设$ f, g\geq 0 $,使得

$ 0<\int_0^\infty f^2(x){\rm d}x<\infty, 0<\int_0^\infty g^2(y){\rm d}y<\infty, $

则有如下著名的Hilbert积分不等式^[1]

(1.1) $ \begin{equation} \int_{0}^\infty\int_{0}^\infty \frac{f(x)g(y)}{x+y}{\rm d}x{\rm d}y<\pi\bigg\{\int_0^\infty f^2(x){\rm d}x\bigg\}^{\frac{1}{2}}\bigg\{\int_0^\infty g^2(y){\rm d}y\bigg\}^{\frac{1}{2}}, \end{equation} $

这里的常数因子$ \pi $是最佳值.不等式(1.1)在分析学和偏微分方程理论中有着重要应用^{[1, 2]}.近年来,杨必成等对式(1.1)进行了一系列的参量化、抽象化和系统化研究,取得了众多成果^[3-13]. 2011年,杨给出了一个具有指数函数核Hilbert型积分不等式^[14]:

(1.2) $ \begin{equation} \int_{0}^\infty\int_{0}^\infty e^{-xy}f(x)g(y){\rm d}x{\rm d}y<\sqrt{\pi}\bigg\{\int_0^\infty f^2(x){\rm d}x\bigg\}^{\frac{1}{2}}\bigg\{\int_0^\infty g^2(y){\rm d}y\bigg\}^{\frac{1}{2}}, \end{equation} $

这里的常数因子$ \sqrt{\pi} $是最佳值.本文引入多个参数$ (\lambda_1, \lambda_2, \lambda_3) $,利用“基于Hardy插值难题”的权函数方法和实分析技巧,借助一些特殊函数刻画不等式中的常数因子,得到了具有混合核$ \frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}}{(\max\{1, xy\})^{\lambda_3}} $的Hilbert型积分不等式和其等价式,证明了它们的常数因子是最佳值,并利用算子理论给出了所得不等式的算子范数表达式.

特殊函数是指实与复分析、数学物理和数论等学科中具有基础理论与重要应用价值的函数.特殊函数论是数学研究的一个重要分支.文献[15-22]对工程技术领域中出现的一些重要特殊函数从定义、性质和主要结论及其应用等方面作了较为详细和精辟的阐述.本文研究涉及了如下一些特殊函数:

1)设Re$ (s)>0 $, Gamma函数$ \Gamma(s) $和不完全Gamma函数$ \Gamma(s, \alpha) $的定义分别为^[15-22]

(2.1) $ \begin{equation} \Gamma(s): = \int_{0}^\infty <italic>e</italic>^{-t}t^{s-1}{\rm d}t. \end{equation} $

(2.2) $ \begin{equation} \Gamma(s, \alpha): = \int_{\alpha}^\infty <italic>e</italic>^{-t}t^{s-1}{\rm d}t. \end{equation} $

2) Beta函数$ B(u, v)(u, v>0) $的定义为^[15-16]

(2.3) $ \begin{equation} B(u, v): = \int_0^1(1-t)^{u-1}t^{v-1}{\rm d}t = \frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}. \end{equation} $

3)合流超几何函数(又称Kummer函数) $ _1F_1(\alpha, \beta, z)(\alpha, \beta, z>0) $的定义为^[15]

(2.4) $ \begin{equation} _1F_1(\alpha, \beta, z): = \sum\limits_{n = 0}^\infty\frac{(\alpha)_n}{n!(\beta)_n}z^n = \frac{\Gamma(\beta)}{\Gamma(\alpha)}\sum\limits_{n = 0}^\infty\frac{z^n\Gamma(n+\alpha)}{n!\Gamma(\beta+n)}, \end{equation} $

记号$ (x)_n = x(x+1)\cdots(x+n-1) = \frac{\Gamma(x+n)}{\Gamma(x)}(x>0) $.由(2.4)式,当$ 0<\lambda_1<4-\lambda_3+\lambda_2 $时,有

(2.5) $ \begin{equation} _1F_1 \bigg(1, 2-\frac{\lambda_1-\lambda_2+\lambda_3}{2}, \lambda_1\bigg) = \Gamma\bigg(2-\frac{\lambda_1-\lambda_2+\lambda_3}{2}\bigg)\sum\limits_{n = 0}^\infty\frac{\lambda^n}{\Gamma(n+2-\frac{\lambda_1-\lambda+\lambda_3}{2})}. \end{equation} $

