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指數(shù)權(quán)的加權(quán)Lebesgue空間中擬齊次核最優(yōu)Hilbert型積分不等式的參數(shù)條件及應(yīng)用

2024-01-01 00:00:00趙茜洪勇孔蔭瑩

摘要: 首先, 引入擬齊次核的概念, 討論權(quán)函數(shù)為指數(shù)函數(shù)的加權(quán)Lebesgue空間中具有擬齊次核的Hilbert型積分不等式; 其次, 利用權(quán)系數(shù)方法及若干分析技巧, 給出最優(yōu)Hilbert型積分不等式的等價(jià)參數(shù)條件, 并獲得了最佳常數(shù)因子的計(jì)算公式; 最后, 討論其在算子理論中的應(yīng)用.

關(guān)鍵詞: 加權(quán)Lebesgue空間; 擬齊次核; Hilbert型積分不等式; 最佳搭配參數(shù); 等價(jià)條件; 有界算子; 算子范數(shù)

中圖分類號: O178""文獻(xiàn)標(biāo)志碼: A""文章編號: 1671-5489(2024)06-1325-09

Parameter Conditions of Optimal Hilbert-Type Integral Inequalitieswith Quasi-homogeneous Kernel in Weighted Lebesgue Spaceswith Exponential Weight and Applications

ZHAO Qian1, HONG Yong1, KONG Yinying2

(1. College of Artificial Intelligence, Guangzhou Huashang College, Guangzhou 511300, China; 2. College of Statistics and Mathematics, Guangdong University of Finance and Ec

onomics, Guangzhou 510320, China)

Abstract: Firstly, the concept of quasi-homogeneous kernel was introduced to discuss Hilbert-type integral inequalities with quasi-homogeneous kernel

in weighted Lebesgue spaces with exponential functions. Secondly, by using the weight coefficient method and several "analysis techniques, equivalent parameter conditio

n for optimal Hilbert-type integral inequalities was given, and the calculation formula for the best constant factor was obtained. Finally, its applications in operator theory were discussed.

Keywords: weighted Lebesgue space; quasi-homogeneous kernel; Hilbert-type integral inequality; optimal matching parameter; equivalent condition; bounded operator; operator norm

0"引"言

設(shè)1p+1q=1(pgt;1), f(x)∈Lp(0,+∞), g(y)∈Lq(0,+∞), 則Hilbert積分不等式[1]

+∞0∫+∞01x+yf(x)g(y)dxdy≤πsin(π/p)‖f‖p‖g‖q,(1)

其中常數(shù)因子πsin(π/p)是最佳值. 定義積分算子

T(f)(y)=∫+∞01x+yf(x)dx,""f(x)∈Lp(0,+∞),

則式(1)可等價(jià)地表示為如下算子T的不等式:

‖T(f)‖p≤πsin(π/p)‖f‖p.

因此T是Lp(0,+∞)中的有界算子, 且算子范數(shù)‖T‖=πsin(π/p).

設(shè)φ(x)gt;0, 引入加權(quán)Lebesgue空間:

Lφ(x)p(a,b)=f(x): ‖f‖p,φ(x)=∫baφ(x)f(x)pdx1/plt;+∞.

一般地, 若1p+1q=1(pgt;1), φ1(x)gt;0, φ2(y)gt;0, 則以K(x,y)為核的不等式

∫ba∫baK(x,y)f(x)g(y)dxdy≤M‖f‖p,φ1(x)‖g‖q,φ2(y)(2)

稱為Hilbert型積分不等式, 若其常數(shù)因子M為最佳值, 則式(2)稱為最優(yōu)不等式.

目前, 關(guān)于Hilbert型不等式的研究已有很多成果[2-24]. 例如: 文獻(xiàn)[2]考慮指數(shù)權(quán)函數(shù)的加權(quán)Lebesgue空間, 得到了一個(gè)具有最佳常數(shù)因子的Hilbert型不等式:

+∞-∞∫+∞-∞f(x)g(y)(1+exey)λdxdy≤Bλ2,λ2‖f‖2,e-λx‖g‖2,e-λy

其中λgt;0, B(u,v)是Beta函數(shù); 文獻(xiàn)[3]在文獻(xiàn)[2]的基礎(chǔ)上引入獨(dú)立參數(shù)σ和σ1, 討論了核為H(eaxeby)的Hilbert型積分不等式:

+∞-∞∫+∞-∞H(eaxeby)f(x)g(y)dxdy≤

M‖f‖p,exp{σapx}‖g‖q,exp{σ1bqy},

并得到了最佳常數(shù)因子的等價(jià)條件. 本文引入一類廣義齊次核:

K(x,y)=G(eλ1x,eλ2y),

其中G(u,v)是λ階齊次非負(fù)函數(shù), 討論指數(shù)函數(shù)加權(quán)Lebesgue空間中最優(yōu)Hilbert型不等式的等價(jià)參數(shù)條件.

