设$ \; \mathbb{C} $表示复平面, $ H(\mathbb{C}) $表示在$ \; \mathbb{C} $上的解析函数全体, $ dA $表示$ \; \mathbb{C} $上的$ Lebesgue $面积测度, $ F^p (\mathbb{C}) $是由$ \; L^p\left(\mathbb{C}, e^{-\frac{p}{2}|z|^2}dA(z)\right) $中的整函数所构成的空间, 即

$ F^p(\mathbb{C}) = \left\{f \in H(\mathbb{C}):||f||_p = \left(\frac {p}{2\pi} \int_\mathbb{C} |f(z)|^p e^{-\frac {p|z|^2}{2}} dA(z)\right)^\frac {1}{p} < \infty\right\}, \; 其中\; 0<p<\infty. $

对任意给定的正整数$ \; m $, Fock-Sobolev空间$ \; F^{{p, m}}=\{f\in H(\mathbb{C}): f^{(m)} \in F^p\} $, Zhu在[1]中证明了$ \; f\in F^{p, m} $当且仅当$ \; z^mf(z)\in F^p. $因此, 在$ \; F^{(p, m)} $中通常使用下述范数:

$ ||f||{^p_{p, m}}=c(p, m){\int_{\mathbb{C}}}|z^mf(z)e^{-|z|^2/2}|^pdA(z), \, \, \, \, \, c(p, m)=\frac{(p/2)^{(mp/2)+1}}{{\pi}\Gamma((mp/2)+1)}. $

若$ \; g\in H(\mathbb{C}) $, 则线性算子$ \; V_gf(z )= \int_0^zf(\zeta)g'(\zeta)\, d\zeta\; $称为Volterra型积分算子. 若$ \; \varphi \in H(\mathbb{C}) $, 则线性算子$ \; C_{\varphi}f(z) = f(\varphi(z))\; $称为复合算子. Volterra型积分算子和复合算子的乘积的定义如下:

$ V{_g^\varphi}f(z)\, =\, V{_g}{\circ}C{_\varphi}f(z)\, =\, {\int_0^z}f({\varphi}(\zeta))g'(\zeta)d\zeta, $

$ C{_\varphi^g}f(z)\, =\, C{_\varphi}{\circ}V{_g}f(z)\, =\, {\int_0^{\varphi(z)}}f(\zeta)g'(\zeta)d\zeta. $

方便起见, 后文使用$ \; T\; $来表示$ \; V_g^\varphi\; $和$ \; C{_\varphi^g}.\; $

人们对于Fock空间和Fock-Sobolev空间已经有了很多研究. Hong在[2]中研究了Fock-Sobolev空间上复合算子有界性和紧性. Constantin在[3]中研究了Fock空间上的Volterra型积分算子. Mengestie在[4]中研究了Fock空间上Volterra型积分算子和复合算子的乘积, 使用Berezin变换刻画了算子的有界性和紧性, 但是他的结果略微复杂, Tien在[5]中给出了更易使用的结果. Cho在[6]中研究了Fock-Sobolev空间上的Toeplitz算子. Hu和Lv在[7]中研究了$ \; F^p(\varphi) $上Topelitz算子的性质, Fock-Sobolev空间可以视为$ \; F^p(\varphi) $的一个特例. Mengestie在[8, 9]中对Fock-Sobolev空间和Fock-Sobolev空间上的Volterra型积分算子进行了许多研究. 因此, 本文受上述启发, 将研究Fock-Sobolev空间上Volterra型积分算子和复合算子乘积的有界性和紧性.

本文通过$ \; M_T(z)\; $来刻画算子T的有界性和紧性, 定义如下:

$ \begin{eqnarray} M_{V{_g^\varphi}}(z) &=& \frac {|g'(z)|(1+|z|)^{(m+1)} e^{\frac {1}{2}|\varphi(z)|^2- \frac {1}{2}|z|^2}} {(1+|z|+\big||z|^2+|z|-m\big|)(1+|\varphi(z)|)^m} \\ &=& \frac {|g'(z)|(1+|z|)e^{\phi(\varphi(z))-\phi(z)}} {(1+|z|+\big||z|^2+|z|-m\big|)}\; , \\ M_{C{_\varphi^g}}(z) &=& \frac {|g'(\varphi(z))\varphi '(z)|(1+|z|)^{(m+1)} e^{\frac {1}{2}|\varphi(z)|^2 - \frac {1}{2}|z|^2}} {(1+|z|+\big||z|^2+|z|-m\big|)(1+|\varphi(z)|)^m} \\ &=& \frac {|g'(\varphi(z))\varphi '(z)|(1+|z|)e^{\phi(\varphi(z))-\phi(z)}} {(1+|z|+\big||z|^2+|z|-m\big|)}\; , \end{eqnarray} $

其中$ \; \phi(z)=|z|^2/2-mlog(1+|z|) $. 后文中若无特殊说明, $ \phi(z) $都为此定义.

下文, 我们用$ \; C\; $表示非负常数且不同处的$ C $可以取值不同. 对于两个非负量$ X $和$ Y $, 用$ \; X\lesssim Y\; $表示存在常数$ \; C > 0\; $使得$ \; X\leq CY, \; $用$ \; X\gtrsim Y\; $表示存在常数$ \; C > 0\; $使得$ \; X\geq CY $. 如果$ \; X\lesssim Y\; $和$ \; X\gtrsim Y, \; $则记$ \; X\simeq Y.\; $本文的主要结论为下述两个定理.

