Functional inequalities for modified Struve functions

0.17 MB·English

by  Árpád Baricz

FUNCTIONAL INEQUALITIES FOR MODIFIED STRUVE FUNCTIONS
A´RPA´DBARICZANDTIBORK.POGA´NY
Abstract. Inthispaper,byusingageneralresultonthemonotonicityofquotientsofpowerseries,our
aim is to prove some monotonicity and convexity results for the modified Struve functions. Moreover,
asconsequencesoftheabovementionedresults,wepresentsomefunctionalinequalitiesaswellaslower
andupperboundsformodifiedStruvefunctions. Ourmainresultscomplementandimprovetheresults
3 ofJoshiandNalwaya[11].
1
0
1. Introduction
2
n Inthelastfewdecadesmanyinequalitiesandmonotonicitypropertiesforspecialfunctions(likeBessel,
a modified Bessel, hypergeometric) and their several combinations have been deduced by many authors,
J motivated by various problems that arise in wave mechanics, fluid mechanics, electrical engineering,

3 quantum billiards, biophysics, mathematical physics, finite elasticity, probability and statistics, special
2
relativity and radar signal processing. Although the inequalities involving the quotients of modified

Bessel functions of the first and second kind are interesting in their own right, recently the lower and
]
A upper bounds for such ratios received increasing attention, since they play an important role in various
C problems of mathematical physics and electrical engineering. For more details, see for example [8] and
. the referencestherein. The modifiedStruve functions arerelatedto modifiedBesselfunctions, thus their
h
properties canbe useful in problems of mathematical physics. In [11] Joshi andNalwayapresentedsome
t
a two-sided inequalities for modified Struve functions and for their ratios. They deduced also some Tur´an
m
and Wronski type inequalities for modified Struve functions by using some generalized hypergeometric
[ function representation of the Cauchy product of two modified Struve functions. Our main motivation

to write this paper is to complement and improve the results of Joshi and Nalwaya [11]. In this paper,
1
v byusing aclassicalresultonthe monotonicityofquotients ofMacLaurinseries,ouraimisto provesome
3 monotonicity and convexity results for the modified Struve functions. Moreover, as consequences of the
2 above mentioned results, we present some functional inequalities as well as lower and upper bounds for
4 modified Struve functions. The paper is organized as follows: Section 2 contains the main results of the
5
paper and their proofs; while Section 3 contains the concluding remarks on the main results of Section
.
1 2. The key tools in the proofs of the main results are the techniques developed in the extensive study
0
of modified Bessel functions of the first and second kind and their ratios. The difficulty in the study
3
of the modified Struve function consists in the fact that the modified Struve differential equation is not
1
: homogeneous, however, as we can see below, the power series structure of modified Struve function is
v
very useful in order to study its monotonicity and convexity properties.
i
X
2. Modified Struve function: Monotonicity patterns and functional inequalities
r
a
We begin with an old result of Biernacki and Krzyz˙ [9], which will be used in the sequel.
Lemma 1. Consider the power series f(x) = a xn and g(x) = b xn, where a R and b > 0
n n n n
∈
n 0 n 0
≥ ≥
for all n 0,1,... , and suppose that both coPnverge on ( r,r), r >P0. If the sequence a /b is
n n n 0
∈ { } − { } ≥
increasing (decreasing), then the function x f(x)/g(x) is increasing (decreasing) too on (0,r).
7→
Fordifferentproofsandvariousapplicationsofthisresulttheinterestedreaderisreferredtotherecent
papers [1, 2, 3, 4, 5, 6, 7, 8, 10, 16] and to the references therein. We note that the above result remains
true if we get even or odd functions, that is, if we have in Lemma 1 the power series f(x) = a x2n
n
n 0
and g(x)= b x2n or f(x)= a x2n+1 and g(x)= b x2n+1. P≥
n n n
n 0 n 0 n 0
≥ ≥ ≥
P P P
File: BP_Struve.tex, printed: 2013-01-24, 2.19
2010 Mathematics Subject Classification. Primary33C10, Secondary39B62.
Keywords and phrases. ModifiedStruvefunction,modifiedBesselfunction,Tura´ntypeinequality.
1
2 A´RPA´DBARICZANDTIBORK.POGA´NY
Our main result is the following theorem.
Theorem 1. The following assertions are true:
a. If ν 1, then the function x I (x)/L (x) is increasing on (0, ).
≥−2 7→ ν+1 ν ∞
b. If ν 3, 1 , then the function x I (x)/L (x) is decreasing on (0, ).
∈ −2 −2 7→ ν+1 ν ∞
c. If µ ν > 3, then the function x 2ν µxµ νL (x)/L (x) is increasing on (0, ).
≥(cid:0) −2 (cid:3) 7→ − − ν µ ∞
d. If ν µ> 3, then the function x 2ν µxµ νL (x)/L (x) is decreasing on (0, ).
