The Newton Polyhedron and positivity of {}_2F_3 hypergeometric functions
TThe Newton polyhedron and positivityof F hypergeometric functions Yong-Kum Cho ∗ Seok-Young Chung † Abstract.
As for the F hypergeometric function of the form F (cid:20) a , a b , b , b (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:21) ( x > , where all of parameters are assumed to be positive, we give sufficient conditions on( b , b , b ) for its positivity in terms of Newton polyhedra with vertices consistingof permutations of ( a , a + 1 / , a ) or ( a , a + 1 / , a ) . As an application, weobtain an extensive validity region of ( α, λ, µ ) for the inequality (cid:90) x ( x − t ) λ t µ J α ( t ) dt ≥ x > . Keywords.
Bessel functions, fractional integrals, Newton polyhedron, p F q hypergeometric functions, sums of squares method, transference principle. This paper concerns primarily the geometric structure of parameters thatensures the positivity of F hypergeometric functions of typeΦ( x ) ≡ F (cid:34) a , a b , b , b (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) ( x > , (1.1) ∗ [email protected]. Department of Mathematics, College of Natural Sciences, Chung-Ang University, 84 Heukseok-Ro, Dongjak-Gu, Seoul 06974, Korea. † [email protected]. Department of Mathematics, University of Central Florida,4393 Andromeda Loop N, Orlando, FL 32816, USA. a r X i v : . [ m a t h . C A ] F e b here all of parameters a i , b j are assumed to be positive.In our previous studies [8], [9], we considered the problem of positivityfor F hypergeometric functions of similar type and described regions ofpositivity in the plane of denominator-parameter pairs by means of Newtonpolyhedra and their hyperbolic extensions (see Theorem 9.2 for a summary).In a consistent manner, we aim at providing sufficient conditions for thepositivity of F hypergeometric functions of type (1.1) in terms of Newtonpolyhedra in the ( b , b , b )-space for each fixed ( a , a ).To state briefly, let A , B be the spatial sets of all permutations of( a , a + 1 / , a ) , ( a , a + 1 / , a ) , respectively. In view of the known formula ([18, 6.2], [26, 5.4]) J ν ( x ) = ( x/ ν [Γ( ν + 1)] F (cid:34) ν + 1 / ν + 1 , ν + 1 (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) , (1.2)where J ν stands for the Bessel function of the first kind of order ν ∈ R , itis evident that Φ is nonnegative for each b = ( b , b , b ) ∈ A ∪ B . By means of fractional integrals, to be explained in detail, it is simpleto observe a transference principle which asserts if Φ is nonnegative for apoint b , then Φ remains strictly positive for all points of the octant b + R except for the corner b . On applying this transference principle, it is thusfound that Φ remains strictly positive for all points of (cid:91) b ∈ A∪B (cid:0) b + R (cid:1) (1.3)excluding A ∪ B where Φ is shown to be nonnegative.As it is natural to ask if this positivity region could be extended further,we shall prove that Φ indeed remains positive for all points of the convex hullcontaining (1.3), so called the Newton polyhedron of
A ∪ B , under certain admissible conditions on a , a . From a geometric view-point, the Newtonpolyhedron of A∪B represents an infinite polygonal region in R surroundedby a hexagonal or triangular bottom face, several side quadrilaterals andinfinite planar faces parallel to one of the coordinate axes.An expansion formula due to Gasper [13] givesΦ( x ) = x − ν ∞ (cid:88) n =0 C n ( ν ) J n + ν ( x ) (1.4)2or any real number ν subject to certain condition. Once each coefficient C n ( ν ) were shown to be nonnegative, we may conclude that Φ is nonnegativefrom this expansion. Often referred to as Gasper’s sums of squares method ,it is this framework that our investigation will be based on.In practice, as each coefficient C n ( ν ) involves either terminating F or F hypergeometric series, we are confronted with the problem of how todetermine the sign of those terminating series. In particular, if ( b , b , b )belongs to one of the aforementioned side quadrilaterals, we must deal withthe terminating F hypergeometric series of the from F (cid:34) − n, n + α , α , α , α β , β , β , β (cid:35) , n = 1 , , · · · , (1.5)with appropriate values of α i , β j expressible in terms of a i , b j , ν. In our previous work [8], we exploited one of Whipple’s transformationformulas and an induction argument to set up a criterion of positivity forthe terminating F hypergeometric series of similar type. Concerning (1.5),however, as an appropriate version of Whipple’s transformation formula isno longer available, we shall make use of a modification of more generaltransformation formulas developed by Fields and Wimp [12] to decomposeit into a finite sum involving products of terminating F hypergeometricseries, which will enable us to obtain a set of positivity criteria.In summary, it turns out that each of Newton polyhedra with vertices A , B , A∪B is available as a positivity region of Φ under different conditionson the pair ( a , a ). For this reason, our positivity results will be presentedin three separate statements (Theorems 6.1, 6.2, 8.1) in each of which ananalytic description of the associated Newton polyhedron will be given forthe sake of practical applications.The problem of positivity for the F hypergeometric function of type(1.1) arises often as a critical issue in various disguises. As an illustration,the famous problem of determining α, λ, µ for validity of (cid:90) x ( x − t ) λ t µ J α ( t ) dt ≥ x >
0) (1.6)is equivalent to the problem of positivity for a special case of (1.1) implicitlyinvolved (see (9.5) for details). In application of our results, we shall specifyan extensive validity region of ( α, λ, µ ) for the inequality (1.6) which will3rovide at once simplified proofs or improvements for a number of knowninequalities and answers to some open conjectures.As usual, we shall define the p F q hypergeometric function by p F q (cid:34) u , · · · , u p v , · · · , v q (cid:12)(cid:12)(cid:12)(cid:12) z (cid:35) = ∞ (cid:88) n =0 ( u ) n · · · ( u p ) n n ! ( v ) n · · · ( v q ) n z n ( z ∈ C ) , (1.7)where parameters u i , v j are real numbers subject to the condition that noneof v j ’s coincides with a negative integer and the coefficients are written inPochhammer’s notation; for α ∈ R and a positive integer n ,( α ) n = α ( α + 1) · · · ( α + n − , ( α ) = 1 . In the special case z = 1 , it is customary to delete the unit argument. Forthe general reference, we refer to Bailey [4] and Luke [18].To simplify notation, we shall use throughout the logical symbol ∧ todenote and ; A ∧ B is true if and only if both A and B are true. In addition,we shall write R + = [0 , ∞ ) and R ∗ + = (0 , ∞ ) distinctively. As observed by many authors in different contexts (see [2], [9], [11], [13]),it is possible to transfer a known positivity region to a much larger regionwith the aid of fractional integrals. We shall denote the beta function by B ( α, β ) = (cid:90) (1 − t ) α − t β − dt ( α > , β > . Proposition 2.1. (transference principle)
Suppose Φ( a , b ; x ) ≡ F (cid:34) a , a b , b , b (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) ≥ x > for some a = ( a , a ) ∈ R and b = ( b , b , b ) ∈ R , where all of thecomponents of a , b are assumed to be positive. (i) For any (cid:15) = ( (cid:15) , (cid:15) , (cid:15) ) ∈ R , if (cid:15) , (cid:15) , (cid:15) are not simultaneously zero,then Φ( a , b + (cid:15) ; x ) > for all x > . (ii) For any δ = ( δ , δ ) ∈ R , if δ , δ are not simultaneously zero and ≤ δ j < a j , j = 1 , , then Φ( a − δ , b ; x ) > for all x > . roof. For (cid:15) > , the readily-verified representationΦ( a , b + (cid:15) , b , b ; x ) = 2 B ( b , (cid:15) ) (cid:90) Φ( a , b ; xt )(1 − t ) (cid:15) − t b − dt implies Φ( a , b + (cid:15) , b , b ; x ) > x > . By permuting the role of b j , (cid:15) j or replacing the kernel appropriately, it is straightforward to verifypart (i) in all possible occasions.In a similar manner, if 0 < δ < a , then the representationΦ( a − δ , a , b ; x ) = 2 B ( δ , a − δ ) (cid:90) Φ( a , b ; xt )(1 − t ) δ − t a − δ − dt implies Φ( a − δ , a , b ; x ) > x > . By permuting the role of a j , δ j or replacing the kernel appropriately, it is also straightforward to verify part(ii) in all possible occasions.By making use of the asymptotic behavior and the above transferenceprinciple, we deduce a set of necessary conditions as follows. Proposition 2.2.
