西南师范大学学报(自然科学版)   2020, Vol. 45 Issue (2): 26-30.  DOI: 10.13718/j.cnki.xsxb.2020.02.005
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  • 调和Fock空间    [PDF全文]
    陈雪 , 黄穗     
    重庆师范大学 数学科学学院, 重庆 401331
    摘要:主要在解析Fock空间中函数的性质的基础上,讨论调和Fock空间中函数的性质结构.首先计算了调和Fock空间的标准正交基、再生核,得到了投影算子的积分表示形式.其次对调和Fock空间中的函数值进行了估计,证明了极值函数的存在性,得到了其基本性质,并且在此基础上讨论了不同调和Fock空间的关系.
    关键词调和Fock空间    再生核    投影算子    极值函数    

    $\mathbb{C}$表示复平面.对于任意α>0,我们考虑Gaussian测度

    $ {\text{d}}{\lambda _\alpha }\left( z \right) = \frac{\alpha }{{\text{\pi }}}{{\text{e}}^{ - \alpha {{\left| z \right|}^2}}}{\text{d}}A\left( z \right) $

    其中dA(z)=dxdy是复平面$\mathbb{C}$上的Lebesgue面积测度.

    L2($\mathbb{C}$,dλα(z))是复平面$\mathbb{C}$上关于Gaussian测度dλα(z)平方可积的Lebesgue可测函数全体.经典Fock空间Fα2是由L2($\mathbb{C}$,dλα(z))中的全体整函数构成的闭子空间;调和Fock空间Fh2是由L2($\mathbb{C}$,dλα(z))中的全体调和函数构成的闭子空间.因此Fα2Fh2是Hilbert空间,其上的内积定义为

    $ \left\langle {f,g} \right\rangle = \int_\mathbb{C} f (z)\bar g(z){\text{d}}{\lambda _\alpha }(z) $

    范数定义为

    $ \left\| f \right\|_{2,\alpha }^2 = \frac{\alpha }{{\text{\pi }}}\int_\mathbb{C} {{{\left| {f(z){{\text{e}}^{ - \frac{\alpha }{2}|z{|^2}}}} \right|}^2}} {\text{d}}A(z) $

    F02={fFα2f(0)=0},经典Fock空间Fα2和调和Fock空间Fh2有如下关系:

    $ F_h^2 = F_\alpha ^2 \oplus \overline {F_0^2} $

    即对∀fFh2,存在f1Fα2f2F02,使得f=f1+$\overline{f_{2}}$.

    与有界区域上经典的Hardy空间、Bergman空间相比,作为定义在无界区域上的整函数空间,对Fock空间的研究相对要滞后一些.为了刻画量子力学中由单粒子或多个全同粒子构成的量子态空间,文献[1]首次构造了一个无界区域上的全纯函数空间来表示Heisenberg群,其后称之为Fock空间.文献[2-3]引入了n维复空间$\mathbb{C}$n上关于Gaussian测度平方可积的整函数全体构成的Hilbert空间作为量子力学中算子的模型空间,并且证明了此空间与Fock空间同构,所以Fock空间又称为Segal-Bargmann空间.在Bargman空间的基础上,不少学者研究讨论了调和Bargman空间的性质结构(如文献[4]).近20年来,Fock空间及其上的算子理论的研究吸引了越来越多学者的关注,研究成果越来越丰富.文献[3, 5-6]系统地研究了这类Fock空间的性质结构.文献[7-9]将其推广成为Fock-Sobolev空间,并且对其性质结构进行了研究.文献[10]引入了一类由某类特殊函数所诱导的加权Fock型空间.文献[5]是关于Fock空间及其相关函数空间与算子理论的第一本专著.文献[11]对调和Fock空间上函数的Berezin变换进行了估计.

    本文主要在解析Fock空间中函数的性质的基础上,讨论调和Fock空间中函数的基本性质.首先计算了调和Fock空间的标准正交基、再生核,得到了投影算子的积分表示形式.其次对调和Fock空间中的函数值进行了估计,证明了极值函数的存在性,得到了其基本性质结构,并且在此基础上讨论了不同调和Fock空间的关系.

    调和Fock空间Fh2L2($\mathbb{C}$,dλα(z))的闭子空间,由其与Fock空间的关系,在Fock空间Fα2的标准正交基的基础上得到Fh2的标准正交基.