≥ −2 7→ − − ν µ ∞
e. The function ν (x) = 2νΓ ν+ 3 x νL (x) is decreasing and log-convex on ( 3, ) for
7→ Lν 2 − ν −2 ∞
all x>0.
(cid:0) (cid:1)
f. The function ν L (x)/L (x) is decreasing on ( 3, ) for all x>0.
7→ ν+1 ν −2 ∞
g. The function x xL (x)/L (x) is increasing on (0, ) for all ν > 3.
7→ ′ν ν ∞ −2
h. The function x 1 +2νΓ ν+ 3 x νL (x) is log-convex on (0, ) for all ν 1.
7→ √π 2 − ν+1 ∞ ≥−2
In particular, for all x>0 the following inequalities are valid:
(cid:0) (cid:1)
2Γ(ν+2) 1
(2.1) L (x)< I (x), ν > ,
ν √πΓ ν+ 3 ν+1 −2
2
(cid:0) (cid:1)
3 3 3
(2.2) 2νΓ ν+ x νL (x)>2µΓ µ+ x µL (x), µ>ν > ,
− ν − µ
2 2 −2
(cid:18) (cid:19) (cid:18) (cid:19)
[L (x)]2 1
(2.3) 0<[L (x)]2 L (x)L (x)< ν , ν > ,
ν − ν−1 ν+1 ν+ 3 −2
2
L (x) coshx 1 1
ν+1
(2.4) < − <1, ν > ,
L (x) sinhx −2
ν
(2.5) 1+ ν2 [L (x)]2 [L (x)]2+ ν+ 12 xν−1Lν(x) >0, ν > 3.
x2 ν − ′ν √π2ν 1Γ ν+ 3 −2
(cid:18) (cid:19) (cid:0) −(cid:1) 2
Moreover, if ν 3, 1 , then (2.1) and the left-hand side(cid:0)of (2.4(cid:1)) are reversed; if ν >µ> 3, then
∈ −2 −2 −2
(2.2) is reversed; equality holds in (2.1) and in the left-hand side of (2.4) if ν = 1; and equality holds
(cid:0) (cid:1) −2
in (2.2) if µ=ν.
Proof. a. &b. LetusrecallthepowerseriesrepresentationsofthemodifiedBesselandStruvefunctions
I and L , which are as follows
ν+1 ν
x 2n+ν+1 x 2n+ν+1
I (x)= 2 and L (x)= 2 .
ν+1 n!Γ(n+ν+2) ν Γ n+ 3 Γ n+ν+ 3
n 0 (cid:0) (cid:1) n 0 (cid:0)2 (cid:1) 2
X≥ X≥
In view of the above representations the quotient I (x)/L (x) can(cid:0)be rew(cid:1)rit(cid:0)ten as (cid:1)
ν+1 ν
α x 2n
Iν+1(x) n 0 ν,n 2
Lν(x) = P≥ βν,n(cid:0)x2(cid:1)2n,
n 0
P≥ (cid:0) (cid:1)
where
1 1
α = and β = .
ν,n n!Γ(n+ν+2) ν,n Γ n+ 3 Γ n+ν+ 3
2 2
Now, if we let
(cid:0) (cid:1) (cid:0) (cid:1)
α Γ n+ 3 Γ ν+n+ 3
q = ν,n = 2 2 ,
n
β Γ(n+1)Γ(ν+n+2)
ν,n (cid:0) (cid:1) (cid:0) (cid:1)
then
q n+ 3 ν+n+ 3
n+1 = 2 2 ,
q (n+1)(ν+n+2)
n (cid:0) (cid:1)(cid:0) (cid:1)
and for each n 0,1,... this is greater (less) than or equal to 1 if ν 1 (ν 1). The assumption
∈{ } ≥−2 ≤−2
that ν > 3 is necessaryin orderto haveallcoefficients ofthe powerseriesofL positive. Now,observe
−2 ν
that the radius of convergence of power series I (x) and L (x) is infinity and applying Lemma 1 the
ν+1 ν
proof of these parts is done.
FUNCTIONAL INEQUALITIES FOR MODIFIED STRUVE FUNCTIONS 3
c. & d. We proceed similarly as above, we shall apply Lemma 1. Observe that
β x 2n
2νx−νLν(x) n 0 ν,n 2
2µx−µLµ(x) = P≥ βµ,n(cid:0)x2(cid:1)2n,
n 0
P≥ (cid:0) (cid:1)
and if we consider the quotient ω = β /β , then ω /ω = µ+n+ 3 / ν+n+ 3 . Conse-
n ν,n µ,n n+1 n 2 2
quently, the sequence ω is increasing (decreasing) if µ ν (µ ν).
n n 0
e. The first part of{the}st≥atement is actually equivalent to≥(2.2),≤s(cid:0)o it is enou(cid:1)gh(cid:0)to prove ((cid:1)2.2). But
this inequality is equivalent to
2νx νL (x) 2νx νL (x) Γ µ+ 3
− ν > lim − ν = 2 ,
2µx−µLµ(x) x→0(cid:20)2µx−µLµ(x)(cid:21) Γ(cid:0)ν+ 32(cid:1)
whichreadilyfollowsfrompartc. Notethat,themonotonicityofν (cid:0) ν(x)(cid:1)canbeprovedalsobyusinga
7→L
differentargument. Forthisconsidertheascendingfactorial(a) =a(a+1)...(a+n 1)=Γ(a+n)/Γ(a)
n −
and consider the sequence γ , defined by
{ ν,n}n 0
≥
3 Γ ν+ 3 1
γ =Γ ν+ β = 2 = .
