尼罗罗非鱼IgM单抗的制备及IgM+B淋巴细胞吞噬能力的研究

吴丽婷; 孔令贺; 王君如; 杨延舰; 尹晓雪; 叶剑敏

doi:10.6054/j.jscnun.2018100

尼罗罗非鱼IgM单抗的制备及IgM+B淋巴细胞吞噬能力的研究

（华南师范大学生命科学学院∥广东省水产健康安全养殖重点实验室,广州 510631）

基金项目:

国家自然科学基金项目（31172432，31472302）;华南师范大学研究生创新计划项目（2017LKXM010）

详细信息

通讯作者:
叶剑敏，教授，国家“千人计划”青年人才项目入选者，Email: yjmying@126.com.

中图分类号: Q813.2
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出版历程
- 收稿日期: 2018-09-29
- 刊出日期: 2018-10-24

The Preparation of McAb Against IgM from Nile Tilapia (Oreochromis niloticus) and the Analysis of the Phagocytosis of Fluorescence Microspheres by mIgM+B Lymphocytes

(School of Life Sciences∥Guangdong Provincial Key Laboratory for Healthy and Safe Aquaculture,South China Normal University,Guangzhou 510631,China)

Funds:

国家自然科学基金项目（31172432，31472302）;华南师范大学研究生创新计划项目（2017LKXM010）

摘要

摘要: 采用免疫亲和层析方法纯化尼罗罗非鱼抗TNP特异性IgM，并以此为抗原免疫Babl/c小鼠进行单克隆抗体制备. 免疫的小鼠脾脏细胞与骨髓瘤细胞(SP2/0)融合成杂交瘤细胞，利用酶联免疫吸附法、蛋白质印迹法(Western blot)和流式细胞术筛选获得两株抗IgM重链的单克隆抗体，分别命名为3B3和9D12. 纯化后的单克隆抗体3B3和9D12在1 mg/mL浓度下其抗体效价分别为740,741和359,712 units/mL. 利用制备的单抗结合激光共聚焦显微镜检测发现，膜结合型IgM位于B细胞膜表面(IgM+ B细胞). 流式细胞术检测尼罗罗非鱼IgM+ B细胞的免疫组织分布，发现IgM+ B细胞分布存在组织差异性，其中在外周血(PBL)所占比例最大，约为37.6%，其次是脾脏(SPL)，占百分比33.7%，头肾(AK)占比例约为23.9%，而后肾(PK)占比例约为20%. IgM+ B细胞荧光微球的吞噬能力分析发现，IgM+ B细胞对0.5 m的荧光微球吞噬能力强于1 m. 另外，IgM+ B细胞的吞噬能力存在免疫组织差异性，其中对于0.5 m荧光微球的吞噬，PBL明显高于其他组织，而AK IgM+ B细胞对1 m荧光微球的吞噬能力最强. 以上结果表明，本研究成功制备了鼠抗尼罗罗非鱼IgM单抗工具，并利用其IgM单抗发现IgM+ B细胞具有吞噬能力且其吞噬能力存在组织差异性，表明IgM+ B细胞在先天性免疫中可能发挥重要作用.
- 尼罗罗非鱼 /
- IgM /
- 单克隆抗体 /
- IgM+B细胞 /
- 吞噬能力
Abstract: In this study, the TNP-specific IgM was purified from Nile tilapia (Oreochromis niloticus) by affinity chromatography and immunized the Babl/c mice for monoclonal antibody (McAb) preparation. The splenocytes from immunized mice were fused with myeloma cells (SP2/0) by hybridoma technique. The positive cells, secreting anti-tilapia IgM antibodies, were screened by enzyme-linked immunosorbent assay (ELISA), Western blotting and flow cytometry. Two hybridomas were identified, named 3B3 and 9D12, and both monoclonal antibodies were prepared. The titers of purified McAb, 3B3 and 9D12 at the concentration of 1 mg/mL, were 740,741 and 359,712 units/mL, respectively. Utilizing the McAb, confocal laser microscopy detection showed that membrane IgM (mIgM) was located on the surface of B cell membrane. Distribution analysis of mIgM+ cell in immune tissues by flow cytometry revealed its tissue-dependent, with the maximum in peripheral blood (PBL) about 37.6%, followed by the spleen (SPL) 33.7%, the anterior kidney (AK) for about 23.9%, and the posterior kidney about 20%. The phagocytosis of mIgM+ B cells uptaking fluorescent microspheres showed that the phagocytic ability of mIgM+ B cells was higher on 0.5 m beads than for 1 m beads. The phagocytosis of mIgM+ B cells varied from immune tissues, with the uptaking 0.5 m beads highest in PBL but AK occupying the maximum for 1 m beads. Taken together, the anti-Nile tilapia IgM McAb was successfully prepared, and the mIgM+ B cells had the tissue-dependent phagocytic activity, indicating that mIgM+ B cells might play an important role in innate immune response.
- Nile tilapia /
- IgM /
- monoclonal antibody /
- IgM+B cells /
- phagocytosis