4) Whittaker函数$ M(k, m, z) $定义为^[15]

(2.6) $ \begin{equation} M(k, m, z): = z^{m+\frac{1}{2}}<italic>e</italic>^{-\frac{z}{2}}\ _1F_1(m-k+\frac{1}{2}, 2m+1, z). \end{equation} $

由(2.5)和(2.6)式,有

(2.7) $ \begin{eqnarray} &&M\bigg(-\frac{\lambda_1-\lambda_2+\lambda_3}{4}, -\frac{\lambda_1-\lambda_2+\lambda_3}{4}+\frac{1}{2}, \lambda_1\bigg) \\ & = &\lambda_1^{-\frac{\lambda_1-\lambda_2+\lambda_3}{4}+1}<italic>e</italic>^{-\frac{\lambda_1}{2}} \Gamma\bigg(2-\frac{\lambda_1-\lambda_2+\lambda_3}{2}\bigg)\sum\limits_{n = 0}^\infty\frac{\lambda_1^n}{\Gamma(n+2-\frac{\lambda_1-\lambda_2+\lambda_3}{2})}. \end{eqnarray} $

设$ 0<\lambda_1<2-\lambda_3+\lambda_2 $,由(2.3)和(2.7)式,有

(2.8) $ \begin{eqnarray} I_1& = &\int_0^1 <italic>e</italic>^{-\lambda_1 t}t^{-\frac{\lambda_1-\lambda_2+\lambda_3}{2}}{\rm d}t \\ & = &e^{-\lambda_1}\int_0^1 e^{\lambda(1- t)}t^{-\frac{\lambda_1-\lambda_2+\lambda_3}{2}}{\rm d}t \\ & = &e^{-\lambda_1}\sum\limits_{n = 0}^\infty\frac{\lambda_1^n}{n!}\int_0^1(1-t)^nt^{-\frac{\lambda_1-\lambda_2+\lambda_3}{2}}{\rm d}t \\ & = &e^{-\lambda_1}\sum\limits_{n = 0}^\infty\frac{\lambda_1^n}{n!}\frac{\Gamma(n+1)\Gamma(1-\frac{\lambda_1-\lambda_2+\lambda_3}{2})}{\Gamma(n+2-\frac{\lambda_1-\lambda_2+\lambda_3}{2})} \\ & = &\frac{\lambda_1^{\frac{\lambda_1-\lambda_2+\lambda_3}{4}-1}e^{-\frac{\lambda_1}{2}}\Gamma (1-\frac{\lambda_1-\lambda_2+\lambda_3}{2})}{\Gamma(2-\frac{\lambda_1-\lambda_2+\lambda_3}{2})} \\ &&\times \bigg[\lambda_1^{-\frac{\lambda_1-\lambda_2+\lambda_3}{4}+1}e^{-\frac{\lambda_1}{2}} \Gamma\bigg(2-\frac{\lambda_1-\lambda_2+\lambda_3}{2}\bigg)\sum\limits_{n = 0}^\infty\frac{\lambda_1^n}{\Gamma(n+2-\frac{\lambda_1-\lambda_2+\lambda_3}{2})}\bigg] \\ & = &\frac{\lambda_1^{\frac{\lambda_1-\lambda_2+\lambda_3}{4}-1}e^{-\frac{\lambda_1}{2}}}{1-\frac{\lambda_1-\lambda_2+\lambda_3}{2}} M\bigg(-\frac{\lambda_1-\lambda_2+\lambda_3}{4}, -\frac{\lambda_1-\lambda_2+\lambda_3}{4}+\frac{1}{2}, \lambda_1\bigg). \end{eqnarray} $

设$ 0<\lambda_1<2-\lambda_2-3\lambda_3 $,令$ \lambda_1 t = u $,由(2.2)式,则有

(2.9) $ \begin{equation} I_2 = \int_1^\infty <italic>e</italic>^{-\lambda_1 t}t^{-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}}{\rm d}t = \lambda_1^{\frac{\lambda_1+\lambda_2+3\lambda_3}{2}-1}\Gamma\bigg(1-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}, \lambda_1\bigg). \end{equation} $

引理2.1  设$ p>1, \frac{1}{p}+\frac{1}{q} = 1, 0<\lambda_1<2-\max\{\lambda_3-\lambda_2, \lambda_2+3\lambda_3\} $,定义如下权函数