定理1  设$ \; 1<p\leq q<\infty, \; g $为整函数且不为常函数, $ \; \varphi(z)=az+b, \; 0<|a|\leq 1 $, 则有:

(a)   算子$ \; T:F^{p, m}(\mathbb{C})\rightarrow F^{q, m}(\mathbb{C}) $有界当且仅当$ M_T(z)\in L^\infty (\mathbb{C}, dA) $, 且

$ ||M_{V{_g^\varphi}}||_{L^\infty} \simeq ||V{_g^\varphi}||, $

$ ||M_{C{_\varphi^g}}||_{L^\infty} \lesssim ||C{_g^\varphi}|| \lesssim (||C_b^g||^q+|a|^{-2} ||M_{V{_g^\varphi}}||^q_{L^\infty})^{\frac {1}{q}}. $

(b)   算子$ \; T:F^{p, m}(\mathbb{C})\rightarrow F^{q, m}(\mathbb{C}) $为紧算子当且仅当$ \lim\limits_{|z| \to \infty} M_T(z)\, =0. $

定理2  设$ \; 1<q< p<\infty, \; g $为整函数且不为常函数, $ \; \varphi(z)=az+b, \; 0<|a|\leq 1 $, 则下列陈述等价:

(i)   算子$ T:F^{p, m}(\mathbb{C})\rightarrow F^{q, m}(\mathbb{C})\; $是有界的.

(ii)   算子$ T:F^{p, m}(\mathbb{C})\rightarrow F^{q, m}(\mathbb{C})\; $是紧的.

(iii)   $ M_T(z) \in L^{\frac {pq}{p-q}}(\mathbb{C}, dA) $. 且有

$ ||M_{V{_g^\varphi}}||_{L^{\frac {pq}{p-q}}} \simeq ||V{_g^\varphi}||, $

$ |a|^{\frac {2(p-q)}{p-q}} ||M_{C{_\varphi^g}}||_{L^{\frac {pq}{p-q}}} \lesssim ||C{_\varphi^g}|| \lesssim (||C_b^g||^q + |a|^{-\frac {2q}{p}} ||M_{C_\varphi ^g}||^q_{L^{\frac {pq}{p-q}}})^{\frac {1}{q}}. $

在本节中, 我们给出一些引理和相关推论. Mensgestie在[9]中给出了Fock-Sobolev空间范数的一个等价表述:

$ ||f||{^p_{p, m}}\simeq {\int_{\mathbb{C}}}|f(z)e^{-\phi(z)}|^pdA(z), \; \phi(z)=|z|^2/2-mlog(1+|z|). $

由上式可得$ ||f||_p \lesssim ||f||_{p, m}. $ Hu在[7]中对这类的加权Fock空间进行了一定探讨. 与Fock空间类似, 我们给出Fock-Sobolev空间中的再生核函数$ \; K_m(z, \omega), \; $后文中将其简记为$ \; K_z(\omega) $. 从[7]我们可以得知, $ \; \forall z, \omega \in \mathbb{C}, \; $存在正常数$ \; \theta, C\; $使得

$ |K_z(\omega)e^{-\phi(z)}e^{-\phi(\omega)}| \leq Ce^{-\theta|z-\omega|} \leq C. $

存在正常数$ \; r, \; $对于$ \; \forall z \in \mathbb{C}, $当$ |\omega - z| < r $时, 有

$ |K_z(\omega)e^{-\phi(z)}e^{-\phi(\omega)}| \geq CK_z(z)e^{-2\phi(z)} \geq C. $

由上面两个不等式, 显然有

$ |K_z(z)| \simeq e^{2\phi(z)}. $

对于每个$ \; z \in \mathbb{C}, \; $定义$ \; k_z(\omega)=\frac {K_z(\omega)}{\sqrt{K_z(z)}}, \; $有$ |k_z(z)|\simeq e^{\phi(z)}, ||k_z||_{p, m}\simeq 1.\; $固定$ \; \omega \in \mathbb{C}, \; $当$ \; z \to \infty\; $时, 有$ \; |k_z(\omega)|\simeq |K_z(\omega)e^{-\phi(z)}| \leq Ce^{-\theta|z-\omega|}e^{\phi(\omega)} \to 0. $

引理2.1  设$ \; p \in (0, \infty), \; f\in F^p(\mathbb{C}). $则对任意$ z\in \mathbb{C}, \; $存在常数$ C $使得

$ |f(z)|e^{-\frac {|z|^2}{2}} \leq C||f||_p. $

当$ \; f\in F^{p, m}(\mathbb{C})\; $时, 有$ \; z^mf \in F^p $, 自然可推出$ \; |f(z)z^m|e^{-\frac {|z|^2}{2}} \leq C||f||_{p, m} $.

引理2.2  设$ \; 0<p<q<\infty, \; $则有$ \; F^{p}(\mathbb{C}) \subset F^{q}(\mathbb{C}), \; $且

$ ||f||_{q} \leq (\frac {q}{p})^{\frac {1}{q}}||f||_{p}, \; \; \forall f \in F^{p}. $

显然可以推出, 若$ \; 0<p<q<\infty, \; $则有$ \; F^{p, m}(\mathbb{C}) \subset F^{q, m}(\mathbb{C}), \; $且$ \; ||f||_{q, m} \leq (\frac {q}{p})^{\frac {1}{q}}||f||_{p, m}. $

由文献[6, 引理3.8] 得到下述紧算子的判别方法:

引理2.3  设$ \; 1<p, q<\infty, \; T:F^{p, m} \to F^{q, m}\; $为紧算子当且仅当对于$ \; F^{p, m}\; $中任意有界且满足在$ \; \mathbb{C}\; $的紧子集上一致收敛于0的序列$ \; \{f_k\}, \; $有$ \; ||Tf_k||_{p, m} \to 0, \; k \to \infty. $由文献[9, 引理2.2] 得到如下引理:

引理2.4  设$ 0<p<\infty, \; f \in F^{p, m}.\; $则

$ ||f||_{p, m}^p \simeq |f(0)|^p + \int_\mathbb{C} \frac {|f'(z)|^p(1+|z|)^p e^{-p\phi(z)}} {(1+|z| + \big| |z|^2+|z|-m \big|)^p}dA(z). $

为了更好的应用引理2.4, 我们构造如下函数:

$ m_{V{_g^\varphi}}(z) = \frac {|g'(z)|(1+|z|)e^{-\phi(z)}} {(1+|z|+\big||z|^2+|z|-m\big|)} = e^{-\phi(\varphi(z))}M_{V{_g^\varphi}}, $

$ m_{C{_\varphi^g}}(z) = \frac {|g'(\varphi(z))\varphi(z)|(1+|z|)e^{-\phi(z)}} {(1+|z|+\big||z|^2+|z|-m\big|)} = e^{-\phi(\varphi(z))}M_{C{_\varphi^g}}. $

在后文中, 使用$ \; m_T\; $来简记上述两个函数.

对文献[8, 推论3.2] 的证明做简单的修改, 可得下述引理:

引理2.5  设$ g, \; \varphi $为整函数, 且$ g $不恒为0. 若满足不等式

$ \sup \limits_{z\in\mathbb{C}} \frac {|g(z)|(1+|z|)^{m+1} e^{\frac {1}{2}|\varphi(z)|^2 -\frac {1}{2}|z|^2}} {(1+|z|+\big||z|^2+|z|-m\big|)(1+|\varphi(z)|)^m} < \infty , $

则有$ \varphi (z)=az+b $, 且$ |a|\leq 1 $.

根据引理2.5, 可以得知如果有$ \sup\limits_{z \in \mathbb{C}}M_T(z) \leq M $, 则$ \varphi(z)=az+b, \; |a|\leq 1 $.

设$ \; 0<p, q<\infty, \; \mu\; $是$ \; \mathbb{C}\; $上的正Borel测度, 若存在正数$ C $使其满足:

$ \left(\int_{\mathbb{C}}|f|^q e^{-q\phi(z)} d\mu (z)\right)^{\frac {1}{q}} \leq C||f||_{p, m}, \, \, \forall f \in F^{p, m}(\mathbb{C}), $

则称$ \mu $是$ \; (p, q)\; $型Fock Sobolev-Carleson测度. 显然, 可以根据$ \; \mu\; $构建恒等算子$ \; i:F^{p, m}(\mathbb{C})\to L^q(\mathbb{C}, \; e^{-q\phi(z)} d\mu ) $, 定义$ \; ||\mu||=||i|| $.

由文献[7, 定理2.8] 得到下述引理:

引理2.6  设$ 0<q<p<\infty, \; \mu $是$ \mathbb{C}\; $上的正Borel测度, 则$ \mu $是$ (p, q) $型的Fock Sobolev-Carleson测度当且仅当$ \widetilde{\mu} \in L^{\frac {p}{p-q}}(\mathbb{C}, dA) $, 其中

$ \widetilde{\mu}(z)=\int_{\mathbb{C}}|k_z(\zeta)|^q e^{-q\phi(\zeta)} d\mu(\zeta), \, \, z\in \mathbb{C} $

且有$ ||\mu||\simeq ||\widetilde{\mu}||^{\frac {1}{q}}_{L^{\frac {p}{p-q}}} $.

本节将刻画Fock Sobolev空间上算子的有界性和紧性. 下面的命题给出一个在后文中多次使用的重要等式.

命题3.1  设$ q\in (0, \infty), \; g $为整函数且不为常函数. 对于任意的整函数$ f $和$ z\in \mathbb{C} $, 有

$ \begin{eqnarray} ||Tf||_{q, m}^{q} &\simeq& |(Tf)(0)|^q + \int_\mathbb{C} |f(\varphi (\zeta))|^q e^{-q\phi(\varphi(\zeta))} M_T(\zeta)^q dA(\zeta) \\ &= & |(Tf)(0)|^q + \int_\mathbb{C} |f(\varphi (\zeta))|^q m_T(\zeta)^q dA(\zeta) \\ &\gtrsim& |(Tf)(0)|^q + |f(\varphi(z))|^q m_T(z)^q, \end{eqnarray} $

(3.1)

和

$ \begin{eqnarray} \int _\mathbb{C} m_T(\zeta)^q dA(\zeta) \lesssim ||T1||_{q, m}^q. \end{eqnarray} $

(3.2)

证  令$ T $为$ \; V_g^{\varphi} $, 注意到$ \; V_g^{\varphi}f(0)=0, \; $由引理2.4, 可得

$ \begin{eqnarray} ||V_g^{\varphi}f||_{q, m}^q &\simeq& |V_g^{\varphi}f(0)|^q + \int_\mathbb{C} \frac {|f(\varphi(\zeta))g'(\zeta)|^q(1+|\zeta|)^q e^{-q\phi(\zeta)}} {(1+|\zeta| + \big| |\zeta|^2+|z|-m \big|)^q}dA(\zeta) \\ &=& \int_\mathbb{C} |f(\varphi (\zeta))|^q e^{-q\phi(\varphi(\zeta))} M_{V_g^{\varphi}}(\zeta)^q dA(\zeta)\\ &=& \int_\mathbb{C} |f(\varphi (\zeta))|^q m_{V_g^{\varphi}}(\zeta)^q dA(\zeta)\\ &\geq& \int_{D(\omega, 1)} |f(\varphi (\zeta))|^q m_{V_g^{\varphi}}(\zeta)^q dA(\zeta)\\ &=& \int_{D(\omega, 1)} \frac {|f(\varphi (\zeta)) g'(\zeta)|^q(1+|\zeta|)^{q(m+1)} e^{- \frac {q}{2}|\zeta|^2}} {(1+|\zeta|+\big||\zeta|^2+|\zeta|-m\big|)^q} dA(\zeta), \end{eqnarray} $