HTML全文

设 $\frac{1}{p}+\frac{1}{q}=1 \quad(p>1)$ ，α, β ∈ $\mathbb{R}$ ，K(x, y)非负可测，若 $L_{p}^{\alpha}(0, +\infty)=\left\{f(x) \geqslant 0:\|f\|_{p, \alpha}=\left(\int_{0}^{+\infty} x^{\alpha} f^{p}(x) \mathrm{d} x\right)^{1 /p} < +\infty\right\}$ , 则称不等式

$\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant M\|f\|_{p, \alpha}\|g\|_{q, \beta}$

为Hilbert型积分不等式. 由于此类不等式与积分算子T：

$T(f)(y)=\int_{0}^{+\infty} K(x, y) f(x) \mathrm{d} x$

有密切的联系，故而Hilbert型积分不等式对于研究算子T的有界性与算子范数有重要意义.

1991年，XU和GAO^[1]首次提出了研究Hilbert型不等式的权系数方法. 该方法的核心是：引入2个搭配参数a、b，利用Hölder不等式，可得到如下形式的不等式：

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\quad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)\left(\int_{0}^{+\infty} x^{\alpha(a, b)} f^{p}(x) \mathrm{d} x\right)^{1 / p} \times \\ &\quad\left(\int_{0}^{+\infty} y^{\beta(a, b)} g^{q}(y) \mathrm{d} x\right)^{1 / q}. \end{aligned}$

(1)

一般地，随意选取的搭配参数a、b并不能使式(1)的常数因子W₁^1/p(b, p)W₂^1/q(a, q)最佳. 已有的相关研究^[2-13]基本上都是凭借丰富的经验和娴熟的分析技巧选取适当的搭配参数a、b，从而获得最佳的Hilbert型不等式.

若选取的搭配参数a、b能够使式(1)的常数因子最佳，则称其为适配参数或适配数. 文献[14]曾讨论了齐次核的Hilbert型级数不等式的适配参数问题，本文将对拟齐次核的Hilbert型积分不等式讨论搭配参数a、b成为适配数的充分必要条件，并讨论其应用.

1. 预备知识

设G(u, v)是λ阶齐次函数，λ₁λ₂>0，则称K(x, y)=G(x^λ₁, y^λ₂)为拟齐次函数. 显然K(x, y)为拟齐次函数等价于：对t>0，有

$\begin{gathered} K(t x, y)=t^{\lambda_{1} \lambda} K\left(x, t^{-\lambda_{1} / \lambda_{2}} y\right), \\ K(x, t y)=t^{\lambda_{2} \lambda} K\left(t^{-\lambda_{2} / \lambda_{1}} x, y\right). \end{gathered}$

下面给出本文证明过程中所需的引理.

引理1 设1/p+1/q=1 (p>1)，a, b, λ∈ $\mathbb{R}$ ，λ₁λ₂>0，G(u, v)是λ阶齐次非负函数，K(x, y)=G(x^λ₁, y^λ₂)，aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，记

$\begin{gathered} W_{1}(b, p)=\int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t, \\ W_{2}(a, q)=\int_{0}^{+\infty} K(t, 1) t^{-a q} \mathrm{~d} t, \end{gathered}$

则W₁(b, p)/λ₁=W₂(a, q)/λ₂，且

$\begin{aligned} &\omega_{1}(b, p, x)=\int_{0}^{+\infty} K(x, y) y^{-b p} \mathrm{~d} y=x^{\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} W_{1}(b, p), \\ &\omega_{2}(a, q, y)=\int_{0}^{+\infty} K(x, y) x^{-a q} \mathrm{~d} x=y^{\lambda_{2}\left(\lambda-a q / \lambda_{1}+1 / \lambda_{1}\right)} W_{2}(a, q) . \end{aligned}$