$ \begin{eqnarray*} \omega(\lambda_1, \lambda_2, \lambda_3, x): = \int_0^\infty \frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}}{(\max\{1, xy\})^{\lambda_3}}\frac{y^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}}{x^{\frac{\lambda_1+\lambda_2+\lambda_3}{2}-1}}{\rm d}y, x\in(0, +\infty), \end{eqnarray*} $

$ \begin{eqnarray*} \omega(\lambda_1, \lambda_2, \lambda_3, y): = \int_0^\infty \frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}}{(\max\{1, xy\})^{\lambda_3}}\frac{x^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}}{y^{\frac{\lambda_1+\lambda_2+\lambda_3}{2}-1}}{\rm d}x, y\in(0, +\infty), \end{eqnarray*} $

则有

$ \omega(\lambda_1, \lambda_2, \lambda_3, x) = \omega(\lambda_1, \lambda_2, \lambda_3, y) = C(\lambda_1, \lambda_2, \lambda_3), $

其中

(2.10) $ \begin{eqnarray} C(\lambda_1, \lambda_2, \lambda_3) & = &\frac{\lambda_1^{\frac{\lambda_1-\lambda_2+\lambda_3}{4}-1}e^{-\frac{\lambda_1}{2}}}{1-\frac{\lambda_1-\lambda_2+\lambda_3}{2}} M\bigg(-\frac{\lambda_1-\lambda_2+\lambda_3}{4}, -\frac{\lambda_1-\lambda_2+\lambda_3}{4}+\frac{1}{2}, \lambda_1\bigg) \\ && +\lambda_1^{\frac{\lambda_1+\lambda_2+3\lambda_3}{2}-1}\Gamma\bigg(1-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}, \lambda_1\bigg). \end{eqnarray} $

证  令$ xy = t $,由(2.8)和(2.9)式,得到

$ \begin{eqnarray*} \omega(\lambda_1, \lambda_2, \lambda_3, x)&: = &\int_0^\infty \frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}}{(\max\{1, xy\})^{\lambda_3}}\frac{y^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}}{x^{\frac{\lambda_1+\lambda_2+\lambda_3}{2}-1}}{\rm d}y\\ & = &\int_0^\infty\frac{e^{-\lambda_1 t}\min\{1, t^{\lambda_2}\}}{\max\{1, t^{\lambda_3}\}}t^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}{\rm d}t\\ & = &\int_0^1 e^{-\lambda_1 t}t^{-\frac{\lambda_1-\lambda_2+\lambda_3}{2}}{\rm d}t+\int_1^\infty e^{-\lambda_1 t}t^{-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}}{\rm d}t\\ & = &I_1+I_2 = C(\lambda_1, \lambda_2, \lambda_3). \end{eqnarray*} $

同理可得$ \omega(\lambda_1, \lambda_2, \lambda_3, y) = C(\lambda_1, \lambda_2, \lambda_3) $.

引理2.2  设$ p>1, \frac{1}{p}+\frac{1}{q} = 1, 0<\lambda_1<2-\max\{\lambda_3-\lambda_2, \lambda_2+3\lambda_3\} $,且$ \varepsilon>0 $充分地小,定义如下两个实函数$ \tilde{f}(x), \tilde{g}(y) $:

$ \tilde{f}(x) = \left\{\begin{array}{ll}0, &x\in(0, 1)\\ x^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}-\frac{\varepsilon}{p}}, & x\in[1, \infty)\end{array} \right., \tilde{g}(y) = \left\{\begin{array}{ll}0, &y\in(1, \infty)\\ y^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}+\frac{\varepsilon}{q}}, &y\in(0, 1]\end{array} \right., $

则有

(2.11) $ \begin{equation} \tilde{J}\cdot\varepsilon = \bigg[\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}\tilde{f}^p(x){\rm d}x\bigg]^{\frac{1}{p}}\bigg[\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}\tilde{g}^q(y){\rm d}y\bigg]^{\frac{1}{q}}\cdot\varepsilon = 1, \end{equation} $

(2.12) $ \begin{eqnarray} \tilde{h}\cdot\varepsilon& = &\varepsilon\int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}\tilde{f}(x)\tilde{g}(y)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x{\rm d}y \\ &>&C(\lambda_1, \lambda_2, \lambda_3)(1-o(1)), \ \ \varepsilon\rightarrow 0^+. \end{eqnarray} $

