关于高维函数方程解析解的一个重要结果
我们研究下面这个(复变量)方程的局部解析解:
\[\begin{equation}\tag{eq-smajdor}
\phi(z)=h(z,\phi(f(z)))
\end{equation}
\]
其中
\[f:\mathbb{C}^n\to \mathbb{C}^n,~~~~h:\mathbb{C}^n\times \mathbb{C}^m\to \mathbb{C}^m
\]
定义在\(z=0\)和\((z,w)=(0,0)\)的某邻域内, 且\(\phi:\mathbb{C}^n\to \mathbb{C}^m\)是未知函数. 定义\(\mathbb{C}^n\)和\(\mathbb{C}^m\)中向量的范数: \(z=(z_1,\cdots,z_n),w=(w^1,\cdots,w^m)\),
\[\begin{equation}\tag{eq-5-1}
\|z\|=\left(\sum_{i=1}^{n}|z_i|\right)^{1/2},~~~
\|w\|_0=\left(\sum_{k=1}^{m}|w^k|^2\right)^{1/2}
\end{equation}
\]
并对于\(U=[u_i^j]_{n\times n},V=[v_l^k]_{m\times m}\), 定义
\[\begin{equation}\tag{eq-5-2}
\|U\|_*=\sup_{|t|=1}\left(\sum_{i=1}^{n}\left|\sum_{j=1}^{n}u_i^jt_j\right|^2\right)^{1/2},~~~
\
\|V\|^*=\left(\sum_{k,l=1}^{m}|v_l^k|^2\right)^{1/2}.
\end{equation}
\]
注记: 为了与后面的保持一致, 在这种写法\([u_i^j]\)中, 规定下标\(i\)表示列指标, 上标\(j\)表示行指标. 那么
\[\|U\|_*=\sup_{|t|=1}\|U^Tt\|.
\]
假设
- (I) \(f(z)=(f_1(z_1,\cdots,f_n(z)),\cdots,
f_n(z_1,\cdots,f_n(z)))\)在\(|z|\le r_0\)解析且在此集合中(即\(|z|\le r_0\))有展开式
\[\begin{equation}\tag{eq-5-3}
f_i(z_1,\cdots,f_n(z))=\sum_{\mu=1}^{\infty}\sum_{j_1,\cdots,j_\mu=1}^{n}
b^i_{j_1\cdots j_\mu}z_{j_1}\cdots z_{j_\mu},~~~~i=1,\cdots,n.
\end{equation}
\]
- (II)\(f\)在\(z=0\)处的Jacobi矩阵\(B=[b_i^j]_{n\times n}\)满足
\[\begin{equation}\tag{eq-5-4}
\|B\|_*<1.
\end{equation}
\]
- (III) \(h(z,w)=(h^1(z_1,\cdots,z_n;w^1,\cdots, w^m),\cdots,h^m(z_1,\cdots,z_n;w^1,\cdots,w^m))\)
在\(|z|\le r_0\),\(\|w\|\le R_0\)解析且有展开式
\[\begin{equation}\tag{eq-5-5}
h^k(z_1,\cdots,z_n;w^1,\cdots,w^m)=
\sum_{\substack{\mu,\nu=0 \\ \mu+\nu\ge 1}}^{\infty}
\sum_{\substack{j_i=0 \\ 1\le i\le \mu}}^{n}
\sum_{\substack{l_t=1 \\ 1\le t\le \nu}}^{n}
a_{j_1\cdots j_\mu;l_1\cdots l_\nu}^k
z_{j_1}\cdots z_{j_\mu}w^{l_1}\cdots w^{l_\nu},~~~~k=1,\cdots,m.
\end{equation}
\]
方程{eq-smajdor}在原点的某邻域内有一个解析解\(\phi(z)\)(\(\phi(0)=0\))的一个必要条件是它在原点的某邻域内有形式级数解
\[\begin{equation}\tag{eq-5-6}
\phi(z)=(\phi^1(z_1,\cdots,z_n),\cdots,\phi^m(z_1,\cdots,z_n)),
\end{equation}
\]
其中
\[\begin{equation}\tag{eq-5-7}
\phi^k(z_1,\cdots,z_n)=\sum_{\mu=1}^{\infty}\sum_{j_1,\cdots,j_\mu=1}^{n}c_{j_1\cdots j_\mu}^k z_{j_1}\cdots z_{j_\mu},~~~k=1,\cdots,m.
\end{equation}
\]
通过对比系数我们得到
\[\begin{equation}\tag{eq-5-8}
\begin{aligned}
c_i^k=&a_{i,0}^k+\sum_{l=1}^{m}\sum_{j=1}^{n}a_{0,l}^k c_j^l b_i^j \\
\cdots &\cdots \cdots\\
c_{i_1,\cdots,i_q}^k=&F_{i_1,\cdots,i_q}^k+\sum_{l=1}^{m}\sum_{j_1,\cdots,j_q=1}^{n}
a_{0,l}^k c_{j_1,\cdots,j_q}^l b_{i_1}^{j_1}\cdots b_{i_q}^{j_q},
\end{aligned}
\end{equation}
\]
对于 \(~k=1,\cdots,m; ~q=2,3,\cdots;i_1,\cdots,i_q=1,\cdots,n;\)其中\(F^k_{i_1,\cdots,i_q}\)是\(c^l_{j_1,\cdots,j_s},l=1,\cdots,i;j_1,\cdots,j_s=1,\cdots,n;s=1\cdots,q-1\) 以及 \(f\)和\(h\)的展开式中的系数(这表明\(c_{i_1,\cdots,i_q}^k\)被递归的确定). 进而, 利用混合偏导相等的性质, 我们可以得到
\[\begin{equation}\tag{eq-5-9}
c_{i_{p_1},\cdots,i_{p_q}}^k=c_{i_{1},\cdots,i_{q}}^k,~~~k=1,\cdots,m;i_1,\cdots,i_q=1,\cdots,n; q=1,2,\cdots
\end{equation}
\]
其中\((p_1,\cdots,p_q)\)是\((1,\cdots,q)\)的任一置换.
{{eq-5-8} 的推导过程如下:
首先把各个函数的展开式代入{eq-smajdor}, 我们有
\[\begin{equation}\tag{eq-5-10}
\begin{aligned}
\phi^k(z_1,\cdots,z_n)
=\sum_{\mu=1}^{\infty}\sum_{j_1,\cdots,j_\mu=1}^{n}c_{j_1\cdots j_\mu}^k z_{j_1}\cdots z_{j_\mu},\\
\end{aligned}
\end{equation}\
\]
\[\begin{equation}\tag{eq-5-11}
\begin{aligned}
&h^k(z,\phi(f(z)))h^k(z_1,\cdots,z_n;\phi^1(f_1(z),\cdots,f_n(z)),\cdots,\phi^m(f_1(z),\cdots,f_n(z)))\\
=&
\sum_{\substack{\mu,\nu=0 \\ \mu+\nu\ge 1}}^{\infty}
\sum_{j_1\cdots j_\mu=1}^{n}
\sum_{l_1\cdots l_\nu=1}^{m}a_{j_1\cdots j_\mu;l_1\cdots l_\nu}^k z_{j_1}\cdots z_{j_\mu} \cdot
\phi^{l_1}(f_{j_1}(z),\cdots,f_{j_\mu}(z))\cdots \phi^{l_\nu}(f_{j_1}(z),\cdots,f_{j_\mu}(z))\\
=&
\sum_{\substack{\mu,\nu=0 \\ \mu+\nu\ge 1}}^{\infty}
\sum_{j_1 \cdots j_\mu=1}^{n}
\sum_{l_1 \cdots l_\nu=1}^{m}a_{j_1\cdots j_\mu;l_1\cdots l_\nu}^k z_{j_1}\cdots z_{j_\mu} \cdot
\\
&\cdot
\left[\sum_{\mu=1}^{\infty}\sum_{j_1\cdots j_\mu=1}^{n}c_{j_1\cdots j_\mu}^{l_1}f_{j_1}(z)
\cdots f_{j_\mu}(z)\right]
\cdots
\left[\sum_{\mu=1}^{\infty}\sum_{j_1\cdots j_\mu=1}^{n}c_{j_1\cdots j_\mu}^{l_\nu}f_{j_1}(z)
\cdots f_{j_\mu}(z)\right]\\
=&
\sum_{\substack{\mu,\nu=0 \\ \mu+\nu\ge 1}}^{\infty}
\sum_{j_1 \cdots j_\mu=1}^{n}
\sum_{l_1 \cdots l_\nu=1}^{m}a_{j_1\cdots j_\mu;l_1\cdots l_\nu}^k z_{j_1}\cdots z_{j_\mu} \cdot
\\
&\cdot
\left\{\sum_{\mu=1}^{\infty}\sum_{j_1\cdots j_\mu=1}^{n}c_{j_1\cdots j_\mu}^{l_1}
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_1}z_{r_1}\cdots z_{r_\mu}\right]
\cdots
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_\mu}z_{r_1}\cdots z_{r_\mu}\right]
\right\}\\
& \cdots \cdots \cdots\\
&\left\{\sum_{\mu=1}^{\infty}\sum_{j_1\cdots j_\mu=1}^{n}c_{j_1\cdots j_\mu}^{l_\nu}
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_1}z_{r_1}
\cdots z_{r_\mu}\right]
\cdots
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_\mu}z_{r_1}
\cdots z_{r_\mu}\right]
\right\}\\
\end{aligned}
\end{equation}
\]
想要从这个非常复杂的等式中看出第\(k\)个分量中的项\(z_{j_1}z_{j_2}\cdots z_{j_p}\)的系数的对应关系,
观键在于处理好\(\sum\)({\color{red}要逐层的拨开\(\sum\)}).
现在, 我们首先观察{eq-5-10}和{eq-5-11}中的一次项中的\(z_i\)的次数:
先找\(\mu=1,\nu=0\)的项, 有
\[a_{i,0}^k
\]
再找\(\mu=0,\nu=1\)的项,
\((\mu,\nu)=(0,1)\)去掉了第一个\(\sum\),
\(\mu=0\)去掉了第二个\(\sum\),
因为是一次的, 所以后面的\(\{\cdot\}\)只能存在一个,
第三\(\sum\)写为\(\sum_{l=1}^{n}\), 有
\[\sum_{l=1}^{n}a_{0,l}^k
\]
再从
\[\left\{\sum_{\mu=1}^{\infty}\sum_{j_1\cdots j_\mu=1}^{n}c_{j_1\cdots j_\mu}^{l_\nu}
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_1}z_{r_1}
\cdots z_{r_\mu}\right]
\cdots
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_\mu}z_{r_1}
\cdots z_{r_\mu}\right]
\right\}
\]
中去掉第一个\(\sum\)(因为要找一次项), 第二个\(\sum\)改为\(\sum_{j=1}^{n}\), 得到
\[\sum_{j=1}^{n}c_j^l
\]
下面的这些\([\cdot]\)中
\[\left[\sum_{\mu=1}^{\infty}
\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_1}z_{r_1}
\cdots z_{r_\mu}\right]
\cdots\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_\mu}z_{r_1}
\cdots z_{r_\mu}\right]
\]
只有一个存在
\[\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j}z_{r_1}
\cdots z_{r_\mu}\right]
\]
要找\(z_i\)的系数(\(\mu=1\)去掉了第一个\(\sum\)), 于是
\[b^j_i.