For < a ≤ a and b j > , j = 1 , , , put Φ( x ) ≡ F (cid:34) a , a b , b , b (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) ( x > . If Φ( x ) ≥ for all x > , then it is necessary that b ∧ b ∧ b ≥ a , b + b + b ≥ a + a + 1 / . (2.1) Moreover, Φ changes sign at least once if these conditions are violated.Proof. Assume first that a − a is not an integer. Setting χ = b + b + b − a − a − / , a well-known asymptotic expansion formula ([18], [21]) givesΦ( x ) = Γ( b )Γ( b )Γ( b )Γ( a )Γ( a ) (cid:26) Γ( a )Γ( a − a )Γ( b − a )Γ( b − a )Γ( b − a ) (cid:16) x (cid:17) − a (cid:2) O (cid:0) x − (cid:1)(cid:3) + Γ( a )Γ( a − a )Γ( b − a )Γ( b − a )Γ( b − a ) (cid:16) x (cid:17) − a (cid:2) O (cid:0) x − (cid:1)(cid:3) + 1 √ π (cid:16) x (cid:17) − χ (cid:104) cos (cid:16) x − πχ (cid:17) + O (cid:0) x − (cid:1)(cid:105)(cid:27) (2.2)5s x → ∞ . In view of the oscillatory nature of the last part, it is evidentthat the condition χ ≥ a , a ) or equivalently b + b + b ≥ a + a + 1 / a − a happens to be an integer. For 0 < δ < min(1 , a ) , it follows from the transference principle of Proposition 2.1 that F (cid:34) a , a − δb , b , b (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) > x > b , b , b must satisfy the condition b + b + b ≥ a + a − δ + 1 / a , a − δ ) . By letting δ → , it is thus found that (2.3) is necessary.It remains to verify the additional requirements b j ≥ a for all j underthe assumption (2.3). By using a limiting argument as above, it suffices todeal with the case b + b + b > a + a + 1 / . Let us assume 0 < b < a . Integrating termwise, it is elementary to deduce F (cid:34) a b , b (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) = 2 B ( b , a − b ) (cid:90) Φ( xt )(1 − t ) a − b − t b − dt > x > , whence b ∧ b > a according to Theorem 9.2.In the case a < a , the asymptotic behavior (2.2) impliesΦ( x ) ∼ Γ( b )Γ( b )Γ( b )Γ( a − a )Γ( a )Γ( b − a )Γ( b − a )Γ( b − a ) (cid:16) x (cid:17) − a as x → ∞ . Replacing b by b + (cid:15) with (cid:15) > , if necessary, we may assume a − < b < a so that Γ( b − a ) < , it implies that Φ must changesign at least once, which contradicts the nonnegativity assumption on Φ.Thus b ≥ a . In the case a = a , replacing a by a − δ, < δ < a , thesame reasonings would lead to b ≥ a − δ and subsequently b ≥ a byletting δ → . By symmetry, it is also necessary that b ≥ a , b ≥ a . Expansions of p F q hypergeometric functions The purpose of this section is to give a list of expansion formulas for p F q hypergeometric functions which will be applied frequently in the sequel.For simplicity, it is often convenient to use the vector notation u p = ( u , · · · , u p ) , ( u p ) n = ( u ) n · · · ( u p ) n so that the p F q hypergeometric function of (1.7) can be written as p F q (cid:34) u p v q (cid:12)(cid:12)(cid:12)(cid:12) z (cid:35) = ∞ (cid:88) n =0 ( u p ) n n ! ( v q ) n z n . An expansion formula due to Fields and Wimp [12, 3.3] states p F q (cid:34) u p v q (cid:12)(cid:12)(cid:12)(cid:12) zw (cid:35) = ∞ (cid:88) n =0 ( α r ) n ( − z ) n n !( n + δ ) n ( β s ) n r F s +1 (cid:34) n + α r n + δ + 1 , n + β s (cid:12)(cid:12)(cid:12)(cid:12) z (cid:35) × p + s +2 F q + r (cid:34) − n, n + δ, β s , u p α r , v q (cid:12)(cid:12)(cid:12)(cid:12) w (cid:35) , (3.1)valid for any nonnegative integers p, q, r, s, real parameters u p , v q , α r , β s , δ, and complex arguments z, w , provided that none of v q , α r , β s , δ coincideswith a negative integer and each of the involved series converges. We referto Luke and Coleman [19] for closely related expansion formulas.In view of the formula (1.2), if we choose r = s = 1 and take α = ν + 1 / , β = ν + 1 , δ = 2 ν, then (3.1) yields Gasper’s sums of squares formula [13, (3.1)] p F q (cid:34) u p v q (cid:12)(cid:12)(cid:12)(cid:12) − z (cid:35) = Γ ( ν + 1) (cid:16) z (cid:17) − ν ∞ (cid:88) n =0 n + 2 νn + 2 ν (2 ν + 1) n n ! × p +3 F q +1 (cid:34) − n, n + 2 ν, ν + 1 , u p ν + 1 / , v q (cid:35) J n + ν ( z ) , (3.2)valid if p ≤ q and none of 2 ν, v q coincides with a negative integer.In the special case when one of numerator parameters happens to bea negative integer, the expansion formula (3.1) gives rise to various kindsof transformation formulas for terminating hypergeometric series with unitargument. Of subsequent importance will be the following formula:7 or each positive integer n , if p ≥ , then p +1 F p (cid:34) − n, u , · · · , u p v , · · · , v p (cid:35) = n (cid:88) k =0 (cid:18) nk (cid:19) ( σ ) k ( u ) k · · · ( u p ) k ( k + δ ) k ( v ) k · · · ( v p ) k F (cid:34) − k, k + δ, u , u σ, v , v (cid:35) × p F p − (cid:34) − n + k, k + σ, k + u , · · · , k + u p k + δ + 1 , k + v , · · · , k + v p (cid:35) . (3.3)The proof is immediate on taking r = p − , s = p − α r = ( − n, σ, u , · · · , u p ) , β s = ( v , · · · , v p ) , z = w = 1 . An advantageous feature in this transformation formula is that there aretwo free parameters δ, σ available. For instance, if we recall [4, 2.2(1)] F (cid:34) − k, k + α , α β , β (cid:35) = ( β − α ) k ( β − α ) k ( β ) k ( β ) k , (3.4)valid under the Saalsch¨utzian condition 1 + α + α = β + β , and choose σ = u , δ = v + v − u − , then formula (3.3) yields p +1 F p (cid:34) − n, u , · · · , u p v , · · · , v p (cid:35) = n (cid:88) k =0 (cid:18) nk (cid:19) ( v − u ) k ( v − u ) k ( u ) k · · · ( u p ) k ( k + v + v − u − k ( v ) k · · · ( v p ) k × p F p − (cid:34) − n + k, k + u , k + u , · · · , k + u p k + v + v − u , k + v , · · · , k + v p (cid:35) , (3.5)which coincides with Gasper’s transformation formula [14, (5.1)]. This section focuses on establishing positivity criteria for terminating F or F hypergeometric series under the Saalsch¨utzian condition, whichwill play crucial roles in later developments.8 .1 Terminating F hypergeometric series Lemma 4.1.
For each positive integer n , put Θ n = F (cid:34) − n, n + α , α , α β , β , β (cid:35) and assume that α > − , α > , α > and α + α + α = β + β + β . (4.1) If parameters α j , β j satisfy either set of the following additional conditionssimultaneously, then Θ n ≥ for all n ≥ . (A1) (cid:40) α ≤ β ≤ α ,α ≤ β ∧ β . (A2) α ≤ β ≤ α ,α ≤ β ∧ β , (1 + α ) α α ≤ β β β . Moreover, the underlined condition may be replaced by β ∧ β > , α − ≤ β ∧ β ≤ α . (4.2) Proof.
Under Saalsch¨utzian condition (4.1), if we apply (3.5) and simplifycoefficients with the aid of Saalsch¨utz’s formula (3.4), then we findΘ n = 1( β ) n n (cid:88) k =0 (cid:18) nk (cid:19) ( β − α ) k ( β − α ) k ( k + α + α − β ) k ( n + α ) k ( α ) k ( β ) k ( β ) k × ( k + 1 + α − β ) n − k ( β − α ) n − k (2 k + 1 + α + α − β ) n − k . (4.3)Since coefficients are all nonnegative under stated conditions of (A1),the nonnegativity of Θ n is obvious. On writing( β − α ) k ( β − α ) k = ( β − α )( β − α )( β − α + 1) k − ( β − α + 1) k − (4.4)for k ≥ , we also deduce the same result when the underlined condition of(A1) is replaced by the alternative condition of (4.2).A proof for the nonnegativity of Θ n under Saalsch¨utzian condition (4.1)and (A2) is given in our previous work [8, Lemma 3.1]. By tracking downthe proof therein, it is not hard to find that the underlined condition of (A2)can be replaced by (4.2) due to the same reason as above.9 emark . By extending (4.4) inductively, we may replace (4.2) by β ∧ β > , α − m ≤ β ∧ β ≤ α − m + 1 , (4.5)valid for any integer 1 ≤ m ≤ n − . In the case n = 1 , we noteΩ = β β β − (1 + α ) α α β β β , which explains in part why the last condition of (A2) is required. Theexpansion (4.3) is equivalent to Whipple’s transformation formula [4, 4.3(4)]. F hypergeometric series Lemma 4.2.
