+2
and from (12) and (14) it then follows that
1 2a+b+2(a+1)a+1(b+1)b+1
f(t )= ln (1 x)a(1+x)b .
0 2 (a+b+1)a+b+1 −
(cid:18) (cid:19)
Thus
2a+b+2(a+1)a+1(b+1)b+1 n/2
enf(t0) (1 x)a(1+x)b
≤ (a+b+1)a+b+1 −
(cid:18) (cid:19)
2a+b+2(a+1)a+1(b+1)b+1 n/2
= (1 x)α/2(1+x)β/2.
(a+b+1)a+b+1 −
(cid:18) (cid:19)
The inequality
Γ(n+1)Γ(n+α+β+1) (a+1)a+1(b+1)b+1 n
Γ(n+α+1)Γ(n+β+1) (a+b+1)a+b+1
(19) (cid:18) (cid:19)
1/2
(n+1)(n+α+β+1)
≤ (n+α+1)(n+β+1)
(cid:18) (cid:19)
willbeshowninLemma4.1. Inserting(17)and(19)intoourdefinitionofg(α,β) we
n
obtain the initial bound
1/4
(n+1)(n+α+β+1)
(20) g(α,β)(x) .
| n |≤ (n+α+1)(n+β+1)
(cid:18) (cid:19)
In particular, since (n +1)(n+α+β +1) (n +α+1)(n+β +1) it follows
≤
that g(α,β)(x) 1 (which could also be seen directly from the fact that g(α,β) is a
n n
| |≤
unitary matrix coefficient of orthonormal vectors).
JACOBI POLYNOMIALS 7
In order to improve the estimate we need to replace the inequality f(t) f(t )
0
≤
by a stronger inequality. In Proposition3.1 below we shall establish the inequality
D
(21) f(t) f(t )+ f (t )(t t )2
≤ 0 1+t2 ′′ 0 − 0
0
fort [ 1,1],with asuitableconstantD >0. Followingthe argumentfrombefore
∈ −
andtaking into accountthe secondtermin (21) we canthen improve(17) with the
extra factor
1 π nD
exp f (t )(cosθ t )2 dθ
π 1+t2 ′′ 0 − 0
Z0 (cid:18) 0 (cid:19)
on the right hand side.
Forthe estimationofthe exponentialintegralwe use Lemma 3.6 below,whichis
applicable since f (t )<0. We let
′′ 0
nD nD
u=t f (t ), v = f (t ),
0s1+t20| ′′ 0 | s1+t20| ′′ 0 |
and observe that u2+v2 =nD f (t ). We thus obtain
′′ 0
| |
(22) |In(α,β)(x)|≤2enf(t0)(nD|f′′(t0)|)−1/4
and hence (20) has been improved to
1/4
(n+1)(n+α+β+1)
|gn(α,β)(x)|≤ (n+α+1)(n+β+1) 2(nD|f′′(t0)|)−1/4.
(cid:18) (cid:19)
From (15), (18) and (14) it follows that
a+b+2 (a+b+1)(a+b+2)
(23) f (t )= = (1 x2),
′′ 0
−2(t t )(t t ) − 2(a+1)(b+1) −
0 1 2 0
− −
and hence
(α+β+n)(α+β+2n)
f (t ) = (1 x2).
′′ 0
| | 2(α+n)(β+n) −
Since
n+α+β+1 n+α+β
(n+α+1)(n+β+1) ≤ (n+α)(n+β)
and
n+1 3
n(2n+α+β) ≤ 2n+α+β+1
for all n 1 and α,β 0, it finally follows that
≥ ≥
g(α,β)(x) C (α+β+2n+1) 1/4(1 x2) 1/4
| n |≤ ′ − − −
where C = 24 6/D = 2√4168 < 8 with the value D = 1/28 from below. This
′
completes the proof of Theorem 1.1 in the integral case (up to the cited results
from below). p (cid:3)
Proposition 3.1. Fix x [ 1,1] and let f(t) and t be as above. Then
0
∈ −
1
f(t) f(t )+ f (t )(t t )2
≤ 0 28(1+t2) ′′ 0 − 0
0
for all t [ 1,1].
∈ −
Proof. We begin the proof by a sequence of lemmas.
Lemma 3.2. The following relation holds
2a2 2b2
(24) (a+b)2+4(a+b+1)t2 = + .