ν,n 2 ν,n Γ n+ 3 Γ n+ν+ 3 Γ n+ 3 ν+ 3
(cid:18) (cid:19) 2(cid:0) (cid:1) 2 2 2 n
Clearly if µ ν > 3, then µ+ 3 ν(cid:0)+ 3 (cid:1) an(cid:0)d consequ(cid:1)ently γ(cid:0) (cid:1)γ(cid:0) for(cid:1)all n 0,1,... .
≥ −2 2 n ≥ 2 n µ,n ≤ ν,n ∈ { }
This in turn implies that
(cid:0) (cid:1) (cid:0) (cid:1)
x 2n+1 x 2n+1
(x)= γ γ = (x)
µ µ,n ν,n ν
L 2 ≤ 2 L
nX≥0 (cid:16) (cid:17) nX≥0 (cid:16) (cid:17)
for all x>0, i.e., indeed the function ν (x) is decreasing. Now, for log-convexity of ν (x), we
ν ν
7→L 7→L
observethatit is enoughto show the log-convexityofeachindividual termandto use the factthat sums
of log-convex functions are log-convex too. Thus, we just need to show that for each n 0,1,... we
∈ { }
have
3 3
∂2[logγ ]/∂ν2 =ψ ν+ ψ ν+n+ 0,
ν,n ′ ′
2 − 2 ≥
(cid:18) (cid:19) (cid:18) (cid:19)
where ψ(x) = Γ(x)/Γ(x) is the so-called digamma function. But, the digamma function ψ is concave
′
and consequently the function ν γ is log-convex on 3, for all n 0,1,... , as we required.
7→ ν,n −2 ∞ ∈ { }
Now, observe that, by definition for all ν ,ν > 3, α [0,1] and x>0 we have
1 2 −2 ∈ (cid:0) (cid:1)
Lαν1+(1−α)ν2(x)≤[Lν1(x)]α[Lν2(x)]1−α.
Now, choosing α = 1, ν = ν 1 and ν = ν +1 in the above inequality, we obtain the Tur´an type
2 1 − 2
inequality
2(x) (x) (x) 0,
Lν −Lν−1 Lν+1 ≤
which is equivalent to the right-hand side of (2.3).
Notealsothatthereverseof (2.2)clearlyfollowsfrompartdofthistheorem. Moreover,theinequality
(2.1)andits reversefollowfrompartsaandb, wejustneedto computethe limitofI (x)/L (x)when
ν+1 ν
x tends to zero, which is [√πΓ ν+ 3 ]/[2Γ(ν+2)].
2
f. By using part c we clearly have
(cid:0) (cid:1)
2νx−νLν(x) ′
0
2µx µL (x) ≥
(cid:20) − µ (cid:21)
when µ ν and x>0. This is equivalent to inequality
≥
x−νLν(x) ′ x−µLµ(x) x−νLν(x) x−µLµ(x) ′ 0,
− ≥
which in view of [15, p.(cid:2)292] (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)(cid:2) (cid:3)
2 ν
x−νLν(x) ′ = √πΓ −ν+ 3 +x−νLν+1(x),
2
(cid:2) (cid:3)
can be rewritten as (cid:0) (cid:1)
2 µx νL (x) 2 νx µL (x)
x ν µ[L (x)L (x) L (x)L (x)] − − ν − − µ .
− − ν+1 µ − ν µ+1 ≥ √πΓ µ+ 3 − √πΓ ν+ 3
2 2
(cid:0) (cid:1) (cid:0) (cid:1)
4 A´RPA´DBARICZANDTIBORK.POGA´NY
But, in view of (2.2), the expression on the right-hand side of the above inequality is positive, and with
this the proof of this part is done. Now, if we choose µ=ν+1 in the inequality
L (x)L (x) L (x)L (x) 0,
ν+1 µ ν µ+1
− ≥
and then we change ν to ν 1, we obtain the left-hand side of the Tur´an type inequality (2.3).
−
Finally, let us focus on the inequality (2.4). By using the particular cases [15, p. 291]
2 2
L (x)= sinhx and L (x)= (coshx 1),
1 1
−2 rπx 2 rπx −
and the result of part f of this theorem, for all x>0 and ν > 1 we have that
−2
Lν+1(x) < L12(x) = coshx−1.