证明由aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，可得- λ₁λ + λ₁bp/λ₂-λ₁/λ₂－1=－aq. 则有

$\begin{aligned} &W_{1}(b, p)=\int_{0}^{+\infty} t^{\lambda_{2} \lambda} K\left(t^{-\lambda_{2} / \lambda_{1}}, 1\right) t^{-b{p}} \mathrm{~d} t= \\ &\ \ \ \ \ \ \frac{\lambda_{1}}{\lambda_{2}} \int_{0}^{+\infty} K(u, 1) u^{-\lambda_{1} \lambda+\lambda_{1} b p / \lambda_{2}-\lambda_{1} / \lambda_{2}-1} \mathrm{~d} u= \\ &\ \ \ \ \ \ \frac{\lambda_{1}}{\lambda_{2}} \int_{0}^{+\infty} K(u, 1) u^{-a q} \mathrm{~d} u=\frac{\lambda_{1}}{\lambda_{2}} W_{2}(a, q), \end{aligned}$

故W₁(b, p)/λ₁=W₂(a, q)/λ₂.

作变换y=x^λ₁/λ₂t，有

$\begin{aligned} &\omega_{1}(b, p, x)=\int_{0}^{+\infty} x^{\lambda_{1} \lambda} K\left(1, x^{-\lambda_{1} / \lambda_{2}} y\right) y^{-b p} \mathrm{~d} y= \\ &\ \ \ \ \ \ x^{\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t=\\ &\ \ \ \ \ \ x^{\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} W_{1}(b, p). \end{aligned}$

同理可证ω₂(a, q, y)=y^{λ₂(λ－aq/λ₁+1/λ₁)}W₂(a, q). 证毕.

2. 适配参数的充分必要条件

定理1 设1/p+1/q=1 (p>1)，a, b, λ∈ $\mathbb{R}$ ，λ₁λ₂>0，G(u, v)是λ阶齐次非负可测函数，K(x, y)=G(x^λ₁, y^λ₂)，W₁(b, p)与W₂(a, q)如引理1所定义. 那么

(1) 若 $\alpha=\lambda_{1}\left[\lambda+\frac{1}{\lambda_{2}}+p\left(\frac{a}{\lambda_{1}}-\frac{b}{\lambda_{2}}\right)\right], \beta=\lambda_{2}\left[\lambda+\frac{1}{\lambda_{1}}+\right.$ $\left.p\left(\frac{b}{\lambda_{2}}-\frac{a}{\lambda_{1}}\right)\right]$ , 则有

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\quad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)\|f\|_{p, \alpha}\|g\|_{q, \beta}, \end{aligned}$

(2)

其中, $f(x) \in L_{p}^{\alpha}(0, +\infty), g(y) \in L_{q}^{\beta}(0, +\infty)$ .

(2) 式(2)中的常数因子W₁^1/p(b, p)W₂^1/q(a, q)是最佳的，当且仅当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，W₁(b, p)和W₂(a, q)都收敛. 当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ时，式(2)化为

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\ \ \ \ \ \ \ \ \frac{W_{0}}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\|f\|_{p, a p q-1}\|g\|_{q, b p q-1}, \end{aligned}$

(3)

其中, W₀=|λ₁|W₂(a, q)= |λ₂|W₁(b, p).

证明 (i)选择a、b为搭配参数. 根据Hölder不等式和引理1，利用权系数方法，有

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y= \\ &\qquad \int_{0}^{+\infty} \int_{0}^{+\infty}\left(\frac{x^{a}}{y^{b}} f(x)\right)\left(\frac{y^{b}}{x^{a}} g(y)\right) K(x, y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\qquad \left(\int_{0}^{+\infty} \int_{0}^{+\infty} \frac{x^{a p}}{y^{b p}} f^{p}(x) K(x, y) \mathrm{d} x \mathrm{d} y\right)^{1 / p} \times \\ &\qquad \left(\int_{0}^{+\infty} \int_{0}^{+\infty} \frac{y^{b q}}{x^{a q}} g^{q}(y) K(x, y) \mathrm{d} x \mathrm{d} y\right)^{1 / q}=\\ &\qquad \left(\int_{0}^{+\infty} x^{a p} f^{p}(x) \omega_{1}(b, p, x) \mathrm{d} x\right)^{1 / p} \times \\ &\qquad \left(\int_{0}^{+\infty} y^{b q} g^{q}(y) \omega_{2}(a, q, y) \mathrm{d} y\right)^{1 / q}= \\ &\qquad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q) \times \\ &\qquad \left(\int_{0}^{+\infty} x^{a p+\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} f^{p}(x) \mathrm{d} x\right)^{1 / p} \times \\ &\qquad \left(\int_{0}^{+\infty} y^{b q+\lambda_{2}\left(\lambda-a q / \lambda_{1}+1 / \lambda_{1}\right)} g^{q}(y) \mathrm{d} x\right)^{1 / q}= \\ &\qquad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)\|f\|_{p, \alpha}\|g\|_{q, \beta}, \end{aligned}$