其中$ D(\omega, 1)=\{\zeta \in \mathbb{C}:|\zeta-\omega|< 1 \}. $当$ \; \zeta\in D(\omega, \; 1)\; $时, 有

$ \begin{equation} \left\{ \begin{aligned} &1+|\zeta| \simeq 1+|\omega| \\ &1+|\zeta| + \big| |\zeta|^2+|\zeta|-m \big| \simeq 1+|\omega| + \big| |\omega|^2+|\omega|-m \big|. \end{aligned} \right.\nonumber \end{equation} $

由次调和性得

$ \int_{D(\omega, 1)} \frac {|f(\varphi (\zeta)) g'(\zeta)|^q(1+|\zeta|)^{q(m+1)} e^{- \frac {q}{2}|\zeta|^2}} {(1+|\zeta|+\big||\zeta|^2+|\zeta|-m\big|)^q} dA(\zeta) \gtrsim |f(\varphi (\omega))|^q m_{V_g^{\varphi}}(\omega)^q, \; $

从而有$ ||V_g^{\varphi}f||_{q, m}^{q} \gtrsim |V_g^{\varphi}f(0)|^q + |f(\varphi (z))|^q m_{V_g^{\varphi}}(z)^q.\; $

令$ T $为$ \; C{_\varphi^g}, \; $由引理2.4, 可得

$ \begin{eqnarray} ||C{_\varphi^g}f||_{q, m}^q &\simeq& |C{_\varphi^g}f(0)|^q + \int_\mathbb{C} \frac {|f(\varphi(\zeta))g'(\varphi (\zeta))\varphi' (\zeta)|^q(1+|\zeta|)^q e^{-q\phi(\zeta)}} {(1+|\zeta| + \big| |\zeta|^2+|\zeta|-m \big|)^q}dA(\zeta) \\ &=& |C{_\varphi^g}f(0)|^q + \int_\mathbb{C} |f(\varphi (\zeta))|^q e^{-q\phi(\varphi(\zeta))} M_{C{_\varphi^g}}(\zeta)^q dA(\zeta)\\ &=& |C{_\varphi^g}f(0)|^q + \int_\mathbb{C} |f(\varphi (\zeta))|^q m_{C{_\varphi^g}}(\zeta)^q dA(\zeta)\\ &\geq& |C{_\varphi^g}f(0)|^q + \int_{D(\omega, 1)} |f(\varphi (\zeta))|^q m_{C{_\varphi^g}}(\zeta)^q dA(\zeta)\\ &=& |C{_\varphi^g}f(0)|^q + \int_{D(\omega, 1)} \frac {|f(\varphi (\zeta)) g'(\varphi (\zeta))\varphi' (\zeta)|^q(1+|\zeta|)^{q(m+1)} e^{- \frac {q}{2}|\zeta|^2}} {(1+|\zeta|+\big||\zeta|^2+|\zeta|-m\big|)^q} dA(\zeta). \end{eqnarray} $

由次调和性得

$ \int_{D(\omega, 1)} \frac {|f(\varphi (\zeta)) g'(\varphi (\zeta))\varphi' (\zeta)|^q(1+|\zeta|)^{q(m+1)} e^{- \frac {q}{2}|\zeta|^2}} {(1+|\zeta|+\big||\zeta|^2+|\zeta|-m\big|)^q} dA(\zeta) \gtrsim |f(\varphi (\omega))|^q m_{C{_\varphi^g}}(\omega)^q, \; $

显然有$ |C{_\varphi^g}f(0)|^q \gtrsim |C{_\varphi^g}f(0)|^q $, 即得$ ||C{_\varphi^g}f||_{q, m}^{q} \gtrsim |C{_\varphi^g}f(0)|^q + |f(\varphi (z))|^q m_{C{_\varphi^g}}(z)^q. $

最后, 取$ \; f = 1, \; $即得$ \int _\mathbb{C} m_T(\zeta)^q dA(\zeta) \lesssim ||T1||_{q, m}^q. $

命题3.2  设$ p, \; q \in (0, \infty), \; g, \; \varphi $为整函数且$ g $不为常函数. 如果$ T:\; F^{p, m}(\mathbb{C})\rightarrow F^{q, m}(\mathbb{C}) $为有界算子, 则$ M_T(z) \in L^\infty (\mathbb{C} , dA), \; \varphi (z)=az+b, |a|\leq 1 $, 且有

$ \begin{eqnarray} M_T(z) \lesssim ||Tk_{\varphi (z)}||_{q, m} \lesssim ||T||, \forall z \in \mathbb{C} . \end{eqnarray} $

(3.3)

证  利用$ ||k_\omega||_{p, m}\simeq 1 $和(3.1) 可得$ ||T||\gtrsim ||Tk_\omega||_{q, m}\gtrsim |k_{\omega}(\varphi (z))|m_T(z) , \forall z \in \mathbb{C}. $取$ \omega = \varphi (z) $, 有

$ \begin{eqnarray} ||T|| \gtrsim |k_{\varphi(z)} (\varphi(z))|m_T(z) \simeq e^{\phi(\varphi(z))}m_T(z) = M_T(z). \end{eqnarray} $

由引理2.5可得$ \varphi(z)=az+b, |a|\leq 1. $

推论3.3  设$ p, \; q \in (0, \infty), \; g $为整函数且不为常函数, $ \varphi(z) = b, \; $则有:

(a)   $ V_g^b:\; F^{p, m}(\mathbb{C}) \to F^{q, m}(\mathbb{C}) $为紧算子当且仅当$ g\in F^{q, m}({\mathbb{C}}) $.