For each positive integer n , put Ω n = F (cid:34) − n, n + α , α , α , α β , β , β , β (cid:35) and assume that α > − , α > , α > , α > and α + α + α + α = β + β + β + β . (4.6) If parameters α j , β j satisfy either set of the following additional conditionssimultaneously, then Ω n ≥ for all n ≥ . (C1) α ≤ β ∧ β ,α ≤ β ∧ β ,α + α ≤ β + β ≤ α + α . (C2) α ≤ β ∧ β ,α ∧ β ∧ β ≤ α , α + α ≤ β + β ≤ α + α ,α α α ≤ β β ( β + β − α − . Moreover, the underlined condition may be replaced by (4.2) .Proof.
We apply Gasper’s transformation formula (3.5) to decomposeΩ n = n (cid:88) k =0 (cid:18) nk (cid:19) ( β − α ) k ( β − α ) k ( n + α ) k ( α ) k ( α ) k ( k + β + β − α − k ( β ) k ( β ) k ( β ) k ( β ) k W k , (4.7)10alid under Saalsch¨utzian condition (4.6), where W k = F (cid:34) − n + k, n + k + α , k + α , k + α k + β + β − α , k + β , k + β (cid:35) with the agreement W n = 1 . Since each W k is Saalsch¨utzian, if we check(A1) of Lemma 4.1 with matching parameters by( α , α , α ) (cid:55)→ (2 k + α , k + α , k + α ) , ( β , β , β ) (cid:55)→ ( k + β , k + β , k + β + β − α ) , then we find W k ≥ ≤ k ≤ n − α ≤ β + β − α ≤ α , α ≤ β ∧ β . In view of (4.7), we may conclude Ω n ≥ n under assumptions (4.6) and (C2), we applythe transformation formula (3.3) to decomposeΩ n = n (cid:88) k =0 (cid:18) nk (cid:19) ( σ ) k ( n + α ) k ( α ) k ( k + δ ) k ( β ) k ( β ) k U k V k , where (4.8) U k = F (cid:34) − k, k + δ, α , α σ, β , β (cid:35) ,V k = F (cid:34) − n + k, n + k + α , k + σ, k + α k + δ + 1 , k + β , k + β (cid:35) with the convention U = V n = 1 . Owing to condition (4.6), we note that U k , V k are Saalsch¨utzian for any pair ( δ, σ ) satisfying σ = 1 + δ + α + α − β − β . (4.9)By checking (A2) of Lemma 4.1 with matching parameters by( α , α , α ) (cid:55)→ ( δ, α , α ) , ( β , β , β ) (cid:55)→ ( β , β , σ ) , we find that U k ≥ ≤ k ≤ n when α ≤ σ ≤ δ, α ≤ β ∧ β , (1 + δ ) α α ≤ β β σ (4.10)11here the second condition may be replaced by (4.2). On the other hand,if we check (A1) of Lemma 4.1 with matching parameters by( α , α , α ) (cid:55)→ (2 k + α , k + α , k + σ ) , ( β , β , β ) (cid:55)→ ( k + β , k + β , k + δ + 1) , then we find that V k ≥ ≤ k ≤ n − α ≤ δ ≤ α , σ ≤ β ∧ β . (4.11)On selecting 1 + δ = α so that σ = β + β − α − , it is simple tofind that (4.9), (4.10), (4.11) amount to (C2). In view of the decomposition(4.8), we may conclude Ω n ≥ n ≥ In dealing with the last condition of (A2), Lemma 4.1, which often causesdifficulties in practice, it will be useful to invoke the following inequality.
Lemma 4.3.
Given < α ≤ α ≤ α , we have β β β ≥ α α α for allpositive real numbers β , β , β satisfying the conditions β + β + β = α + α + α , α ≤ β ∧ β ∧ β ≤ α . Proof.
We may assume α < α with no loss of generality. Setting f ( x , x ) = x x ( α + α + α − x − x ) ,D = (cid:8) ( x , x ) ∈ [ α , α ] × [ α , α ] : α + α ≤ x + x ≤ α + α (cid:9) , it suffices to prove that f ( x , x ) ≥ α α α for all ( x , x ) ∈ D. As readily verified, ∇ f ( x , x ) = (0 ,
0) only when x = x = α + α + α f (cid:18) α + α + α , α + α + α (cid:19) = (cid:18) α + α + α (cid:19) ≥ α α α .
12t remains to prove f ( x , x ) ≥ α α α for all ( x , x ) ∈ ∂D, where theboundary ∂D consists of six line segments ∂D = (cid:8) α ≤ x ≤ α , x = α (cid:9) ∪ (cid:8) α ≤ x ≤ α , x + x = α + α (cid:9) ∪ (cid:8) x = α , α ≤ x ≤ α (cid:9) ∪ (cid:8) α ≤ x ≤ α , x = α (cid:9) ∪ (cid:8) α ≤ x ≤ α , x + x = α + α (cid:9) ∪ (cid:8) x = α , α ≤ x ≤ α (cid:9) . For ( x , x ) ∈ (cid:8) α ≤ x ≤ α , x = α (cid:9) , the first line segment, if wewrite x = (1 − λ ) α + λα , ≤ λ ≤ , then the AM-GM inequality gives f ( x , x ) = α x ( α + α − x )= α (cid:2) (1 − λ ) α + λα (cid:3)(cid:2) λα + (1 − λ ) α (cid:3) ≥ α α α . On the second line segment of ∂D , we put x = (1 − λ ) α + λα , ≤ λ ≤ f ( x , x ) = α x ( α + α − x )= α (cid:2) (1 − λ ) α + λα (cid:3)(cid:2) λα + (1 − λ ) α (cid:3) ≥ α α α . By carrying out similar reasonings for other line segments of ∂D , it is notdifficult to find that the same conclusion continues to hold true. In what follows, we shall denote by Φ the F hypergeometric functiondefined in (1.1) unless specified otherwise. As it will become clearer in thesequel, our regions of positivity for Φ will be built on the basis of hexagonsor triangles whose vertices consist of permutations of ( a , a + 1 / , a ) . In order to represent such a hexagon or triangle in such a way that each oftheir vertices always indicates a fixed location, as observed in Figures 1, 2,it is effective to introduce the following arrangement.
Definition 5.1.
For a > , a > , we denote by ξ , ξ , ξ the arrangementof a , a + 1 / , a in ascending order of magnitude so that ξ = min (cid:0) a , a + 1 / , a (cid:1) , ξ = max (cid:0) a , a + 1 / , a (cid:1) A ⊂ R the set of all permutations of ( ξ , ξ , ξ ) labelled by A = (cid:8) A ( ξ , ξ , ξ ) , A ( ξ , ξ , ξ ) , A ( ξ , ξ , ξ ) ,A ( ξ , ξ , ξ ) , A ( ξ , ξ , ξ ) , A ( ξ , ξ , ξ ) (cid:9) . Partly due to our way of handling those terminating hypergeometricseries considered in the preceding section, it is essential for our investigationof positivity to impose the following admissible conditions on ( a , a ). Definition 5.2.
We denote by Λ the set of ( a , a ) ∈ ( R ∗ + ) satisfying a ≤ (cid:18) a + 12 (cid:19) or 1 ≤ a ≤ a + 1 if a ≤ / ,a ≤ max (cid:20) a + 32 , (cid:18) a + 12 (cid:19)(cid:21) if a ≥ / . Our principal result in this section reads as follows.
Proposition 5.1.