故式(2)成立.

(ii) 充分性：设aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，W₁(b, p)和W₂(a, q)收敛. 由引理1，有W₁(b, p)/λ₁=W₂(a, q)/λ₂，故

$W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} W_{1}(b, p)=\frac{W_{0}}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}},$

且α=apq－1，β=bpq－1，于是式(2)可化为式(3).

设式(3)的最佳常数因子为M₀，则M₀≤W₀/(|λ₁^1/q|λ₂|^1/p)，且用M₀取代式(3)中的常数因子后，式(3)仍然成立.

取充分小的ε>0及δ>0，令

$\begin{aligned} f(x) =&\left\{\begin{array}{l} x^{\left(-a p q-\left|\lambda_{1}\right| \varepsilon\right) / p}\quad (x \geqslant 1), \\ 0 \quad (0<x<1); \end{array} \right.\\ g(y) = &\begin{cases}y^{\left(-b p q-\left|\lambda_{2}\right| \varepsilon\right) / q}\quad (y \geqslant \delta) ,\\ 0 \quad (0<y<\delta).\end{cases} \end{aligned}$

则

$\begin{array}{c} \begin{aligned} &\|f\|_{p, a p q-1}\|g\|_{q, b p q-1}= \\ &\quad\left(\int_{1}^{+\infty} x^{-1-\left|\lambda_{1}\right| \varepsilon} \mathrm{d} x\right)^{1 / p}\left(\int_{\delta}^{+\infty} y^{-1-\left|\lambda_{2}\right| \varepsilon} \mathrm{d} y\right)^{1 / q}= \\ &\quad\left(\frac{1}{\left|\lambda_{1} \varepsilon\right|}\right)^{1 / p}\left(\frac{1}{\left|\lambda_{2}\right| \varepsilon} \delta^{-\left|\lambda_{2}\right| \varepsilon}\right)^{1 / q}= \\ &\quad\frac{1}{\varepsilon\left|\lambda_{1}\right|^{1 / p}\left|\lambda_{2}\right|^{1 / q}} \delta^{-\left|\lambda_{2}\right| \varepsilon / q}, \end{aligned}\\ \begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y= \\ &\quad\int_{1}^{+\infty} x^{-a q-\left|\lambda_{1}\right| \varepsilon / p}\left(\int_{\delta}^{+\infty} y^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(x, y) \mathrm{d} y\right) \mathrm{d} x= \\ &\quad\int_{1}^{+\infty} x^{-a q-\left|\lambda_{1}\right| \varepsilon / p+\lambda \lambda_{1}}\left(\int_{\delta}^{+\infty} y^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K\left(1, x^{-\lambda_{1} / \lambda_{2}} y\right) \mathrm{d} y\right) \mathrm{d} x= \\ &\quad\int_{1}^{+\infty} x^{-1-\left|\lambda_{1}\right| \varepsilon}\left(\int_{x^{-\lambda_{1} / \lambda_{2}} \delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t\right) \mathrm{d} x \geqslant \\ &\quad\int_{1}^{+\infty} x^{-1-\left|\lambda_{1}\right| \varepsilon}\left(\int_{\delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t\right) \mathrm{d} x= \\ &\quad\frac{1}{\left|\lambda_{1}\right| \varepsilon} \int_{\delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t. \end{aligned} \end{array}$

于是

$\frac{1}{\left|\lambda_{1}\right|} \int_{\delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t \leqslant \frac{M_{0}}{\left|\lambda_{1}\right|^{1 / p}\left|\lambda_{2}\right|^{1 / q}} \delta^{-\left|\lambda_{2}\right| \varepsilon / q}.$

先令ε→0⁺，再令δ→0⁺，得

$W_{1}(b, p)=\int_{0}^{+\infty} t^{-b p} K(1, t) \mathrm{d} t \leqslant \frac{M_{0}}{\left|\lambda_{1}\right|^{1 / p}\left|\lambda_{2}\right|^{1 / q}}.$

再根据引理1，可得到W₀/(|λ₁|^1/q|λ₂|^1/p)≤M₀. 所以式(3)的最佳常数因子M₀=W₀/(|λ₁|^1/q|λ₂|^1/p).

