R E S E A R C H
Open Access
A Hilbert-type integral inequality in the
whole plane related to the kernel of exponent
function
Yanru Zhong
1*, Meifa Huang
2and Bicheng Yang
3*Correspondence:
1School of Computer Science and
Information Security, Guilin University of Electronic Technology, Guilin, P.R. China
Full list of author information is available at the end of the article
Abstract
By using real analysis and weight functions, we obtain a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the kernel of exponent function with intermediate variables. The constant factor related to the gamma function is proved to be the best possible. We also consider some particular cases and the operator expressions.
MSC: 26D15; 31A10
Keywords: Hilbert-type integral inequality; Weight function; Intermediate variable; Equivalent statement; Operator; Gamma function
1 Introduction
If 0 <0∞f2(x)dx<∞and 0 <∞
0 g2(y)dy<∞, then we have the following well-known
Hilbert integral inequality (see [1]):
∞
0
∞
0
f(x)g(y)
x+y dx dy<π
∞
0
f2(x)dx
∞
0
g2(y)dy
1 2
, (1)
where the constant factorπis the best possible. In 1925, by introducing the pair of con-jugate exponents (p,q) (p> 1, p1 +1q = 1), Hardy et al. gave an extension of (1) (see [1], Theorem 316). Recently, by means of weight functions, some new extensions of (1) and the Hardy’s work were given by Yang [2,3] and in [4–9]. Most of them are built in the quarter plane of the first quadrant.
In 2007, Yang [10] provided a Hilbert-type integral inequality in the whole plane with the exponent function and intermediate variables as follows:
∞
–∞
∞
–∞
f(x)g(y)
(1 +ex+y)λdx dy<B
λ
2,
λ
2
∞
–∞e
–λxf2(x)dx
∞
–∞e
–λyg2(y)dy
1 2
, (2)
where the constant factorB(λ2,λ2) is the best possible (λ> 0,B(u,v) is the beta function). He et al. [11–19] proved some new Hilbert-type integral inequalities in the whole plane with the best possible constant factors.
In 2017, Hong [20] gave two equivalent statements between Hilbert-type inequalities with general homogenous kernel and a few parameters. A few authors continue to study this topic (see [21–25]).
In this paper, by using real analysis and weight functions we obtain a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the exponent function with intermediate variables. The constant factor related to the gamma function is proved to be the best possible. We also consider some particular cases and operator expressions.
2 Some lemmas
Forγ,ρ,σ> 0, settingh(u) :=e–ρuγ
(u> 0), we find
kρ(γ)(σ) :=
∞
0
h(u)uσ–1du=
∞
0
e–ρuγuσ–1du v=ρuγ
= 1
γρσ/γ
∞
0
e–vv(σ/γ)–1dv=(σ/γ)
γρσ/γ ∈R+= (0,∞), (3)
where(s) :=0∞e–vvs–1dv(Res> 0) is the gamma function (see [26]).
Forδ∈ {–1, 1},α,β∈(–1, 1), we set
xα:=|x|+αx, yβ:=|y|+βy
x,y∈R= (–∞,∞),
Eδ:=
t∈R;|t|δ≥1 , E
–δ=
t∈R;|t|δ≤1 . Lemma 1 For c> 0,θ=α,β∈(–1, 1),we have
Eδ
tθ–cδ–1dt=
1
c
1 (1 +θ)cδ+1 +
1 (1 –θ)cδ+1
, (4)
E–δ
tcθδ–1dt=
1
c
1 (1 +θ)–cδ+1 +
1 (1 –θ)–cδ+1
; (5)
and for c≤0,we have
Eδ
t–cδ–1
θ dt=
E–δ
tcδ–1
θ dt=∞.