For ( a , a ) ∈ Λ , let H A denote the closed polygon withvertices A . If ( b , b , b ) ∈ H A , then Φ( x ) ≥ for all x > with strictinequality unless ( b , b , b ) ∈ A . We observe that the polygon H A lies on the plane b + b + b = ξ + ξ + ξ = 3 a + a + 1 / • If ξ , ξ , ξ are distinct, then H A represents the cyclic hexagon withvertices A , A , A , A , A , A in the clockwise order and center ξ + ξ + ξ , , . • If either ξ = ξ < ξ or ξ < ξ = ξ , then H A represents the trianglewith vertices A , A , A or A , A , A in the clockwise order. • In the case ξ = ξ = ξ , which occurs only when a = 1 / , a = 1 , then H A reduces to the singleton (cid:8) (1 , , (cid:9) . x ) = Γ ( a + 1 / (cid:16) x (cid:17) − a +1 × ∞ (cid:88) n =0 L n n + 2 a − n + 2 a − a ) n n ! J n + a − / ( x ) , (5.2)valid for each point ( b , b , b ) lying on the plane (5.1), where L n = F (cid:34) − n, n + 2 a − , a + 1 / , a b , b , b (cid:35) (5.3)for each n ≥ L ≡ . This expansion results from Gasper’s sums ofsquares formula (3.2), where the free parameter ν is chosen so that each L n satisfies the Saalscht¨utzian condition, that is,2 ν + 1 = b + b + b − a − a − / a . Once each L n were shown to be nonnegative, it is obvious from (5.1)that Φ ≥ , ∞ ). Furthermore, in view of the known fact ([26, 15.22])that those positive zeros of Bessel functions J ν , J ν +1 are interlaced for any ν ∈ R , we observe that Φ is strictly positive in the case when at least one ofthe L n ’s is strictly positive. In other words, if L n ≥ n ≥ , thenΦ is strictly positive unless (5.1) reduces toΦ( x ) = Γ ( a + 1 / (cid:16) x (cid:17) − a +1 J a − / ( x )= F (cid:34) a a + 1 / , a (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) , (5.4)which happens only when ( b , b , b ) ∈ A . By making use of positivity criteria established in Lemma 4.1, we nowproceed to prove the nonnegativity of each L n for ( b , b , b ) ∈ H A and( a , a ) ∈ Λ. In preparation for the proof, let us introduce Q j = (cid:8) ( b , b , b ) ∈ H A : ξ ≤ b j ≤ ξ (cid:9) , j = 1 , , . (5.5)As illustrated in Figure 1, if H A happens to be a hexagon, for instance, then Q , Q , Q represent separately quadrilaterals A A A A , A A A A , A A A A .
15n addition, we denote by T the subregion of H A bounded by three linesegments A A , A A , A A , which represents the triangle with vertices( ξ + ξ − ξ , ξ , ξ ) , ( ξ , ξ + ξ − ξ , ξ ) , ( ξ , ξ , ξ + ξ − ξ )unless H A reduces to a point. As readily observed, the decomposition H A = Q ∪ Q ∪ Q ∪ T (5.6)holds in any case. We note T ⊂ Q j for each j only if ξ + ξ − ξ ≥ ξ . Figure 1: A decomposition of hexagon in the case ξ + ξ > ξ . Our proof will be divided into two large cases.
Case 1. a ≥ / . We note a + 1 / ≤ a and we shall deal with threepossible occasions separately depending on the location of a . Subcase 1.1. a +1 / ≤ a ≤ a . Since ξ + ξ − ξ = 3 a +1 / − a , Q j = (cid:8) ( b , b , b ) ∈ H A : a ≤ b j ≤ a (cid:9) , j = 1 , , , and T has vertices ( c, a , a ) , ( a , c, a ) , ( a , a , c ) with c = 3 a + 1 / − a . By Lemma 4.1 applied with (A1), we find L n ≥ a ≤ b ≤ a , b ∧ b ≥ a + 1 / , L n ≥ b , b , b ) ∈ Q . By permuting the b j ’s, we findthe same also holds true for each point of Q , Q . (i) If c ≥ a , that is, 2 a ≤ a + 1 , then T ⊂ Q j for each j and wemay conclude L n ≥ b , b , b ) ∈ H A . (ii) In the opposite occasion 2 a > a + 1 , we apply Lemma 4.1 with(A1) and (4.2) to deduce that L n ≥ a + 1 / ≤ b ≤ a , a − ≤ b ∧ b ≤ a . If we further restrict c ≥ a − , that is, 2 a ≤ a + 3 / , then T ⊂ (cid:2) a + 1 / , a (cid:3) × (cid:2) a − , a (cid:3) × (cid:2) a − , a (cid:3) . It follows that L n is nonnegative for each point of T and hence for anypoint of the entire hexagon H A in this occasion.In summary, we have proved L n ≥ b , b , b ) ∈ H A when a ≥ / , a + 1 / ≤ a ≤ max (cid:20) a , (cid:18) a + 12 (cid:19)(cid:21) . (5.7) Subcase 1.2. a + 1 / ≤ a ≤ a . We note H A ⊂ (cid:2) a + 1 / , a (cid:3) × (cid:2) a + 1 / , a (cid:3) × (cid:2) a + 1 / , a (cid:3) ≡ I. By Lemma 4.1 applied with (A2) and (4.2), we deduce L n ≥ a + 1 / ≤ b ≤ a + 1 , a − ≤ b ∧ b ≤ a , a ( a + 1 / a ≤ b b b . (5.8)If we further assume a − ≤ a + 1 / , that is, a ≤ a + 3 / , whichholds only if 1 / < a ≤ / , then a ≤ a + 1 so that I ⊂ (cid:2) a + 1 / , a + 1 (cid:3) × (cid:2) a − , a (cid:3) × (cid:2) a − , a (cid:3) and hence L n ≥ α = a + 1 / , α = 2 a , α = a and β j = b j ,it is plain to verify the last condition of (5.8).In summary, we have proved L n ≥ b , b , b ) ∈ H A when1 / ≤ a ≤ / , a ≤ a ≤ a + 3 / . (5.9)17 ubcase 1.3. a ≤ a + 1 / ≤ a . Since ξ = a , ξ = a + 1 / , ξ = 2 a , ξ + ξ − ξ = a + a − / ≡ c, Q j = (cid:8) ( b , b , b ) ∈ H A : a + 1 / ≤ b j ≤ a (cid:9) , j = 1 , , , and vertices of triangle T are given by( c, a + 1 / , a + 1 / , ( a + 1 / , c, a + 1 / , ( a + 1 / , a + 1 / , c ) . By applying Lemma 4.1 with (A1), we find L n ≥ a + 1 / ≤ b ≤ a , b ∧ b ≥ a , which implies L n ≥ b , b , b ) ∈ Q . By permuting the b j ’s, we findthe same also holds true for each point of Q , Q . (i) If c ≥ a + 1 / , that is, a ≥ , then T ⊂ Q j for all j = 1 , , L n ≥ b , b , b ) ∈ H A . (ii) In the case c ≤ a + 1 / a ≤ , we apply Lemma 4.1 with (A1)and (4.2) to find that L n ≥ a ≤ b ≤ a , a − / ≤ b ∧ b ≤ a + 1 / . Since a − / ≤ c, it is evident that T ⊂ (cid:2) a , a (cid:3) × (cid:2) a − / , a + 1 / (cid:3) × (cid:2) a − / , a + 1 / (cid:3) , which implies L n ≥ T and hence for any point ofthe entire hexagon H A .In summary, we have proved L n ≥ b , b , b ) ∈ H A when a ≥ / , < a ≤ a + 1 / . (5.10) Case 2. < a < / . We note 2 a < a + 1 / a . Since the method is basically same as above, our proof will be sketchy.18 ubcase 2.1. a ≤ a ≤ a + 1 / . In decomposition (5.6), we note Q j = (cid:8) ( b , b , b ) ∈ H A : a ≤ b j ≤ a + 1 / (cid:9) , j = 1 , , , and T has ( c, a , a ) , ( a , c, a ) , ( a , a , c ) , with c ≡ a + 1 / − a . ByLemma 4.1, we find L n ≥ a ≤ b ≤ a + 1 , a − / ≤ b ∧ b ≤ a + 1 / ,b ∧ b > , a ( a + 1 / a ≤ b b b . (5.11)Since 2 a +1 > a +1 / , a − / < a and the last condition can be verifiedby Lemma 4.3, we conclude L n ≥ b , b , b ) ∈ Q . By permuting the b j ’s, we find the same also holds for each point of Q , Q . In the case c ≥ a , that is, 2 a ≤ a + 1 , we note T ⊂ Q j for each j and hence L n ≥ b , b , b ) ∈ H A . (Unfortunately, positivitycriteria of Lemma 4.1 are not suitable for dealing with the case c ≤ a . )In summary, we have proved L n ≥ b , b , b ) ∈ H A when0 < a < / , a ≤ a ≤ (cid:18) a + 12 (cid:19) . (5.12) Subcase 2.2. a < a + 1 / ≤ a . By interchanging a + 1 / , a and proceeding as in the preceding subcase, we find that if a ≤ a + 1 , then L n ≥ b , b , b ) ∈ Q j , where Q j denotes Q j = (cid:8) ( b , b , b ) ∈ H A : a + 1 / ≤ b j ≤ a (cid:9) , j = 1 , , . In regards to the triangle T having vertices( c, a + 1 / , a + 1 / , ( a + 1 / , c, a + 1 / , ( a + 1 / , a + 1 / , c ) , where c ≡ a + a − / , if c ≥ a + 1 / , then T ⊂ Q j for each j and L n ≥ b , b , b ) ∈ H A . We remark once again that positivity criteriaof Lemma 4.1 are not suitable for dealing with the case c ≤ a + 1 / . In summary, we have proved L n ≥ b , b , b ) ∈ H A when0 < a < / , ≤ a ≤ a + 1 . (5.13)19 ubcase 2.3. a ≤ a < a + 1 / . As H A ⊂ (cid:2) a , a + 1 / (cid:3) × (cid:2) a , a + 1 / (cid:3) × (cid:2) a , a + 1 / (cid:3) , it is immediate to confirm that the positivity region obtained by (5.11) coversthe whole of H A . Thus L n ≥ b , b , b ) ∈ H A when0 < a < / , < a ≤ a . (5.14) The end of proof for Proposition 5.1.