\]
综上得到
\[c^k_i=a_{i,0}^k+\sum_{l=1}^{n}a_{0,l}^k\left(\sum_{j=1}^{n}c_j^l\right)b^j_i
=a_{i,0}^k+\sum_{l=1}^{n}\sum_{j=1}^{n}a_{0,l}^kc_j^l b^j_i,
\]
这就是{eq-5-8}的第一个等式.
我们继续推导{eq-5-8}的第二个等式(通过对比\(z_{i_1}z_{i_2}\cdots z_{i_p}\)的系数来完成).
从{eq-5-11}中找\(z_{i_1}z_{i_2}\cdots z_{i_p}\)的系数:
我们把所有\(c_{j_1\cdots j_\mu}^{k}\)的下标\(\mu<q\)的项都全部整合在一起, 记为\(F_{i_1\cdots i_q}^k\),
只看能出现\(c_{j_1\cdots j_q}^{k}\)的项.
{eq-5-11}最后一个等号后面的第一行中\(\mu=0,\nu=1\),
去掉{eq-5-11}最后一个等号后面的第一和第二个\(\sum\),第三个\(\sum\)写为\(\sum_{l=1}^{m}\),
有
\[\sum_{l=1}^{m}a_{0,l}^k
\]
后面的\(\{\cdot\}\)中只有一个:
\[\left\{\sum_{\mu=1}^{\infty}\sum_{j_1\cdots j_\mu=1}^{n}c_{j_1\cdots j_\mu}^{l}
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_1}z_{r_1}
\cdots z_{r_\mu}\right]
\cdots
\left[\sum_{\mu=1}^{\infty}\sum_{r_1\cdots r_\mu=1}^{n}b_{r_1\cdots r_\mu}^{j_\mu}z_{r_1}
\cdots z_{r_\mu}\right]
\right\}
\]
要从它中找\(z_{i_1}z_{i_2}\cdots z_{i_p}\)的系数, \(\mu=p\)去掉第一个\(\sum\), 上式变为
\[\left\{\sum_{j_1\cdots j_p=1}^{n}c_{j_1\cdots j_p}^{l}
\left[\sum_{r_1\cdots r_p=1}^{n}b_{r_1\cdots r_p}^{j_1}z_{r_1}
\cdots z_{r_p}\right]
\cdots
\left[\sum_{r_1\cdots r_p=1}^{n}b_{r_1\cdots r_p}^{j_p}z_{r_1}
\cdots z_{r_p}\right]
\right\}
\]
由于\(j_s\ge1\), 所以只能从每个\([\cdot]\)中找到一个\(z_{i_s}\),
设从第\(r\)个\([\cdot]\)中找到\(z_{i_{s_r}}\), 于是
\[\sum_{j_1\cdots j_p=1}^{n}c_{j_1\cdots j_p}^{l}b_{i_{s_1}}^{j_1}\cdots b_{i_{s_p}}^{j_p}
\]
其中\((i_{s_1},\cdots,i_{s_p})\)是\((1,\cdots,p)\)的一个置换.
因为\(j_1,\cdots,j_p\)每一个从\(1,\cdots,n\)是"遍历的", 于是上式可以写为
\[\sum_{j_1\cdots j_p=1}^{n}c_{j_1\cdots j_p}^{l}b_{i_1}^{j_1}\cdots b_{i_p}^{j_p}.
\]
综上得到
\[c_{i_1\cdots i_p}^k=F_{i_1\cdots i_p}^k+\sum_{l=1}^{m}\sum_{j_1\cdots j_p=1}^{n}a_{0,l}^kc_{j_1\cdots j_p}^l b_{i_1}^{j_1}\cdots b_{i_p}^{j_p}.
\]
因此, 如果存在{eq-5-8}的解\(\{c_{i_1,\cdots,i_q}^k\}(k=1,\cdots,m;i_1,\cdots,i_q=1,\cdots,n;q=1,2,\cdots)\)
满足{eq-5-9}, 那么方程{eq-smajdor}在原点的某邻域内有一个形式级数解{eq-5-6}.
下面我们假设
- (IV) {eq-smajdor}有一个形式级数解{eq-5-6}.
我们引入一列向量空间\(W^{(q)},q=0,1,2,\cdots:W^{(0)}=C^m,W^{(q)}(q>0)\)是\(m\times n^q\)-tuples
\(w^{(q)}=\{w^k_{i_1\cdots i_q}\}(k=1,\cdots,m;i_1,\cdots,i_q=1,\cdots,n)\)的集合,
其中\(w^k_{i_1\cdots i_q}\in \mathbb{C}\), 具有通常的数的加法和乘法.
定义\(W^{(q)}\)中的范数
\[\begin{equation}\tag{eq-5-12}
\|w^{(q)}\|_q=\left(\sum_{k=1}^{m}\sum_{i_1\cdots i_q=1}^{n}|w^k_{i_1\cdots i_q}|^2\right)^{1/2}.
\end{equation}
\]
设正整数\(p\)满足
\[\|[a_{0,l}^k]\|^*(\|B\|_*)^p<1.
\]
这样的\(p\)存在是因为\(\|B\|_*<1\).
因此存在一个常数\(\theta<1\)使得
\[\begin{equation}\tag{eq-5-13}
\|[a_{0,l}^k]\|^*(\|B\|_*)^p<\theta.
\end{equation}
\]
利用函数\(\frac{\partial h^k(z,w)}{\partial w^l}\)和\(\frac{\partial f_j(z)}{\partial z_i}\)的连续性,
我们可以得到: 存在\(r_1\)和\(R\) {\color{red}(\(r_1\le r_0,R\le R_0\))}使得
\[\begin{equation}\tag{eq-5-14}
\left|\left|\frac{\partial h^k(z,w)}{\partial w^l}\right|\right|^*
\left(\left|\left| \left[\frac{\partial f_j(z)}{\partial z_i}\right]\right|\right|_*\right)^p<\theta,
~~~~||z||\le r_1,~\|w\|_0\le R.
\end{equation}
\]
引理(lemma-5-1)
设\([\alpha_l^k],[\beta_i^j]\)是两个\(m\times m\)和\(n\times n\)的矩阵,且
\(\gamma^{(p)}=\{\gamma^l_{j_1\cdots j_p}\}(l=1,\cdots,m;j_1,\cdots,j_p=1,\cdots,n)\)
是空间\(W^{(p)}\)中的一个向量.
那么
\[\begin{equation}\tag{eq-5-15}
\|[\alpha_l^k]\gamma^{(p)} [\beta_i^j]^p\|^2
=
\sum_{k=1}^{m} \sum_{i_1,\cdots,i_p=1}^{n} \left|\sum_{l=1}^{m}\sum_{j_1\cdots j_p=1}^{n}\alpha_l^k
\gamma_{j_1\cdots j_p}^{l}\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}\right|^2
\le (\|[\alpha_l^k]\|^*)^2(\|[\beta_i^j]\|_*)^{2p}(\|\gamma^{(p)}\|_p)^2.
\end{equation}
\]
where \([\beta_i^j]^q=\underbrace{ [\beta_i^j]\bigotimes \cdots \bigotimes [\beta_i^j]}_p\)
在证明这个引理之前, 我们先看看
\[[\alpha_l^k]\gamma^{(p)} [\beta_i^j]^p
=
([\alpha_l^k]_{m\times m}\cdot [\gamma^l_{j_1\cdots j_p}]_{m\times n^p}
\cdot [\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}]_{n^p\times n^p})_{m\times n^p}.
\]
的计算规则.
在\([a_l^k]\)中\(k\)是行指标, \(l\)是列指标; 在\(\gamma^l_{j_1\cdots j_p}\)中\(l\)是行指标,
\(j_1,\cdots,j_p\)是列指标;
在\([\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}]\) 中\(j_1,\cdots,j_p\)是行指标,\(i_1,\cdots,i_p\) 是列行指标.
\(j_1\cdots j_p\)对应的一个数是\((j_1j_2\cdots j_p)\): \((j_1,\cdots,j_p)\)在由\(1,\cdots,n\)组成的\(p\)元数组,
\((j_1j_2\cdots j_p)\)是这些数组在字典序下,
数组\((j_1,j_2\cdots, j_p)\)的位置(即在字典序下, 这个数组\((j_1,\cdots,j_p)\)是第\((j_1j_2\cdots j_p)\)个),
比如\((1,1,\cdots,1)\)是第一个,它对于数\(1\); \((1,1,\cdots,1,n)\)是第\(n\)个,它对应数\(n\).
在这个矩阵\([\alpha_l^k]\gamma^{(p)}\)中, \((k,(j_1\cdots j_p))\)位置的元素是
\[\sum_{l=1}^{m}\alpha_l^k \gamma^l_{j_1\cdots j_p}.
\]
进而, 在\([\alpha_l^k]\gamma^{(p)} [\beta_i^j]^p\)中, \((k,(i_1\cdots i_p))\)位置的元素是
\[\sum_{j_1\cdots j_p}^{n}
\left(\sum_{l=1}^{m}\alpha_l^k\gamma^l_{j_1\cdots j_p}\right)
(\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}).