0 1 x 1+x
−
8 HAAGERUPANDSCHLICHTKRULL
Proof. Using (16) we obtain
(a b+(a+b)x)2
4(a+b+1)t2 = − .
0 1 x2
−
On the other hand
2a2 2b2 2(a2+b2+(a2 b2)x)
+ = − .
1 x 1+x 1 x2
− −
Hence (24) follows from the identity
(a+b)2(1 x2)+(a b+(a+b)x)2 =2(a2+b2+(a2 b2)x),
− − −
which is straightforward. (cid:3)
Lemma 3.3. We have
(a+1)(b+1)
1 x2 16 (1+t2)
− ≤ (a+b+2)2 0
for all x [ 1,1].
∈ −
Proof. Note first that if we replace the triple (a,b,x) by (b,a, x), then t ,t ,t
1 0 2
−
are replaced by t , t , t and hence the asserted inequality is unchanged. We
2 0 1
− − −
may thus assume that a b.
≤
It follows from Lemma 3.2 that
2b2
(a+b)2+4(a+b+1)t2
0 ≥ 1+x
and therefore
2b2
1+x .
≥ (a+b)2+4(a+b+1)t2
0
Hence
2b2 a2+2ab+4(a+b+1)t2
1 x 2 =2 0.
− ≤ − (a+b)2+4(a+b+1)t2 (a+b)2+4(a+b+1)t2
0 0
and
a2+2ab+4(a+b+1)t2
1 x2 2(1 x) 4 0.
− ≤ − ≤ (a+b)2+4(a+b+1)t2
0
Since the right hand side is an increasing function of t2 we have for t2 1 that
0 0 ≤
a2+2ab+4(a+b+1) (a+1)(b+1)
1 x2 4 16 ,
− ≤ (a+b)2+4(a+b+1) ≤ (a+b+2)2
where in the last step we used that a b implies a2+2ab 4ab. For t2 1 we
≤ ≤ 0 ≥
obtain similarly
(a2+2ab)t2+4(a+b+1)t2 (a+1)(b+1)
1 x2 4 0 0 16 t2.
− ≤ (a+b)2+4(a+b+1) ≤ (a+b+2)2 0
This completes the proof of Lemma 3.3. (cid:3)
Lemma 3.4. We have
1 1
(25) t t and t t .
2− 0 ≥ 4(1+t2)1/2 0− 1 ≥ 4(1+t2)1/2
0 0
Proof. It follows from (14) and Lemma 3.3 that
(a+b+2)2
t t (1+t2) 1/2,
2− 1 ≥ 4 (a+1)(b+1)(a+b+1) 0 −
and hence by (18) p
√a+1(a+b+2)
t t (1+t2) 1/2.
2− 0 ≥ 4 (b+1)(a+b+1) 0 −
p
JACOBI POLYNOMIALS 9
Using (b+1)(a+b+1) (a+b+2)2 and√a+1 1weobtainthe firstinequality
in (25). The second one≤is analogous. ≥ (cid:3)
Lemma 3.5. We have
(26) (u t )(t u) 14(1+t2)(t t )(t t )
− 1 2− ≤ 0 0− 1 2− 0
for all u [t ,t ] for which 1 u t or t u 1.
1 2 0 0
∈ − ≤ ≤ ≤ ≤
Proof. We first assume a b. Then by (18)
≤
a+b+2
(27) u t t t = (t t ) 2(t t ).
1 2 1 0 1 0 1
− ≤ − b+1 − ≤ −
In order to estimate t u we first note that u t 1+ t and hence
2 0 0
− | − |≤ | |
t u t t + t u t t +1+ t .
2 2 0 0 2 0 0
− ≤ − | − |≤ − | |
By Lemma 3.4
1+ t √2(1+t2)1/2 4√2(1+t2)(t t )
| 0|≤ 0 ≤ 0 2− 0
and hence
(28) t u (1+4√2)(1+t2)(t t ) 7(1+t2)(t t ).