L (x) L (x) sinhx
ν 1
−2
Moreover, the above inequality is reversed if ν 3, 1 and x > 0; the equality is attained when
∈ −2 −2
ν = 1. For the right-hand side inequality in (2.4) observe that the function x L (x)/L (x) is
−2 (cid:0) (cid:1) 7→ 21 −12
increasing on (0, ) since
∞
L1(x) ′ coshx 1
2 = − >0
"L 1(x)# sinh2x
−2
for allx>0. Onthe otherhand, L (x)/L (x) tends to 1, asx tends to infinity, andthis completes the
1 1
2 −2
proof of (2.4).
g. Observe that the quotient xL (x)/L (x) can be rewritten as
′ν ν
δ x 2n
xLLν′ν((xx)) = nP≥0βνν,,nn(cid:0)2x2(cid:1)2n,
n 0
P≥ (cid:0) (cid:1)
where δ = (2n+ν +1)β . Since the sequence δ /β is increasing, by applying Lemma 1,
ν,n ν,n ν,n ν,n n 0
the function x xL (x)/L (x) is clearly increasing{on (0, )}fo≥r all ν > 3.
7→ ′ν ν ∞ −2
Now,recallthatthemodifiedStruvefunctionL isaparticularsolutionofthemodifiedStruveequation
ν
[15, p. 288]
xν+1
x2y (x)+xy (x) (x2+ν2)y(x)= ,
′′ ′ − √π2ν 1Γ ν+ 1
− 2
and consequently
(cid:0) (cid:1)
ν2 1 xν 1
(2.6) L (x)= 1+ L (x) L (x)+ − .
′ν′ x2 ν − x ′ν √π2ν 1Γ ν+ 1
(cid:18) (cid:19) − 2
Thus, by using part g of this theorem we obtain that
(cid:0) (cid:1)
(2.7) x1 [Lν(x)]2 xLL′ν((xx)) ′ = 1+ νx22 [Lν(x)]2−[L′ν(x)]2+ √νπ+2ν21 1xΓν−ν1L+ν(3x),
(cid:20) ν (cid:21) (cid:18) (cid:19) (cid:0) −(cid:1) 2
and this is positive for all x>0 and ν > 3, as we required. (cid:0) (cid:1)
−2
h. First consider the integral representation [15, p. 292]
2 x ν π2
(2.8) L (x)= 2 sinh(xcost)(sint)2νdt,
ν √πΓ ν+ 1
(cid:0) (cid:1) 2 Z0
which holds for all ν > 1. Applying the(cid:0)change(cid:1)of variable cost=s in the above integral, we get
−2
Lν(x)= √π2Γ x2ν+ν 1 1 1−s2 ν−12 sinh(xs)ds.
(cid:0) (cid:1) 2 Z0
(cid:0) (cid:1)
Integrating by parts we obtain
(cid:0) (cid:1)
Lν(x)=−√πΓx2 νν−+11 + 2 √νπ−Γ21ν+x21ν−1 1 1−s2 ν−23 scosh(xs)ds,
(cid:0) (cid:1) 2 (cid:0) (cid:1)(cid:0) (cid:1)2 Z0
(cid:0) (cid:1)
and replacing ν by ν+1 one obtains
(cid:0) (cid:1) (cid:0) (cid:1)
Lν+1(x)=−√πΓx2νν+ 3 + 2√νπΓ+ 21ν+x23 ν 1 1−s2 ν−12 scosh(xs)ds,
(cid:0) (cid:1) 2 (cid:0) (cid:1)(cid:0) 2(cid:1) Z0
(cid:0) (cid:1)
(cid:0) (cid:1) (cid:0) (cid:1)
FUNCTIONAL INEQUALITIES FOR MODIFIED STRUVE FUNCTIONS 5
where ν > 3. Thus, we get the following integral representation
−2
2νΓ ν+ 3 x−νLν+1(x)= 1 + 2 ν+ 21 1 1 s2 ν−12 scosh(xs)ds.