\]
证明
下面我们约定\(0/0=1/\sqrt{n}\).
\[\begin{align*}
&\sum_{k=1}^{m} \sum_{i_1,\cdots,i_p=1}^{n}
\left|\sum_{l=1}^{m}\sum_{j_1\cdots j_p=1}^{n}\alpha_l^k
\gamma_{j_1\cdots j_p}^{l}\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}\right|^2
\\
\le & \sum_{k=1}^{m} \sum_{i_1,\cdots,i_p=1}^{n}
\left|\sum_{l=1}^{m}\alpha_l^k \sum_{j_1\cdots j_p=1}^{n}\gamma_{j_1\cdots j_p}^{l}\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}
\right|^2,
~~~
(T^l(i_1,\cdots,i_p):=\sum_{j_1\cdots j_p=1}^{n}\gamma_{j_1\cdots j_p}^{l}
\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p})\\
=&\sum_{k=1}^{m} \sum_{i_1,\cdots,i_p=1}^{n} \left|\sum_{l=1}^{m}\alpha_l^k T^l(i_1,\cdots,i_p)\right|^2
\le \sum_{k=1}^{m} \sum_{i_1,\cdots,i_p=1}^{n}
\left|\left(\sum_{l=1}^{m}|\alpha_l^k|^2\right)
\left(\sum_{l=1}^{m}|T^l(i_1,\cdots,i_p)|^2\right)\right|\\
=&\left(\sum_{k=1}^{m}\sum_{l=1}^{m}|\alpha_l^k|^2\right)
\left( \sum_{i_1,\cdots,i_p=1}^{n} \sum_{l=1}^{m}|T^l(i_1,\cdots,i_p)|^2\right)
=(\|[\alpha_l^k]\|^*)^2 \left( \sum_{i_1,\cdots,i_p=1}^{n} \sum_{l=1}^{m}
\left|
\sum_{j_1\cdots j_p=1}^{n}\gamma_{j_1\cdots j_p}^{l}\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}\right|^2\right)\\
=&(\|[\alpha_l^k]\|^*)^2 \left( \sum_{i_1,\cdots,i_p=1}^{n} \sum_{l=1}^{m}
\left|\sum_{j_p=1}^{n}
\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\beta_{i_p}^{j_p}\right|^2
\right)\\
=&
(\|[\alpha_l^k]\|^*)^2 \left( \sum_{i_1,\cdots,i_p=1}^{n} \sum_{l=1}^{m}
\left|\sum_{j_p=1}^{n}\beta_{i_p}^{j_p}
\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]
\right|^2
\right)\\
=&
(\|[\alpha_l^k]\|^*)^2
\left\{\sum_{i_1,\cdots,i_p=1}^{n}
\sum_{l=1}^{m}
{\color{red}
\left(\sum_{s_p=1}^{n}
\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]^2
\right)
}
\right.
\\
&\left.\left|\sum_{j_p=1}^{n}\beta_{i_p}^{j_p}
\frac{\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]}{\color{red}\left(\sum_{s_p=1}^{n}\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]^2\right)^{1/2}}\right|^2
\right\}
\\
=&(\|[\alpha_l^k]\|^*)^2
\left\{\sum_{i_1,\cdots,i_{p-1}=1}^{n}
\sum_{l=1}^{m}
{\color{red}
\left(\sum_{s_p=1}^{n}
\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]^2
\right)
}\cdot
\right.
\\
&\left.\cdot
\left(
{\color{blue}\sum_{i_p=1}^{n}\left|\sum_{j_p=1}^{n}\beta_{i_p}^{j_p}
\frac{\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]}{\color{red}\left(\sum_{s_p=1}^{n}\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]^2\right)^{1/2}}\right|^2
}
\right)
\right\}
\\
=&(\|[\alpha_l^k]\|^*)^2 \left\{\sum_{i_1,\cdots,i_{p-1}=1}^{n}
\sum_{l=1}^{m}
{\color{red}
\left(\sum_{s_p=1}^{n}
\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\left(\gamma_{j_1\cdots j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}\right)
\right]^2
\right)
}
(\|[\beta_i^j]\|_*)^2
\right\}
\end{align*}
\]
于是
\[\begin{align*}
\tag{eq-5-16}
&\sum_{k=1}^{m} \sum_{i_1,\cdots,i_p=1}^{n}
\left|\sum_{l=1}^{m}\sum_{j_1\cdots j_p=1}^{n}\alpha_l^k
\gamma_{j_1\cdots j_p}^{l}\beta_{i_1}^{j_1}\cdots \beta_{i_p}^{j_p}\right|^2
\\
\le & (\|[\alpha_l^k]\|^*)^2
(\|[\beta_i^j]\|_*)^2
\left\{
\sum_{l=1}^{m}\sum_{i_1,\cdots,i_{p-1}=1}^{n}
{\color{red}
\left(\sum_{s_p=1}^{n}
\left[\sum_{j_1\cdots j_{p-1}=1}^{n}
\gamma_{j_1\cdots j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-1}}^{j_{p-1}}
\right]^2
\right)
}
\right\}
\\
=
& (\|[\alpha_l^k]\|^*)^2(\|[\beta_i^j]\|_*)^2
\left\{
\sum_{l=1}^{m}\sum_{i_1,\cdots,i_{p-1}=1}^{n}
{\color{red}
\left(\sum_{s_p=1}^{n}
\left[\sum_{j_{p-1}}^{n}\beta_{i_{p-1}}^{j_{p-1}}
\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]
\right]^2
\right)
}
\right\}
\\
=\cdot \cdot
&
\left\{
\sum_{l=1}^{m}\sum_{i_1,\cdots,i_{p-1}=1}^{n}
{\color{red}
\left(\sum_{s_p=1}^{n}
\left[\sum_{j_{p-1}}^{n}\beta_{i_{p-1}}^{j_{p-1}}
\frac{\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]}
{\left(\sum_{s_{p-1}=1}^{n}\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, s_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]^2\right)^{1/2}}
\right]^2
\right)
}\cdot
\right.
\\
&\left.\cdot
\left(\sum_{s_{p-1}=1}^{n}\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, s_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]^2\right)
\right\}
\\
=& \cdot\cdot
\left\{
\sum_{l=1}^{m}\sum_{i_1,\cdots,i_{p-2}=1}^{n} \sum_{s_p=1}^{n}
{\color{red}
\left(\sum_{i_{p-1}=1}^{n}
\left[\sum_{j_{p-1}}^{n}\beta_{i_{p-1}}^{j_{p-1}}
\frac{\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, j_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]}
{\left(\sum_{s_{p-1}=1}^{n}\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, s_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]^2\right)^{1/2}}
\right]^2
\right)
}\cdot
\right.
\\
&\left.\cdot
\left(\sum_{s_{p-1}=1}^{n}\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, s_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]^2\right)
\right\}\\
=&(\|[\alpha_l^k]\|^*)^2(\|[\beta_i^j]\|_*)^4
\left\{
\sum_{l=1}^{m}\sum_{i_1,\cdots,i_{p-2}=1}^{n} \sum_{s_p=1}^{n}
\cdot
\sum_{s_{p-1}=1}^{n}\left[\sum_{j_1\cdots j_{p-2}=1}^{n}
\gamma_{j_1\cdots j_{p-2}, s_{p-1}s_p}^{l} \beta_{i_1}^{j_1}\cdots \beta_{i_{p-2}}^{j_{p-2}}
\right]^2
\right\}\\
&\cdots\cdots\\
\le &(\|[\alpha_l^k]\|^*)^2(\|[\beta_i^j]\|_*)^{2p}
\left\{\sum_{l=1}^{m}\sum_{s_1,\cdots,s_{p}=1}^{n}
\left|
\gamma_{s_1\cdots s_p}^{l}
\right|^2
\right\}\\
=&(\|[\alpha_l^k]\|^*)^2(\|[\beta_i^j]\|_*)^{2p}(\|\gamma^{(p)}\|_p)^2.
\end{align*}
\]
因此
\[\|[\alpha_l^k]\gamma^{(p)} [\beta_i^j]^p\|
\le (\|[\alpha_l^k]\|^*)(\|[\beta_i^j]\|_*)^{p}\|\gamma^{(p)}\|_p.
\]
证毕.
递归地, 定义\(H_{i_1\cdots i_q}^k(z,w,w^{(1)},\cdots, w^{(q)})\)如下: \(\left(w^{(r)}=(w^l_{j_1\cdots j_r})\right)\)
\[\begin{equation}
\tag{eq-5-17}
\begin{aligned}
H_{i_1}^k(z,w,w^{(1)})&=\frac{\partial h^k}{\partial z_{i_1}}
+\sum_{l=1}^{m}\sum_{j_1=1}^{n}\frac{\partial h^k}{\partial w^l}
w_{j_1}^{l}\frac{\partial f_{j_1}}{\partial z_{i_1}},\\
H_{i_1\cdots i_{q+1}}^k(z,w,w^{(1)},\cdots, w^{(q+1)})
&=\frac{\partial H_{i_1\cdots i_{q}}^k}{\partial z_{i_{q+1}}}+
\sum_{s=0}^{q}\sum_{l=1}^{m}\sum_{j_1\cdots j_{s+1}=1}^{n}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots j_s}^l} w_{j_1\cdots j_{s+1}}^{l}
\frac{\partial f_{j_{s+1}}}{\partial z_{i_{q+1}}}.
\end{aligned}
\end{equation}
\]
在{eq-5-17}中, 当\(s=0\)时, \(w_{j_1\cdots j_{s}}^{l}=w^{l}\).
引理(lemma-5-2)
假设\(f,h\)满足前面的假设{\rm (I)(III)}. 设\(\psi(z)=h(z,\phi[f(z)])\),
其中\(\phi\)在\(z=0\)的一个邻域内有定义并且解析.
那么
\[\begin{equation}\tag{eq-5-18}
\frac{\partial^q \psi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}
=H_{i_1\cdots i_q}^k(z,\phi[f(z)],\phi^{(1)}[f(z)],\cdots,\phi^{(q)}[f(z)]),
\end{equation}
\]
\(k=1,\cdots,m;i_1,\cdots,i_q=1,\cdots,n;q=1\cdots,p;\)
其中
\[\phi^{(q)}[f(z)]=\left\{\frac{\partial ^s}{\partial z_{i_1}\cdots z_{i_s}}\phi^k(z)\right\}, ~~k=1,\cdots,m;~
i_1,\cdots,i_s=1,\cdots,n;~s=0,1,\cdots,p.
\]
证明
我们用数学归纳法来证明此引理.
首先, 对
(\(h(z,w)=h(z_1,\cdots,z_n;w^1,\cdots,w^m)\))
\[\begin{aligned}
\psi(z)=h(z,\phi[f(z)])
=&h(z_1,\cdots,z_n;\phi^1(f_1(z_1,\cdots,z_n),\cdots,
f_n(z_1,\cdots,z_n)),
\cdots,\\
&\phi^m(f_1(z_1,\cdots,z_n),\cdots,f_n(z_1,\cdots,z_n))),
\end{aligned}
\]
两边对\(z_{i_1}\)求导, 有
\[\begin{equation*}
\begin{aligned}
\frac{\partial \psi^k(z)}{\partial z_{i_1}}
=&\frac{\partial h^k(z,\phi[f(z)])}{\partial z_{i_1}}
+
\sum_{l=1}^{m}\frac{\partial h^k}{\partial w^l}\sum_{j_1=1}^{n}
\frac{\partial \phi^l}{\partial z_{j_1}}\frac{\partial f_{j_1}}{\partial z_{i_1}}\\
=&\frac{\partial h^k(z,\phi[f(z)])}{\partial z_{i_1}}
+
\sum_{l=1}^{m}\sum_{j_1=1}^{n}\frac{\partial h^k}{\partial w^l}
\frac{\partial \phi^l}{\partial z_{j_1}}\frac{\partial f_{j_1}}{\partial z_{i_1}}\\
=&H_{i_1}^k(z,\phi[f(z)],\phi^{(1)}[f(z)]).