    引理1  对任意非负整数n,令$e_{n}(z)=\sqrt{\frac{\alpha^{n}}{n!}} z^{n}$$\bar{e}_{n}(z)=\sqrt{\frac{\alpha^{n}}{n!}} \bar{z}^{n}$,则{en}n=0+∞∪{en}n=1+∞Fh2的规范正交基.

      {en}n=0+∞∪{en}n=1+∞显然是Fh2中的规范正交集.下面证{en}n=0+∞∪{en}n=1+∞Fh2上是完全的.设fFh2,〈fen〉=0,〈fen〉=0,因为f是调和函数,所以可以表示为

    $ f(z) = \sum\limits_{n = 0}^{ + \infty } {{a_n}} {z^n} + \sum\limits_{m = 1}^{ + \infty } {{b_m}} {{\bar z}^m} $

    通过直接计算得

    $ \begin{gathered} \left\langle {f,{e_k}} \right\rangle = \left\langle {\sum\limits_{n = 0}^{ + \infty } {{a_n}} {z^n} + \sum\limits_{n = 1}^{ + \infty } {{b_m}} {{\bar z}^m},{e_k}} \right\rangle = \left\langle {\sum\limits_{n = 0}^{ + \infty } {{a_n}} {z^n},{e_k}} \right\rangle + \left\langle {\sum\limits_{n = 1}^{ + \infty } {{b_m}} {{\bar z}^m},{e_k}} \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 0}^{ + \infty } {{a_n}} \left\langle {{z^n},{e_k}} \right\rangle + 0 = \sum\limits_{n = 0}^{ + \infty } {{a_n}} \left\langle {{z^n},\sqrt {\frac{{{\alpha ^k}}}{{k!}}} } \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;{a_k}\left\langle {{z^k},\sqrt {\frac{{{\alpha ^k}}}{{k!}}} } \right\rangle = 0 \hfill \\ \end{gathered} $

    ak=0.又因为

    $ \begin{gathered} \left\langle {f,{e_k}} \right\rangle = \left\langle {\sum\limits_{n = 0}^{ + \infty } {{a_n}} {z^n} + \sum\limits_{m = 1}^{ + \infty } {{b_m}} {{\bar z}^m},{e_k}} \right\rangle = \left\langle {\sum\limits_{n = 0}^{ + \infty } {{a_n}} {z^n},{e_k}} \right\rangle + \left\langle {\sum\limits_{m = 1}^{ + \infty } {{b_m}} {{\bar z}^m},{e_k}} \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;0 + \left\langle {\sum\limits_{m = 1}^{ + \infty } {{b_m}} {{\bar z}^m},{e_k}} \right\rangle = \sum\limits_{m = 1}^{ + \infty } {{b_m}} \left\langle {{{\bar z}^m},\sqrt {\frac{{{\alpha ^k}}}{{k!}}} } \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;{b_k}\left\langle {{{\bar z}^k},\sqrt {\frac{{{\alpha ^k}}}{{k!}}} } \right\rangle = 0 \hfill \\ \end{gathered} $

    bk=0,从而f=0.因此{en}n=0+∞∪{en}n=1+∞Fh2上是完全的.证明完毕.

    对固定的ω$\mathbb{C}$,由于映射$f \longmapsto f(\omega)$Fh2上的有界线性泛函,所以由Riesz表示定理,存在唯一的函数HωFh2,使得对∀fFh2f(ω)=〈fHω〉,函数H(zω)=Hz(ω)叫作Fh2的再生核.下面通过标准正交基计算出调和Fock空间的再生核.

    定理1  Fh2的再生核为Hz(ω)=eαzω+eαzω-1.

      设fFh2,则f=f1+f2,其中f1Fα2f2F02.由于Fh2是Hilbert空间,故

    $ \begin{gathered} {H_z}(\omega ) = \sum\limits_{n = 0}^{ + \infty } {{e_n}} (z)\overline {{e_n}(\omega )} + \sum\limits_{n = 1}^{ + \infty } {\overline {{e_n}(z)} } \overline{\overline {{e_n}(\omega )}} = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 0}^{ + \infty } {{e_n}} (z)\overline {{e_n}(\omega )} + \sum\limits_{n = 1}^{ + \infty } {\overline {{e_n}(z)} } \overline{\overline {{e_n}(\omega )}} - \overline {{e_0}(z)} \overline{\overline {{e_0}(\omega )}} = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 0}^{ + \infty } {\frac{{{\alpha ^n}{z^n}{{\bar \omega }^n}}}{{n!}}} + \sum\limits_{n = 0}^{ + \infty } {\frac{{{\alpha ^n}{{\bar z}^n}{\omega ^n}}}{{n!}}} - 1 = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;{{\text{e}}^{\alpha z\bar \omega }} + {{\text{e}}^{\alpha z\bar \omega }} - 1 \hfill \\ \end{gathered} $

    由于Fh2L2($\mathbb{C}$,dλα(z))的闭子空间,由Hilbert空间的性质结构,存在从L2($\mathbb{C}$,dλα(z))到Fh2的唯一正交投影Qα.