证  容易得到

$ \begin{eqnarray*} \tilde{J}\cdot\varepsilon& = &\bigg[\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}\tilde{f}^p(x){\rm d}x\bigg]^{\frac{1}{p}}\bigg[\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}\tilde{g}^q(y){\rm d}y\bigg]^{\frac{1}{q}}\cdot\varepsilon\\ & = &\bigg[\int_1^\infty x^{-1-\varepsilon}{\rm d}x\bigg]^{\frac{1}{p}}\bigg[\int_0^1 y^{-1+\varepsilon}{\rm d}y\bigg]^{\frac{1}{q}}\cdot\varepsilon = 1. \end{eqnarray*} $

函数$ H(t): = e^{-\lambda_1 t}t^{1-\frac{\lambda_2+3\lambda_3}{2}} $在$ [x, \infty)(x\geq1) $内连续,且用洛比达(L'Hospital)法则^[23]可得$ { }\lim_{t\rightarrow\infty}H(t) = 0 $.因此存在一个常数$ T>0 $,使得$ H(u)\leq T $.令$ xy = t $,由Fubini's定理^[24], (2.8)和(2.9)式可得

$ \begin{eqnarray*} \tilde{h}\cdot\varepsilon& = &\varepsilon\int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}\tilde{f}(x)\tilde{g}(y)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x{\rm d}y\\ & = &\varepsilon\int_1^\infty x^{-1-\varepsilon}{\rm d}x\int_0^{x}\frac{e^{-\lambda_1 t}t^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}+\frac{\varepsilon}{q}}(\min\{1, t\})^{\lambda_2}}{(\max\{1, t\})^{\lambda_3}}{\rm d}t\\ & = &\varepsilon\int_1^\infty x^{-1-\varepsilon}{\rm d}x\bigg[\int_0^1e^{-\lambda_1t}t^{-\frac{\lambda_1-\lambda_2+\lambda_3}{2}+\frac{\varepsilon}{q}}{\rm d}t+\int_1^xe^{-\lambda_1t}t^{-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}+\frac{\varepsilon}{q}}{\rm d}t\bigg]\\ & = &\varepsilon\int_1^\infty x^{-1-\varepsilon}{\rm d}x\bigg[\int_0^1e^{-\lambda_1t}t^{-\frac{\lambda_1-\lambda_2+\lambda_3}{2}+\frac{\varepsilon}{q}}{\rm d}t +\int_1^\infty e^{-\lambda_1t}t^{-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}+\frac{\varepsilon}{q}}{\rm d}t\\ &&- \int_x^\infty e^{-\lambda_1t}t^{-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}+\frac{\varepsilon}{q}}{\rm d}t\bigg]\\ & = &\frac{\lambda_1^{\frac{\lambda_1-\lambda_2+\lambda_3}{4}-\frac{\varepsilon}{2q}-1}e^{-\frac{\lambda_1}{2}}}{1-\frac{\lambda_1-\lambda_2+\lambda_3}{2}+\frac{\varepsilon}{q}} M\big(-\frac{\lambda_1-\lambda_2+\lambda_3}{4}+\frac{\varepsilon}{2q}, -\frac{\lambda_1-\lambda_2+\lambda_3}{4}+\frac{\varepsilon}{2q}+\frac{1}{2}, \lambda_1\big)\\ &&+ \lambda_1^{\frac{\lambda_1+\lambda_2+3\lambda_3}{2}-\frac{\varepsilon}{q}-1}\Gamma\big(1-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}+\frac{\varepsilon}{q}, \lambda_1\big)\\ &&- \varepsilon\int_1^\infty x^{-1-\varepsilon}{\rm d}x\int_x^\infty e^{-\lambda_1t}t^{-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}+\frac{\varepsilon}{q}}{\rm d}t\\ &>&\frac{\lambda_1^{\frac{\lambda_1-\lambda_2+\lambda_3}{4}-\frac{\varepsilon}{2q}-1}e^{-\frac{\lambda_1}{2}}}{1-\frac{\lambda_1-\lambda_2+\lambda_3}{2}+\frac{\varepsilon}{q}} M\big(-\frac{\lambda_1-\lambda_2+\lambda_3}{4}+\frac{\varepsilon}{2q}, -\frac{\lambda_1-\lambda_2+\lambda_3}{4}+\frac{\varepsilon}{2q}+\frac{1}{2}, \lambda_1\big)\\ &&+ \lambda_1^{\frac{\lambda_1+\lambda_2+3\lambda_3}{2}-\frac{\varepsilon}{q}-1}\Gamma\big(1-\frac{\lambda_1+\lambda_2+3\lambda_3}{2}+\frac{\varepsilon}{q}, \lambda_1\big)- \varepsilon T\int_1^\infty x^{-1}{\rm d}x\int_x^\infty t^{-\frac{\lambda_1}{2}-1+\frac{\varepsilon}{q}}{\rm d}t\\ & = &C(\lambda_1, \lambda_2, \lambda_3)+o_1(1)-\frac{T\varepsilon}{(\frac{\lambda_1}{2}-\frac{\varepsilon}{q})^2}\\ & = &C(\lambda_1, \lambda_2, \lambda_3)(1-o(1))\ \ (\varepsilon\rightarrow 0^+). \end{eqnarray*} $