\end{aligned}
\end{equation*}
\]
假设{eq-5-18}对\(q\)成立, 我们证明它对\(q+1\)也成立, 即证明:
\[\begin{equation}\tag{eq-5-19}
\begin{aligned}
\frac{\partial^{q+1}\psi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_{q+1}}}
=&H_{i_1\cdots i_{q+1}}^k(z,\phi[f(z)],\phi^{(1)}[f(z)],\cdots,\phi^{(q+1)}[f(z)])\\
=&\frac{\partial H_{i_1\cdots i_{q}}^k}{\partial z_{i_{q+1}}}+
\sum_{s=0}^{q}\sum_{l=1}^{m}\sum_{j_1\cdots j_{s+1}=1}^{n}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots j_s}^l}
\frac{\partial^{s+1}\phi^l}{\partial z_{j_{1}}\cdots \partial z_{j_{s+1}}}
\frac{\partial f_{j_{s+1}}}{\partial z_{i_{q+1}}}.
\end{aligned}
\end{equation}
\]
{eq-5-18}对\(q\)成立说明
\[\begin{equation*}
\begin{aligned}
\frac{\partial^q\psi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}
=&H_{i_1\cdots i_q}^k(z,\phi[f(z)],\phi^{(1)}[f(z)],\cdots,\phi^{(q)}[f(z)])\\
=&
{\color{red}
{H_{i_1\cdots i_q}^k\left(z_1,\cdots,z_n;
\left(\frac{\partial^s \phi^l[f(z)]}{\partial z_{j_1}\cdots \partial z_{j_s}}:l=1\cdots,m;
j_1,\cdots,j_s=1,\cdots,n;s=0,\cdots,q\right)
\right)}
}
\end{aligned}
\end{equation*}
\]
这是把
\[H_{i_1\cdots i_q}^k(z_1,\cdots,z_n;(w^l_{j_1\cdots j_s}:l=1\cdots,m;j_1,\cdots,j_s=1,\cdots,n;
s=0,1,\cdots,q))
\]
中的\(w^l_{j_1\cdots j_s}\)替换为
\[\frac{\partial^s \phi^l}{\partial z_{j_1}\cdots \partial z_{j_s}}
\]
的结果.
因此(对上面红色部分微分)
\[\begin{equation*}
\begin{aligned}
\frac{\partial^{q+1}\psi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q+1}}
=&\frac{\partial H_{i_1\cdots i_q}^k}{\partial z_{i_{q+1}}}+
\sum_{s=0}^{q}\sum_{l=1}^{m}\sum_{j_1,\cdots,j_{s+1}=1}^{n}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots i_{s}}^l}
\frac{\partial }{\partial z_{j_{s+1}}}
\left(
\frac{\partial^{s}
\phi^l}{\partial z_{j_1}\cdots \partial z_{j_{s}}}
\right)
\frac{\partial f_{j_{s+1}}}{\partial z_{i_{q+1}}}\\
=&\frac{\partial H_{i_1\cdots i_q}^k}{\partial z_{i_{q+1}}}+
\sum_{s=0}^{q}\sum_{l=1}^{m}\sum_{j_1,\cdots,j_{s+1}=1}^{n}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots i_{s}}^l}
\frac{\partial^{s+1}
\phi^l}{\partial z_{j_1}\cdots \partial z_{j_{s+1}}}
\frac{\partial f_{j_{s+1}}}{\partial z_{i_{q+1}}}\\
=&H_{i_1\cdots i_{q+1}}^k(z,\phi[f(z)],\phi^{(1)}[f(z)],\cdots,\phi^{(q+1)}[f(z)]).
\end{aligned}
\end{equation*}
\]
至此我们完成了整个证明.
引理(lemma-5-3)
在假设{\rm (I)(III)(IV)}下, 我们有
\[\begin{equation}\tag{eq-5-20}
c_{i_1\cdots i_q}^k=H_{i_1\cdots i_q}^k(0,0,c^{(1)},\cdots,c^{(q)}),~~
k=1,\cdots,m;i_1,\cdots,i_q=1,\cdots,n;q=1,2,\cdots,p,
\end{equation}
\]
其中
\[c^{(q)}=\{c_{i_1\cdots i_q}^k\}(k=1,\cdots,m;i_1,\cdots,i_q=1,\cdots,n).
\]
证明
由上面的红色部分我们知道
\[\begin{equation*}
\begin{aligned}
&\frac{\partial^q\phi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}
=H_{i_1\cdots i_q}^k(z,\phi[f(z)],\phi^{(1)}[f(z)],\cdots,\phi^{(q)}[f(z)])\\
=&
{H_{i_1\cdots i_q}^k\left(z_1,\cdots,z_n;
\left(\frac{\partial^s \phi^l}{\partial z_{j_1}\cdots \partial z_{j_s}}:l=1\cdots,m;
j_1,\cdots,j_s=1,\cdots,n;s=0,\cdots,q\right)
\right)}
\end{aligned}
\end{equation*}
\]
由{eq-5-7}, 我们知道\(c_{i_1\cdots i_q}^k\)是
\(\frac{\partial^q\phi^k}{\partial z_{i_1}\cdots z_{i_q}}\)
这个函数在\(z=0\)处的值.
上式中令\(z=0\), 我们得到
\[\begin{equation}\tag{eq-5-21-0}
c_{i_1\cdots i_q}^k=H_{i_1\cdots i_q}^k(0,0,c^{(1)},\cdots,c^{(q)}).
\end{equation}
\]
证毕.
引理(lemma-5-4)
在假设{\rm (I)(III)}下, 我们有
\[\begin{equation}\tag{eq-5-21}
H_{i_1\cdots i_q}^k(z,w,w^{(1)},\cdots, w^{(q)}),~k=1,\cdots,m;i_1,\cdots,i_q=1,\cdots,n;q=1,2,\cdots,p,
\end{equation}
\]
是关于\(z,w,w^{(1)},\cdots, w^{(q)}\)在\(\|z\|\le r_0,\|w\|_0\le R_0,
\|w^{(1)}\|\in W^{(1)},\cdots,\|w^{(q)}\|\in W^{(q)}\)上的解析函数.
并且,
\[\begin{equation}
\tag{eq-5-22}
\begin{aligned}
&H_{i_1\cdots i_q}^k(z,w,w^{(1)},\cdots, w^{(q)})\\
=&G_{i_1\cdots i_q}^k(z,w,w^{(1)},\cdots, w^{(q-1)})+
\sum_{l=1}^{m}\sum_{j_1\cdots j_{q}=1}^{n}
\frac{\partial h^k(z,w)}{\partial w^l}w_{j_1\cdots j_{q}}^{l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{q}}}{\partial z_{i_q}},
~~q=1,2,\cdots,p,
\end{aligned}
\end{equation}
\]
其中\(G_{i_1\cdots i_q}^k(z,w,w^{(1)},\cdots, w^{(q-1)})\)是关于
\(z,w,w^{(1)},\cdots, w^{(q-1)}\)在$|z|\le r_0,|w|_0\le R_0, $
\(\|w^{(1)}\|\in W^{(1)},\cdots,\)
\(\|w^{(q-1)}\|\in W^{(q-1)}\)上的解析函数.
证明
由{eq-5-17}, 我们知道\(H_{i_1\cdots i_q}^k\)是解析的.
下面我们利用数学归纳法来证明{eq-5-22}.
对于\(q=1\)的情况, {eq-5-17}表明
\[H_{i_1}^k(z,w,w^{(1)})=\frac{\partial h^k}{\partial z_{i_1}}+
\sum_{l=1}^{m}\sum_{j_1=1}^{n}\frac{\partial h^k}{\partial w^l} w_{j_1}^{l} \frac{\partial f_{j_1}}{\partial z_{i_1}}
=G_{i_1}^k(z,w)+\sum_{l=1}^{m}\sum_{j_1=1}^{n}\frac{\partial h^k}{\partial w^l} w_{j_1}^{l} \frac{\partial f_{j_1}}{\partial z_{i_1}},
\]
where
\[G_{i_1}^k(z,w)=\frac{\partial h^k(z,w)}{\partial z_{i_1}}.
\]
这就证明了\(q=1\)的情况.
假设{eq-5-22}对\(q\)成立, 我们证明它对\(q+1\)也成立, 即证明:
\[\begin{equation}\tag{eq-5-23}
\begin{aligned}
&H_{i_1\cdots i_{q+1}}^k(z,w,w^{(1)},\cdots, w^{(q+1)})\\
=&G_{i_1\cdots i_{q+1}}^k(z,w,w^{(1)},\cdots, w^{(q)})+
\sum_{l=1}^{m}\sum_{j_1\cdots j_{q+1}=1}^{n}
\frac{\partial h^k(z,w)}{\partial w^l}w_{j_1\cdots j_{q+1}}^{l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{q+1}}}{\partial z_{i_{q+1}}}.
\end{aligned}
\end{equation}
\]
利用{eq-5-17}我们有
\[\begin{equation}\tag{eq-5-24}
\begin{aligned}
&H_{i_1\cdots i_{q+1}}^k(z,w,w^{(1)},\cdots, w^{(q+1)})
=\frac{\partial H_{i_1\cdots i_q}^k}{\partial z_{i_{q+1}}}+
\sum_{s=0}^{q}\sum_{l=1}^{m}\sum_{j_1\cdots j_{s+1}=1}^{n}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots j_s}^l}
w_{j_1\cdots j_{s+1}}^{l}
\frac{\partial f_{j_{s+1}}}{\partial z_{i_{q+1}}}\\
=&\left[\frac{\partial H_{i_1\cdots i_q}^k}{\partial z_{i_{q+1}}}+
\sum_{s=0}^{q-1}\sum_{l=1}^{m}\sum_{j_1\cdots j_{s+1}=1}^{n}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots j_s}^l}
w_{j_1\cdots j_{s+1}}^{l}
\frac{\partial f_{j_{s+1}}}{\partial z_{i_{q+1}}}\right]+\sum_{l=1}^{m}\sum_{j_1\cdots j_{q+1}=1}^{n}
{\color{red}\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots j_q}^l} }
w_{j_1\cdots j_{q+1}}^{l}
\frac{\partial f_{j_{q+1}}}{\partial z_{i_{q+1}}}.
