管状铜掺杂ZnO双晶结构材料的形貌控制合成及荧光性质研究

梁晓韵; 梁威威; 李旭仙; 许海珊; 万霞; 铁绍龙

doi:10.6054/j.jscnun.2014.12.009

摘要: 采用直接沉淀法合成了具有六棱柱体形貌的Cu2+掺杂ZnO双晶结构材料，研究了Cu2+的存在对ZnO双晶的形成及形貌的影响，发现溶液中存在的Cu2+浓度越高，获得的Zn(Cu)O材料粒径越大，形貌从细长棒形逐步变为短粗六方柱体，长径比也从10:1变到1.2:1。采用简单的碱腐蚀法获得了管状结构的Cu2+掺杂Zn(Cu)O材料，并探讨了管状结构的形成机理。Cu掺杂使得Zn(Cu)O样品的绿光发射由550nm蓝移至520nm附近，且强度大幅增加。形成管状结构使绿光发射进一步增强，该发射由Cu2+掺杂引起的样品内部的Cu2+与Cu+之间的相互转变引起。

关键词:

光致荧光

Abstract: The Cu-doped ZnO twinned structure materials with hexagonal prism morphology were synthesized by direct precipitation and the effects of Cu2+ ion on the formation of twinned structure and morphology were investigated. With the increase of Cu2+ concentration in reaction solution the particle sizes of as-prepared ZnO materials increased following the change of morphologies from slender rod to stumpy hexagonal prism and the decrease of length-diameter ratios from 10:1 to 1.2:1. The tubular Cu-doped ZnO materials were obtained via corroding process by concentrated aqueous alkali and the formation mechanism was proposed. The photoluminescence properties of Zn(Cu)O materials were studied and the Cu-doped ZnO showed broad and strong green emission at 520 nm, attributed to the transition of copper ion between Cu2+ and Cu+ in interior of ZnO crystal. The formation of tubular construction enhance the intensity of green emission.

Keywords:

photoluminescence

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.

[1] Klingshirn C, ZnO: Material, Physics and Applications [J].ChemPhysChem, 2007, 8:782–803
[2] Jin D H, Kim D, Seo Y, et al, Morphology-controlled synthesis of ZnO crystals with twinned structures and the morphology dependence of their antibacterial activities[J].Materials Letters, 2014, 115:205–207
[3] Kim Y, Seshadri R, Optical Properties of Cation-Substituted Zinc Oxide[J].Inorganic Chemistry, 2008, 47: 8437–8443
[4] Ferhat M, Zaoui A, Ahuja R, Magnetism and band gap narrowing in Cu-doped ZnO[J].Applied Physics Letters, 2009, 94, 142502–1-3
[5] 邢伯阳, 牛世峰, Cu掺杂ZnO纳米粒子的光学性能及铁磁性[J].材料科学与工程学报, 2012, 30(3):333–336
[6] Mahanti M, Basak D, Cu/ZnO nanorods′ hybrid showing enhanced photoluminescence properties due to surface plasmon resonance[J].Journal of Luminescence, 2014, 145:19–24
[7] Hu H M, Deng C H, Huang X H, Hydrothermal growth of center-hollow multigonal star-shaped ZnO architectures assembled by hexagonal conic nanotubes[J].Materials Chemistry and Physics, 2010, 121: 364–369
[8] 黄明，万霞，铁绍龙，管状氧化锌的碱溶法合成及光催化降解亚甲基蓝的性能[J].华南师范大学学报(自然科学版)，2014, 46（5）：0-0(即将正式出版)
[9] 李汶军, 施尔畏, 负离子配位多面体生长基元的理论模型与晶粒形貌[J].人工晶体学报, 1999, 28(2): 117–125
[10] Li W J, Shi E W, Zhong W Z, et al, Growth mechanism and growth habit of oxide crystals[J].Journal of Crystal Growth, 1999, 203:186–196
[11]王步国, 仲维卓, 施尔畏, 等.ZnO晶体的极性生长习性与双晶的形成机理[J].人工晶体学报, 1997, 26(2):102-107
[12] Chu D, Masuda Y, Ohji T, et al, Formation and photocatalytic application of ZnO nanotubes using aqueous solution[J].Langmuir, 2010, 26(4):2811–2815
[13] Gladyshchuk A A, Gurskii A L, Nikitenko V A, et al, Luminescence of ZnO monocrystals at excitation by streamer discharges and laser radiation[J].Journal of Luminescence, 1988, 42(1):49–55
[14] Xu P S, Sun Y M, Shi C S, et al, Electronic structure of ZnO and its defects[J].Science in China (A), 2001, 44(9):1174–1181
[15]. Vanheusden K, Seager C H, Warren W L, et al, Correlation between photoluminescence and oxygen vacancies in ZnO phosphors[J], Applied Physics Letters, 1996, 68:403–405
[16] Dijken A V, Meulenkamp E A, Vanmaekelbergh D, et al, The Kinetics of the Radiative and Nonradiative Processes in Nanocrystalline ZnO Particles upon Photoexcitation[J].Journal Physics Chemistry B, 2000, 104:1715–1723
[17]. Djuri?i? A B, Leung Y H, Tam K H, Green, yellow, and orange defect emission from ZnO nanostructures:Influence of excitation wavelength[J], Applied Physics Letters, 2006, 88:103107-1-3
[18] Liang G F, Hu L W, Feng W, et al, Enhanced photocatalytic performance of ferromagnetic ZnO:Cu hierarchical microstructures[J].Applied Surface Science, 2014, 296:158–162
[19] Garces N Y, Wang L, Bai L, et al, Role of copper in the green luminescence from ZnO crystals[J].Applied Physics Letters, 2002, 81:622–624

期刊类型引用(1)

1.

肖世伟，李承凯，杨美娜，冯祥虎，孙国萃，杜军. MIPS指令集的流水线CPU模型机设计. 单片机与嵌入式系统应用. 2023(02): 15-18 .

百度学术