1"引"理

引理1"設(shè)G(u,v)是λ階齊次非負(fù)函數(shù), λ1λ2gt;0, K(x,y)=G(eλ1x,eλ2y), 則

K(x,y)=eλλ1xK0,y-λ1λ2x=eλλ2yKx-λ2λ1y,0.

特別地, 有

K(0,t)=eλλ2tK-λ2λ1t,0,""K(t,0)=eλλ1tK0,-λ1λ2t.

證明: 根據(jù)G(u,v)是λ階齊次函數(shù), 有

K(x,y)=G(eλ1x,eλ2y)=eλλ1xG(1,eλ2y-λ1x)eλλ1xGe0λ1,eλ2\=eλλ1xK0,y-λ1λ2x.

同理可證K(x,y)=eλλ2yKx-λ2λ1y,0.

引理2"設(shè)1p+1q=1(pgt;1), λ1λ2gt;0, a,b∈, G(u,v)是λ階齊次非負(fù)函數(shù),

K(x,y)=G(eλ1x,eλ2y), 記

W1(s)=∫+∞-∞K(0,t)estdt,""W2(s)=∫+∞-∞K(t,0)estdt,

則有

ω1(b,p,x)=∫+∞-∞K(x,y)e-bpydy=eλ1\1λ1aq+1λ2bp=λ, 則1λ1W1(-bp)=1λ2W2(-aq).

證明: 根據(jù)引理1, 有

ω1(b,p,x)= "eλλ1x+∞-∞K0,y-λ1λ2xe-bpydy

=eλλ1x+∞-∞K(0,t)e-bp\"eλ1\

類似可證ω2(a,q,y)=eλ2\1λ1aq+1λ2bp=λ, 則

W1(-bp)= "∫+∞-∞K(0,t)e-bptdt=∫+∞-∞K-λ2λ1t,0e

(λλ2-bp)tdt= "λ1λ2∫+∞-∞K(u,0)e"λ1λ2∫+∞-∞K(u,0)e-aqudu=λ1λ2W2(-aq),

故1λ1W1(-bp)=1λ2W2(-aq).

引理3"設(shè)1p+1q=1(pgt;1), K(x,y)≥0, φ1(x)gt;0, φ2(y)gt;0, Ω是一個(gè)可測集, 積分算子T為

T(f)(y)=∫ΩK(x,y)f(x)dx,

則Hilbert型不等式

∫Ω∫ΩK(x,y)f(x)g(y)dxdy≤M‖f‖p,φ1(x)‖g‖q,φ2(y)(3)

等價(jià)于算子不等式

‖T(f)‖p,φ1-p2(y)≤M‖f‖p,φ1(x).(4)

證明: 若式(4)成立, 則根據(jù)H?lder不等式, 有

∫Ω∫ΩK(x,y)f(x)g(y)dxdy= "∫Ωg(y)T(f)(y)dy= "∫Ωφ1/q2(y)g(y)φ-1/q2(y)T(f)(y)dy

≤ "∫Ωφ2(y)g(y)qdy1/q∫Ωφ-p/q2(y)T(f)(y)pdy1/p

= "‖g‖q,φ2(y)∫Ωφ1-p2(y)T(f)(y)pdy1/p

= "‖g‖q,φ2(y)‖T(f)‖p,φ1-p2(y)≤M‖f‖p,φ1(x)‖g‖q,φ2(y),

故式(4)成立.

反之, 若式(3)成立, 則

‖T(f)‖pp,φ1-p2(y)= "∫Ωφ1-p2(y)T(f)(y)pdy= "∫Ωφ1-p2(y)∫ΩK(x,y)f(x)dxpdy

≤ "∫Ωφ1-p2(y)∫ΩK(x,y)f(x)dxp-1∫ΩK(x,y)f(x)dxdy.

令g(y)=φ1-p2(y)∫ΩK(x,y)f(x)dxp-1, 則

‖T(f)‖pp,φ1-p2(y)≤ "∫Ωg(y)∫ΩK(x,y)f(x)dxdy= "∫Ω∫ΩK(x,y)f(x)g(y)dxdy

≤ "M‖f‖p,φ1(x)‖g‖q,φ2(y)=M‖f‖p,φ1(x)‖g‖q,φ2(y)= "M‖f‖p,φ1(x)∫Ωφ2(y)

φ1-p2(y)∫ΩK(x,y)f(x)dxp-1qdy1/q=

M‖f‖p,φ1(x)∫Ωφ1-p2(y)∫ΩK(x,y)f(x)dxpdy1/q

= "M‖f‖p,φ1(x)∫Ωφ1-p2(y)T(f)(y)pdy1/q= "M‖f‖p,φ1(x)‖T(f)‖p-1

p,φ1-p2(y),

從而可得式(4).