\end{aligned}
\end{equation}
\]
由{eq-5-22}对\(q\)的情况, 我们有
\[\begin{equation*}
\begin{aligned}
&H_{i_1\cdots i_q}^k(z,w,w^{(1)},\cdots, w^{(q)})\\
=&G_{i_1\cdots i_q}^k(z,w,w^{(1)},\cdots, w^{(q-1)})+
\sum_{l=1}^{m}\sum_{j_1\cdots j_{q}=1}^{n}
\frac{\partial h^k(z,w)}{\partial w^l}w_{j_1\cdots j_{q}}^{l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{q}}}{\partial z_{i_q}},
~~q=1,2,\cdots,p,
\end{aligned}
\end{equation*}
\]
利用上式我们可以得到
\[\begin{equation*}
\begin{aligned}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots j_q}^l}
=\frac{\partial h^k(z,w)}{\partial w^l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{q}}}{\partial z_{i_q}}
\end{aligned}
\end{equation*}
\]
把它带入{eq-5-24}, 我们得到
\[\begin{equation*}
\begin{aligned}
&H_{i_1\cdots i_{q+1}}^k(z,w,w^{(1)},\cdots, w^{(q+1)})\\
=&\left[\frac{\partial H_{i_1\cdots i_q}^k}{\partial z_{i_{q+1}}}+
\sum_{s=0}^{q-1}\sum_{l=1}^{m}\sum_{j_1\cdots j_{s+1}=1}^{n}
\frac{\partial H_{i_1\cdots i_q}^k}{\partial w_{j_1\cdots j_s}^l}
w_{j_1\cdots j_{s+1}}^{l}
\frac{\partial f_{j_{s+1}}}{\partial z_{i_{q+1}}}\right]\\
&+\sum_{l=1}^{m}\sum_{j_1\cdots j_{q+1}=1}^{n}
\frac{\partial h^k(z,w)}{\partial w^l}w_{j_1\cdots j_{q+1}}^{l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{q}}}{\partial z_{i_q}}
\frac{\partial f_{j_{q+1}}}{\partial z_{i_{q+1}}}\\
=&G_{i_1,\cdots,i_{q+1}}^k(z,w,\cdots,w^{(q)})+
\sum_{l=1}^{m}\sum_{j_1\cdots j_{q+1}=1}^{n}
\frac{\partial h^k(z,w)}{\partial w^l}w_{j_1\cdots j_{q+1}}^{l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{q}}}{\partial z_{i_q}}
\frac{\partial f_{j_{q+1}}}{\partial z_{i_{q+1}}}.
\end{aligned}
\end{equation*}
\]
证毕.
用\(H_p(z,w,w^{(1)},\cdots, w^{(p)})\)表示空间\(W^{(p)}\)中的向量
\(
\left(\frac{\partial ^p\phi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_p}}\in W^{(p)}\right)\):
\[\{H_{i_1\cdots i_p}^k(z,w,w^{(1)},\cdots, w^{(p)})\},~k=1,\cdots,m;i_1,\cdots,i_p=1,\cdots,n.
\]
引理(lemma-5-5)
在假设{\rm (I)(II)(III)}下, 存在一个常数\(L>0\)(独立于\(z\))使得对于任意的 \((z,\hat{w},\hat{w}^{(1)},\)
\(\cdots,\hat{w}^{(p)})\), \((z,\hat{\hat{w}},\hat{\hat{w}}^{(1)},\cdots,\hat{\hat{w}}^{(p)})\in
Z\), 我们有
\[\begin{equation}\tag{eq-5-25}
\|H_p(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p)})-H_p(z,\hat{\hat{w}},\hat{\hat{w}}^{(1)},\cdots, \hat{\hat{w}}^{(p)})\|
\le L\sum_{q=0}^{p-1}+\|\hat{w}^{(q)}-\hat{\hat{w}}^{(q)}\|_q+\theta \|\hat{w}^{(p)}-\hat{\hat{w}}^{(p)}\|_p,
\end{equation}
\]
其中\(\theta\)是前面出现过的一个小于\(1\)的常数并且
\[Z=\{z:\|z\|\le r_0\}\times \{w:\|w\|_0\le R\}\times
K(\mathring{w}^{(1)},\rho_1)\times \cdots \times K(\mathring{w}^{(p-1)},\rho_{p-1})\times
K(\mathring{w}^{(p)},\rho_p),
\]
这里\(\mathring{w}^{(q)}\)是\(W^{(q)}\)中的任意固定的元素, \(\rho_q\)是任意固定的正数, 并且
\[K(\mathring{w}^{(q)},\rho_q)=\{w^{(q)}:\|w^{(q)}-\mathring{w}^{(q)}\|_q\le \rho_q\},~q=1,\cdots,p.
\]
证明
\[\begin{aligned}
H^k_{i_1\cdots i_p}(z,w,w^{(1)},\cdots, w^{(p-1)})
=&G^k_{i_1\cdots i_p}(z,w,w^{(1)},\cdots, w^{(p-1)})+
\sum_{l=1}^{m}\sum_{j_1\cdots j_{p}=1}^{n}
\frac{\partial h^k(z,w)}{\partial w^l}w_{j_1\cdots j_{p}}^{l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{p}}}{\partial z_{i_p}}\\
=&G^k_{i_1\cdots i_p}(z,w,w^{(1)},\cdots, w^{(p-1)})+N^k_{i_1\cdots i_p}(z,w, w^{(p)})
\end{aligned}
\]
其中
\[N^k_{i_1\cdots i_p}(z,w,w^{(p)})=\sum_{l=1}^{m}\sum_{j_1\cdots j_{p}=1}^{n}
\frac{\partial h^k(z,w)}{\partial w^l}w_{j_1\cdots j_{p}}^{l}
\frac{\partial f_{j_{1}}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_{p}}}{\partial z_{i_p}}.
\]
记
\[\begin{aligned}
H_p(z,w,w^{(1)},\cdots, w^{(p)})&=\{H^k_{i_1\cdots i_p}(z,w,w^{(1)},\cdots, w^{(p)})\},\\
G_p(z,w,w^{(1)},\cdots, w^{(p-1)})&=\{G^k_{i_1\cdots i_p}(z,w,w^{(1)},\cdots, w^{(p-1)})\},\\
N_p(z,w,w^{(p)})&=\{N^k_{i_1\cdots i_p}(z,w,w^{(p)})\}.
\end{aligned}
\]
{\color{red}\bf 为了把项\(w^l_{j_1\cdots j_p}\)单独分离出来, }
我们写
\[\begin{aligned}
&\|H_p(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p)})-
H_p(z,\hat{\hat{w}},\hat{\hat{w}}^{(1)},\cdots, \hat{\hat{w}}^{(p)})\|_p\\
\le & \|H_p(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p-1)},\hat{w}^{(p)})-
H_p(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p-1)},\hat{\hat{w}}^{(p)})\|_p\\
&+\|H_p(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p-1)},\hat{\hat{w}}^{(p)})-
H_p(z,\hat{\hat{w}},\hat{\hat{w}}^{(1)},\cdots,\hat{\hat{w}}^{(p-1)},\hat{\hat{w}}^{(p)})\|_p\\
\le & \|N_p(z,\hat{w},\hat{w}^{(p)})-N_p(z,\hat{w},\hat{\hat{w}}^{(p)})\|_p\\
&+\|H_p(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p-1)},\hat{\hat{w}}^{(p)})-
H_p(z,\hat{\hat{w}},\hat{\hat{w}}^{(1)},\cdots,\hat{\hat{w}}^{(p-1)},\hat{\hat{w}}^{(p)})\|_p.