Proof SettingE+
δ :={t∈R+;t
δ≥1},E–
δ :={–t∈R+; (–t)
δ≥1}, we findE
δ=E+δ ∪E–δ and
Eδ
tθ–cδ–1dt=
E+δ
(1 +θ)t–cδ–1dt+
E– δ
(1 –θ)(–t)–cδ–1dt
=
1 (1 +θ)cδ+1 +
1 (1 –θ)cδ+1
E+δ
t–cδ–1dt.
Settingt=u1δ, we find
E+ δ
t–cδ–1dt= 1 |δ|
∞
1
u1δ(–cδ–1)u1δ–1du=
∞
1
Hence, forc> 0, (4) follows, and forc≤0,Eδt–cδ–1
θ dt=∞. Since, forc> 0,
E–δ
tcθδ–1dt=
E(–δ)
t–c(–θ δ)–1dt=
1 (1 +θ)–cδ+1+
1 (1 –θ)–cδ+1
1
0
uc–1du,
we have (5), and forc≤0,E–δtcδ–1
θ dt=∞.
The lemma is proved.
In the following, We further assume that p> 1, 1p +1q = 1,δ∈ {–1, 1}, α,β∈(–1, 1),
γ,ρ,σ> 0,σ1∈R,kρ(γ)(σ) is given by (3), and
Kα(γ,β)(σ) := 2k
(γ)
ρ (σ)
(1 –α2)1/q(1 –β2)1/p. (6)
Forn∈N={1, 2, . . .},E–1= [–1, 1],x∈Eδ, we define:
I(–)(x) :=
0
–1
e–ρ(xδ αyβ)γyσ+
1
qn–1
β dy, I
(+)(x) :=
1
0
e–ρ(xδ αyβ)γyσ+
1
qn–1
β dy,
I(x) :=I(–)(x) +I(+)(x) =
E–1
e–ρ(xδαyβ)γyσ+ 1
qn–1
β dy.
Foryβ= (sgn(y) +β)y, where
sgn(y) := ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–1, y< 0,
0, y= 0,
1, y> 0,
xδ α=
1 +αsgn(x)δ|x|δ≥ min δ∈{–1,1}
1± |α|δ (x∈Eδ),
and 1 –|α| ≤(1 +|α|)–1≤1 +|α| ≤(1 –|α|)–1, we have
(1±β)xδα≥mα,β:=
1 –|β|1 –|α|> 0 (x∈Eδ). (7)
For fixedx∈Eδ, settingu=xδαyβ, we find
I(–)(x) =x
–δ(σ+qn1)
α
1 –β
(1–β)xδ α 0
e–ρuγuσ+qn1–1du≥x
–δ(σ+qn1)
α
1 –β
mα,β
0
e–ρuγuσ+qn1–1du,
I(+)(x) = x
–δ(σ+qn1) α
1 +β
(1+β)xδα
0
e–ρuγuσ+qn1–1du≥x
–δ(σ+qn1) α
1 +β
mα,β
0
e–ρuγuσ+qn1–1du,
I(x) =x–δ(σ+
1
qn) α
1 1 –β
(1–β)xδ α 0
e–ρuγuσ+qn1–1du+ 1 1 +β
(1+β)xδ α 0
e–ρuγuσ+qn1–1du
≥2x –δ(σ+qn1)
α
1 +β
mα,β
0
e–ρuγuσ+qn1–1
Forn∈N={1, 2, . . .},x∈E–δ, we define:
J(–)(x) :=
–1
–∞e
–ρ(xδαyβ)γyσ– 1
qn–1
β dy,
J(+)(x) :=
∞
1
e–ρ(xδαyβ)γyσ– 1
qn–1
β dy,
J(x) :=J(–)(x) +J(+)(x) =
E–1
e–ρ(xδ αyβ)γyσ–
1
qn–1
β dy.