2 −√π √π −
(cid:18) (cid:19) (cid:0) (cid:1)Z0
(cid:0) (cid:1)
Now, we shall use the classical Ho¨lder-Rogers inequality for integrals [14, p. 54], that is,
1/p 1/q
b b b
(2.9) f(t)g(t)dt f(t)pdt g(t)qdt ,
Za | | ≤"Za | | # "Za | | #
where p > 1, 1/p+1/q = 1, f and g are real functions defined on [a,b] and f p, g q are integrable
| | | |
functions on [a,b]. Using (2.9) and the fact that the hyperbolic cosine is log-convex,we obtain that
1
1 s2 ν−21 scosh((αx+(1 α)y)s)ds
− −
Z0
(cid:0) 1 (cid:1)
1 s2 ν−21 s[cosh(xs)]α[cosh(ys)]1−αds
≤ −
Z0
= 1(cid:0) 1 s(cid:1)2 ν−21 scosh(xs) α 1 s2 ν−12 scosh(ys) 1−αds
− −
Z0 1h(cid:0)1 s2(cid:1)ν−21 scosh(xs)idshα(cid:0) 1 (cid:1)1 s2 ν−21 scoish(ys)ds 1−α,
≤ − −
(cid:20)Z0 (cid:21) (cid:20)Z0 (cid:21)
(cid:0) (cid:1) (cid:0) (cid:1)
where α [0,1], x,y > 0 and ν > 3. In other words, the function x 1 1 s2 ν−12 scosh(xs)ds
∈ −2 7→ 0 −
is log-convex on (0, ) for all ν > 3, and consequently x 1 + 2νΓ ν+ 3 x νL (x) is also
∞ −2 7→ √π R (cid:0) 2 −(cid:1) ν+1
log-convex on (0, ) for all ν 1. (cid:3)
∞ ≥−2 (cid:0) (cid:1)
3. Concluding remarks and further results
In what follows we present some consequences of Theorem 1 and we present new proofs for some of
the results contained therein. For example, by using different approach than in the proof of Theorem 1,
we show that the Tur´an type inequality (2.3) is valid for all ν > 3 and x>0.
−2
1. It is worth to mention here that it is possible to improve the left-hand side of the inequality (2.4)
by using part f of the above theorem. However, the obtained bound will be more complicated. Namely,
for all x>0 and ν > 1 we have that
2
Lν+1(x) < L32(x) < L12(x) = coshx−1.
L (x) L (x) L (x) sinhx
ν 1 1
2 −2
Moreover,theleft-handsideoftheaboveinequalityisreversedifν 3,1 andx>0;andtheequality
∈ −2 2
is attained when ν = 1. By using the particular case [15, p. 291]
2 (cid:0) (cid:1)
x 2 2 coshx
L (x)= 1 + sinhx ,
32 −r2π (cid:18) − x2(cid:19) rπx(cid:18) − x (cid:19)
the above inequality becomes
(3.1) Lν+1(x) < L32(x) = 1−coshx+xsinhx− x22.
L (x) L (x) xcoshx x
ν 1
2 −
2. Now,we wouldlike to presentsome consequencesofthe inequality (2.4). For this, firstwe combine
the recurrence relations [15, p. 292]
2ν x ν
(3.2) L (x) L (x)= L (x)+ 2 ,
ν−1 − ν+1 x ν √πΓ ν+ 3
(cid:0) (cid:1) 2
x(cid:0) ν (cid:1)
Lν−1(x)+Lν+1(x)=2L′ν(x)− √πΓ(cid:0)2ν(cid:1)+ 23
and we obtain
(cid:0) (cid:1)
(3.3) xL (x)+νL (x)=xL (x).
′ν ν ν 1
−
6 A´RPA´DBARICZANDTIBORK.POGA´NY
Now,inviewof (2.4)itisclearthatL (x)<L (x)forallν 1,andbyusingtherecurrencerelation
ν+1 ν ≥−2
(3.3) we get
L (t) ν
(3.4) ′ν >1
L (t) − t
ν
forallν 1 andt>0.Integratingbothsidesof (3.4)on[x,y],where0<x<y,weobtaintheinequality
≥ 2
L (x) y ν
(3.5) ν <ex y ,
−
L (y) x
ν
(cid:16) (cid:17)
which holds for all ν 1 and 0 < x < y. Note that this inequality was proved also by Joshi and
≥ 2
Nalwaya [11, p. 51] for ν > 1, but just for 0 < x < y < 2ν +1. For similar inequalities on modified
−2
Bessel functions of the first kind we refer to the paper of Laforgia [13], see also [8] for a survey on this
kind of inequalities for modified Bessel functions. Moreover, we mention that it is possible to improve
the inequalities (3.4) and (3.5). Namely, by using the left-hand side of (2.4) we obtain the following
improvement of (3.4)
L (t) sinht ν
(3.6) ′ν
L (t) ≥ cosht 1 − t
ν
−
forallν 1 andt>0.Integratingbothsidesof (3.4)on[x,y],where0<x<y,weobtaintheinequality
≥ 2
L (x) coshx 1 y ν
ν
(3.7) − ,
L (y) ≤ coshy 1 x
ν (cid:18) − (cid:19)(cid:16) (cid:17)
whichholdsforallν 1 and0<x<y.Observethatclearly(3.7)improves(3.5). Finally,letusmention
≥ 2
alsothat(3.5)and(3.7)actuallymeanthatthefunctionsx xνe xL (x)andx xν(coshx 1) 1L (x)
− ν − ν
7→ 7→ −
are increasing on (0, ) for all ν 1. Furthermore,according to the fact that the left-hand side of (2.4)
∞ ≥ 2
isreversedwhenν 3, 1 ,theinequalities(3.6)and(3.7)arereversedwhen ν < 1.Inotherwords,
∈ −2 −2 | | 2
the function x xν(coshx 1) 1L (x) is decreasing on (0, ) for all ν < 1.