\end{aligned}
\]
在上式中,
\[\begin{aligned}
&\|N_p(z,\hat{w},\hat{w}^{(p)})-N_p(z,\hat{w},\hat{\hat{w}}^{(p)})\|_p
=\left[\sum_{k=1}^{m}\sum_{i_1\cdots i_p=1}^{n}\left|N^k_{i_1\cdots i_p}(z,\hat{w},\hat{w}^{(p)})
-N^k_{i_1\cdots i_p}(z,\hat{w},\hat{\hat{w}}^{(p)})\right|^2\right]^{1/2}\\
=&\left[\sum_{k=1}^{m}\sum_{i_1\cdots i_p=1}^{n}
\left|\sum_{l=1}^{m}\sum_{j_1\cdots j_p=1}^{n}
\frac{\partial N^k_{i_1\cdots i_p}(z,\hat{w},\hat{w}^{(p)})}{\partial w_{j_1\cdots j_p}^l}
\left[\hat{w}_{j_1\cdots j_p}^{l}-\hat{\hat{w}}_{j_1\cdots j_p}^{l}\right]
\right|^2\right]^{1/2}\\
=&\left[\sum_{k=1}^{m}\sum_{i_1\cdots i_p=1}^{n}
\left|\sum_{l=1}^{m}\sum_{j_1\cdots j_p=1}^{n}
\frac{\partial h^k}{\partial w^l}\frac{\partial f_{j_1}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_p}}{\partial z_{i_p}}
\left[\hat{w}_{j_1\cdots j_p}^{l}-\hat{\hat{w}}_{j_1\cdots j_p}^{l}\right]
\right|^2\right]^{1/2}\\
=&\left[\sum_{k=1}^{m}\sum_{i_1\cdots i_p=1}^{n}
\left|\sum_{l=1}^{m}\sum_{j_1\cdots j_p=1}^{n}
\frac{\partial h^k}{\partial w^l}
\left[\hat{w}_{j_1\cdots j_p}^{l}-\hat{\hat{w}}_{j_1\cdots j_p}^{l}\right]
\frac{\partial f_{j_1}}{\partial z_{i_1}}\cdots \frac{\partial f_{j_p}}{\partial z_{i_p}}
\right|^2\right]^{1/2}~~~(\text{利用引理}{lemma-5-1})\\
\le &\left(\left|\left|\frac{\partial h^k(z,w)}{\partial w^l}\right|\right|^*\right)
\left(\left|\left|\frac{\partial f_j(z)}{\partial z_i}\right|\right|_*\right)^p
\|\hat{w}^{(p)}-\hat{\hat{w}}^{(p)}\|_p ~~~(\text{利用引理}{eq-5-13})\\
\le &\theta \|\hat{w}^{(p)}-\hat{\hat{w}}^{(p)}\|_p
\end{aligned}
\]
并且
\[\begin{aligned}
&\|H_p(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p-1)},\hat{\hat{w}}^{(p)})-
H_p(z,\hat{\hat{w}},\hat{\hat{w}}^{(1)},\cdots,\hat{\hat{w}}^{(p-1)},\hat{\hat{w}}^{(p)})\|_p\\
=&\left(
\sum_{k=l}^{m}\sum_{i_1\cdots i_p=1}^{n}
\left|
H^k_{i_1\cdots i_p}(z,\hat{w},\hat{w}^{(1)},\cdots, \hat{w}^{(p-1)},\hat{\hat{w}}^{(p)})
- H^k_{i_1\cdots i_p}(z,\hat{\hat{w}},\hat{\hat{w}}^{(1)},\cdots,\hat{\hat{w}}^{(p-1)},\hat{\hat{w}}^{(p)})
\right|^2
\right)^{1/2}\\
\le &\left(
\sum_{k=l}^{m}\sum_{i_1\cdots i_p=1}^{n}
\left|
\sum_{q=0}^{p-1}
\sum_{l=1}^{m}
\sum_{j_1\cdots j_q=1}^{n}
\frac{\partial H^k_{i_1\cdots i_p}}{\partial w^l_{j_1\cdots j_q}}
\left[\hat{w}^l_{j_1\cdots j_q}-\hat{\hat{w}}^l_{j_1\cdots j_q}\right]
\right|^2
\right)^{1/2}
~~~~~(w^l_{j_1\cdots j_q}=w^l~\text{for}~q=0)
\\
&(\text{利用}\|\hat{w}^l_{j_1\cdots j_q}-\hat{\hat{w}}^l_{j_1\cdots j_q}\|\le
\|\hat{w}^{(q)}-\hat{\hat{w}}^{(q)}\|_q)\\
\le &\left(
\sum_{k=l}^{m}\sum_{i_1\cdots i_p=1}^{n}
\left|
\sum_{q=0}^{p-1}\|\hat{w}^{(q)}-\hat{\hat{w}}^{(q)}\|_q
\sum_{l=1}^{m}
\sum_{j_1\cdots j_q=1}^{n}
\left|\frac{\partial H^k_{i_1\cdots i_p}}{\partial w^l_{j_1\cdots j_q}}\right|
\right|^2
\right)^{1/2}~~~
(\text{假设} \sup_{Z}\left|\frac{\partial H^k_{i_1\cdots i_p}}{\partial w^l_{j_1\cdots j_q}}\right|\le L_1)\\
\le &\left(L_1^2m n^p \left(\sum_{q=0}^{p-1}\|\hat{w}^{q}-\hat{\hat{w}}^{(q)}\|_q\right)^2\right)^{1/2}
\le L_1 m^{1/2} n^{p/2} \sum_{q=0}^{p-1}\|\hat{w}^{q}-\hat{\hat{w}}^{(q)}\|_q\\
=&L \sum_{q=0}^{p-1}\|\hat{w}^{q}-\hat{\hat{w}}^{(q)}\|_q,~~~~L=L_1 m^{1/2} n^{p/2}.\\
\end{aligned}
\]
证毕.
定理 (theorem-5-1)
假设 {\rm (I)(II)(III)(IV)} 成立，则对于由 {eq-5-20} 确定的每一组 \(c^{(1)}, \cdots, c^{(p)}\)，恰好存在唯一的一个函数 \(\phi\) 在 \(z=0\) 的一个邻域内有定义且解析，满足方程 {eq-smajdor}，使得：
\[\begin{equation}\tag{eq-5-26}
\phi(0) = 0, \quad \phi^{(q)} = c^{(q)}, \quad q = 1, 2, \cdots, p.
\end{equation}
\]
证明:
利用压缩映像原理证明.
固定一个正数\(K>0\)以及\(\vartheta\in (\theta,1)\). 取\(r>0\)满足下面的不等式:
\[\begin{equation}\tag{eq-determine-r}
\begin{aligned}
&\sup_{\|z\|\le r}\left|\left|\frac{\partial f_j(z)}{\partial z_i}\right|\right|_*<1,\\
&r<\min(1,r_1),\\
&m^{1/2} n^{p} r (c_{1}+ c_{2}+\cdots+c_{p-1}+c_p+K)<R,\\
&L \sum_{q=0}^{p-1} m^{1/2} n^{p-q/2} r \left(\sum_{t=q+1}^{p} c_{t}+K\right)\le \frac{1}{2}(1-\theta)K, \\
&\|H_p(z,0,c^{(1)},\cdots,c^{(p)})-H_p(0,0,c^{(1)},\cdots,c^{(p)})\|_p\le \frac{1}{2} (1-\theta)K,~~~\forall \|z\|\le r,\\
& L\sum_{q=0}^{p-1}m^{1/2} n^{p-q/2} r^{p-q} +\theta \le \vartheta.
\end{aligned}
\end{equation}
\]
为了利用引理{lemma-5-5}, 我们取\(\mathring{w}^{(q)}=c^{(q)},q=1\cdots,p\), 且
\[\rho_q=m^{1/2} n^{p-q/2} (c_{q+1}+ c_{q+2}+\cdots+c_{p-1}+c_p+K).
\]
Step 1. 构造度量空间\(\mathfrak{R}\)为满足下面条件的函数\(\phi\)的集合:
\[\phi^k(z)=\sum_{q=0}^{p}c^k_{i_1\cdots i_q}z_{i_1}\cdots z_{i_q}
+\sum_{q=p+1}^{\infty}d^k_{i_1\cdots i_q}z_{i_1}\cdots z_{i_q},~~~\|z\|\le r
\]
其中\(c^k_{i_1\cdots i_q}\)已确定, \(d^k_{i_1\cdots i_q}\)是任意的复数.
\[\|\phi^{(p)}(z)-c^{(p)}\|_p\le K,~~~~\|z\|\le r.
\]
在空间\(\mathfrak{R}\)中定义度量\(\rho\):
\[\rho(\phi,\hat{\phi})=\sup_{\|z\|\le r}\|\phi^{(p)}(z)-\hat{\phi}^{(p)}(z)\|_p
\]
Step 2. 证明空间\((\mathfrak{R},\rho)\)是完备的.
设\(\{\phi_n|n\ge 1\}\)是空间\((\mathfrak{R},\rho)\)中的任一Cauchy序列, 即: 对于任意的\(\varepsilon>0\),
存在一个正整数\(N=N(\varepsilon)\)使得
\[\rho(\phi_n,\phi_m)=\sup_{\|z\|\le r}
\left(\sum_{k=1}^{m}\sum_{i_1\cdots i_p=1}^{n}
\left|\frac{\partial^p\phi_n^k(z)}{\partial z_{i_1}\cdots z_{i_p}}
-\frac{\partial^p\phi_m^k(z)}{\partial z_{i_1}\cdots z_{i_p}}
\right|^2\right)^{1/2}
<\varepsilon,~~~\forall n>m\ge N.
\]
因此, 对于固定的\(1\le k\le m\)和\(1\le i_1,\cdots i_p\le n\),
\[\left\{\frac{\partial^p\phi_n^k(z)}{\partial z_{i_1}\cdots z_{i_p}} \middle| n\ge 1\right\}
\]
一致收敛. 因此存在\(\phi_{(p)}(z)\)使得
\[\phi_n^{(p)}(z)\rightrightarrows \phi_{(p)}(z) ,~~~~ \text{on}~~ \|z\|\le r.
\]
利用Taylor公式:
\[\begin{equation}\tag{Taylor-formula}
\begin{aligned}
\frac{\partial^q\phi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}
=&c_{i_1\cdots i_q}^k+
\sum_{j_1=1}^{n}c^{k}_{i_1\cdots i_q j_1}z_{j_1}
+\frac{1}{2!}\sum_{j_1,j_2=1}^{n}c^{k}_{i_1\cdots i_q j_1j_2} z_{j_1} z_{j_2}\\
&+\cdots+
\frac{1}{(p-q-1)!}\sum_{j_1,j_2,\cdots, j_{p-q-1}=1}^{n}c^{k}_{i_1\cdots i_q j_1j_2}z_{j_1} z_{j_2}\cdots z_{j_{p-q-1}}\\
&+\frac{1}{(p-q)!}\sum_{j_1,j_2,\cdots, j_{p-q}=1}^{n}
\frac{\partial^p\phi^{k}(\xi)}{\partial z_{i_1}\cdots
\partial z_{i_q}\partial z_{j_1}\cdots \partial z_{j_{p-q}}}z_{j_1} z_{j_2}\cdots z_{j_{p-q}}.
\end{aligned}
\end{equation}
\]
进而,
\[\left|\frac{\partial^q\phi_n^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}
-\frac{\partial^q\phi_m^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}
\right|
\le \frac{1}{(p-q)!} r^{p-q} \rho(\phi_m,\phi_n)
\]
并且
\[\|\phi_m^{(q)}(z)-\phi_n^{(q)}(z)\|_q\le\frac{1}{(p-q)!} m^{1/2}n^{(p-q)/2} r^{p-q}\rho(\phi_m,\phi_n).
\]
有了它, 利用与上面类似的方式证明: 存在\(\phi_{(q)},q=0,1\cdots,p-1\)使得
\[\phi_n^{(q)}(z)\rightrightarrows \phi_{(q)}(z) ~~\text{on}~\|z\|\le r.
\]
进而
\[\phi_{0}^{q}(z)=\phi_{(q)}(z),~~q=1,2,\cdots,p.
\]
记\(\phi_{(0)}(z)=\phi(z)\).
我们论证\(\phi(z)\in \mathfrak{R}\).
\(1^\circ\) \(\phi_n\)在\(\|z\|\)上是连续的且一致收敛于\(\phi\)从而\(\phi\)在\(\|z\|\)上也是连续的;
\(\phi_n\)在\(\|z\|<r\)内是解析的且一致收敛于\(\phi\), Weierstrass定理保证了\(\phi\)也是解析的在\(\|z\|<r\)内.
\(2^\circ\) \(\phi_n\)在\(\|z\|<r\)上是解析的, 从而有形式级数展开; 此外, 对于\(q=1,\cdots,p\),
\[\phi_n^{(q)}(0)\to \phi^{(q)}(0),~~n\to \infty,
\]
因为\(\phi_n^{(q)}(0)=c^{(q)}\), 所以\(\phi^{(q)}(0)=c^{(q)}\).
同理, \(\phi(0)=0\).