Since, forx∈E–δ,
xδα=
1 +αsgn(x)δ|x|δ≤ max
δ∈{–1,1}
1± |α|δ =1 –|α|–1,
we have
Mα,β:=
1 +|β|1 –|α|–1≥1± |β|xδ
α (x∈E–δ). (9)
For fixedx∈E–δ, settingu=xδαyβ, we find
J(–)(x) =x
–δ(σ–qn1)
α
1 –β
∞
(1–β)xδα
e–ρuγ
uσ–qn1–1du≥x
–δ(σ–qn1)
α
1 –β
∞
Mα,β
e–ρuγ
uσ–qn1–1du,
J(+)(x) = x
–δ(σ–qn1) α
1 +β
∞
(1+β)xδα
e–ρuγuσ–qn1–1du≥x
–δ(σ–qn1) α
1 +β
∞
Mα,β
e–ρuγuσ–qn1–1du,
J(x) =x–δ(σ–
1
qn) α
1 1 –β
∞
(1–β)xδα
e–ρuγ
uσ–qn1–1du+ 1 1 +β
∞
(1+β)xδα
e–ρuγ
uσ–qn1–1du
≥2x –δ(σ–qn1)
α
1 +β
∞
Mα,β
e–ρuγuσ–qn1–1du. (10)
In view of (8) and (10), we have the following:
Lemma 2 We have the following inequalities:
I1:=
Eδ
I(x)xδ(σ1–
1
pn)–1
α dx
≥ 2
1 –β2
Eδ
x–δ(σ–σ1+
1
n)–1
α dx
mα,β
0
e–ρuγuσ+qn1–1du, (11)
J1:=
E–δ
J(x)xδ(σ1+
1
pn)–1
α dx
≥ 2
1 –β2
E–δ
xδ(σ1–σ+
1
n)–1
α dx
∞
Mα,β
Lemma 3 If there exists a constant M such that,for any nonnegative measurable functions f(x)and g(y)inR,
I:=
∞
–∞
∞
–∞e
–ρ(xδαyβ)γf(x)g(y)dx dy
≤M
∞
–∞x p(1–δσ1)–1
α fp(x)dx
1
p ∞
–∞y q(1–σ)–1
β gq(y)dy
1
q
, (13)
then we haveσ1=σ.
Proof Ifσ1>σ, then forn≥σ1–1σ (n∈N), we define the functions:
fn(x) :=
⎧ ⎨ ⎩
xδ(σ1–
1
pn)–1
α , x∈Eδ,
0, x∈R\Eδ,
gn(y) :=
⎧ ⎨ ⎩
yσ+
1
qn–1
β , y∈E–1,
0, y∈R\E–1,
and by (4) and (5) it follows that
˜
J1:=
∞
–∞x
p(1–δσ1)–1
α fnp(x)dx
1
p ∞
–∞y q(1–σ)–1
β gnq(y)dy
1
q
=
Eδ
x–
δ
n–1
α dx
1
p
E–1
y
1
n–1 β dy
1
q
=n
1
(1 +α)nδ+1
+ 1
(1 –α)δn+1 1
p 1
(1 +β)–1n+1
+ 1
(1 –β)–1n+1 1
q <∞.
By (11) and (13) (forf =fn,g=gn) we have
2 1 –β2
Eδ
x–δ(σ–σ1+
1
n)–1
α dx
mα,β
0
e–ρuγuσ+qn1–1du
≤I1=
∞
–∞
∞
–∞e –ρ(xδ
αyβ)γf
n(x)gn(y)dx dy≤M˜J1<∞.
Since for anyn≥σ1–1σ,σ–σ1+1n≤0, by Lemma1it follows that
Eδx
–δ(σ–σ1+1n)–1
α dx=∞.
In view of 0mα,βe–ρuγ
uσ+qn1–1du> 0, we find that∞ ≤M˜J
1<∞, which is a
contradic-tion.