On collecting all of the abovecase-by-case results summarized in (5.7), (5.9), (5.10), (5.12), (5.13), (5.14),it is simple to find that each L n is nonnegative for ( b , b , b ) ∈ H A provided( a , a ) ∈ Λ . In view of the expansion (5.2) and our analysis on the strictpositivity, the proof is now complete.
As standard, the Newton polyhedron of a set E ⊂ R refers to theconvex hull of the union of octants with corners at all points of E , (cid:91) x ∈ E (cid:0) x + R (cid:1) , which will be denoted by Γ + ( E ) (see [25] for instance).According to the transference principle of Proposition 2.1, if Φ wereshown to be nonnegative for a single point b = ( b , b , b ) , then Φ remainsstrictly positive for all points of octant b + R except for the corner b . Asa consequence, we obtain the following extension of Proposition 5.1 whichconstitutes one of our main positivity results in this paper. Theorem 6.1.
For ( a , a ) ∈ Λ , if ( b , b , b ) ∈ Γ + ( A ) , then Φ( x ) ≥ for all x > and strict inequality holds true unless ( b , b , b ) ∈ A . Remark . For practical applications, it would be desirable to characterize Γ + ( A ) explicitly. On inspecting boundary planes both analytically andgeometrically, it is not difficult to find that Γ + ( A ) consists of all points( b , b , b ) satisfying the following conditions simultaneously: b ∧ b ∧ b ≥ ξ , ( b + b ) ∧ ( b + b ) ∧ ( b + b ) ≥ ξ + ξ ,b + b + b ≥ ξ + ξ + ξ . (6.1)20igure 2: A Newton polyhedron with a hexagonal face.As readily observed, Γ + ( A ) has the following geometric features. • If ξ , ξ , ξ are distinct, as illustrated in Figure 2, Γ + ( A ) representsan infinite polygonal region having 6 vertices at A , 7 faces includingthe hexagon H A , and 12 edges comprised of 6 boundary lines of H A and 6 rays emanating from A parallel to the coordinate axes. • If ξ = ξ < ξ or ξ < ξ = ξ , then Γ + ( A ) represents an infinitepolygonal region having 3 vertices at A , 4 faces including the triangle H A , and 6 edges comprised of 3 boundary lines of H A and 3 raysemanating from A parallel to the coordinate axes. • If ξ = ξ = ξ , which happens only when a = 1 / , a = 1 , then Γ + ( A ) = + R , where = (1 , , . If a = a = a, it is Theorem 6.1 that we wish to establish ultimatelyin this paper. Since ( a, a ) ∈ Λ for any a > , with no further restrictions,it may be noteworthy to single out the result in this special occasion.21 orollary 6.1. For any a > , suppose (cid:40) b ∧ b ∧ b ≥ a, b + b + b ≥ a + 1 / , ( b + b ) ∧ ( b + b ) ∧ ( b + b ) ≥ a + 1 / − σ, (6.2) where σ = max (cid:0) a + 1 / , a (cid:1) . If we define φ ( x ) = F (cid:34) a, ab , b , b (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) ( x > , then φ ( x ) ≥ for all x > and strict inequality holds true unless ( b , b , b ) coincides with a permutation of ( a, a + 1 / , a ) . In the case when a , a are distinct, however, there is another Newtonpolyhedron available for the positivity of Φ whose vertices are comprisedof permutations of ( a , a + 1 / , a ). To state more precisely, we shallrearrange these vertices for the same purpose as before. Definition 6.1.
For a > , a > , we denote by η , η , η the arrangementof a , a + 1 / , a in ascending order of magnitude so that η = min (cid:0) a , a + 1 / , a (cid:1) , η = max (cid:0) a , a + 1 / , a (cid:1) and by B ⊂ R the set of all permutations of ( η , η , η ) labelled by B = (cid:8) B ( η , η , η ) , B ( η , η , η ) , B ( η , η , η ) ,B ( η , η , η ) , B ( η , η , η ) , B ( η , η , η ) (cid:9) . Let Λ ∗ be the reflection of Λ across the line a = a , that is,Λ ∗ = { ( a , a ) : ( a , a ) ∈ Λ } . Applying Proposition 5.1 with roles of a , a interchanged, we find that for any pair ( a , a ) ∈ Λ ∗ , if ( b , b , b ) ∈ H B , the closed polygon withvertices B , then Φ( x ) ≥ for all x > and strict inequality holds trueunless ( b , b , b ) ∈ B . Since { ( a , a ) : 0 < a < a } ⊂ Λ ∗ , an extension ofthis positivity result by the Newton polyhedron of B can be summarized asfollows, which constitutes our second main positivity result. Theorem 6.2.
For < a < a , if ( b , b , b ) ∈ Γ + ( B ) , then Φ( x ) ≥ for all x > and strict inequality holds true unless ( b , b , b ) ∈ B . emark . We should emphasize that no additional conditions on the pair( a , a ) are required for validity. In analogue with (6.1), Γ + ( B ) consists ofall points ( b , b , b ) satisfying the following conditions simultaneously: b ∧ b ∧ b ≥ η , ( b + b ) ∧ ( b + b ) ∧ ( b + b ) ≥ η + η ,b + b + b ≥ η + η + η , (6.3)where η = a , η + η + η = 3 a + a + 1 / Γ + ( A ) , Γ + ( B ) are available as positivityregions of Φ, it is natural to ask whether the Newton polyhedron Γ + ( A ∪ B )also gives rise to a region of positivity. What matters are side polygonalregions between H A , H B which we shall now investigate. Note Γ + ( A ) ∪ Γ + ( B ) ⊂ Γ + ( A ∪ B ) . In order to facilitate our geometric interpretations, we shall consider both H A , H B as if they were hexagons and hence the interconnecting polygonalregion would consist of six side quadrilaterals. Since any of H A , H B inother occasions may be regarded as continuous deformations of hexagons inan obvious way, such a presumption would be harmless.Due to our arrangements, it is simple to observe that each of six sidequadrilaterals connecting the two hexagons has vertices A i , A j , B j , B i forsome i, j ∈ { , , } with i (cid:54) = j , if enumerated in a fixed direction, whichwill be denoted by S (cid:2) A i A j B j B i (cid:3) . It is convenient to interpret S (cid:2) A i A j B j B i (cid:3) = (cid:91) λ ∈ [0 , L λ , (7.1)the collection of all line segments L λ joining two points(1 − λ ) A i + λB i , (1 − λ ) A j + λB j , ≤ λ ≤ . By parameterizing in an elementary way, it is immediate to characterizethe line segment L λ in two exemplary cases:23i) For S (cid:2) A A B B (cid:3) , each L λ consists of all ( b , b , b ) satisfying b = (1 − λ ) ξ + λη ,b + b = (1 − λ )( ξ + ξ ) + λ ( η + η ) , (1 − λ ) ξ + λη ≤ b ∧ b ≤ (1 − λ ) ξ + λη . (7.2)(ii) For S (cid:2) A A B B (cid:3) , each L λ consists of all ( b , b , b ) satisfying b = (1 − λ ) ξ + λη ,b + b = (1 − λ )( ξ + ξ ) + λ ( η + η ) , (1 − λ ) ξ + λη ≤ b ∧ b ≤ (1 − λ ) ξ + λη . (7.3)As readily verified, by characterizing each L λ in the same way or bygeometric considerations, it turns out that the other four side quadrilateralsarise as mirror images of the above two quadrilaterals across either the plane b = b or b = b in the senses specified as follows. Lemma 7.1.
