超声微波辅助共沉淀法制备Li1.2Ni0.2Mn0.6O2正极材料及其性能

赵瑞瑞; 梁家星; 杨子莲; 梁昌铖; 关熊聪; 高爱梅; 陈红雨

doi:10.6054/j.jscnun.2017104

摘要: 通过超声微波共沉淀法制备一系列纳米正极材料Li1.2Ni0.2Mn0.6O2。通过X射线衍射，扫描电镜，X射线光电子能谱和电化学方法研究超声时间对合成材料的影响。结果表明反应时间为2h时，材料表现出最优异的电化学性能，其在0.1C和2C倍率下的容量分别为265mAh.g-1 和180mAh.g-1。材料优异的电化学性能取决于其均一的颗粒粒径，理想的元素分布和高反应活性的氧化还原电对。超声微波系统可以在富锂材料的实际生产中起到良好的辅助作用，它使用方便，且能够节约时间。

关键词:

共沉淀

Abstract: A series of nanocrystalline lithium-rich cathode materials (Li1.2Ni0.2Mn0.6O2) have been prepared via an ultrasonic/microwave-assisted co-precipitation method. The effects of the ultrasonic time are emphasis investigated by using X-ray diffraction, scanning electron microscopy, X-ray photoelectron spectroscopy and electrochemical measurements. The optimum reaction time is 2h, while the sample can exhibit the best electrochemical properties with an initial discharge capacity of 265mAh.g-1 at 0.1C and 180mAh-1 at 2C after 90 cycles, respectively. The superior electrochemical performance of this material can be attributed to the uniform particles, desired element distribution and high activities of the redox couples in the bulk material. This study also suggests that the ultrasonic/microwave system can be a good assistant in the practical Li-rich material production, while it can be easy to use and time-saving.

Keywords:

Co-precipitation

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.

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