Ifσ1<σ, then forn≥σ–1σ1 (n∈N), we define the functions:
˜
fn(x) :=
⎧ ⎨ ⎩
xδ(σ+
1
pn)–1
α , x∈E–δ,
0, x∈R\E–δ,
˜
gn(y) :=
⎧ ⎨ ⎩
yσ–
1
qn–1
β , y∈E1,
and by (4) and (5) it follows that
˜
J2:=
∞
–∞x
p(1–δσ1)–1
α ˜fnp(x)dx
1
p ∞
–∞y q(1–σ)–1
β g˜
q n(y)dy
1
q
=
E–δ
x
δ
n–1
α dx
1
p
E1
y
–1
n–1
β dy
1
q
=n
1
(1 +α)–δn+1
+ 1
(1 –α)–δn+1 1
p 1
(1 +β)1n+1
+ 1
(1 –β)–n+1 1
q <∞.
By (12) and (13) (forf =f˜n,g=g˜n) we have
2 1 –β2
E–δ
xδ(σ1–σ+
1
n)–1
α dx
∞
Mα,β
e–ρuγuσ–qn1–1du
≤J1=
∞
–∞
∞
–∞e
–ρ(xδαyβ)γf˜
n(x)g˜n(y)dx dy≤MJ˜2<∞.
Since for anyn≥ σ–1σ
1,σ1–σ+ 1
n≤0, by Lemma1it follows that
E–δx
δ(σ1–σ+1n)–1
α dx=
∞. In view of M∞
α,βe
–ρuγuσ–qn1–1
du> 0, we have∞ ≤M˜J2<∞, which is a
contradic-tion.
Hence we conclude thatσ1=σ.
The lemma is proved.
Lemma 4 If there exists a constant M such that,for any nonnegative measurable functions f(x)and g(y)inR,
∞
–∞
∞
–∞e
–ρ(xδαyβ)γf(x)g(y)dx dy
≤M
∞
–∞x p(1–δσ)–1
α fp(x)dx
1
p ∞
–∞y q(1–σ)–1
β gq(y)dy
1
q
, (14)
then we have Kα(γ,β)(σ)≤M.
Proof Forσ1=σ, by (8) we have
I1=
Eδ
I(x)xδ(σ–
1
pn)–1
α dx=I1(–)+I1(+),
I1(–)=
Eδ
I(–)(x)xδ(σ–
1
pn)–1
α dx, I1(+):=
Eδ
I(+)(x)xδ(σ–
1
pn)–1
α dx.
In view of the presented results, we find
I1(–)= 1 1 –β
Eδ
x–
δ
n–1 α
(1–β)xδ α 0
e–ρuγ
uσ+qn1–1du dx
= 1
1 –β
Eδ
x–
δ
n–1 α
∞
0
e–ρuγuσ+qn1–1du–
∞
(1–β)xδ α
e–ρuγuσ+qn1–1du
= n 1 –β
1
(1 +α)δn+1
+ 1
(1 –α)δn+1
k(ργ)
σ+ 1
qn
– 1
1 –β
Eδ
x–
δ
n–1 α
∞
(1–β)xδ α
e–ρuγuσ+qn1–1du dx. (15)
Sincee–ρuγu2σ is continuous in (0,∞), ande–ρuγu2σ→0 (u→ ∞), there exists a positive
constantM1such thate–ρu γ
u2σ ≤M
1(u∈[mα,β,∞)). By (4) it follows that
0 <
Eδ
x–
δ
n–1 α
∞
(1–β)xδα
e–ρuγuσ+qn1–1du dx
≤M1
Eδ
x–
δ
n–1 α
∞
(1–β)xδ α
u–σ+qn1–1du
dx=M1
Eδx
–δ(σ+pn1)–1
α dx
(1 –β)σ–qn1
=(σ+
1 pn)
–1M 1
(1 –β)σ–qn1
1
(1 +α)δ(σ+pn1)+1
+ 1
(1 –α)δ(σ+pn1)+1
,
so that
1 1 –β
Eδ
x–
δ
n–1 α
∞
(1–β)xδ α
e–ρuγuσ+qn1–1du dx=O(1).