2"主要結(jié)果

定理1"設(shè)1p+1q=1(pgt;1), a,b∈, λ1λ2gt;0, G(u,v)是λ階齊次非負(fù)函數(shù),

K(x,y)=G(eλ1x,eλ2y), 則下列結(jié)論成立:

1) 記φ1(x)=expλ1paλ1-bλ2+λx,

φ2(y)=expλ2qbλ2-aλ1+λy, 則

A(K,f,g)=∶ "∫+∞-∞∫+∞-∞K(x,y)f(x)g(y)dxdy

≤ "W1/p1(-bp)W1/q2(-aq)‖f‖p,φ1(x)‖g‖q,φ2(y),(5)

其中f(x)∈Lφ1(x)p(-∞,+∞), g(y)∈Lφ2(y)q(-∞,+∞).

2) 若W1(-bp)lt;+∞, W2(-aq)lt;+∞, 則W1/p1(-bp)W1/q2(-aq)是式(5)最佳常數(shù)因子的充分必要條件是1λ1aq+1λ2bp=λ, 且當(dāng)1λ1aq+1λ2bp=λ時(shí), 式(5)化為

A(K,f,g)≤λ2λ11/qW1(-bp)‖f‖p,exp{apqx}‖g‖q,exp{bpqy},(6)

其中λ2λ11/qW1(-bp)是最佳常數(shù)因子.

證明: 1) 選取a,b為搭配參數(shù), 利用H?lder不等式及引理2, 有

A(K,f,g)= "∫+∞-∞∫+∞-∞(eax-byf(x))(eby-axg(y))K(x,y)dxdy≤

+∞-∞∫+∞-∞e(ax-by)pf(x)pK(x,y)dxdy1/p

× "∫+∞-∞∫+∞-∞e(by-ax)qg(y)qK(x,y)dxdy1/q

= "∫+∞-∞eapxf(x)pω1(b,p,x)dx1/p

+∞-∞ebqyg(y)qω2(a,q,y)dy1/q= "W1/p1(-bp)W1/q2(-aq)

+∞-∞expλ1paλ1-bλ2+λxf(x)pdx1/p× "∫+∞-∞expλ2qbλ2-

aλ1+λyg(y)qdx1/q= "W1/p1(-bp)W1/q2(-aq)‖f‖p,φ1(x)‖g‖q,φ2(y).

故式(5)成立.

2) 充分性. 設(shè)1λ1aq+1λ2bp=λ, 則計(jì)算可得φ1(x)=exp{apqx}, φ2(y)=exp{bpqy}.

又根據(jù)引理2, 有1λ1W1(-bp)=1λ2W2(-aq), 于是式(5)可化為式(6). 從而只需證明式(6)的常數(shù)因子最佳即可.

若式(6)的常數(shù)因子不是最佳的, 則存在常數(shù)M0lt;λ2λ11/qW1(-bp), 使得

A(K,f,g)≤M0‖f‖p,exp{apqx}‖g‖q,exp{bpqy}.

取充分小的εgt;0及足夠大的正數(shù)N, 令

f(x)=exp{(-apq-λ1ε)x/p},x≥N,0,xlt;N,

g(y)=exp{(-bpq-λ2ε)y/q},y≥0,0,ylt;0,

則有

‖f‖p,exp{apqx}‖g‖q,exp{bpqy}= "∫+∞Ne-λ1εxdx1/p

+∞0e-λ2εydy1/q= "1λ1εe-λ1εN1/p1λ2ε1/q= "1ελ11/pλ21/q

e-λ1εN/p.

根據(jù)引理1, 有

A(K,f,g)= "∫+∞Nexp-aq-λ1εpx

+∞0K(x,y)exp-bp-λ2εqydydx

= "∫+∞Nexp-aq-λ1εpx×

+∞0K0,y-λ1λ2xe

λλ1xexp-bp-λ2εqydydx=

+∞Nexpλλ1-aq-λ1εpx×

+∞-(λ1/λ2)xK(0,t)exp-bp-λ2εqt+λ1λ2xdtdx

= "∫+∞Nexpλλ1-aq-λ1εp-λ1λ2bp+

λ1εqx× "∫+∞-(λ1/λ2)xK(0,t)exp-bp-λ2εqtdtdx= "∫+∞Neλ1εx

+∞-(λ1/λ2)xK(0,t)exp-bp-λ2εqtdtdx

≥ "∫+∞Neλ1εxdx∫+∞-(λ1/λ2)NK(0,t)exp-bp-λ

2εqtdt= "1ελ1e-λ1εN

+∞-(λ1/λ2)NK(0,t)exp-bp-λ2εqtdt.