7→ (cid:0) −(cid:1) − ν ∞ | | 2
3. We also mention that the results of remark 2 can be improved too by using the inequality (3.1):
the inequality
L (t) tcosht t ν
(3.8) ′ν −
Lν(t) ≥ 1 cosht+tsinht t2 − t
− − 2
is valid for all ν 3 and t>0. Integrating both sides of (3.8) on [x,y], where 0<x<y, we obtain the
≥ 2
inequality
L (x) 1 coshx+xsinhx x2 y ν
(3.9) ν − − 2 ,
Lν(y) ≤ 1 coshy+ysinhy y2 ! x
− − 2 (cid:16) (cid:17)
which holds for all ν 3 and 0 < x < y. Observe that clearly (3.8) improves (3.6), and (3.9) improves
≥ 2
(3.7), when ν 3. Finally, since (3.1) is reversed when ν 3,1 , the inequalities (3.8) and (3.9) are
≥ 2 ∈ −2 2
reversed when ν 1,3 . In other words, the function x xν(1 coshx+xsinhx x2) 1L (x) is
∈ −2 2 (cid:0)7→ (cid:1)− − 2 − ν
increasing on (0, ) when ν 3 and is decreasing on (0, ) for all ν 1,3 .
∞(cid:0) (cid:1) ≥ 2 ∞ ∈ −2 2
4. Observe that
xL (x) 8 (cid:0) (cid:1)
′ν =ν+1+ x2+...
L (x) 3(2ν+3)
ν
and by using part g of the above theorem, we get that xL (x)/L (x)>ν+1 for all x>0 and ν > 3.
′ν ν −2
This inequality in view of (3.3) is equivalent to L (x)/L (x)>(2ν+1)/x, where x>0 and ν > 3.
Note that the later inequality was obtained also byν−J1oshianνd Nalwaya[11, p. 52] for x>0 and ν >−21.
−2
In what follows we show that the above inequality can be used to prove the left-hand side of the Tur´an
type inequality (2.3) for ν > 3 and x>0. For this, we combine first the recurrence relations (3.2) and
−2
(3.3), and we obtain
ν x ν
(3.10) L (x)=L (x) L (x) 2 .
ν+1 ′ν − x ν − √πΓ ν+ 3
(cid:0) (cid:1) 2
Now, by using (3.3) and (3.10) we obtain (cid:0) (cid:1)
ν2 xνL (x)
∆ν(x)=[Lν(x)]2−Lν−1(x)Lν+1(x)= 1+ x2 [Lν(x)]2−[L′ν(x)]2+ √π2νΓν−ν1+ 3
(cid:18) (cid:19) 2
(cid:0) (cid:1)
FUNCTIONAL INEQUALITIES FOR MODIFIED STRUVE FUNCTIONS 7
and consequently
∆ν(x)− x1 [Lν(x)]2 xLL′ν((xx)) ′ = 2ν√xπνΓLνν(x+) 3 LLν−(1x(x)) − 2νx+1 >0
(cid:20) ν (cid:21) 2 (cid:20) ν (cid:21)
forallν > 3 andx>0.Thus,byusingpartgofTheore(cid:0)m1,w(cid:1)econcludethattheTur´antypeinequality
−2
(2.3) is valid for all ν > 3 and x>0. We note that Joshi and Nalwaya [15, p. 55] proved also (2.3) for
−2
ν > 3 and x>0, by using a closed form expression of the Cauchy product L (x)L (x).
−2 ν µ
Finally, we mention that integrating on [x,y] both sides of the inequality
L (t) ν+1
′ν > ,
L (t) t
ν
we obtain the inequality
L (x) x ν+1
ν
< ,
L (y) y
ν (cid:18) (cid:19)
where ν > 3 and 0 < x < y. This inequality was proved also by Joshi and Nalwaya [15, p. 52] for
−2
ν > 1 and 0<x<y.
−2
5. We also mention that it is possible to deduce different upper bounds for the modified Struve
function L than in (2.1). For example, by using part e of our main theorem, we can deduce that ν
ν
7→
2νΓ(ν+1)x νL (x) is decreasingon ( 1, ) as the product of the functions ν 2νΓ ν+ 3 x νL (x)
− ν − ∞ 7→ 2 − ν
and ν q(ν)=Γ(ν+1)/Γ ν+ 3 . Here we used that for each ν > 1 we have
7→ 2 − (cid:0) (cid:1)
(cid:0) (cid:1)q′(ν) 3
=ψ(ν+1) ψ ν+ <0
q(ν) − 2
(cid:18) (cid:19)
since the gamma function is log-convex,that is, the digamma function is increasing. Thus, by using the
monotonicity of ν 2νΓ(ν+1)x νL (x) we obtain
− ν
7→
xνL (x)
(3.11) L (x) 0
ν ≤ 2νΓ(ν+1)
for all ν 0 and x > 0. Moreover, the above inequality is reversed when ν ( 1,0) and x > 0.