必要性：设式(2)的常数因子W₁^1/p(b, p)W₂^1/q(a, q)是最佳的，则W₁(b, p)和W₂(a, q)是收敛的. 下证aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ.

记 $\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p-\left(\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}+\lambda\right)=c, a_{1}=a-\frac{\lambda_{1} c}{p q}, b_{1}=$ $b-\frac{\lambda_{2} c}{p q}$ ，则

$\begin{gathered} \alpha=\lambda_{1}\left[\lambda+\frac{1}{\lambda_{2}}+p\left(\frac{a_{1}}{\lambda_{1}}-\frac{b_{1}}{\lambda_{2}}\right)\right]=\alpha_{1}, \\ \beta=\lambda_{2}\left[\lambda+\frac{1}{\lambda_{1}}+p\left(\frac{b_{1}}{\lambda_{2}}-\frac{a_{1}}{\lambda_{1}}\right)\right]=\beta_{1}, \\ W_{2}(a, q)=\int_{0}^{+\infty} K(t, 1) t^{-a q} \mathrm{~d} t=\frac{\lambda_{2}}{\lambda_{1}} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t . \end{gathered}$

于是可知式(2)等价于

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\quad W_{1}^{1 / p}(b, p)\left(\frac{\lambda_{2}}{\lambda_{1}} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q}\|f\|_{p, \alpha_{1}}\|g\|_{q, \beta_{1}} . \end{aligned}$

又经计算有a₁q/λ₁+b₁p/λ₂=1/λ₁+1/λ₂+ λ，α₁=a₁pq－1，β₁=b₁pq－1，故式(2)进一步等价于

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\qquad W_{1}^{1 / p}(b, p)\left(\frac{\lambda_{2}}{\lambda_{1}} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q} \times \\ &\qquad\|f\|_{p, a_{1} p q-1}\|g\|_{q, b_{1} p q-1} . \end{aligned}$

(4)

根据假设，式(4)的最佳常数因子是W₁^1/p(b, p)× ${\left({\frac{{{\lambda _2}}}{{{\lambda _1}}}\int_0^{ + \infty } K (1, t){t^{ - bp + {\lambda _2}c}}{\rm{d}}t} \right)^{1/q}}$ . 又由 $\frac{1}{{{\lambda _1}}}{a_1}q + \frac{1}{{{\lambda _2}}}{b_1}p = \frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}} + \lambda$ 及充分性的证明，可知式(4)的最佳常数因子为

$\begin{gathered} \frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left(\left|\lambda_{2}\right| \int_{0}^{+\infty} K(1, t) t^{-b_{1} p} \mathrm{~d} t\right)= \\ \left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c / q} \mathrm{~d} t \end{gathered}$

于是得到

$\begin{aligned} &\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c / q} \mathrm{~d} t= \\ &\quad W_{1}^{1 / p}(b, p)\left(\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q}. \end{aligned}$

(5)

对于1和t^λ₂c/q，应用Hölder不等式，有

$\begin{aligned} &\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c / q} \mathrm{~d} t=\int_{0}^{+\infty} t^{\lambda_{2} c / q} K(1, t) t^{-b p} \mathrm{~d} t \leqslant \\ &\qquad\left(\int_{0}^{+\infty} 1^{p} K(1, t) t^{-b p} \mathrm{~d} t\right)^{1 / p}\left(\int_{0}^{+\infty} t^{\lambda_{2} c} K(1, t) t^{-b p} \mathrm{~d} t\right)^{1 / q}= \\ &\qquad W_{1}^{1 / p}(b, p)\left(\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q}. \end{aligned}$

(6)

根据式(5)，可知式(6)取等号. 又根据Hölder不等式取等号的条件，可得t^λ₂c/q=常数，故c=0，即aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ₁. 证毕.

注1 定理1表明: 当且仅当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ时，搭配参数a、b是适配参数. 因此，只要选取a、b满足aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，就可以得到各种各样的具有最佳常数因子的Hilbert型积分不等式.