(b)   $ C_b^g:\; F^{p, m}(\mathbb{C}) \to F^{q, m}(\mathbb{C}) $是紧算子.

证  $ (a) $  必要性, 令$ f=1, \; V_g^bf(z)=g(z)-g(0) $, 则$ g\in F^{q, m}({\mathbb{C}}) $.

充分性, 显然$ V_g^b $是一个有限秩算子, 因此只需证明其有界即可. 有

$ ||Tf||_{q, m}=|f(b)|\; ||g-g(0)||_{q, m} \leq e^{\frac {|b|^2}{2}} ||g-g(0)||_{q, m} ||f||_p \lesssim ||g-g(0)||_{q, m} ||f||_{p, m}, $

即得$ V_g^b $是紧算子.

(b)   设$ f\in F^{p, m}, \; C_b^gf(z)=\int_0^bf(\zeta)g'(\zeta)d\zeta, \; $有

$ ||C_b^gf||_{q, m}=\left|\int_0^b f(\zeta)g'(\zeta)d\zeta\right| \leq |b| \sup\limits_{\zeta\in[0, b]}|f(\zeta)g'(\zeta)| . $

由引理2.1可知$ \forall z \in \mathbb{C}, \; |f(z)|\lesssim ||f||_p $, 又因为$ \; g'\; $是整函数, 所以$ g'(\zeta)\; $在$ \; [0, b]\; $上有界, 可得

$ |b| \sup\limits_{\zeta\in[0, b]}|f(\zeta)g'(\zeta)| \lesssim ||f||_p \lesssim ||f||_{p, m}. $

显然$ \; C_b^g\; $是秩为1的有限秩算子, 从而可知$ \; C_b^g\; $是紧算子.

当$ \varphi(z)=az+b, \; 0<a\leq1 $时, 情形更为复杂, 我们分$ 1<p\leq q<\infty $和$ 1<q<p<\infty $来进行讨论, 先研究$ 1<p\leq q<\infty $的情况.

定理1的证明  (a)  首先证明必要性, 由命题3.2可知$ M_T(z)\lesssim ||T||, $即得$ ||M_T||_{L^\infty} \leq ||T||. $

再证充分性, 设$ M_T(z)\in L^\infty (\mathbb{C}, dA) $. 由命题3.1可得

$ \begin{eqnarray} ||Tf||_{q, m}^{q} &\simeq& |Tf(0)|^q + \int_\mathbb{C} |f(\varphi (\zeta))|^q e^{-q\phi(\varphi(\zeta))} M_T(\zeta)^q dA(\zeta)\\ &\leq & |Tf(0)|^q + ||M_T||_{L^\infty}^q \int_\mathbb{C} |f(\varphi (\zeta))|^q e^{-q\phi(\varphi(\zeta))}dA(\zeta)\\ &\lesssim & |Tf(0)|^q + |a|^{-2} ||M_T||_{L^\infty}^q \int_\mathbb{C} |f(\zeta)|^q e^{-q\phi(\zeta)}dA(\zeta)\\ &\lesssim& |Tf(0)|^q + |a|^{-2} ||M_T||^q_{L^\infty} ||f||^q_{q, m} \\ &\lesssim& |Tf(0)|^q + |a|^{-2} ||M_T||^q_{L^\infty} ||f||^q_{p, m}. \end{eqnarray} $

令$ T $为$ \; V_g^{\varphi} $, 注意到$ \; V_g^{\varphi}f(0)=0, \; $由上式可得

$ ||V{_g^\varphi}f||_{q, m}^{q} \lesssim |a|^{-2} ||M_{V{_g^\varphi}}||^q_{L^\infty} ||f||^q_{p, m}. $

即得$ V{_g^\varphi} $为有界算子, 再由命题3.2即可得$ ||M_{V{_g^\varphi}}||_{L^\infty} \simeq ||V{_g^\varphi}||. $

令$ T $为$ \; C_{\varphi}^g $, 注意到$ C_{\varphi}^gf(0)=C_b^gf $, 由推论3.3可知$ C_b^g $是有界算子, 得

$ \begin{eqnarray} ||C_{\varphi}^gf||_{q, m}^{q} &\lesssim& |C_b^gf|^q + |a|^{-2} ||M_{V{_g^\varphi}}||^q_{L^\infty} ||f||^q_{p, m}\\ &\lesssim& ||C_b^g||^q\; ||f||_{p, m}^q + |a|^{-2} ||M_{V{_g^\varphi}}||^q_{L^\infty} ||f||^q_{p, m}\\ &=& (||C_b^g||^q+|a|^{-2} ||M_{V{_g^\varphi}}||^q_{L^\infty})||f||^q_{p, m}. \end{eqnarray} $

即得$ C_{\varphi}^g $为有界算子, 且有$ ||M_{C{_\varphi^g}}||_{L^\infty} \lesssim ||C{_g^\varphi}|| \lesssim (||C_b^g||^q + |a|^{-\frac {2}{q}}||M_{C{_\varphi^g}}||_{L^\infty}^q)^{\frac {1}{q}}. $

(b)   首先证明必要性, 设$ z_n\to \infty $, 则有$ \varphi(z_n)\to \infty $, $ \{k_{\varphi(z_n)}\} $满足$ F^{p, m} $中有界且在$ \; \mathbb{C}\; $的紧子集上一致收敛于0. 由引理2.3可知$ ||Tk_{\varphi(z_n)}||_{q, m} \to 0 $, 再由命题3.2即得$ z\to \infty $时, $ M_T(z)\to 0 $.