    推论1  投影算子QαL2($\mathbb{C}$,dλα)→Fh2可以表示为

    $ {Q_\alpha }f\left( z \right) = \int_\mathbb{C} {f\left( \omega \right)\overline {{H_z}\left( \omega \right)} {\text{d}}{\lambda _\alpha }\left( \omega \right)} $

    其中fL2($\mathbb{C}$,dλα).

      设fL2($\mathbb{C}$,dλα),有

    $ {Q_\alpha }f\left( z \right) = \left\langle {{Q_\alpha }f,{H_z}} \right\rangle = \left\langle {f,{H_z}} \right\rangle = \int_\mathbb{C} f (\omega )\overline {{H_z}(\omega )} {\text{d}}{\lambda _\alpha }(\omega ) $

    其中z$\mathbb{C}$,这就证明了投影算子Qα是以Hz(ω)为核的积分算子.

    现在将Fh2推广到Fhp(1≤p < +∞).对∀α>0,p>0,用Lαp表示所有复平面$\mathbb{C}$上满足$f(z) e^{-\frac{a|z|^{2}}{2}}$Lp($\mathbb{C}$,dA)的Lebesgue可测函数全体.对fLαp,定义

    $ \left\| f \right\|_{p,\alpha }^p = \frac{{p\alpha }}{{2{\text{\pi }}}}\int_0 {{{\left| {f\left( z \right){{\text{e}}^{ - \frac{{\alpha |z{|^2}}}{2}}}} \right|}^p}} {\text{d}}A(z) $

    类似地,对α>0,p=+∞,用Lα来表示所有复平面$\mathbb{C}$上满足

    $ {\left\| f \right\|_{\infty ,\alpha }} = esssup \left\{ {\left| {f\left( z \right)} \right|{{\text{e}}^{ - \frac{{a{{\left| z \right|}^2}}}{2}}}:z \in \mathbb{C}} \right\} < + \infty $

    的函数全体.

    α>0,1≤p≤+∞,定义Fhp是由所有调和函数fLαp构成的空间,称之为调和Fock空间.显然FhpLαp的闭子空间.关于Fhp的结构,有以下结果:

    定理2  设fFhp(1≤p < +∞),fr(z)=f(rz).则:

    (ⅰ)‖fr-fpα0,r→1-

    (ⅱ)存在调和多项式{gn},使得‖gn-fpα0,n+∞.

      (ⅰ)设fFhp,则存在f1Fαpf2F0p,使得f=f1+$\overline{f_{2}}$,那么fr=f1r+$\overline{f_{2r}}$.由解析Fock空间Fαp的性质可得

    $ {\left\| {{f_{1r}} - {f_1}} \right\|_{p,\alpha }} \to 0\;\;\;\;r \to {1^ - } $
    $ {\left\| {{f_{2r}} - {f_2}} \right\|_{p,\alpha }} \to 0\;\;\;\;r \to {1^ - } $

    因此

    $ {\left\| {{f_r} - f} \right\|_{p,\alpha }} \to 0\;\;\;\;r \to {1^ - } $

    (ⅱ)由于解析多项式在Fock空间Fαp中稠密,故存在pnFαpqmFαp,使得

    $ {\left\| {{p_n} - {f_1}} \right\|_{p,\alpha }} \to 0\;\;\;\;n \to + \infty $
    $ {\left\| {{q_m} - {f_2}} \right\|_{p,\alpha }} \to 0\;\;\;\;m \to + \infty $

    $g_{n, m}=p_{n}+\overline{q_{m}}$,则

    $ {\left\| {{g_{n,m}} - f} \right\|_{p,\alpha }} \to 0\;\;\;\;n,m \to + \infty $

    由定理2的证明,可知调和多项式在Fhp中稠密.

    下面将对函数值进行估计.