证毕.

定理3.1  设$ p>1, \frac{1}{p}+\frac{1}{q} = 1, 0<\lambda_1<2-\max\{\lambda_3-\lambda_2, \lambda_2+3\lambda_3\}, f(x), g(y)\geq0 $,使得

$ 0<\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x)){\rm d}x<\infty, 0<\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g^q(y){\rm d}y<\infty, $

则有

(3.1) $ \begin{eqnarray} &&\int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}f(x)g(y)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x{\rm d}y \\ & <&C(\lambda_1, \lambda_2, \lambda_3)\bigg\{\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x){\rm d}x\bigg\}^{\frac{1}{p}}\bigg\{\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g^q(y){\rm d}y\bigg\}^{\frac{1}{q}}, \end{eqnarray} $

这里的常数因子$ C(\lambda_1, \lambda_2, \lambda_3) $ (同(2.10)式)是最佳值.

证  由带权Hölder不等式^[25]和引理2.1,可得到

(3.2) $ \begin{eqnarray} &&\int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}f(x)g(y)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x{\rm d}y \\ & = &\int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}f(x)g(y)}{(\max\{1, xy\})^{\lambda_3}}\bigg[\frac{y^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2p}}}{x^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2q}}}\bigg] \bigg[\frac{x^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2q}}}{y^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2p}}}\bigg]{\rm d}x{\rm d}y \\ &\leq&\bigg\{\int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}f^p(x)}{(\max\{1, xy\})^{\lambda_3}}\frac{y^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}}{x^{-\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2q}}}{\rm d}x{\rm d}y\bigg\}^{\frac{1}{p}} \\ &&\times \bigg\{\int_0^\infty\int_0^\infty \frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}g^p(y)}{(\max\{1, xy\})^{\lambda_3}}\frac{x^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}}{y^{-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2p}}}{\rm d}x{\rm d}y\bigg\}^{\frac{1}{q}} \\ & = &\bigg\{\int_0^\infty \omega(\lambda_1, \lambda_2, \lambda_3, x)x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x){\rm d}x \bigg\}^{\frac{1}{p}} \\ &&\times \bigg\{\int_0^\infty \omega(\lambda_1, \lambda_2, \lambda_3, y)y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g^q(y){\rm d}y \bigg\}^{\frac{1}{q}} \\ & = &C(\lambda_1, \lambda_2, \lambda_3)\bigg\{\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x){\rm d}x\bigg\}^{\frac{1}{p}}\bigg\{\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g^q(y){\rm d}y\bigg\}^{\frac{1}{q}}. \end{eqnarray} $

现在假设式(3.2)中的等号成立,则存在不全为零的常数$ A $和$ B $,使得

$ A\frac{y^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}}{x^{-\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2q}}}f^p(x) = B\frac{x^{-\frac{\lambda_1+\lambda_2+\lambda_3}{2}}}{y^{-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2p}}}g^q(y) $

在$ (0, \infty)\times(0, \infty) $内几乎处处成立,于是存在不为零的常数$ C $,使得

$ Ax^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}}f^p(x) = By^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}}g^q(y) = C $

在$ (0, \infty)\times(0, \infty) $内几乎处处成立.不妨设$ A\neq0 $,则有$ x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x) = \frac{C}{Ax} $于$ (0, \infty) $内几乎处处成立.因为广义积分$ \int_0^\infty\frac{C}{Ax}{\rm d}x $是发散的,与条件$ 0<\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x)){\rm d}x<\infty $矛盾.因此(3.2)式只能取严格不等号.若(3.1)式中的常数因子$ C(\lambda_1, \lambda_2, \lambda_3) $不是最佳值,则存在一个正数$ K<C(\lambda_1, \lambda_2, \lambda_3) $,使得用$ K $代替$ C(\lambda_1, \lambda_2, \lambda_3) $时(3.1)式仍然成立.于是由(2.11)和(2.12)式有$ C(\lambda_1, \lambda_2, \lambda_3)(1-o(1))<K $.令$ \varepsilon\rightarrow0^{+} $,就得到$ K\geq C(\lambda_1, \lambda_2, \lambda_3) $,这与$ K<C(\lambda_1, \lambda_2, \lambda_3) $矛盾,所以(3.1)式中的常数因子$ C(\lambda_1, \lambda_2, \lambda_3) $是最佳值.