2− ≤ 0 2− 0 ≤ 0 2− 0
Now (27) and (28) together imply (26). The proof for a b is analogous. (cid:3)
≥
WecannowproveProposition3.1. Lett [ 1,1]. Itfollowsfrom(15), (26)and
∈ −
(23) that
f (u) a+b+2
′
=
u t −2(u t )(t u)
0 1 2
− − −
a+b+2 f (t )
′′ 0
=
≤−28(1+t2)(t t )(t t ) 14(1+t2)
0 0− 1 2− 0 0
for all u R between t and t . Hence
0
∈
t
f(t)=f(t )+ f (u)du
0 ′
Zt0
f (t ) t f (t )
f(t )+ ′′ 0 (u t )du=f(t )+ ′′ 0 (t t )2. (cid:3)
≤ 0 14(1+t2) − 0 0 28(1+t2) − 0
0 Zt0 0
Lemma 3.6. Let u,v R with u2+v2 >0. Then
∈
(29) 1 πe−(u+vcoss)2ds 2
π ≤ (u2+v2)1/4
Z0
Proof. We will show (29) with the slightly stronger bound
√2
.
max u, v
{| | | |}
The statement is invariant underpthe map (u,v) ( u, v) and, using the substi-
7→ − −
tution s π s, also under v v. Hence, it is sufficient to show
7→ − 7→−
1 πe (u vcoss)2ds √2
− −
π ≤ max u,v
Z0 { }
for u 0, v 0. p
≥ ≥
Supposefirst0 u v,thenv =0. Letσ [0,π]be suchthatcosσ = u. Then
≤ ≤ 6 ∈ 2 v
s+σ s σ
u vcoss=v(cosσ coss)=2vsin( )sin( − ).
− − 2 2
10 HAAGERUPANDSCHLICHTKRULL
Note thatsin(s+σ) sin(s σ) becausesin2(s+σ) sin2(s σ)=sinssinσ 0 for
2 ≥| −2 | 2 − −2 ≥
s [0,π] and σ [0,π]. Using also that sint 2 t for t π, it follows that
∈ ∈ 2 | |≥ π| | | |≤ 2
1 πe−(u−vcoss)2ds= 1 πe−4v2sin2(s+2σ)sin2(s−2σ)ds
π π
Z0 Z0
1 πe 4v2π−4(s σ)4ds
− −
≤ π
Z0
1 ∞ e 4v2π−4s4ds 2 ,
−
≤ π ≤ √2v
Z−∞
where we used that 0∞e−t4dt=Γ(54)≤1.
Suppose next 0 v u 2v. Then u vcoss v(1 coss) = 2vsin2(s).
≤R ≤ ≤ − ≥ − 2
Hence,
1 πe−(u−vcoss)2ds 1 πe−4v2sin4(2s)ds
π ≤ π
Z0 Z0
1 πe−4v2π−4s4ds 1 1
≤ π ≤ √2v ≤ √u
Z0
using again 0∞e−t4dt≤1.
Suppose finally 0 2v u. Then u vcoss u and hence
R ≤ ≤ − ≥ 2
1 πe−(u−vcoss)2ds e−u42 1
π ≤ ≤ √u
Z0
where we used that xe x4 1 for all x 0. (cid:3)
− ≤ √2 ≥
4. Some inequalities with gamma functions
In this section we prove some inequalities which were used in the preceding
section. We assume that α,β are real and non-negative.
Lemma 4.1. Let n,α,β 0. Then
≥
Γ(n+1)Γ(n+α+β+1)
Γ(n+α+1)Γ(n+β+1)
(30)
nn(α+β+n)α+β+n (n+1)(n+α+β+1) 1/2
.
≤ (α+n)α+n(β+n)β+n (n+α+1)(n+β+1)
(cid:18) (cid:19)
Proof. We have for x,y,z 0
≥
Γ(x+1)Γ(x+y+z+1) y z
(31) ln = (lnΓ) (x+s+t+1)dtds.
′′
Γ(x+y+1)Γ(x+z+1)
Z0 Z0
We claim that
1 1
(32) (lnΓ) (u+1)
′′
≤ u − 2(u+1)2
for all u>0. The asserted inequality (30) follows easily from (31) and (32).
In order to prove (32) we recall that
∞ 1 ∞
(lnΓ) (u+1)= = A(u+k),
′′
(u+k)2
k=1 k=0
X X
where
1
A(u)= .
(u+1)2
For the other side of (32) we use the telescoping series
1 ∞ 1 ∞
= B(u+k), = C(u+k),
u 2(u+1)2
k=0 k=0
X X