    定理3  设fFhp(1≤p < +∞),则sup{|f(z)|:‖fpα≤1}=$e^{\frac{\alpha|z|^{2}}{2}}$.

      由于fFhp,则由|f|p的次调和性可得

    $ {\left| {f\left( 0 \right)} \right|^p} \leqslant \frac{{p\alpha }}{{2{\text{\pi }}}}\int_\mathbb{C} {{{\left| {f\left( \omega \right){{\text{e}}^{ - \frac{{\alpha |\omega {|^2}}}{2}}}} \right|}^p}} {\text{d}}A\left( \omega \right) = \left\| f \right\|_{p,\alpha }^p $

    τz(ω)=z-ω,从而

    $ \begin{gathered} {\left| {f\left( z \right)} \right|^p} = \left| {f \circ {\tau _z}\left( 0 \right)} \right| \leqslant \int_\mathbb{C} {{{\left| {f \circ {\tau _z}\left( 0 \right){{\text{e}}^{ - \frac{{\alpha |\omega {|^2}}}{2}}}} \right|}^p}} {\text{d}}A\left( \omega \right) = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\int_\mathbb{C} {{{\left| {f\left( {z - \omega } \right){{\text{e}}^{ - \frac{{\alpha |\omega {|^2}}}{2}}}} \right|}^p}} {\text{d}}A\left( \omega \right) = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\int_\mathbb{C} {{{\left| {f\left( {\omega '} \right){{\text{e}}^{ - \frac{{\alpha {{\left| {z - \omega '} \right|}^2}}}{2}}}} \right|}^p}} {\text{d}}A\left( \omega \right) = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{\int_\mathbb{C} {\left| {f\left( \omega \right)} \right|} ^p}{{\text{e}}^{ - \frac{{\alpha p}}{2}\left( {{{\left| z \right|}^2} + {{\left| \omega \right|}^2} - \bar z\omega - \bar z\omega } \right)}}{\text{d}}A\left( \omega \right) = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{{\text{e}}^{ - \frac{{\alpha p{{\left| z \right|}^2}}}{2}}}{\int_\mathbb{C} {\left| {f\left( \omega \right)} \right|} ^p}{{\text{e}}^{ - \frac{{\alpha p}}{2}\left( {|\omega {|^2} - \bar z\omega - \bar z\omega } \right)}}{\text{d}}A(\omega ) = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{{\text{e}}^{ - \frac{{\alpha p{{\left| z \right|}^2}}}{2}}}{\int_\mathbb{C} {\left| {f\left( \omega \right){{\text{e}}^{ - \frac{{\alpha {{\left| \omega \right|}^2}}}{2}}}} \right|} ^p}{{\text{e}}^{ - \frac{{\alpha p}}{2}\left( {|\omega {|^2} - \bar z\omega - \bar z\omega } \right)}}{\text{d}}A(\omega ) \hfill \\ \end{gathered} $

    F(ω)=f(ω)$\mathrm{e}^{-\frac{\alpha \bar{z} \omega}{2}-\frac{\alpha \bar{\omega} z}{2}}$,则

    $ {\left| {f(z){{\text{e}}^{ - \alpha |z{|^2}}}} \right|^p} \leqslant {{\text{e}}^{ - \frac{{ap|z{|^2}}}{2}}}\int_\mathbb{C} {{{\left| {f(\omega ){{\text{e}}^{ - \frac{{\alpha |\omega {|^2}}}{2}}}} \right|}^p}} {\text{d}}A(\omega ) $

    那么

    $ {\left| {f(z)} \right|^p} \leqslant {{\text{e}}^{\frac{{\alpha p|z{|^2}}}{2}}}\int_\mathbb{C} {{{\left| {f(\omega ){{\text{e}}^{ - \frac{{\alpha |\omega {|^2}}}{2}}}} \right|}^p}} {\text{d}}A(\omega ) $

    f(ω)=${\rm{e}}^{-\alpha \bar{z} \omega-\left(\frac{\alpha|z|^{2}}{2}\right)+i \theta}$时,不等式取等号,从而有sup{|f(z)|:‖fpα≤1}=${\rm{e}}^{\frac{a|z|^{2}}{2}}$.

    推论2  设fFhp,1≤p≤+∞,则|f(z)|≤‖fpα${\rm{e}}^{\frac{a|z|^{2}}{2}}$(∀z$\mathbb{C}$).