For < a < a , the following hold true. S (cid:2) A A B B (cid:3) = (cid:110) ( b , b , b ) : ( b , b , b ) ∈ S (cid:2) A A B B (cid:3)(cid:111) , S (cid:2) A A B B (cid:3) = (cid:110) ( b , b , b ) : ( b , b , b ) ∈ S (cid:2) A A B B (cid:3)(cid:111) , S (cid:2) A A B B (cid:3) = (cid:110) ( b , b , b ) : ( b , b , b ) ∈ S (cid:2) A A B B (cid:3)(cid:111) , S (cid:2) A A B B (cid:3) = (cid:110) ( b , b , b ) : ( b , b , b ) ∈ S (cid:2) A A B B (cid:3)(cid:111) . We are now in a position to prove the positivity of Φ when ( b , b , b )belongs to those side quadrilaterals. In view of the symmetry specified asabove, it suffices to deal with any two side quadrilaterals of type S (cid:2) A A B B (cid:3) , S (cid:2) A A B B (cid:3) . (cid:2) A A B B (cid:3) . As to three side quadrilaterals of this type, it turns out that admissible conditions for ( a , a ) are independent of Λ and given as follows. Definition 7.1.
We denote by Σ the set of ( a , a ) ∈ ( R ∗ + ) satisfying (cid:40) a < a ≤ min (cid:0) a , / (cid:1) if a < / ,a < a ≤ a if a ≥ / . roposition 7.1. For ( a , a ) ∈ Σ , if ( b , b , b ) ∈ S (cid:2) A A B B (cid:3) , then Φ( x ) ≥ for all x > and strict inequality holds true unless it coincideswith one of vertices. By symmetry, the same holds true for S (cid:2) A A B B (cid:3) , S (cid:2) A A B B (cid:3) . We apply Gasper’s sums of squares formula (3.2) to expandΦ( x ) = Γ ( ν + 1) (cid:16) x (cid:17) − ν ∞ (cid:88) n =0 K n n + 2 νn + 2 ν (2 ν + 1) n n ! J n + ν (cid:16) x (cid:17) , where ν is an arbitrary real number subject to the condition that 2 ν doesnot coincide with a negative integer, K n = F (cid:34) − n, n + 2 ν, ν + 1 , a , a ν + 1 / , b , b , b (cid:35) (7.4)for each n ≥ K ≡ . By the same reason as mentioned in the proofof Theorem 5.1, it will be sufficient to prove K n ≥ n ≥ . In accordance with (7.1), we shall write S (cid:2) A A B B (cid:3) = (cid:91) λ ∈ [0 , L λ and deduce the nonnegativity of K n on L λ by applying positivity criteriafor the terminating F hypergeometric series established in Lemma 4.2.For each ( b , b , b ) ∈ L λ , it follows from the characterization (9.10) thatthe Saalscht¨utzian condition for K n amounts to the choice2 ν + 1 = b + b + b − a − a − /
2= 2 (cid:2) (1 − λ ) a + λa (cid:3) . (7.5)Since 2 ν + 1 > λ ∈ [0 , , this choice would be legitimate.We shall divide our proof into three cases. Case 1. / ≤ a < a ≤ a + 1 / ≤ a . Due to the correspondence( ξ , ξ , ξ ) = ( a , a + 1 / , a ) , ( η , η , η ) = ( a , a + 1 / , a ) , ν + 1 / − λ ) ξ + λη ,ν + 1 = (1 − λ ) ξ + λη , ν + 1 = (1 − λ ) ξ + λη . Concerning the positivity criterion (C1) of Lemma 4.2, if we match( α , α , α , α ) (cid:55)→ (2 ν, ξ , ν + 1 , η ) , ( β , β , β , β ) (cid:55)→ ( b , b , b , ν + 1 / , then it follows readily from the above expression and (9.10) α ≤ β ∧ β , α ≤ β ∧ β , α + α ≤ β + β ≤ α + α , whence we may conclude K n ≥ n ≥ Case 2. / ≤ a < a + 1 / ≤ a ≤ a . Due to the correspondence( ξ , ξ , ξ ) = ( a + 1 / , a , a ) , ( η , η , η ) = ( a , a + 1 / , a ) , the Saalscht¨utzian condition (7.5) allows us to express ν + 1 / − λ ) η + λξ ,ν + 1 = (1 − λ ) ξ + λη , ν + 1 = (1 − λ ) ξ + λη . We shall deduce the nonnegativity of K n based on the criterion (C2) ofLemma 4.2. On matching parameters by( α , α , α , α ) (cid:55)→ (2 ν, ν + 1 , η , ξ ) , ( β , β , β , β ) (cid:55)→ ( ν + 1 / , b , b , b ) , it is straightforward to verify the first three conditions of (C2) α ≤ β ∧ β , α ∧ β ∧ β ≤ α , α + α ≤ β + β ≤ α + α . For the last condition of (C2), we use the fact ξ − η = η − ξ = 1 / β β ( β + β − α − (cid:2) (1 − λ ) η + λξ (cid:3)(cid:2) (1 − λ ) ξ + λη (cid:3)(cid:2) (1 − λ ) ξ + λη (cid:3) ≥ (cid:2) (1 − λ ) η + λξ (cid:3) (cid:8) η ξ + (cid:2) (1 − λ ) ξ + λη (cid:3) / (cid:9) ≥ (cid:8)(cid:2) (1 − λ ) η + λξ (cid:3) + 1 / (cid:9) η ξ = α α α and hence we may conclude K n ≥ n ≥ ase 3. a < a ≤ / , a ≤ a ≤ a + 1 / . We note the correspondence( ξ , ξ , ξ ) = ( a , a , a + 1 / , ( η , η , η ) = ( a , a , a + 1 / ν + 1 / − λ ) η + λξ ,ν + 1 = (1 − λ ) ξ + λη , ν + 1 = (1 − λ ) ξ + λη . In this case, we apply the positivity criterion (C1) of Lemma 4.2 withthe alternative condition (4.2) and parameter-matchings( α , α , α , α ) (cid:55)→ (2 ν, ξ , ν + 1 , η ) , ( β , β , β , β ) (cid:55)→ ( b , b , b , ν + 1 / . While the other conditions can be verified easily, we use (9.10) to observe ν ≤ b ∧ b ≤ (1 − λ )( a + 1 /
2) + λ ( a + 1 /
2) = ν + 1 , which verifies (4.2) and hence K n ≥ n ≥ Remark . Each of quadrilaterals S (cid:2) A A B B (cid:3) , S (cid:2) A A B B (cid:3) , S (cid:2) A A B B (cid:3) lies separately on the plane b + b + b = 3 a + 12 + a + 2( a − a ) a − ξ ( b − ξ ) ,b + b + b = 3 a + 12 + a + 2( a − a ) a − ξ ( b − ξ ) ,b + b + b = 3 a + 12 + a + 2( a − a ) a − ξ ( b − ξ ) . (7.6) (cid:2) A A B B (cid:3) . A geometric inspection reveals that three side quadrilaterals of this typeare part of Newton polyhedra of their neighboring side quadrilaterals andhence the positivity of Φ follows from the preceding results.27 roposition 7.2.
For ( a , a ) ∈ Λ ∩ Σ , if ( b , b , b ) ∈ S (cid:2) A A B B (cid:3) , then Φ( x ) ≥ for all x > and strict inequality holds true unless it coincideswith one of vertices. By symmetry, the same also holds true for S (cid:2) A A B B (cid:3) , S (cid:2) A A B B (cid:3) . Proof.
Let { e , e , e } be the standard basis of R and Π the union of infinitevertical strips bounded below by line segments A A , A B , A B ,Π = { b + (cid:15) e : (cid:15) ≥ , b ∈ A A ∪ A B ∪ A B } . (7.7)Since ( a , a ) ∈ Λ ∩ Σ , it follows from Propositions 5.1, 7.1, together withthe transference principle, that Φ remains strictly positive for each point ofΠ except for vertices A , A , B , B where Φ is nonnegative.As readily observed, S (cid:2) A A B B (cid:3) lies on the plane b + b + b = 3 a + 1 / a + 2( a − a ) η − ξ ( b − ξ ) . (7.8)(i) If ξ + ξ = η + η or equivalently η − ξ = 2( a − a ) , then theplane becomes vertical with its equation b + b = ξ + ξ so that S (cid:2) A A B B (cid:3) ⊂ Π , whence the result follows immediately.(ii) In the case ξ + ξ < η + η , which may happen only when0 < a < a ≤ min (cid:0) a , / (cid:1) , (7.8) becomes b + b − b = a + a − / b + b = 2 a + a , b + b = 2 a + 2 a , b + 2 b = 2 a + 2 a . By considering the inner product between normal vectors of theseplanes, it is simple to find, for instance, that S (cid:2) A A B B (cid:3) and Πare positively distant apart with respect to the b -direction in thesense that any point P on this quadrilateral, excluding the commonedges, can be written as P = Q + (cid:15) e with unique Q ∈ Π and (cid:15) > . Consequently, the result follows by the transference principle.(Alternatively, one may use the inclusion S (cid:2) A A B B (cid:3) ⊂ Γ + (cid:0)(cid:8) A , A , A , A , B , B , B , B (cid:9)(cid:1) . Main positivity result
On combining Theorem 5.1, Lemma 7.1 and Propositions 7.1, 7.2, withthe aid of the transference principle, we may conclude that the Newtonpolyhedron of
A ∪ B is a positivity region of Φ for any pair ( a , a ) ∈ Λ ∩ Σ , what we aimed to establish ultimately. Theorem 8.1.