By (15) it follows that
1
nI
(–) 1 =
k(ργ)(σ+qn1)
1 –β
1
(1 +α)δn+1
+ 1
(1 –α)δn+1
–O(1)
n . (16)
In the same way, we have
1
nI
(+) 1 =
kρ(γ)(σ+qn1)
1 +β
1
(1 +α)δn+1
+ 1
(1 –α)nδ+1
–O˜(1)
n . (17)
By (14) (forf =fn,g=gn), we have
1
nI1=
1
n
I1(–)+I1(+)≤1 nMJ˜1.
Forn→ ∞, by Fatou lemma (see [27]), (16), and (17) we find
2 1 –β2
2kρ(γ)(σ)
1 –α2 ≤M
2 1 –α2
1
p 2
1 –β2
1
q ,
so thatKα(γ,β)(σ) =
2k(γρ)(σ)
(1–α2)1/q(1–β2)1/p≤M.
The lemma is proved.
Lemma 5 We define the following weight functions:
ωδ(σ,y) :=yσβ
∞
–∞e –ρ(xδ
αyβ)γxδσ–1
δ(σ,x) :=xδσα
∞
–∞e
–ρ(xδαyβ)γyσ–1
β dy (x∈R). (19)
Then we have
1 –α2
2 ωδ(σ,y) = 1 –β2
2 δ(σ,x) =k
(γ)
ρ (σ)
x,y∈R\{0}. (20)
Proof For fixedy∈R\{0}, settingu=xδ
αyβ, we find
ωδ(σ,y) =yσβ
0
–∞e –ρ(xδ
αyβ)γ(1 –α)(–x)δσ–1dx
+yσ β
∞
0
e–ρ(xδ
αyβ)γ(1 +α)xδσ–1dx
=
1
1 –α +
1 1 +α
∞
0
e–ρuγ
uσ–1du= 2
1 –α2k (γ)
ρ (σ);
for fixedx∈R\{0}, settingu=xδ
αyβ, it follows that
δ(σ,x) =xδσα
0
–∞e –ρ(xδ
αyβ)γyσ–1
β dy+x δσ α
∞
0
e–ρ(xδαyβ)γyσ–1
β dy
= 2
1 –β2
∞
0
e–ρuγuσ–1du= 2 1 –β2k
(γ)
ρ (σ).
Hence we have (20).
The lemma is proved.
3 Main results
Theorem 1 If M is a constant,then the following statements(i), (ii),and(iii)are equiva-lent:
(i) For anyf(x)≥0,we have:
J:=
∞
–∞y pσ–1
β
∞
–∞e –ρ(xδ
αyβ)γf(x)dx
p
dy
1
p
≤M
∞
–∞x p(1–δσ1)–1
α f
p(x)dx
1
p
. (21)
(ii) For anyf(x),g(y)≥0,we have:
I=
∞
–∞
∞
–∞e –ρ(xδ
αyβ)γf(x)g(y)dx dy
≤M
∞
–∞x p(1–δσ1)–1
α fp(x)dx
1
p ∞
–∞y q(1–σ)–1
β gq(y)dy
1
q
. (22)
Proof (i)=>(ii). By Hölder’s inequality (see [28]) we have
I=
∞
–∞
yσ–
1
p β
∞
–∞e –ρ(xδ
αyβ)γf(x)dxy–σ+ 1
p β g(y)
dy
≤J
∞
–∞y q(1–σ)–1
β g
q(y)dy
1
q
. (23)
Then by (21) we have (22).
(ii)=>(iii). By Lemma1we haveσ1=σ. Then by Lemma2we haveKα(γ,β)(σ)≤M.