\(3^\circ\)
\(\phi_n\in \mathfrak{R}\Rightarrow \|\phi_n^{(p)}(z)-c^{(p)}\|_p \le K\)对于\(\|z\|\le r\),
这意味着对于任意固定的\(z:\|z\|\le r\), \(\|\phi_n^{(p)}(z)-c^{(p)}\|_p \le K,\)
从而对于任意固定的\(z:\|z\|\le r\), \(\|\phi^{(p)}(z)-c^{(p)}\|_p \le K,\)
即\(\|\phi^{(p)}(z)-c^{(p)}\|_p \le K\)对于\(\|z\|\le r.
\)
至此证明完成了Step 3.
Step 3. 我们定义算子\(T\)如下:
\[\begin{aligned}
T(\phi)(z)=&h(z,\phi[f(z)]),~~~\|z\|\le r.
\end{aligned}
\]
我们说明\(T\)是良定义的, 即如果\(\phi\in \mathfrak{R}\), 则\(\|\phi[f(z)]\|\le R\)对于\(\|z\|\le r\)成立.
首先, 利用\(\|B\|_*<1\), 我们有
{\color{red}[这一部分要放在{eq-determine-r}中!]
\[\|f(z)\|\le \sup_{\|z\|\le r}\left|\left|\frac{\partial f_j(z)}{\partial z_i}\right|\right|_*\|z\|\le \|z\|\le r, ~~~\|z\|\le r,
\]
即\(f(z)\in K(0,r)\)当\(z\in K(0,r)\).
}
下面说明\(\phi[f(z)]\in K(0,R)\)当\(\phi\in \mathfrak{R}\)且\(z\in K(0,r)\).
由{Taylor-formula}, 我们有
\[\begin{aligned}
&|\frac{\partial^q\phi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}-c^k_{i_1\cdots i_q}|\\
\le & r c_{q+1}+\frac{1}{2!}r^2 c_{q+2}+\cdots+\frac{1}{(p-q-1)!}r^{p-q-1}c_{p-1}+\frac{1}{(p-q)!}r^{p-q}\|
\sup \|\phi^{(p)}(z)\|_p\\
& (\text{利用}3^\circ: \|\phi^{(p)}(z)-c^{(p)}\|_p\le K,\forall z\in K(0,r))\\
\le &n r c_{q+1} +n^2\frac{1}{2!}r^2 c_{q+2}+\cdots+
n^{p-q-1}\frac{1}{(p-q-1)!}r^{p-q-1}c_{p-1}+n^{p-q}\frac{1}{(p-q)!}r^{p-q}(c_p+K)\\
\le & r n^{p-q} (c_{q+1}+ c_{q+2}+\cdots+c_{p-1}+c_p+K)\\
\end{aligned}
\]
其中
\[c_s=\max_{\substack{1\le k\le m, \\ 1\le i_1,\cdots,i_s\le n}}|c^k_{i_1\cdots i_s}|.
\]
因此, 我们有
\[\begin{equation}\tag{eq-5-27}
\begin{aligned}
\|\phi^{(q)}(z)-c^{(q)}\|_q\le & m^{1/2} n^{q/2} r n^{p-q} (c_{q+1}+ c_{q+2}+\cdots+c_{p-1}+c_p+K)\\
=&m^{1/2} n^{p-q/2} r (c_{q+1}+ c_{q+2}+\cdots+c_{p-1}+c_p+K).\\
(\text{从这里可以导出}&\rho_q=m^{1/2} n^{p-q/2} (c_{q+1}+ c_{q+2}+\cdots+c_{p-1}+c_p+K))
\end{aligned}
\end{equation}
\]
对于\(q=0\), 我们有
\[\|\phi(z)\|\le m^{1/2} n^{p} r (c_{1}+ c_{2}+\cdots+c_{p-1}+c_p+K)<R.
\]
{\color{red}[这要放在{eq-determine-r}中!]}
因此对于\(\phi\in \mathfrak{R}\)和\(z\in K(0,r)\), 我们有\(\phi[f(z)]\in K(0,R)\).
由此, \(T\)是良定义的.
Step 4. 证明\(T(\mathfrak{R})\subset \mathfrak{R}.\)
对于\(\forall \phi\in \mathfrak{R}\),
\[T(\phi)(z)=h(z,\phi[f(z)]).
\]
\(1^\circ\) \(\phi\)在\(\|z\|<r\)解析, \(f\)在\(\|z\|<r\)解析, \(h\)在\(\|z|\le r\)和\(\|w\|_0\le R\)解析,
\(\|\phi[f(z)]\|_0\le R\)对于\(\|z\|\le r\), 于是\(h(z,\phi[f(z)])\)在\(\|z\|<r\)解析.
在\(\|z\|\le r\)上, \(\phi\)是连续的, \(f\)是连续的, \(h\)在\(\|z|\le r\)和\(\|w\|_0\le R\)上连续,
\(\|\phi[f(z)]\|_0\le R\)对于\(\|z\|\le r\), 于是\(h(z,\phi[f(z)])\)在\(\|z\|\le r\)上连续.
\(2^\circ\) 既然\(h(z,\phi[f(z)])\)在\(\|z\|<r\)上解析, 它有级数展开; 我们计算
\[T(\phi)(0)=h(0,0)=0, (\text{see}~{eq-5-5})
\]
\[T(\phi)^{q}(0)=H_q(0,0,c^{(1)},\cdots,c^{(q)})=c^{(q)},~~q=1,2,\cdots,p.
\]
\(3^\circ\)
\[\begin{aligned}
&\|T(\phi)^{(p)}(z)-c^{(p)}\|_p
=\|H_p(z,\phi[f(z)],\phi^{(1)}[f(z)],\cdots,\phi^{(p)}[f(z)])-H_p(0,0,c^{(1)},\cdots,c^{(p)})\|_p\\
\le & \|H_p(z,\phi[f(z)],\cdots,\phi^{(p)}[f(z)])-H_p(z,0,c^{(1)},\cdots,c^{(p)})\|_p\\
&+\|H_p(z,0,c^{(1)},\cdots,c^{(p)})-H_p(0,0,c^{(1)},\cdots,c^{(p)})\|_p, (\text{利用引理}{lemma-5-5})\\
\le & L \sum_{q=0}^{p-1}\|\phi^{(q)}[f(z)]-c^{(q)}\|_q+\theta \|\phi^{(p)}[f(z)]-c^{(p)}\|_p
+\|H_p(z,0,c^{(1)},\cdots,c^{(p)})-H_p(0,0,c^{(1)},\cdots,c^{(p)})\|_p\\
& (\text{利用}3^\circ: \|\phi^{(p)}[f(z)]-c^{(p)}\|_p\le K;\text{and}~~(eq-5-27))\\
\le & L \sum_{q=0}^{p-1} m^{1/2} n^{p-q/2} r \left(\sum_{t=q+1}^{p} c_{t}+K\right)+\theta K
+ \|H_p(z,0,c^{(1)},\cdots,c^{(p)})-H_p(0,0,c^{(1)},\cdots,c^{(p)})\|_p\\
\end{aligned}
\]
为了让上式\(\le K\), 我们可以让
\[\begin{aligned}
&L \sum_{q=0}^{p-1} m^{1/2} n^{p-q/2} r \left(\sum_{t=q+1}^{p} c_{t}+K\right)\le \frac{1}{2}(1-\theta)K, \\
&\|H_p(z,0,c^{(1)},\cdots,c^{(p)})-H_p(0,0,c^{(1)},\cdots,c^{(p)})\|_p\le \frac{1}{2} (1-\theta)K.
\end{aligned}
\]
{\color{red}[这两个都要放到{eq-determin-r}中!]}
我们证明了\(T(\mathfrak{R})\subset \mathfrak{R}\).
Step 5. 证明\(T\)是压缩映射.
对于任意的\(\phi,\hat{\phi}\in \mathfrak{R}\), 我们有: \(\forall \|z\|\le r\),
\[\begin{equation*}
\begin{aligned}
&\|T(\phi)^{(p)}(z)-T(\hat{\phi})^{(p)}(z)\|_p\\
=&\|H_p(z,\phi[f(z)],\phi^{(1)}[f(z)],\cdots,\phi^{(p)}[f(z)])-
H_p(z,\hat{\phi}[f(z)],\hat{\phi}^{(1)}[f(z)],\cdots,\hat{\phi}^{(p)}[f(z)])\|_p\\
\le & L\sum_{q=0}^{p-1}\|\phi^{(q)}[f(z)]-\hat{\phi}^{(q)}[f(z)]\|_q+\theta \|\phi^{(p)}[f(z)]-\hat{\phi}^{(p)}[f(z)]\|_p\\
\end{aligned}
\end{equation*}
\]
利用{Taylor-formula}, 我们可以得到
\[\begin{aligned}
&\left|\frac{\partial^q\phi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}-
\frac{\partial^q\hat{\phi}^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}}
\right|\\
=&\frac{1}{(p-q)!}\sum_{j_1\cdots j_{p-q}=1}^{n}\sup_{\|z\|\le r}
\left|
\frac{\partial^{(p)}\phi^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}\partial z_{j_1}\cdots \partial z_{j_{p-q}}}
-\frac{\partial^{(p)} \hat{\phi}^k(z)}{\partial z_{i_1}\cdots \partial z_{i_q}\partial z_{j_1}\cdots \partial z_{j_{p-q}}}
\right| \|z_{j_1}\cdots z_{j_{p-q}}\|\\
\le & \frac{1}{(p-q)!} r^{p-q} n^{p-q}\rho(\phi,\hat{\phi})\le r^{p-q} n^{p-q}\rho(\phi,\hat{\phi}).
\end{aligned}
\]
于是
\[\|\phi^{(q)}(z)-\hat{\phi}^{(q)}(z)\|_q\le m^{1/2} n^{p-q/2} r^{p-q} \rho(\phi,\hat{\phi}).
\]
进而
\[\begin{equation*}
\begin{aligned}
&\|T(\phi)^{(p)}(z)-T(\hat{\phi})^{(p)}(z)\|_p
\le L\sum_{q=0}^{p-1}m^{1/2} n^{p-q/2} r^{p-q} \rho(\phi,\hat{\phi})
+\theta \rho(\phi,\hat{\phi})\\
\le & \vartheta \rho(\phi,\hat{\phi}),
~~\left[ L\sum_{q=0}^{p-1}m^{1/2} n^{p-q/2} r^{p-q}+\theta\le \vartheta\right].
\end{aligned}
\end{equation*}
\]
{\color{red}(这一部分\([ L\sum_{q=0}^{p-1}m^{1/2} n^{p-q/2} r^{p-q}+\theta\le \vartheta]\)也需放在
{eq-determin-r}中!)}
证毕.