Suppose that a > , a > and a < a ≤ min (cid:0) a , / (cid:1) if a < / ,a < a ≤ min (cid:20) a , (cid:18) a + 12 (cid:19)(cid:21) if a ≥ / . (8.1) If ( b , b , b ) ∈ Γ + ( A ∪ B ) , then Φ( x ) ≥ for all x > and strict inequal-ity holds true unless ( b , b , b ) ∈ A ∪ B . Remark . By exploiting the symmetry specified in Lemma 7.1 and (7.6),it is not simple but elementary to give an analytic expression for the Newtonpolyhedron of
A ∪ B in terms of a , a as follows.(P1) For 1 / ≤ a < a ≤ a + 1 / ≤ a , Γ + ( A ∪ B ) consists of all points( b , b , b ) satisfying the following conditions simultaneously: b i ≥ a , b i + b j ≥ a + a + 1 / i (cid:54) = j ) ,b + b + b ≥ a + a + 1 / ,b + b + 3 b ≥ a + 3 a + 1 / ,b + 3 b + b ≥ a + 3 a + 1 / , b + b + b ≥ a + 3 a + 1 / . (8.2)(P2) For 1 / ≤ a < a + 1 / ≤ a ≤ a , Γ + ( A ∪ B ) consists of all points( b , b , b ) satisfying the following conditions simultaneously: b i ≥ a , b i + b j ≥ a + a + 1 / i (cid:54) = j ) ,b + b + b ≥ a + a + 1 / ,b + b + b ≥ a + 3 a + 1 / − a − a )( b − a ) ,b + b + b ≥ a + 3 a + 1 / − a − a )( b − a ) ,b + b + b ≥ a + 3 a + 1 / − a − a )( b − a ) . (8.3)(Note that (8.3) reduces to (8.2) in the case a = a + 1 / . )29P3) For 0 < a < a ≤ min (cid:0) a , / (cid:1) , Γ + ( A ∪ B ) consists of all points( b , b , b ) satisfying the following conditions simultaneously: b i ≥ a , b i + b j ≥ a + a , (2 b i + b j ) ∧ ( b i + 2 b j ) ≥ a + 2 a ( i (cid:54) = j ) ,b + b + b ≥ a + a + 1 / ,b + b + 3 b ≥ a + 3 a + 1 / ,b + 3 b + b ≥ a + 3 a + 1 / , b + b + b ≥ a + 3 a + 1 / . (8.4)Figure 3: A Newton polyhedron of case (P1) or (P2) with a hexagonal face.A close inspection reveals that ξ + ξ = η + η in both cases of (P1),(P2), whereas ξ + ξ < η + η in case of (P3). From a geometric view-point,if H A is a hexagon, it means that the Newton polyhedra corresponding to(P1), (P2) are alike with the same number of faces, whereas the Newtonpolyhedron corresponding to (P3) has six more faces. For convenience, weillustrate two kinds of Newton polyhedra in Figures 3, 4.30igure 4: A Newton polyhedron of case (P3) with a hexagonal face.We recall from (2.1) of Proposition 2.2 that R = (cid:26) b ∈ a + R : b · ≥ a + a + 12 (cid:27) , (8.5)with = (1 , , , provides a necessity region for Φ. In addition, if b liesoutside R , then Φ oscillates in sign at least once on (0 , ∞ ). As it is evidentthat the Newton polyhedron Γ + ( A ∪ B ) of any case is properly contained in R , we have thus covered all cases of interest in the first octant of R exceptthe missing region R \ Γ + ( A ∪ B ) under the assumption (8.1).
This section deals with an application to the problem of positivity for aclass of Riemann-Liouville fractional integrals of Bessel functions. Of criticalimportance in this subject will be the following result which improves ourearlier version [7, Theorem 3.1] considerably.31 emma 9.1.
For a > , b > , c > − , we have F (cid:34) a, a + 1 / c + 1 , a + b, a + b + 1 / (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) ≥ x >
0) (9.1) when a, b, c satisfy the following case assumptions. (i) If − < c ≤ − / , then < a ≤ min (cid:0) c + 1 , b + c (cid:1) , b ≥ / . (ii) If c ≥ − / , then b ≥ / and < a ≤ min (cid:18) , b + c − (cid:19) or ≤ a ≤ min (cid:18) c + 1 , b + c, b + c , b c , b + c − (cid:19) . Moreover, the inequality of (9.1) is strict unless a = 1 / , b = 1 , c = − / or a = 1 , b = c = 1 / . (9.2) Proof.
In the case when a ≥ / , we note that Theorem 8.1 is applicable.More concretely, the pair ( a, a + 1 /
2) falls under the scope of (P1) and itfollows from the characterization (8.2) of the associated Newton polyhedronthat (9.1) holds when parameters a, b, c satisfy12 ≤ a ≤ min (cid:18) c + 1 , b + c, b + c , b c , b + c − (cid:19) and b ≥ . (9.3)On the other hand, Theorem 6.2 is applicable without any additionalcondition on a . Since ( η , η , η ) = ( a, a + 1 , a + 1) , it follows from (6.3)that (9.1) holds when parameters a, b, c satisfy0 < a ≤ min (cid:18) c + 1 , b + c, b + c − (cid:19) and b ≥ . (9.4)On combining (9.3), (9.4), we confirm immediately validity for the in-equality (9.1) under stated conditions. For the strict positivity, it is elemen-tary to find that the tuple of denominator-parameters coincides with one ofpermutations of ( a + 1 / , a + 1 / , a ) or ( a, a + 1 , a + 1) only in the caseof (9.2), whence the assertion follows by Theorems 6.2, 8.1.32e consider the fractional integral of the form (cid:90) x ( x − t ) λ t µ J α ( t ) dt = B ( λ + 1 , α + µ + 1)2 α Γ( α + 1) x α + λ + µ +1 × F (cid:34) ( α + µ + 1) / , ( α + µ + 2) / α + 1 , ( α + λ + µ + 2) / , ( α + λ + µ + 3) / (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) , (9.5)where x > α > − , λ > − , α + µ + 1 > α, λ, µ for the positivity of the above integralis historic and we refer to [13] for an extensive survey of earlier resultscontributed by numerous authors (see also [1], [2]). Despite many partialresults, however, it appears that a significant progress has not been madeyet for a complete description of parameters for positivity.On recognizing that the F hypergeometric function on the right sideof (9.5) is of class (9.1) with parameter matchings a = α + µ + 12 , b = λ + 12 , c = α, it is a matter of algebra to deduce from Lemma 9.1 the following. Theorem 9.1.