定理3.2  在与定理3.1相同的条件下,还可得到

(3.3) $ \begin{eqnarray} &&\int_0^\infty y^{\frac{p}{q}[1-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}]}\bigg\{\int_0^\infty\frac{e^{-\lambda_1xy}(\min\{1, xy\})^{\lambda_2}f(x)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x\bigg\}^p{\rm d}y \\ & <&C^p(\lambda_1, \lambda_2, \lambda_3)\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x){\rm d}x, \end{eqnarray} $

这里的常数因子$ C^p(\lambda_1, \lambda_2, \lambda_3) $是最佳值,且不等式(3.3)和不等式(3.1)等价.

证  定义函数$ [f(x)]_n: = \min\{n, f(x)\} $.因为$ 0<\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x){\rm d}x<\infty $,则存在$ n_0\in\mathbb{N} $使得$ 0<\int_{\frac{1}{n}}^n x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x){\rm d}x<\infty $ ($ n\geq n_0 $).令

$ g_n(y): = y^{\frac{p}{q}[1-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}]}\bigg[\int _ {\frac{1}{n}}^n\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}[f(x)]_n}{(\max\{1, xy\} )^{\lambda_3}}{\rm d}x\bigg]^{\frac{p}{q}}\ (\frac{1}{n}<y<n, n\geq n_0), $

当$ n\geq n_0 $时,由(3.1)式有

(3.4) $ \begin{eqnarray} 0&<&\int_{\frac{1}{n}}^n y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g_n^q(y){\rm d}y \\ & = &\int_{\frac{1}{n}}^n y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g_n^{q-1}(y)g_n(y){\rm d}y \\ & = &\int_{\frac{1}{n}}^n \int_{\frac{1}{n}}^n\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}[f(x)]_ng_n(y)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x{\rm d}y \\ &<&C(\lambda_1, \lambda_2, \lambda_3)\bigg\{\int_{\frac{1}{n}}^n x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}[f(x)]_n^p{\rm d}x\bigg\}^{\frac{1}{p}}\bigg\{\int_{\frac{1}{n}}^n y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g_n^{q}(y){\rm d}y\bigg\}^{\frac{1}{q}}. \end{eqnarray} $

进一步,由(3.4)式有

(3.5) $ \begin{eqnarray} 0&<&\int_{\frac{1}{n}}^n y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g_n^q(y){\rm d}y \\ & = &\int_{\frac{1}{n}}^n y^{\frac{p}{q}[1-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}]}\bigg\{\int_{\frac{1}{n}}^n\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}[f(x)]_n}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x\bigg\}^p {\rm d}y \\ &<&C^p(\lambda_1, \lambda_2, \lambda_3)\int_{\frac{1}{n}}^n x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}[f(x)]_n^p{\rm d}x<\infty. \end{eqnarray} $

让$ n\rightarrow\infty $,有$ 0<\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g^q_\infty(y))<\infty, 0<0<\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x))<\infty $.由(3.1)式,可知(3.4)和(3.5)式取严格不等号的条件均满足,因此不等式(3.3)成立.

另一方面,设(3.3)式成立,由Hölder's不等式,可得

$ \begin{eqnarray*} &&\int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}f(x)g(y)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x{\rm d}y\\ & = &\int_0^\infty\bigg[y^{\frac{1-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}}{q}}\int_0^\infty \frac{e^{-\lambda_1xy}(\min\{1, xy\})^{\lambda_2}f(x)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x\bigg] \bigg[y^{\frac{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}{q}}g(y)\bigg]{\rm d}y\\ &\leq&\bigg\{\int_0^\infty y^{\frac{p}{q}[1-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}]}\bigg[\int_0^\infty \frac{e^{-\lambda_1xy}(\min\{1, xy\})^{\lambda_2}f(x)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x\bigg]^p{\rm d}y \bigg\}^{\frac{1}{p}}\\ &&\times \bigg\{\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)-1}{2}}g^q(y){\rm d}y\bigg\}^{\frac{1}{q}}\\ &<&C(\lambda_1, \lambda_2, \lambda_3)\bigg\{\int_0^\infty x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}f^p(x){\rm d}x\bigg\}^{\frac{1}{p}}\bigg\{\int_0^\infty y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1}g^q(y){\rm d}y\bigg\}^{\frac{1}{q}}. \end{eqnarray*} $

即(3.1)式成立,所以(3.3)与(3.1)式是等价的.