再证充分性, 由$ (a) $可知, 此时$ T $是有界算子. 令$ \{f_n\} $为$ F^{p, m} $中的任意序列, 满足$ \{||f_n||\} $有界且$ f_n $在$ \mathbb{C} $的任意紧子集上一致收敛于0. 由命题3.1可得

$ \begin{eqnarray} ||Tf_n||_{q, m}^{q} &\simeq & |Tf_n(0)|^q + \int_{|z|\leq R} |f_n(\varphi (z))|^q m_T(z)^q dA(z) \\ &&+\int_{|z|>R} |f_n(\varphi (z))|^q e^{-q\phi(\varphi (z))} M_T(z)^qdA(z)\\ &\lesssim& |Tf_n(0)|^q + (\sup\limits_{|z|\leq R}|f_n(\varphi (z))|^q )\int_{|z|\leq R} m_T(z)^q dA(z) \\ &&+ (\sup\limits_{|z|>R}M_T(z)^q) \int_{|z|>R} |f_n(\varphi (z))|^q e^{-q\phi(\varphi (z))}dA(z) \\ &\lesssim& |Tf_n(0)|^q + ||T1||_{q, m}^{q} (\sup\limits_{|z|\leq R}|f_n(\varphi (z))|^q) + |a|^{-2}||f_n||_{q, m}^q (\sup\limits_{|z|>R}M_T(z)^q) \\ &\lesssim& |Tf_n(0)|^q + ||T1||_{q, m}^{q} (\sup\limits_{|z|\leq R}|f_n(\varphi (z))|^q) + |a|^{-2}||f_n||_{p, m}^q (\sup\limits_{|z|>R}M_T(z)^q). \end{eqnarray} $

我们有$ V_g^{\varphi}f_n(0)=0 $和$ C_{\varphi}^gf_n(0)=C_b^gf_n=\int_0^b f_n(\zeta)g'(\zeta)d\zeta $. 因为$ \{f_n\} $在紧集$ [0, b] $上一致收敛于0, 所以$ n\to \infty $时, $ C_{\varphi}^gf_n(0) \to 0 $. 再结合$ \{||f_n||\} $有界和$ T $是有界算子即得$ ||Tf_n||_{q, m} \to 0 $, 由引理2.3可知$ T $是紧算子.

接下来证明$ 1<q<p<\infty $的情况, 我们先给出两个$ \mathbb{C} $上测度$ \mu_{T, q} $和$ \lambda_{T, q} $, 定义如下:

$ \begin{eqnarray*} \mu_{T, q} (B)=\int_{\varphi^{-1}(B)} m_T(z)^q dA(z), \quad d\lambda_{T, q}(z) = e^{q\phi(z)}d\mu_{T, q}(z). \end{eqnarray*} $

定理2的证明  (ii) $ \to $ (i) 是显然的.

(i) $ \to $ (iii). 设$ T:F^{p, m}(\mathbb{C})\rightarrow F^{q, m}(\mathbb{C})\; $有界, 由(3.1) 式可得

$ \begin{eqnarray} ||T||^q\; ||f||_{p, m}^q &\geq& ||Tf||_{q, m}^q \simeq |Tf(0)|^q + \int_{\mathbb{C}} |f(\varphi (z))|^q m_T(z)^q dA(z)\\ &\geq& \int_{\mathbb{C}} |f(\varphi (z))|^q m_T(z)^q dA(z) \geq\int_{\mathbb{C}} |f(z)|^q d\mu_{T, q} (z) = \int_{\mathbb{C}} |f(z)|^q e^{-q\phi(z)} d\lambda_{T, q}(z). \end{eqnarray} $

由上述不等式可知$ \lambda_{T, q} $是一个Fock Sobolev-Carleson测度, 由引理2.6可得

$ \widetilde{\lambda_{T, q}}(\omega)=\int_{\mathbb{C}}|k_\omega(\zeta)|^q e^{-q\phi(\zeta)} d\lambda_{T, q}(\zeta)\, \in L^{\frac {p}{p-q}}(\mathbb{C}, dA). $

再使用命题3.1, 有

$ \begin{eqnarray} \int_{\mathbb{C}}|k_\omega(\zeta)|^q e^{-q\phi(\zeta)} d\lambda_{T, q}(\zeta) &=& \int_{\mathbb{C}}|k_{\omega}(\zeta)|^q d\mu_{T, q}(\zeta) =\int_{\mathbb{C}}|k_{\omega}(\varphi(\zeta))|^q m_T(\zeta)^q dA(\zeta) \\ &\gtrsim& |k_{\omega}(\varphi(z))|^q m_T(z)^q. \end{eqnarray} $

令$ \omega=\varphi(z) $, 得到$ \widetilde{\lambda_q}(\varphi(z))\gtrsim e^{q\phi(\varphi(z))}m_T(z)^q = M_T(z)^q $, 从而有