(iii)=>(i). Forσ1=σ, by Hölder’s inequality with weight (see [28]) and (18) we have
∞
–∞e –ρ(xδ
αyβ)γf(x)dx
p
=
∞
–∞e –ρ(xδαyβ)γ
y(σ–1)/p
β f(x)
x(αδσ–1)/q
x(αδσ–1)/q
y(βσ–1)/p
dx
p
≤
∞
–∞e –ρ(xδ
αyβ)γy
σ–1
β fp(x)
x(αδσ–1)p/q
dx
∞
–∞e –ρ(xδ
αyβ)γ x
δσ–1
α
y(βσ–1)q/p
dx
p/q
=ωδ(σ,y)yq(1–β σ)–1
p–1 ∞ –∞e
–ρ(xδαyβ)γy
σ–1
β fp(x)
x(αδσ–1)p/q
dx
=
2kρ(γ)(σ)
1 –α2
p–1
y–pβ σ+1
∞
–∞e
–ρ(xδαyβ)γy
σ–1
β fp(x)
x(αδσ–1)p/q
dx. (24)
By Fubini’s theorem (see [27]), (24), and (19) we have
J≤
2kρ(γ)(σ)
1 –α2
1
q ∞
–∞
∞
–∞e
–ρ(xδαyβ)γy
σ–1
β fp(x)
x(αδσ–1)p/q
dx dy
1
p
=
2k(ργ)(σ)
1 –α2
1
q ∞
–∞δ(σ,x)x p(1–δσ)–1
δ fp(x)dx
1
p
=Kα(γ,β)(σ)
∞
–∞x p(1–δσ)–1
δ f
p(x)dx
1
p .
ForKα(γ,β)(σ)≤M, we have (21) (whenσ1=σ).
Therefore, statements (i), (ii), and (iii) are equivalent.
The theorem is proved.
Theorem 2 If M is a constant,then the following statements(i), (ii),and(iii)are equiva-lent:
(i) For anyf(x)≥0satisfying0 <–∞∞ xαp(1–δσ)–1fp(x)dx<∞,we have:
∞
–∞y pσ–1
β
∞
–∞e –ρ(xδ
αyβ)γf(x)dx
p
dy
1
p
<M
∞
–∞x p(1–δσ)–1
α fp(x)dx
1
p
(ii) For anyf(x)≥0satisfying0 <–∞∞ xαp(1–δσ)–1fp(x)dx<∞,andg(x)≥0satisfying
0 <–∞∞ yq(1–β σ)–1gq(y)dy<∞,we have:
∞
–∞
∞
–∞e –ρ(xδ
αyβ)γf(x)g(y)dx dy
<M
∞
–∞x p(1–δσ)–1
α fp(x)dx
1
p ∞
–∞y q(1–σ)–1
β gq(y)dy
1
q
. (26)
(iii) Kα(γ,β)(σ)≤M.
Moreover,if statement(iii)holds,then the constant factor M=Kα(γ,β)(σ)in(25)and(26)
is the best possible.
In particular, (1)forδ= 1,M=Kα(γ,β)(σ),we have the following equivalent inequalities with nonhomogeneous kernel:
∞
–∞y pσ–1
β
∞
–∞e –ρ(xαyβ)γ
f(x)dx
p
dy
1
p
<Kα(γ,β)(σ)
∞
–∞x p(1–σ)–1
α fp(x)dx
1
p
, (27)
∞
–∞
∞
–∞e
–ρ(xαyβ)γf(x)g(y)dx dy
<Kα(γ,β)(σ)
∞
–∞x p(1–σ)–1
α f
p(x)dx
1
p ∞
–∞y q(1–σ)–1
β g
q(y)dy
1
q
, (28)
where Kα(γ,β)(σ)is the best possible constant factor;
(2)forδ= –1,M=Kα(γ,β)(σ),we have the following equivalent inequalities with homoge-neous kernel of degree0:
∞
–∞y pσ–1
β
∞
–∞e
–ρ(yβ/xα)γf(x)dx
p
dy
1
p
<Kα(γ,β)(σ)
∞
–∞x p(1+σ)–1
α f
p(x)dx
1
p
, (29)
∞
–∞
∞
–∞e –ρ(yβ/xα)γ
f(x)g(y)dx dy
<Kα(γ,β)(σ)
∞
–∞x p(1+σ)–1
α fp(x)dx
1
p ∞
–∞y q(1–σ)–1
β g
q(y)dy
1
q
, (30)
where Kα(γ,β)(σ)is the best possible constant factor.