綜上可得

1λ1e-λ1εN+∞-(λ1/λ2)NK(0,t)exp-bp-λ

2εqtdt≤M01λ11/pλ21/qe-λ1εN/p.

令ε→0+, 利用Fatou引理[25], 有

1λ1∫+∞-(λ1/λ2)NK(0,t)e-bptdt≤M01λ11/pλ21/q

再令N→+∞, 得

1λ1W1(-bp)=1λ1∫+∞-∞K(0,t)e-bptdt

≤M01λ11/pλ21/q

于是有λ2λ11/qW1(-bp)≤M0, 與M0lt;λ2λ11/qW1(-bp)矛盾,

故λ2λ11/qW1(-bp)是式(6)的最佳常數(shù)因子.

必要性. 設(shè)W1/p1(-bp)W1/q2(-aq)是式(5)的最佳常數(shù)因子, 記1λ1aq+1λ2bp-λ=c, a1=a-λ1cpq,

b1=b-λ2cpq, 則

1λ1a1q+1λ2b1p=1λ1aq+1λ2bp-cp-cq=λ+c-c=λ,

φ1(x)=expλ1paλ1-bλ2+λx

=expλ1pa1λ1-b1λ2+λx=exp{a1pqx},

φ2(y)=expλ2qbλ2-aλ1+λy

=expλ2qb1λ2-a1λ1+λy=exp{b1pqy},

W2(-aq)= "∫+∞-∞K(t,0)e-aqtdt=∫+∞-∞K0,-λ1λ2te

λλ1-aq)tdt= "λ2λ1∫+∞-∞K(0,u)e(-bp+λ2c)udu=λ2λ1∫+∞-∞K(0,t)e(-bp+λ2c)tdt,

于是式(5)可化為

A(K,f,g)=W1/p1(-bp)λ2λ1∫+∞-∞K(0,t)e(-bp+λ2c)tdt1/q

‖f‖p,exp{a1pqx}‖g‖q,exp{b1pqy}.(7)

由于式(5)的常數(shù)因子是最佳的, 故式(7)的常數(shù)因子為

λ2λ11/qW1/p1(-bp)∫+∞-∞K(0,t)e(-bp+λ2c)tdt1/q

因?yàn)?λ1a1q+1λ2b1p=λ, 故根據(jù)充分條件的證明可知, 式(7)的最佳常數(shù)因子應(yīng)為

λ2λ11/qW1(-bp)=λ2λ11/q

+∞-∞K(0,t)exp-bpt+λ2cqtdt.

于是

+∞-∞K(0,t)exp-bpt+λ2cqtdt=W1/p1(-bp)∫+∞-∞K(0,t)exp{(-bp+λ2c)t}d

t1/q.(8)

利用H?lder不等式, 有

+∞-∞K(0,t)exp-bp+λ2cqtdt= "∫+∞-∞1·

expλ2cqtK(0,t)e-bptdt≤

+∞-∞1pK(0,t)e-bptdt1/p+∞-∞eλ2ctK(0,t)e-bptdt1/q

= "W1/p1(-bp)∫+∞-∞K(0,t)e(-bp+λ2c)tdt1/q.(9)

由式(8)知式(9)取等號, 再根據(jù)H?lder不等式取等號的條件, 可得eλ2ct=常數(shù), 從而c=0, 即1λ1aq+1λ2bp=λ.

3"應(yīng)"用

根據(jù)引理3和定理1, 可得下面關(guān)于算子理論的定理.

定理2"設(shè)1p+1q=1(pgt;1), a,b∈, λ1λ2gt;0, G(u,v)是λ階齊次非負(fù)可測函數(shù),

K(x,y)=G(eλ1x,eλ2y), W1(-bp)lt;+∞, W2(-aq)lt;+∞, 積分算子T為

T(f)(y)=∫+∞-∞K(x,y)f(x)dx.

1) 記φ1(x)=expλ1paλ1-bλ2+λx,

φ2(y)=expλ2qbλ2-aλ1+λy,

則T是從Lφ1(x)p(-∞,+∞)到Lφ1-p2(y)p(-∞,+∞)的有界算子, 且T的算子范數(shù)‖T‖≤W1/p1(-bp)W1/q2(-aq).