≥ ∈ −
Surprisingly,the inequality (3.11)can be deduced alsoby using the Chebyshev integralinequality [14, p.
40], which is as follows: if f,g : [a,b] R are integrable functions, both increasing or both decreasing
and p:[a,b] R is a positive integrab→le function, then
→
b b b b
(3.12) p(t)f(t)dt p(t)g(t)dt p(t)dt p(t)f(t)g(t)dt.
≤
Za Za Za Za
Note that if one of the functions f or g is decreasing and the other is increasing, then (3.12) is reversed.
Now, we shall use (3.12) and (2.8) to prove (3.11). For this consider the functions p,f,g : 0,π R,
2 →
defined by p(t) = 1, f(t) = sinh(xcost) and g(t) = (sint)2ν. Observe that f is decreasing and g is
(cid:2) (cid:3)
increasing (decreasing) if ν 0 (ν 0). On the other hand, we have
≥ ≤
π2 π π2 √πΓ ν+ 1
sinh(xcost)dt= L (x) and (sint)2νdt= 2 ,
0
2 2Γ(ν+1)
Z0 Z0 (cid:0) (cid:1)
and applying the Chebyshev inequality (3.12) we get the inequality (3.11) for all ν 0, and its reverse
≥
when ν 1,0 .
∈ −2
6. We also note that if we combine the inequality (2.1) with [8, p. 583]
(cid:0) (cid:1)
xν x2
Iν(x)≤ 2νΓ(ν+1)e4(ν+1), ν >−1, x>0,
then we obtain the upper bound
x2
xν+1e4(ν+2)
L (x)< ,
ν √π2νΓ ν+ 3
2
where ν > −21 and x > 0. A similar upper bound for th(cid:0)e mod(cid:1)ified Struve function can be obtained by
combining the inequality (3.11) with (2.1). In this way we obtain that
xν+1ex82
L (x)< ,
ν π2ν 1Γ(ν+1)
−
8 A´RPA´DBARICZANDTIBORK.POGA´NY
where ν 0 and x>0. Another inequality of this kind can be obtained from parts c and d of the main
≥
theorem. Namely, if take µ = 1 in parts c and d of Theorem 1, then clearly x (x)/ (x) =
−2 7→ Lν L−12
√π (x)/sinhx is increasing on (0, ) if ν 3, 1 , and decreasing on (0, ) if ν > 1. Conse-
Lν ∞ ∈ −2 −2 ∞ −2
quently, for all ν > 1 and x>0 we have
−2 (cid:0) (cid:1)
xνsinhx
L (x)< ,
ν √π2νΓ ν+ 3
2
and this inequality is reversed if ν 3, 1 ; equalit(cid:0)y is ob(cid:1)tained when ν = 1. Note also that the
∈ −2 −2 −2
above inequality is a particular case of the fact that the function ν (x) is decreasing on 3,
(cid:0) (cid:1) 7→ Lν −2 ∞
for all x>0, conform part e of Theorem 1.
(cid:0) (cid:1)
Moreover,if we consider the quotient
(x)2n+1
2
ν(x) n 0Γ(n+32)(ν+32)n
Qν(x)= sinLh x = P≥ (x)2n+1 ,
2ν+3 2
(2n+1)!(ν+3)2n+1
n 0 2
≥
P
thenwecangetalowerboundforthemodifiedStruvefunction. Namely,themonotonicityofthesequence
λ , defined by
ν,n n 0
{ } ≥ (2n+1)! ν+ 3 2n+1
λ = 2 ,
ν,n Γ n+ 3 ν+ 3
2(cid:0) (cid:1)2 n
will generate an inequality for the modified Struve function. For this observe that the inequality
(cid:0) (cid:1)(cid:0) (cid:1)
λ (2n+2)(2n+3) ν+ 3 2
ν,n+1 = 2 >1
λ n+ 3 ν+n+ 3
ν,n 2 (cid:0) 2 (cid:1)
is equivalent to
(cid:0) (cid:1)(cid:0) (cid:1)
78 3 15
4(ν+1)(ν+2)n2+ 10ν2+29ν+ n+ ν+ 6ν+ >0,
4 2 2
(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
which is valid for all ν > 1 and n 0,1,... . Thus, the sequence λ is increasing and in view
ν,n n 0
− ∈{ } { } ≥
of Lemma 1 the function Q is increasing too on (0, ) for all ν > 1. Consequently for all ν > 1 and
ν
∞ − −
x>0 we have Q (x)> limQ (x), that is,
ν ν
x 0
→
xνsinh x
L (x)> 2ν+3 .