Proof Forσ1=σ, under and the assumption of statement (i), if (24) takes the form of
equality fory∈R\{0}, then there exist constantsAandBsuch that they are not both zero and (see [28])
A y
σ–1
β
x(αδσ–1)p/q
fp(x) =B x
δσ–1
α
We suppose thatA= 0 (otherwise,B=A= 0). Then it follows that
xp(1–δσ)–1
α f
p(x) =yq(σ–1)
β
B Axα
a.e. inR.
Since–∞∞ x–1
α dx=∞, this contradicts the fact that 0 <
∞
–∞x p(1–δσ)–1
α fp(x)dx<∞. Hence
(24) takes the form of strict inequality, and so does (21). Hence (25) and (26) are valid. In view of Theorem1, we still can conclude that statements (i), (ii), and (iii) in Theorem2
are equivalent.
When statement (iii) holds, namely,Kα(γ,β)(σ)≤M, if there exists a constantM(≤Kα(γ,β)(σ)) such that (26) is valid, thenM=Kα(γ,β)(σ), and we can conclude that the constant factor
M=Kα(γ,β)(σ) in (26) is the best possible.
The constant factorM=Kα(γ,β)(σ) in (25) is still the best possible. Otherwise, by (23) (for
σ1=σ), we would get a contradiction that the constant factorM=K(
γ)
α,β(σ) in (26) is not
the best possible.
The theorem is proved.
4 Operator expressions
We set the following functions:ϕ(x) :=xp(1–α δσ)–1(x∈R), andψ(y) :=yq(1–β σ)–1, where from ψ1–p(y) :=ypβσ–1(y∈R). Define the following real normed linear spaces:
Lp,ϕ(R) :=
f; f p,ϕ:=
∞
–∞ϕ(x)
f(x)pdx
1
p <∞
,
Lq,ψ(R) :=
g; g q,ψ=
∞
–∞ψ(y)
g(y)qdy
1
q <∞
,
Lp,ψ1–p(R) :=
h; h p,ψ1–p=
∞
–∞ψ
1–p(y)h(y)pdy
1
p <∞
.
In view of Theorem2, forf ∈Lp,ϕ(R), setting
h1(y) :=
∞
–∞e
–ρ(yβ/xα)γf(x)dx (y∈R),
by (25) we have
h1 p,ψ1–p=
∞
–∞ψ 1–p(y)h
1(y) p
dy
1
p
≤M f p,ϕ<∞. (31)
Definition 1 Define the Hilbert-type integral operatorT :Lp,ϕ(R)→Lp,ψ1–p(R) as fol-lows: For anyf ∈Lp,ϕ(R), there exists a unique representationTf =h1∈Lp,ψ1–p(R), satis-fying for anyy∈R,Tf(y) =h1(y).
In view of (31), it follows that Tf p,ψ1–p= h1 p,ψ1–p≤M f p,ϕ, and then the operator
T is bounded and satisfies
T = sup f(=θ)∈Lp,ϕ(R)
Tf p,ψ1–p
f p,ϕ
If we define the formal inner product ofTf andgas
(Tf,g) :=
∞
–∞
∞
–∞e
–ρ(yβ/xα)γf(x)dx
g(y)dy,
then we can rewrite Theorem2as follows.
Theorem 3 If M is a constant,then the following statements(i), (ii),and(iii)are equiva-lent:
(i) For anyf(x)≥0,f∈Lp,ϕ(R), f p,ϕ> 0,we have:
Tf p,ψ1–p<M f p,ϕ. (32) (ii) Forf(x),g(y)≥0,f ∈Lp,ϕ(R),g∈Lq,ψ(R), f p,ϕ, g q,ψ> 0,we have:
(Tf,g) <M f p,ϕ g q,ψ. (33)
(iii) Kα(γ,β)(σ)≤M.