$ \begin{eqnarray} \int_\mathbb{C} M_T(z)^{\frac {pq}{p-q}} dA(z) \lesssim \int_\mathbb{C} |\widetilde{\lambda_q}(\varphi(z))|^{\frac {p}{p-q}} dA(z) =|a|^{-2} \int_\mathbb{C} |\widetilde{\lambda_q}(\omega)|^{\frac {p}{p-q}} dA(\omega). \end{eqnarray} $

由引理2.5我们还可知$ ||\widetilde{\lambda_{T, q}}||_{L^{\frac {p}{p-q}}}^{\frac {1}{q}}\simeq||\lambda_{T, q}||\leq||T||, $从而有$ ||M_T||_{L^{\frac {pq}{p-q}}} \lesssim |a|^{-\frac {2(p-q)}{pq}}||T|| $.

(iii) $ \to $ (ii). 设$ M_T(z) \in L^{\frac {pq}{p-q}} $, $ f \in F^{p, m}(\mathbb{C}) $. 注意到$ \frac {1}{p} + \frac {p-q}{pq} = \frac {1}{q} $, 使用命题3.1和Hölder不等式得

$ \begin{eqnarray} ||Tf||_{q, m}^{q} &\simeq& |Tf(0)|^q + \int_\mathbb{C} |f(\varphi (\zeta))|^q e^{-\frac {q|\phi(\varphi (\zeta))|^2}{2}} M_T(\zeta)^q dA(\zeta) \\ &\leq & |Tf(0)|^q + \left(\int_\mathbb{C} |f(\varphi (\zeta))|^p e^{-\frac {p|\phi(\varphi (\zeta))|^2}{2}}dA(\zeta)\right)^{\frac {q}{p}} \times \left( \int_\mathbb{C} M_T(\zeta)^{\frac {pq}{p-q}}dA(\zeta)\right)^{\frac {p-q}{p}} \\ &\lesssim& |Tf(0)|^q + |a|^{-\frac {2q}{p}} ||f||_{p, m}^q ||M_T||_{L^{\frac {pq}{p-q}}}^q. \end{eqnarray} $

有$ V_g^{\varphi}f(0)=0 $和$ C_{\varphi}^gf(0)=C_b^gf, $从而可推出$ T $有界, 且

$ \begin{eqnarray} ||V_g^{\varphi}|| \lesssim |a|^{-\frac {2}{p}} ||M_{V_g^{\varphi}}||_{L^{\frac {pq}{p-q}}}, \quad ||C_{\varphi}^g|| \lesssim (||C_b^g||^q + |a|^{-\frac {2q}{p}} ||M_{C_\varphi ^g}||^q_{L^{\frac {pq}{p-q}}})^{\frac {1}{q}}. \end{eqnarray} $

接下来我们再证$ T $的紧性. 令$ \{f_n\} $为$ F^{p, m} $中的任意序列, 满足$ \{||f_n||_{p, m}\} $有界且$ f_n $在$ \mathbb{C} $的任意紧子集上一致收敛于0. 由命题3.1和Hölder不等式可得

$ \begin{eqnarray} ||Tf_n||_{q, m}^{q} &\simeq& |Tf_n(0)|^q + \int_\mathbb{C} |f_n(\varphi (\zeta))|^q e^{-\frac {q|\phi(\varphi (\zeta))|^2}{2}} M_T(\zeta)^q dA(\zeta) \\ &=& |Tf_n(0)|^q + \int_{|z|\leq R} |f_n(\varphi (z))|^q m_T(z)^q dA(z) \\ &&+\int_{|z|>R} |f_n(\varphi (z))|^q e^{-q\phi(\varphi(z))} M_T(z)^qdA(z)\\ &\lesssim& |Tf_n(0)|^q + \sup\limits_{|z|\leq R}|f_n(\varphi (z))|^q \int_{|z|\leq R} m_T(z)^q dA(z) \\ &\, &+ \left(\int_{|z|> R} |f_n(\varphi (z))|^p e^{-p\phi(\varphi(z))}dA(z)\right)^{\frac {q}{p}} \times \left(\int_{|z|> R} M_T(z)^{\frac {pq}{p-q}}dA(z)\right)^{\frac {p-q}{p}} \\ &\lesssim& |Tf_n(0)|^q + ||T1||_{q, m}^{q} \sup\limits_{|z|\leq R}|f_n(\varphi (z))|^q \\ &&+ |a|^{-\frac {2q}{p}} \sup\limits_{n}||f_n||_{p, m}^q \left(\int_{|z|> R} M_T(z)^{\frac {pq}{p-q}}dA(z)\right)^{\frac {p-q}{p}}. \end{eqnarray} $

由$ \; T\; $有界可知$ ||T1||_{q, m}< \infty $, 再注意到$ M_T(z) \in L^{\frac {pq}{p-q}} $, $ V_g^{\varphi}f_n(0)=0 $和当$ n\to \infty $时, $ C_{\varphi}^gf_n(0)=C_b^gf_n \to 0 $. 所以我们可推出当$ n\to \infty $时, 有$ ||Tf_n||_{q, m}\to 0 $, 再通过引理2.3即可得出$ T $是紧算子.

利用本文的结论, 我们可以对$ \; V_g^{\varphi}, \; C_{\varphi}^g\; $在Fock sobolev空间中的拓扑性质做些简单的探讨, Mengestie在[10]中对Fock空间中$ \; V_g^{\varphi}\; $的拓扑结构做了更深入的研究.