假设(3.3)式中的常数因子不是最佳值,则由上面(3.3)式导出(3.1)式的常数因子也不是最佳值,这与定理3.1的结论矛盾,所以(3.3)式的常数因子$ C^p(\lambda_1, \lambda_2, \lambda_3) $是最佳值.

设$ p>1, $$ \frac{1}{p}+\frac{1}{q} = 1, $$ 0<\lambda_1<2-\max\{\lambda_3-\lambda_2, \lambda_2+3\lambda_3\} $, $ \varphi(x) = x^{\frac{p(\lambda_1+\lambda_2+\lambda_3)}{2}-1}, $$ \psi(y) = y^{\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}-1} (x, y>0) $,显然$ \psi^{1-p}(y) = y^{\frac{p}{q}[1-\frac{q(\lambda_1+\lambda_2+\lambda_3)}{2}]}$.定义如下赋范线性空间:

$ L_\varphi^p(0, \infty): = \bigg\{f: \|f\|_{p, \varphi} = \bigg[\int_0^\infty\varphi(x)|f(x)|^p{\rm d}x\bigg]^{\frac{1}{p}}<\infty\bigg\}, $

$ L_\psi^q(0, \infty): = \bigg\{g: \|g\|_{p, \psi} = \bigg[\int_0^\infty\psi(y)|g(y)|^q{\rm d}y\bigg]^{\frac{1}{q}}<\infty\bigg\}, $

$ L_{\psi^{1-p}}^p(0, \infty): = \bigg\{h: \|h\|_{p, \psi^{1-p}} = \bigg[\int_0^\infty\psi(y)^{1-p}|h(y)|^p{\rm d}y\bigg]^{\frac{1}{p}}<\infty\bigg\}. $

设$ f\in L_\varphi^p(0, \infty) $,定义一个奇异积分算子$ T: L_\varphi^p(0, \infty)\rightarrow L_{\psi^{1-p}}^p(0, \infty) $,

$ \begin{eqnarray*} T(f)(y): = \int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}}{(\max\{1, xy\})^{\lambda_3}}f(x){\rm d}x, y\in(0, \infty). \end{eqnarray*} $

因为$ f\in L_\varphi^p(0, \infty), g\in L_\psi^q(0, \infty) $,定义$ Tf $和$ g $的形式内积如下:

$ \begin{eqnarray*} (Tf, g): = \int_0^\infty\int_0^\infty\frac{e^{-\lambda_1 xy}(\min\{1, xy\})^{\lambda_2}f(x)g(y)}{(\max\{1, xy\})^{\lambda_3}}{\rm d}x{\rm d}y, \end{eqnarray*} $

由(3.3)式,有

(3.6) $ \begin{equation} \|T(f)\|_{p, \psi^{1-p}}^p = \int_0^\infty\psi^{1-p}(y)|T(f)|^p{\rm d}y<C^p(\lambda_1, \lambda_2, \lambda_3)\|f\|_{p, \varphi}^p<\infty. \end{equation} $

根据(3.6)式,算子$ T $是有界的,即

$ \begin{eqnarray*} \|T\|: = \sup\limits_{f(\neq0)\in L^p_{\varphi^{1-p}}(0, \infty)}\frac{\|T(f)\|_{p, \varphi^{1-p}}}{\|f\|_{p, \varphi}}\leq C(\lambda_1, \lambda_2, \lambda_3). \end{eqnarray*} $

又因为常数因子$ C(\lambda_1, \lambda_2, \lambda_3) $是最佳值,故有$ \|T\| = C(\lambda_1, \lambda_2, \lambda_3) $.