For α > − , the inequality (cid:90) x ( x − t ) λ t µ J α ( t ) dt ≥ x >
0) (9.6) holds true under the following case assumptions on λ, µ. (i) If − < α ≤ − / , then λ ≥ − / , − α − < µ ≤ min (cid:0) α + 1 , λ + α (cid:1) . (ii) If α ≥ − / , then λ ≥ − / and − α − < µ ≤ min (cid:18) − α, λ − (cid:19) or − α ≤ µ ≤ min (cid:20) α + 1 , λ + α, λ + 12 , λ + 12 , (cid:18) λ + α + 12 (cid:19)(cid:21) . Moreover, the inequality of (9.6) is strict unless α = − / , λ = 1 , µ = 1 / or α = 1 / , λ = 0 , µ = 1 / nd the integral in both exceptional cases reduces to (cid:114) π (cid:90) x ( x − t ) cos t dt = (cid:114) π (cid:90) x sin t dt = 2 (cid:114) π sin (cid:16) x (cid:17) . (9.8) Remark . By the necessary condition (2.1) for the positivity of the F hypergeometric function defined on the right side of (9.5), it is simple tofind that the inequality (5.7) fails to hold when µ > λ + 1 / α + 1 < µ ≤ λ + 1 / . (9.9)A number of special cases are noteworthy. To describe, let us denote by P α , α > − , the region of validity for (9.6) in the ( λ, µ )-plane determined bythe above case assumptions, which represents an infinite polygonal subregionof strip [ − / , ∞ ) × [ − α − , α + 1] . Figure 5: The validity region P α in the ( λ, µ ) plane for the case α ≥ / , where P = ( α − / , α ) , Q = ( α + 3 / , α + 1).As readily observed in Figure 5, a part of the line µ = λ + 1 / P α only when α ≥ / ≤ λ ≤ α − / , which leads to34S1) If α > / , ≤ λ ≤ α − / , then (cid:90) x ( x − t ) λ t λ +1 / J α ( t ) dt > x > . This inequality was proposed by Gasper [13, (1.5)] as a limiting case ofa conjecture regarding positive sums of Jacobi polynomials in Askey andGasper [3] and proved by himself in a rather complicated way.In a similar manner, it is elementary to observe that a segment of theline µ = λ − / P α only when α ≥ − / (cid:40) − α − / < λ ≤ α + 3 / − / ≤ α ≤ , − / ≤ λ ≤ α + 3 / α > . (9.10)(S2) For any pair ( α, λ ) satisfying (9.10) , the inequality (cid:90) x ( x − t ) λ t λ − / J α ( t ) dt > x > . holds true unless α = − / , λ = 1 . We remark that this inequality was proved by Gasper [13, (2.10)] butonly in the range 1 ≤ λ ≤ α + 3 / , α > − / . For example, if we take α = 1 / , λ = 0 , not available in Gasper’s validity region, (S2) is equivalentto the well-known positivity of sine integral (cid:90) x sin tt dt > x > . A line segment of µ = ( λ + α + 1 / / P α in the range (cid:40) − α + 1 / ≤ λ ≤ α + 3 / | α | ≤ / ,α − / ≤ λ ≤ α + 3 / α ≥ / , (9.11)where the latter case corresponds to the line segment P Q in Figure 5, andintersects the horizontal line µ = α + β at λ = α + 2 β − / . As a conse-quence, we find that the intersection point (cid:0) α + 2 β − / , α + β (cid:1) ∈ P α if2 α + 2 β ≥ , β ≤ | α | ≤ / ≤ β ≤ α ≥ / , which leadsto the following inequality after simplifying.35S3) For any pair ( α, λ ) satisfying (9.11) , the inequality (cid:90) x ( x − t ) λ t ( λ + α + ) J α ( t ) dt > x > . holds true unless α = − / , λ = 1 or α = 1 / , λ = 0 . As a particularcase, if α + β ≥ / , ≤ β ≤ , then (cid:90) x ( x − t ) α +2 β − / t α + β J α ( t ) dt ≥ x > with strict inequality unless α = − / , β = 1 or α = 1 / , β = 0The second inequality was proved by Gasper [13], [14] (the equality casein [14, Theorem 7] must be corrected out as above). The particular case β = 0 was proved earlier by Fields and Ismail [11].On fixing λ = 1 and inspecting the corresponding vertical line segmentcontained in P α case by case, it is immediate to deduce the following.(S4) The inequality (cid:90) x ( x − t ) t µ J α ( t ) dt > x > holds true for any α, µ satisfying the condition − α − < µ ≤ min (cid:20) α + 1 , (cid:18) α + 32 (cid:19) , (cid:21) , α > − except for the case α = − / , µ = 1 / . As observed by Misiewicz and Richards [20], this implies (cid:90) x (cid:0) x δ − t δ (cid:1) λ t µ J α ( t ) dt > x >
0) (9.13)for any 0 < δ ≤ ≤ λ under the condition (9.12) on α, µ , which arises as ageneralization of Kuttner’s problem [17] concerning the positivity of Rieszmeans of Fourier series. The inequality under the present condition (9.12)was proved by ourselves ([7, Theorem 4.1]) in a different manner and appliedto extend Buhmann’s classes of radial basis functions [5].In terms of Fourier cosine or sine transforms, two special cases α = ± / λ (cid:55)→ α − , µ (cid:55)→ β − / . The inequalities (i) (cid:90) cos( xt ) (1 − t ) α − t β − dt ≥ , ( α, β ) ∈ ∆ c , (ii) (cid:90) sin( xt ) (1 − t ) α − t β − dt ≥ , ( α, β ) ∈ ∆ s hold true for all x > , where ∆ c = (cid:110) ( α, β ) : α > , < β ≤ min (1 , α − (cid:111) , ∆ s = (cid:110) ( α, β ) : α ≥ / and − < β ≤ min (0 , α − or ≤ β ≤ min (cid:2) , ( α + 1) / , α − (cid:3)(cid:111) . Moreover, the first inequality is strict unless α = 2 , β = 1 and thesecond inequality is strict unless α = β = 1 . We remark that ∆ c ⊂ ∆ s and more generally P α increases monotoni-cally as α increases. Since the associated beta density f ( t ) = 1 B ( α, β ) (1 − t ) α − t β − ( α > , β > α, β ), these results indicate whyit is so difficult to prove or disprove positivity of Fourier transforms by usingonly intrinsic characters of densities such as convexity or monotonicity, aswas suggested by Tuck [24] or P´olya [22] for instance.For negative results, we note that if β > α or β ≤ α, β > , thenthe Fourier cosine transform in (i) changes sign infinitely often according toProposition 2.2. Likewise, if β > α or β ≤ α, β > , then the Fourier sinetransform in (ii) changes sign. In the special case 0 < α < , β = 1 , werefer to Koumandos [16] for the estimate of those positive zeros.The second inequality extends the result of Williamson [27] (cid:90) sin( xt ) (1 − t ) t dt > , which was used decisively in his simplified proof of Royall’s theorem [23] thatLaplace transforms of 3-times monotone functions on (0 , ∞ ) are univalent in37he right-half plane excluding the imaginary axis. In addition, he considereda family of Fourier sine transforms (cid:0) g α (cid:1) α> defined by g α ( x ) = (cid:90) sin( xt ) (1 − t ) α − t dt ( x >
0) (9.14)and conjectured that there exists α (cid:48) with 2 < α (cid:48) < g α remainsnonnegative for α ≥ α (cid:48) but changes sign for 0 < α < α (cid:48) . If this conjecturewere true, as he pointed out, then Royall’s theorem could be extended tothe class of α -monotone functions by the same method he employed. Proposition 9.1.
Williamson’s conjecture is false in that g α remains strictlypositive on (0 , ∞ ) for α ≥ but changes sign for < α < . The proof is immediate. On taking β = 2 and inspecting ∆ s , it is simpleto find that g α remains strictly positive on (0 , ∞ ) for α ≥ . On the otherhand, it follows from (9.5) that g α ( x ) = c α x G α ( x ) , where G α ( x ) = F (cid:34) α + 3) / , ( α + 4) / (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) and c α is a positive constant. Setting a = 2 , b = ( α + 3) / , c = ( α + 4) / , it follows from the necessity part (i) of Theorem 9.2 that if b + c < a + 1 / , that is, α < , then G α changes sign on (0 , ∞ ) and so does g α . Appendix . The following is a summary of our work [8], [9] concerningthe positivity of F hypergeometric functions of similar type (see also [6],[7] for relevant applications and [15] for a probabilistic approach). Theorem 9.2.
For a > , b > , c > , put Ψ( x ) = F (cid:34) ab, c (cid:12)(cid:12)(cid:12)(cid:12) − x (cid:35) ( x > . (i) If b ≤ a or c ≤ a or b + c < a + 1 / , then Ψ changes sign. (ii) If ( b, c ) ∈ P ∗ a , then Ψ( x ) ≥ for all x > and strict inequality holdstrue unless it belongs to S = (cid:8) ( a + 1 / , a ) , (2 a, a + 1 / (cid:9) , where P ∗ a = (cid:26) ( b, c ) : b > a, c ≥ max (cid:104) a + 1 / − b, a + a b − a ) (cid:105)(cid:27) .
38s illustrated in Figure 6, P ∗ a represents an infinite hyperbolic region in R containing the Newton polyhedron of S characterized by Γ + ( S ) = { ( b, c ) : b ∧ c ≥ min ( a + 1 / , a ) , b + c ≥ a + 1 / } . Figure 6: The positivity region P ∗ a (red-colored) in the case a ≥ / , whichcontains the Newton polyhedron of S = ( a + 1 / , a ) , S = (2 a, a + 1 / . If ( b, c ) belongs to the grey-colored region, then Ψ oscillates in sign.
Acknowledgements . Yong-Kum Cho is supported by the Basic Sci-ence Research Program through the National Research Foundation of Ko-rea (NRF) funded by the Ministry of Education (2018R1D1A1A09083148).Seok-Young Chung is supported by the Chung-Ang University GraduateResearch Scholarship in 2019.
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