推论3.4  设$ \; 1<p\leq q<\infty, \; g_1, \; g_2 $为整函数且不为常函数, $ \; \varphi_1(z)=a_1z+b_1, \; \varphi_2(z)=a_2z+b_2, \; 0<|a_1|, |a_2|\leq 1.\; V_{g_1}^{\varphi_1}, \; V_{g_2}^{\varphi_2}, $是$ F^{p, m}(\mathbb{C})\rightarrow F^{q, m}(\mathbb{C}) $上的有界算子, 如果$ \; V_{g_1}^{\varphi_1}-V_{g_2}^{\varphi_2}\; $是紧算子, 则有$ V_{g_1}^{\varphi_1} $和$ \; V_{g_2}^{\varphi_2} $都为紧算子或者$ a_1=a_2 $.

证  出于方便起见, 用$ T_1, \; T_2 $来代表$ \; V_{g_1}^{\varphi_1} $和$ V_{g_2}^{\varphi_2}\; $. 若$ T_1 $和$ T_2 $中某一个为紧算子, 则通过$ \; T_1-T_2 $为紧算子可以推出$ T_1, \; T_2 $都为紧算子.

若$ \; T_1, T_2\; $都不为紧算子, 我们使用反证法来证明$ \; a_1=a_2 $. 先假设有$ a_1 \neq a_2 $. 对$ (T_1-T_2)f(z)=\int_0^z f({\varphi_1}(\zeta))g_1'(\zeta)d\zeta-\int_0^zf({\varphi_2}(\zeta))g_2'(\zeta)d\zeta $使用引理2.4得

$ \begin{eqnarray} ||(T_1-T_2)f||_{q, m}^q &\simeq& |\big((T_1-T_2)f\big)(0)|^q \\ &&+ \int_\mathbb{C} \frac {|f({\varphi_1}(\zeta))g_1'(\zeta)-f({\varphi_2}(\zeta))g_2'(\zeta)|^q(1+|\zeta|)^q e^{-q\phi(\zeta)}} {(1+|\zeta| + \big| |\zeta|^2+|\zeta|-m \big|)^q}dA(\zeta)\\ &\gtrsim& |(T_1-T_2)f(0)|^q \\ &&+ \frac {|f({\varphi_1}(z))g_1'(z)-f({\varphi_2}(z))g_2'(z)|^q(1+|z|)^q e^{-q\phi(z)}} {(1+|z| + \big| |z|^2+|z|-m \big|)^q}dA(z). \end{eqnarray} $

将$ k_\omega $代入上式, 得

$ \begin{eqnarray} ||(T_1-T_2)k_\omega||_{q, m}^q &\gtrsim& |\big((T_1-T_2)k_\omega\big)(0)|^q \\ &&+ \frac {|k_\omega({\varphi_1}(z))g_1'(z)-k_\omega({\varphi_2}(z))g_2'(z)|^q(1+|z|)^q e^{-q\phi(z)}} {(1+|z| + \big| |z|^2+|z|-m \big|)^q}dA(z)\\ &\gtrsim& |\big((T_1-T_2)k_\omega\big)(0)|^q \\ &&+ \frac {(1+|z|)^q e^{-q\phi(z)}} {(1+|z| + \big| |z|^2+|z|-m \big|)^q}(|k_\omega({\varphi_1}(z))g_1'(z)|-|k_\omega({\varphi_2}(z))g_2'(z)|)^q . \end{eqnarray} $

令$ \; \omega=\varphi_1(z) $, 则存在正数C使得

$ \begin{eqnarray} M_{T_1}(z) &\simeq& \frac{(1+|z|) e^{-\phi(z)}} {(1+|z| + \big| |z|^2+|z|-m \big|)} |k_{\varphi_1(z)}{\varphi_1(z)}| \\ &\leq& \big(C||(T_1-T_2)k_{\varphi_1(z)}||_{q, m}^q - |\big((T_1-T_2)k_{\varphi_1(z)}\big)(0)|^q \big)^{\frac {1}{q}}\\ &&+ M_{T_2}(z)e^{-\phi(\varphi_2(z))}|k_{\varphi_1(z)}(\varphi_2(z))| \\ &\lesssim& \big(C||(T_1-T_2)k_{\varphi_1(z)}||_{q, m}^q - |\big((T_1-T_2)k_{\varphi_1(z)}\big)(0)|^q \big)^{\frac {1}{q}} + M_{T_2}(z)e^{-\theta|\varphi_1(z)-\varphi_2(z)|} . \end{eqnarray} $

通过$ T_1-T_2 $的紧性和$ \; a_1 \neq a_2 $, 可以得出当$ z \to \infty $时, 有$ ||(T_1-T_2)k_{\varphi_1(z)}||_{q, m} \to 0 $和$ e^{-\theta|\varphi_1(z)-\varphi_2(z)|}\to 0 $.又有$ |\big((T_1-T_2)k_{\varphi_1(z)}\big)(0)|=0 $和$ M_{T_2} $有界, 我们可以推出$ \; z \to \infty $时, $ M_{T_1}(z) \to 0 $. 根据定理1, 此时$ \; T_1 $为紧算子, 导出矛盾, 即得此时有$ a_1=a_2 $.

从上述的论证中, 注意到当$ z \to \infty $时, $ \; \big|\big((C_{\varphi_1}^{g_1}-C_{\varphi_2}^{g_2})\big) k_{\varphi_1(z)}\big)(0)\big|\to 0 $. 即可知此推论的结果对于$ \; C_{\varphi_1}^{g_1}-C_{\varphi_2}^{g_2}\; $同样成立.