    上面对单个函数值的估计是最精确的,也就是说${\rm{e}}^{\frac{a|z|^{2}}{2}}$是一个极值函数.在此估计上定义以下函数空间:用fh表示由调和函数f(z)构成的空间,其中f(z)满足$\lim\limits_{z \rightarrow \infty} f(z) \mathrm{e}^{-\frac{a|z|^{2}}{2}}=0$.显然,fhFh的闭子空间.实际上关于{Fhp}p≥1有以下关系:

    定理4  设1≤p < q < +∞,则FhpFhqFh.

      对fFhp,通过计算,得

    $ \begin{gathered} \left\| f \right\|_{q,\alpha }^q = \frac{{q\alpha }}{{2{\text{\pi }}}}{\int_\mathbb{C} {\left| {f(z)} \right|} ^p}{\left| {f(z)} \right|^{q - p}}{{\text{e}}^{ - \frac{{q\alpha {{\left| z \right|}^2}}}{2}}}{\text{d}}A(z) \leqslant \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\frac{{q\alpha }}{{2{\text{\pi }}}}\left\| f \right\|_{q,\alpha }^{p - q}\int_\mathbb{C} {\left| {f(z){{\text{e}}^{ - \frac{{q\alpha {{\left| z \right|}^2}}}{2}}}} \right|{\text{d}}A(z)} = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\frac{q}{p}\left\| f \right\|_{p,\alpha }^q \hfill \\ \end{gathered} $

    $\|f\|_{q \cdot \alpha} \leqslant\left(\frac{q}{p}\right)^{\frac{1}{q}}\|f\|_{p \cdot a}$,因此可得FhpFhq.由推论2有|f(z)|≤‖fpα$\mathrm{e}^{-\frac{a|z|^{2}}{2}}$,故$|f(z) | \mathrm{e}^{-\frac{a|z|^{2}}{2}}$≤‖fpα,因此‖f∞,α≤‖fpα,从而有FhpFh.

    实际上,我们还有以下结果:

    推论3  设1≤p < +∞,则Fhpfh.

      由于调和多项式P包含在fh中,并且调和多项式PFhp中稠密,所以Fhpfh.

    由于Fα2L2($\mathbb{C}$,dλα)的闭子空间,故存在唯一的投影算子PαL2($\mathbb{C}$,dλα)→Fα2,其表示为

    $ {P_\alpha }(z) = \left\langle {f,{K_z}} \right\rangle = \int_\mathbb{C} f (\omega )\overline {{K_z}(\omega )} {\text{d}}{\lambda _\alpha }(\omega ) $

    文献[5]已证明了解析Fock空间Fα2上的投影Pα是有界的,因此,由Pα与Qα的关系有以下结果:

    定理5  投影算子QαL2($\mathbb{C}$,dλα)→Fh2有界.

      由于投影算子PαL2($\mathbb{C}$,dλα)→Fα2有界,且

    $ \begin{gathered} {Q_\alpha }f(z) = \left\langle {f,{H_z}} \right\rangle = \left\langle {f,{K_z} + \overline {{K_z}} - 1} \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\left\langle {f,{H_z}} \right\rangle + \left\langle {f,{H_z}} \right\rangle - \left\langle {f,1} \right\rangle = \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{P_\alpha }f} \right)(z) + \overline {(P\bar f)(z)} - Pf(0) \hfill \\ \end{gathered} $

    所以

    $ \begin{array}{*{20}{c}} {\left\| {{Q_\alpha }f} \right\| \leqslant \left\| {{P_\alpha }f} \right\| + \left\| {\overline {{P_\alpha }\bar f} } \right\| + \left\| f \right\| \leqslant } \\ {\left( {2\left\| {{P_\alpha }} \right\| - 1} \right)\left\| f \right\|} \end{array} $

    因此Qα有界.

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    On Harmonic Fock Space
    CHEN Xue , HUANG Sui     
    School of Mathematical Sciences, Chongqing Normal University, Chongqing 401331, China
    Abstract: In this paper, the structure of harmonic Fock space has been studied on the basis of analytic Fock space. Firstly, the orthonormal basis of space has been calculated, and kernel and integration form of orthogonal projection been reproduced. Secondly, the value of functions and obtain extremal functions have been estimated. Moreover, the relations among different harmonic Fock Spaces have also been discussed.
    Key words: harmonic Fock space    reproducing kernel    projection operator    the extremal functions    
    X