定理3.3  由定理3.1和定理3.2,不等式(3.1)和(3.3)可表示成如下算子范数形式：

(3.7) $ \begin{equation} (Tf, g)<\|T\|\cdot\|f\|_{p, \varphi}\cdot\|g\|_{q, \psi}, \end{equation} $

(3.8) $ \begin{equation} \|T(f)\|_{p, \psi^{1-p}}^p<\|T\|^p\cdot\|f\|_{p, \varphi}^p. \end{equation} $

在(3.1)和(3.3)式中选取一些合适的参数值,借助Maple数学软件计算$ C(\lambda_1, \lambda_2, \lambda_3) $的值,并结合(3.7)和(3.8)式的表示方法,可得到一些新的且形式简美的Hilbert型积分不等式及其等价式.选取参数$ \lambda_1, \lambda_2, \lambda_3 $的方法是先选定参数$ \lambda_1\ (>0) $,然后在由不等式组

$ \Bigg\{\begin{array}{ll}2+\lambda_2-\lambda_3>\lambda_1, \\ \lambda_2+3\lambda_3<2-\lambda_1\end{array} $

所确定的区域内选取合适的参数$ \lambda_2, \lambda_3 $.

例1  取$ \lambda_1 = 1, \lambda_2 = \lambda_3 = 0, p = q = 2 $,计算式(2.10),得到$ C(1, 0, 0) = \sqrt{\pi} $,此时$ \varphi(x) = 1 $.设$ f, g>0, 0<\|f\|_{2}, \|g\|_{2}<\infty $,则可得到(1.2)式和它的等价式:

(4.1) $ \begin{equation} \int_0^\infty\bigg[\int_0^\infty e^{-xy}f(x){\rm d}x\bigg]^2{\rm d}y<\pi\|f\|_2^2, \end{equation} $

这里的常数因子$ \pi $是最佳值.

例2  取$ \lambda_1 = 1, \lambda_2 = 0, \lambda_3 = -1, p = q = 2 $,计算(2.10)式,得到$ C(1, 0, -1) = 1+e^{-1} $,此时$ \varphi(x) = \frac{1}{x} $.设$ f, g>0, 0<\|f\|_{2, \varphi}, \|g\|_{2, \varphi}<\infty $,则有如下等价不等式:

(4.2) $ \begin{equation} \int_0^\infty\int_0^\infty\frac{\max\{1, xy\}f(x)g(y)}{e^{xy}}{\rm d}x{\rm d}y<\frac{1+e}{e}\|f\|_{2, \varphi}\|g\|_{2, \varphi}, \end{equation} $

(4.3) $ \begin{equation} \int_0^\infty y\bigg[\int_0^\infty\frac{\max\{1, xy\}f(x)}{e^{xy}}{\rm d}x\bigg]^2{\rm d}y<\Big(\frac{1+e}{e}\Big)^2\|f\|_{2, \varphi}^2, \end{equation} $

这里的常数因子$ \frac{1+e}{e}, \big(\frac{1+e}{e}\big)^2 $是最佳值.

例3  取$ \lambda_1 = 1, \lambda_2 = \lambda_3 = -\frac{1}{2}, p = q = 2 $,计算(2.10)式,得到

$ C\Big(1, -\frac{1}{2}, -\frac{1}{2}\Big) = \frac{\sqrt{\pi}(erf(1)+1)}{2}+\frac{1}{e} = 2.000930499^+, $

此时$ \varphi(x) = \frac{1}{x} $.设$ f, g>0, 0<\|f\|_{2, \varphi}, \|g\|_{2, \varphi}<\infty $,则有如下等价不等式:

(4.4) $ \begin{equation} \int_0^\infty\int_0^\infty e^{-xy}\sqrt{\frac{\max\{1, xy\}}{\min\{1, xy\}}}f(x)g(y){\rm d}x{\rm d}y<\bigg(\frac{\sqrt{\pi}(erf(1)+1)}{2}+\frac{1}{e}\bigg)\|f\|_{2, \varphi}\|g\|_{2, \varphi}, \end{equation} $

(4.5) $ \begin{equation} \int_0^\infty y\bigg[\int_0^\infty e^{-xy}\sqrt{\frac{\max\{1, xy\}}{\min\{1, xy\}}}f(x){\rm d}x\bigg]^2{\rm d}y<\bigg(\frac{\sqrt{\pi}(erf(1)+1)}{2}+\frac{1}{e}\bigg)^2\|f\|_{2, \varphi}^2, \end{equation} $

这里的常数因子$ \frac{\sqrt{\pi}(erf(1)+1)}{2}+\frac{1}{e}, \big(\frac{\sqrt{\pi}(erf(1)+1)}{2}+\frac{1}{e}\big)^2 $是最佳值,其中$ erf(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}{\rm d}t $为误差函数.