推论1 设1/p+1/q=1 (p>1)，λ₁λ₂>0，λ >0，1/r+1/s=1 (r>1)，α=p(1－ λλ₁/r)－1，β=q(1－ λλ₂/s)－1，则

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} \frac{f(x) g(y)}{\left(x^{\lambda_{1}}+y^{\lambda_{2}}\right)^{\lambda}} \mathrm{d} x \mathrm{d} y \leqslant \\ &\qquad\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}} \mathrm{~B}\left(\frac{\lambda}{r}, \frac{\lambda}{s}\right)\|f\|_{p, \alpha}\|g\|_{q, \beta}, \end{aligned}$

(7)

其中的常数因子是最佳的，f(x)∈L_p^α(0, +∞)，g(y)∈L_q^β(0, +∞).

证明记K(x, y)=G(x^λ₁, y^λ₂)=1/(x^λ₁+y^λ₂)^λ，则G(u, v)是－λ阶齐次非负函数. 选取搭配参数a= $\frac{1}{q}\left(1-\frac{\lambda \lambda_{1}}{r}\right), b=\frac{1}{p}\left(1-\frac{\lambda \lambda_{2}}{s}\right)$ , 可得

$\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p=\frac{1}{\lambda_{1}}\left(1-\frac{\lambda \lambda_{1}}{r}\right)+\frac{1}{\lambda_{2}}\left(1-\frac{\lambda \lambda_{2}}{s}\right)=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}-\lambda,$

故a、b是适配参数. 又因为apq－1=p(1－ λλ₁/r)－1=α，bpq－1=q(1－ λλ₂/s)－1=β，且

$\begin{aligned} W_{0}=&\left|\lambda_{2}\right| W_{1}(b, p)=\left|\lambda_{2}\right| \int_{0}^{+\infty} \frac{1}{\left(1+t^{\lambda_{2}}\right)^{\lambda}} t^{\lambda \lambda_{2} / s-1} \mathrm{~d} t=\\ & \int_{0}^{+\infty} \frac{1}{(1+u)^{\lambda}} u^{\lambda / s-1} \mathrm{~d} u=\mathrm{B}\left(\frac{\lambda}{s}, \lambda-\frac{\lambda}{s}\right)=\mathrm{B}\left(\frac{\lambda}{r}, \frac{\lambda}{s}\right). \end{aligned}$

根据定理1，式(7)成立，且其常数因子是最佳的. 证毕.

3. 在求积分算子范数中的应用

根据Hilbert型不等式与相应积分算子的关系理论，由定理1可得如下定理.

定理2 设1/p+1/q=1 (p>1)，a, b, λ ∈ $\mathbb{R}$ ，λ₁λ₂>0，α=apq－1，β=bpq－1，G(u, v)是λ阶齐次非负可测函数，K(x, y)=G(x^λ₁, y^λ₂)，且

$\begin{aligned} &W_{1}(b, p)=\int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t<+\infty, \\ &W_{2}(a, q)=\int_{0}^{+\infty} K(t, 1) t^{-a q} \mathrm{~d} t<+\infty, \end{aligned}$

则当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ时，积分算子T：

$T(f)(y)=\int_{0}^{+\infty} K(x, y) f(x) \mathrm{d} x, f(x) \in L_{p}^{\alpha}(0,+\infty)$

是从L_p^α(0, +∞)到L_p^β(1－p)(0, +∞)的有界算子，且T的算子范数为

$\|T\|=\frac{\left|\lambda_{2}\right| W_{1}(b, p)}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t.$

推论2 设1/p+1/q=1 (p>1)，λ₁λ₂>0，－1 < λ < min{1±4/λ₁, 1±4/λ₂}，α=p[1+ λ₁(λ －1)/2]－1，β=p[1+ λ₂(λ －1)/2]－1，则积分算子T：

$T(f)(y)=\int_{0}^{+\infty} \frac{\left|x^{\lambda_{1}}-y^{\lambda_{2}}\right|^{\lambda}}{\max \left\{x^{\lambda_{1}}, y^{\lambda_{2}}\right\}} f(x) \mathrm{d} x, f(x) \in L_{p}^{\alpha}(0,+\infty)$