2) 當(dāng)且僅當(dāng)1λ1aq+1λ2bp=λ時(shí), T的算子范數(shù)‖T‖=W1/p1(-bp)W1/q2(-aq). 當(dāng)1λ1aq+

1λ2bp=λ時(shí), 有φ1(x)=exp{apqx}, φ2(y)=exp{bpqy}, 且

‖T‖=W1/p1(-bp)W1/q2(-aq)=λ2λ11/qW1(-bp).

推論1"設(shè)1p+1q=1(pgt;1), λgt;0, λ1λ2gt;0, k1gt;0, k2gt;0, 則積分算子

T(f)(y)=∫+∞-∞f(x)(k1eλ1x+k2eλ2y)λdx

是從Lexp{-λλ1x}p(-∞,+∞)到Lexp{-λλ2(1-p)y}p(-∞,+∞)的有界算子, 且T的算子范數(shù)為

‖T‖=1λ11/qλ21/pkλ/p1

kλ/q2Bλp,λq,

其中B(s,t)是Beta函數(shù).

證明: 令G(u,v)=1(k1u+k2v)λ, 則G(u,v)是-λ階齊次函數(shù), 且

K(x,y)=G(eλ1x,eλ2y)=1(k1eλ1x+k2eλ2y)λ.

取a=-λλ1pq, b=-λλ2pq, 則

1λ1aq+1λ2bp=-λ,"exp{apqx}=exp{-λλ1x},"exp{bpq(1-p)y}=exp{-λλ2(1-p)y},

W1(-bp)= "∫+∞-∞K(0,t)e-bptdt=∫+∞-∞1(k1+k2eλ2t)λe

(λλ2/q)tdt= "1kλ1∫+∞-∞1\e(λλ2/q)tdt= "1kλ1

+∞-∞1(1+u)λk1k2uλ/q1λ2u-1du

= "1λ21kλ/p1kλ/q2∫+∞-∞1(1+u)λuλ/q-1du=

1λ2kλ/p1kλ/q2Bλq,λ-λq

=1λ2kλ/p1kλ/q2Bλp,λq.

于是可得

λ2λ11/qW1(-bp)=1λ11/qλ21/pkλ

/p1kλ/q2Bλp,λq.

綜上并根據(jù)定理2知結(jié)論成立.

推論2"設(shè)1p+1q=1(pgt;1), λ1λ2gt;0, 0lt;σlt;λ, k1gt;0, k2gt;0, 則積分算子

T(f)(y)=∫+∞-∞f(x)(k1+k2eλ1x-λ2y)λdx

是從Lexp{-λ1σpx}p(-∞,+∞)到Lexp{-λ2σpy}p(-∞,+∞)的有界算子, 且T的算子范數(shù)為

‖T‖=kσ-λ1k2λ11/qλ21/pB(σ,λ-σ).

證明: 令G(u,v)=1(k1+k2u/v)λ, 則G(u,v)是0階齊次函數(shù), 且

K(x,y)=G(eλ1x,eλ2y)=1(k1+k2eλ1x-λ2y)λ.

取a=-λ1σq, b=λ2σp, 則

1λ1aq+1λ2bp=0,"exp{apqx}=exp{-λ1σpx},"exp{bpq(1-p)y}=exp{-λ2σpy},

W1(-bp)= "∫+∞-∞K(0,t)e-bptdt=∫+∞-∞1(k1+k2e-λ2t)λe

-λ2σtdt= "1kλ1∫+∞-∞1\e-λ2σtdt= "1λ2kσ-λ1k2

+∞-∞1(1+u)λuσ-1du=1λ2kσ-λ1k2B(σ,λ-σ),

于是可得

λ2λ11/qW1(-bp)=kσ-λ1k2λ11/qλ

21/pB(σ,λ-σ).

綜上并根據(jù)定理2知結(jié)論成立.

若在推論2中取σ=1p, λ=1, 則可得:

推論3"設(shè)1p+1q=1(pgt;1), λ1λ2gt;0, k1gt;0, k2gt;0, 則積分算子

T(f)(y)=∫+∞-∞f(x)k1+k2eλ1x-λ2ydx

是從Lexp{-λ1x}p(-∞,+∞)到Lpexp{-λ2y}(-∞,+∞)的有界算子, 且T的算子范數(shù)為

‖T‖=k-1/q1k-1/p2λ11/qλ21/pB1p,1-1p

=1λ11/qλ21/pk1/q1k1/p2πsin(π/p).

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(責(zé)任編輯: 趙立芹)

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