Moreover,if statement(iii)holds,then the constant factor M=Kα(γ,β)(σ)in(32)and(33)
is the best possible,namely, T =Kα(γ,β)(σ).
Remark1 (1) In particular, forα=β= 0 in (27) and (28), we have the following equivalent inequalities:
∞
–∞|y| pσ–1
∞
–∞e
–ρ|xy|γf(x)dx
p
dy
1
p
<2(σ/γ)
γρσ/γ
∞
–∞|x|
p(1–σ)–1fp(x)dx
1
p
, (34)
∞
–∞
∞
–∞e –ρ|xy|γ
f(x)g(y)dx dy
<2(σ/γ)
γρσ/γ
∞
–∞|x|
p(1–σ)–1fp(x)dx
1
p ∞
–∞|y|
q(1–σ)–1gq(y)dy
1
q
, (35)
where 2γρ(σσ/γ/γ) is the best possible constant factor.
Iff(–x) =f(x),g(–y) =g(y) (x,y∈R+), then we have the following equivalent inequali-ties:
∞
0
ypσ–1
∞
0
e–ρ(xy)γf(x)dx
p
dy
1
p
<(σ/γ)
γρσ/γ
∞
0
xp(1–σ)–1fp(x)dx
1
p
, (36)
∞
0
∞
0
e–ρ(xy)γf(x)g(y)dx dy
<(σ/γ)
γρσ/γ
∞
0
xp(1–σ)–1fp(x)dx
1
p ∞
0
yq(1–σ)–1gq(y)dy
1
q
, (37)
(2) Forα=β= 0 in (29) and (30), we have the following equivalent inequalities:
∞
–∞|y| pσ–1
∞
–∞e
–ρ|y/x|γf(x)dx
p
dy
1
p
<2(σ/γ)
γρσ/γ
∞
–∞|x|
p(1+σ)–1fp(x)dx
1
p
, (38)
∞
–∞
∞
–∞e –ρ|y/x|γ
f(x)g(y)dx dy
<2(σ/γ)
γρσ/γ
∞
–∞|x|
p(1+σ)–1fp(x)dx
1
p ∞
–∞|y|
q(1–σ)–1gq(y)dy
1
q
, (39)
where 2γρ(σσ/γ/γ) is the best possible constant factor.
Iff(–x) =f(x),g(–y) =g(y) (x,y∈R+), then we have the following equivalent inequali-ties:
∞
0
ypσ–1
∞
0
e–ρ(y/x)γf(x)dx
p
dy
1
p
<(σ/γ)
γρσ/γ
∞
0
xp(1+σ)–1fp(x)dx
1
p
, (40)
∞
0
∞
0
e–ρ(y/x)γ
f(x)g(y)dx dy
<(σ/γ)
γρσ/γ
∞
0
xp(1+σ)–1fp(x)dx
1
p ∞
0
yq(1–σ)–1gq(y)dy
1
q
, (41)
where γρ(σσ/γ/γ)is the best possible constant factor.
5 Conclusions
In this paper, by using real analysis and weight functions we obtain a few equivalent state-ments of a Hilbert-type integral inequality in the whole plane related to the kernel of expo-nent function with the intermediate variables (Theorem1). The constant factor related to the gamma function is proved to be the best possible in Theorem2. We also consider some
particular cases and the operator expressions in Remark1and Theorem3. The lemmas
and theorems provide an extensive account of this type of inequalities.
Funding
This work is supported by the National Natural Science Foundation (Nos. 61562016 and 51765012) and Science and Technology Planning Project Item of Guangzhou City (No. 201707010229). We are grateful for this help.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. YZ and MH participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
Author details
Publisher’s Note
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Received: 14 July 2018 Accepted: 2 September 2018 References
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