是从L_p^α(0, +∞)到L_p^β(1－p)(0, +∞)的有界算子，且T的算子范数为

$\begin{aligned} &\|T\|=\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\mathrm{~B}\left(\lambda+1, \frac{1-\lambda}{2}-\frac{2}{\lambda_{2}}\right)+\right. \\ &\ \ \ \ \ \ \left.\mathrm{B}\left(\lambda+1, \frac{1-\lambda}{2}+\frac{2}{\lambda_{2}}\right)\right] . \end{aligned}$

证明记K(x, y)=G(x^λ₁, y^λ₂)= |x^λ₁－y^λ₂|^λ/max{x^λ₁, y^λ₂}，则G(u, v)是λ －1阶齐次函数. 取a= $\frac{1}{q}\left[1+\frac{\lambda_{1}}{2}(\lambda-1)\right], b=\frac{1}{p}\left[1+\frac{\lambda_{2}}{2}(\lambda-1)\right]$ ，则

$\begin{aligned} &\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p=\frac{1}{\lambda_{1}}\left[1+\frac{\lambda_{1}}{2}(\lambda-1)\right]+\frac{1}{\lambda_{2}}\left[1+\frac{\lambda_{2}}{2}(\lambda-1)\right]= \\ &\ \ \ \ \ \ \frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}+\lambda-1, \end{aligned}$

故a、b是适配参数. 又 $a p q-1=p\left[1+\frac{\lambda_{1}}{2}(\lambda-1)\right]-1=\alpha$ ， $b p q-1=q\left[1+\frac{\lambda_{2}}{2}(\lambda-1)\right]-1=\beta$ . 则

$\begin{aligned} &\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t= \\ &\qquad\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} \frac{\left|1-t^{\lambda_{2}}\right|^{\lambda}}{\max \left\{1, t^{\lambda_{2}}\right\}} t^{-\left[1+\lambda_{2}(\lambda-1) / 2\right]} \mathrm{d} t= \\ &\qquad\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\mathrm{~B}\left(\lambda+1, \frac{1-\lambda}{2}-\frac{2}{\lambda_{2}}\right)+\right. \\ &\qquad\left.\mathrm{B}\left(\lambda+1, \frac{1-\lambda}{2}+\frac{2}{\lambda_{2}}\right)\right]<+\infty . \end{aligned}$

根据定理2，知推论2成立. 证毕.

推论3 设1/p+1/q=1 (p>1)，1/r+1/s=1 (r>1)，λ₁λ₂>0，α=p(1－ λ₁/r)－1，β=q(1－ λ₂/s)－1. 则积分算子T：

$T(f)(y)=\int_{0}^{+\infty} \frac{\ln \left(x^{\lambda_{1}} / y^{\lambda_{2}}\right)}{x^{\lambda_{1}}-y^{\lambda_{2}}} f(x) \mathrm{d} x, f(x) \in L_{p}^{\alpha}(0,+\infty)$

是从L_p^α(0, +∞)到L_p^β(1－p)(0, +∞)的有界算子，且T的算子范数为

$\|T\|=\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\zeta\left(2, \frac{1}{r}\right)+\zeta\left(2, \frac{1}{s}\right)\right],$

其中ζ(t, a)是Riemann函数.

证明记

$K(x, y)=G\left(x^{\lambda_{1}}, y^{\lambda_{2}}\right)=\frac{\ln \left(x^{\lambda_{1}} / y^{\lambda_{2}}\right)}{x^{\lambda_{1}}-y^{\lambda_{2}}},$

则G(u, v)是－1阶齐次非负函数.

取搭配参数 $a=\frac{1}{q}\left(1-\frac{\lambda_{1}}{r}\right), b=\frac{1}{p}\left(1-\frac{\lambda_{2}}{s}\right)$ ，则

$\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p=\frac{1}{\lambda_{1}}\left(1-\frac{\lambda_{1}}{r}\right)+\frac{1}{\lambda_{2}}\left(1-\frac{\lambda_{2}}{s}\right)=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}-1,$

故a、b是适配参数. 又apq－1=p(1－ λ₁/r)－1=α，bpq－1=q(1－ λ₂/s)－1=β，且

$\begin{gathered} \left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} \frac{\ln \left(t^{-\lambda_{2}}\right)}{1-t^{\lambda_{2}}} t^{\lambda_{2} / s-1} \mathrm{~d} t= \\ \frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\zeta\left(2, \frac{1}{r}\right)+\zeta\left(2, \frac{1}{s}\right)\right]<+\infty \end{gathered}$

根据定理2，知推论3成立. 证毕.

参考文献(1)