ν √π2ν 1Γ ν+ 3
− 2
7. Now, let us focus on the Tur´antype inequality (2.3)(cid:0). In w(cid:1)hat follows, we show that this result can
be deduced also by using some recent results from [12]. In [12, Theorem 1] Kalmykov and Karp stated
that if the non-trivial and non-negative sequence a is log-concave without internal zeros, then
n n 0
{ } ≥
a xn
n
ν f(ν,x)=
7→ n!Γ(ν+n)
n 0
X≥
is strictly log-concave on (0, ) for all x > 0. Note that, following the proof of [12, Theorem 1], it can
∞
be proved that the function
a xn
n
ν
7→ Γ n+ 3 Γ ν+n+ 3
n 0 2 2
X≥
is strictly log-concave on 3, for all x > (cid:0)0, prov(cid:1)ide(cid:0)d that the(cid:1)sequence a is as above. Now,
since ν x ν+1 is log-l−in2ea∞r, from the above result (by choosing a ={1n}fonr≥0all n 0,1,... )
7→ 2 (cid:0) (cid:1) n ∈ { }
we immediately have that the function ν L (x) is strictly log-concave on 3, for all x > 0.
(cid:0) (cid:1) 7→ ν −2 ∞
Consequently, by definition for all ν ,ν > 3, ν =ν , α (0,1) and x>0 we have
1 2 −2 1 6 2 ∈ (cid:0) (cid:1)
Lαν1+(1−α)ν2(x)>[Lν1(x)]α[Lν2(x)]1−α.
Now, choosing α = 1, ν = ν 1 and ν = ν +1 in the above inequality, we obtain the left-hand side
2 1 − 2
of the Tur´an type inequality (2.3) for all ν > 1 and x > 0. It is worth to mention also here that the
−2
strict log-concavity of ν L (x) is stronger than part f of Theorem 1. Namely, since ν log[L (x)]
ν ν
7→ 7→
is strictly concave on 3, it follows that the function ν log[L (x)] log[L (x)] is strictly
−2 ∞ 7→ ν+a − ν
decreasingon 3, for all a>0 andx>0. Thus, choosinga=1, we obtainthatindeedthe function
−2 ∞ (cid:0) (cid:1)
(cid:0) (cid:1)
FUNCTIONAL INEQUALITIES FOR MODIFIED STRUVE FUNCTIONS 9
ν L (x)/L (x) is strictly decreasing on 3, for all x >0. This result can be obtained also by
7→ ν+1 ν −2 ∞
using[12,Theorem2],whichsaysthatifthenon-trivialandnon-negativesequence b islog-concave
n n 0
(cid:0) (cid:1) { } ≥
without internal zeros, then the function
b xn
n
ν g(ν,x)=
7→ Γ(ν+n)
n 0
X≥
is strictly discrete Wright log-concave on (0, ) for all x>0, that is, we have
∞
g(ν+1,x)g(ν+a,x) g(ν,x)g(ν+a+1,x) 0
− ≥
for all a,x,ν > 0. Now, choosing the log-concave sequence b , defined by b = 1/Γ n+ 3 , we
{ n}n≥0 n 2
get g ν+ 32,x42 = 2ν+1x−ν−1Lν(x), and hence the function ν 7→ 2ν+1x−ν−1Lν(x) is stric(cid:0)tly dis(cid:1)crete
Wrigh(cid:16)t log-conca(cid:17)ve on 3, for all x>0. Consequently, the function
−2 ∞
(cid:0) (cid:1) 2ν+2x−ν−2Lν+1(x) 2Lν+1(x)
ν =
7→ 2ν+1x ν 1L (x) xL (x)
− − ν ν
isstrictlydecreasingon 3, forallx>0.Finally,wementionthatifthenon-trivialandnon-negative
−2 ∞
sequence b is log-concavewithout internal zeros,then the function ν g(ν,x) satisfies the Tur´an
n n 0
{ } ≥ (cid:0) (cid:1) 7→
type inequality [12, Remark 3]
b2 1
0 [g(ν,x)]2 g(ν 1,x)g(ν+1,x) [g(ν,x)]2, ν,x>0,
ν[Γ(ν)]2 ≤ − − ≤ ν
which in our case reduces to
π x 2ν+2 [L (x)]2
4 2 <[L (x)]2 L (x)L (x)< ν ,
ν+ 23 (cid:0) Γ(cid:1) ν+ 32 2 ν − ν−1 ν+1 ν+ 23
where ν > 3 and (cid:0)x > 0.(cid:1)O(cid:2) b(cid:0)serve t(cid:1)h(cid:3)at the left-hand side of the above inequality is better than the
−2
left-hand side of (2.3).
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Departmentof Economics, Babe¸s-BolyaiUniversity, 400591Cluj-Napoca, Romania
E-mail address: [email protected]
Facultyof Maritime Studies, University of Rijeka,Rijeka 51000,Croatia
E-mail address: [email protected]