利用一个基本的手法我们可以弱化假设 (II)，用下面的假设 (V) 替换假设 (II)：
- (V) \(\lambda_0 := \max_{1 \le i \le n} \|\lambda_i\| < 1.\)
其中 \(\lambda_i\) 是 \(B\) 的特征值。
首先我们说明 (V) 是比 (II) 弱的：一方面，利用 \(\|B\|_* \ge \lambda_0\)，我们知道 \(\|B\|_* < 1\) 蕴含 \(\lambda_0 < 1\)；另一方面，我们可以很容易找到例子说明 \(\lambda_0 < 1\) 但 \(\|B\|_* > 1\)，例如：
\[B = \begin{pmatrix}
\frac{1}{2} & 10\\
0 & \frac{1}{2}
\end{pmatrix}.
\]
我们有
\[B^T (1,0)^T = \begin{pmatrix}
\frac{1}{2} & 0\\
10 & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
1\\
0
\end{pmatrix}
= \begin{pmatrix}
\frac{1}{2}\\
10
\end{pmatrix},
\]
\[\Rightarrow \|B^T(1,0)\|_2 = \sqrt{\frac{1}{4}+100} > 1 \Rightarrow \|B\|_* = \sup_{\|t\|=1,t\in \mathbb{R}^2} \|B^T t\|_2 > 1.
\]
下面我们证明定理{theorem-5-1} 的弱化版本：这里我们取正整数 \(p\) 满足
\[\begin{equation}\tag{eq-determine-p}
\|[a_{0,l}^k]\|^* \lambda_0^p < 1.
\end{equation}
\]
定理 (theorem-5-2)
假设 (I)(III)(IV)(V) 成立，则对于由 {eq-5-20} 确定的每一组 \(c^{(1)}, \cdots, c^{(p)}\)，恰好存在唯一的一个函数 \(\phi\) 在 \(z=0\) 的一个邻域内有定义且解析，满足方程 {eq-smajdor}，使得
\[\begin{equation}\tag{eq-5-28}
\phi(0) = 0, ~ \phi^{(q)} = c^{(q)}, ~ q = 1,2,\cdots,p.
\end{equation}
\]
证明：
利用引理{lem:1.1}，对于一个固定的 \(\varepsilon > 0\) 满足 \(\lambda_0 + \varepsilon < 1\)，我们可以找到一个非奇异矩阵 \(T\) 使得
\[\|T^{-1} BT\|_* < 1.
\]
查看原方程 {eq-smajdor}：
\[\phi(z) = h(z, \phi[f(z)]).
\]
我们可以做如下的变形：
\[\phi(Tz) = h(Tz, \phi[f(Tz)]) = h(Tz, \phi[TT^{-1} f(Tz)]).
\]
令 \(\psi(z) = \phi(Tz)\)，\(\hat{h}(z, w) = h(Tz, w)\)，\(\hat{f}(z) = T^{-1} f(Tz)\)，我们有
\[\begin{equation}\tag{eq-5-29}
\psi(z) = \hat{h}(z, \psi[\hat{f}(z)]).
\end{equation}
\]
我们将对上述方程应用定理 {theorem-5-1}。为此，我们需要验证 (I)(II)(III)(IV) 成立。
(I) \(\hat{f}\) 使得 (III) 成立的变量范围是
\[z \in T^{-1}(K(0, r_0)).
\]
(II) 对 \(\hat{f}\) 变为
\[\left|\left| \frac{\partial \hat{f}_j(z)}{\partial z_i} \bigg|_{z=0}\right|\right|_* = \left|\left| \frac{\partial (T^{-1} f(Tz))_j}{\partial z_i} \bigg|_{z=0}\right|\right|_* = \left|\left| T^{-1} B T \right|\right|_* < 1.
\]
(III) \(\hat{h}\) 使得 (I) 成立的变量范围是
\[z \in T^{-1}(K(0, r_0)),~~ w \in K(0, R_0.
\]
(IV) 因为假设 (IV) 对于关于 \(\phi\) 的方程{eq-smajdor} 成立，即存在\(\phi(z) = (\phi^1(z), \cdots, \phi^m(z))\) 满足 {eq-smajdor}，其中
\[\phi^k(z) = \sum_{\mu=1}^{\infty} \sum_{i_1 \cdots i_\mu=1}^{n} c^k_{i_1 \cdots i_{\mu}} z_{i_1} \cdots z_{i_{\mu}}.
\]
我们可以很容易看到 \(\psi(z) = (\psi^1(z), \cdots, \psi^m(z))\) 满足上面的{eq-5-29}，其中
\[\begin{equation}
\tag{eq-5-30}
\begin{aligned}
\psi^k(z) &= \sum_{\mu=1}^{\infty} \sum_{j_1 \cdots j_\mu=1}^{n} \hat{c}^k_{j_1 \cdots j_{\mu}} z_{j_1} \cdots z_{j_{\mu}}, \\
\hat{c}^k_{j_1 \cdots j_{\mu}} &= \sum_{i_1 \cdots i_\mu=1}^{n} c_{i_1 \cdots i_{\mu}}^k t^{i_1}_{j_1} \cdots t^{i_\mu}_{j_\mu}.
\end{aligned}
\end{equation}
\]
一个简单的推导：按照之前的约定 \([t_i^j]\) 的上标是行标，下标是列标。
\[(\hat{z}_1, \cdots, \hat{z}_n)^T = [t_i^j](z_1,\cdots,z_n)^T = \left(\sum_{i=1}^{n} t_i^1 z_i, \cdots, \sum_{i=1}^{n} t_i^n z_i\right)^T,
\]
\[\begin{aligned}
\psi^k(z) &= \phi^k(Tz) = \sum_{\mu=1}^{\infty} \sum_{i_1 \cdots i_\mu=1}^{n} c^k_{i_1 \cdots i_{\mu}} \left(\sum_{j_1=1}^{n} t_{j_1}^{i_1} z_{j_1}\right) \cdots \left(\sum_{j_\mu=1}^{n} t_{j_\mu}^{i_\mu} z_{j_\mu}\right) \\
&= \sum_{\mu=1}^{\infty} \sum_{i_1 \cdots i_\mu=1}^{n} c^k_{i_1 \cdots i_{\mu}} \left[\sum_{j_1 \cdots j_\mu=1}^{n} t_{j_1}^{i_1} \cdots t_{j_\mu}^{i_\mu} z_{j_1} \cdots z_{j_\mu}\right] \\
&= \sum_{\mu=1}^{\infty} \sum_{j_1 \cdots j_\mu=1}^{n} \left[\sum_{i_1 \cdots i_\mu=1}^{n} c^k_{i_1 \cdots i_{\mu}} t_{j_1}^{i_1} \cdots t_{j_\mu}^{i_\mu}\right] z_{j_1} \cdots z_{j_\mu}.
\end{aligned}
\]
由此，
\[\begin{equation}
\tag{eq-5-31}
\hat{c}^k_{j_1 \cdots j_{\mu}} = \sum_{i_1 \cdots i_\mu=1}^{n} c_{i_1 \cdots i_{\mu}}^k t^{i_1}_{j_1} \cdots t^{i_\mu}_{j_\mu}.
\end{equation}
\]
因为 \(T\) 是可逆的线性变换，所以由 {eq-5-31} 建立的 \(\hat{c}^k_{j_1 \cdots j_{\mu}}\) 与 \(c^k_{j_1 \cdots j_{\mu}}\) 之间的对应是一一对应.
验证了 {\rm (I)(II)(III)(IV)}，我们可以应用定理{theorem-5-1}，从而得到方程{eq-5-29} 在 \(z=0\) 的一个邻域内有解析解 \(\psi\)，满足：
\[\psi(0) = 0, ~ \psi^{(q)} = \hat{c}^{(q)}, ~ q = 1, 2, \cdots, p.
\]
由此，我们得到了方程{eq-smajdor} 在 \(z=0\) 的某邻域内有唯一的局部解析解 \(\phi = \psi \circ T^{-1}\)，满足
\[\phi(0) = 0, ~ \phi^{(q)} = c^{(q)}, ~ q = 1, 2, \cdots, p.
\]
定理得证.
下面是定理{theorem-5-2} 的一个应用：
假设 \(h(z,w)\) 是下面类型的函数：
\[h: \mathbb{C}^n \times \mathbb{C}^m \to \mathbb{C}
\]
它在 \((z,w) = (0,0)\) 的某个邻域内解析 [这意味着假设 {\bf (III)} 满足]，并设 \(f(z)\) 满足假设 {\bf (I)(V)}。
一个 \(m\) 阶的方程
\[\begin{equation}\tag{eq-5-32}
\varphi(z) = h(z, \varphi[f(z)], \cdots, \varphi[f^m(z)]),
\end{equation}
\]
其中的未知函数 \(\varphi(z)\) 是这种类型的：
\[\varphi: \mathbb{C}^n \to \mathbb{C},
\]
可以被约化为一个由 \(m\) 个 \(1\) 阶方程组成的方程组：
\[\begin{equation}\tag{eq-5-33}
\left\{
\begin{aligned}
\varphi_1(z) &= h_1(z, \varphi_1[f(z)], \cdots, \varphi_{m-1}[f(z)]), \\
\varphi_2(z) &= \varphi_1[f(z)], \\
\cdots \\
\varphi_m(z) &= \varphi_{m-1}[f(z)],
\end{aligned}
\right.
\end{equation}
\]
它是一个形式为{eq-smajdor} 的方程：
\[\hat{\varphi}(z) =
\begin{pmatrix}
\varphi_1(z) \\ \varphi_2(z) \\ \cdots \\ \varphi_m(z)
\end{pmatrix}
=
\begin{pmatrix}
h_1(z, \varphi_1[f(z)], \cdots, \varphi_m[f(z)]) \\ \varphi_1[f(z)] \\ \cdots \\ \varphi_{m-1}[f(z)]
\end{pmatrix}
= h(z, \varphi_1[f(z)], \cdots, \varphi_m[f(z)]).
\]
由此, 对应于{eq-smajdor}的\(h\)是：
\[h(z, w^1, \cdots) =
\begin{pmatrix}
h_1(z, w^1, \cdots, w^{m}) \\
w^1 \\
\cdots \\
w^{m-1}
\end{pmatrix}.
\]
从而，如果令
\[[a_{0,l}^k] = \frac{\partial h_k(0,0)}{\partial w^l},
\]
那么
\[a_{0,l}^1 = \frac{\partial h_1(0,0)}{\partial w^l}, ~~ a_{0,l}^k = \frac{\partial w^{k-1}}{\partial w_l} = \delta_{k-1,l} \quad (\text{Kronecker's delta}), ~ k = 2, \cdots, m.
\]
假设方程{eq-5-32} 有一个形式级数解：
\[\begin{equation}\tag{eq-5-34}
\varphi(z) = \sum_{\mu=1}^{\infty} \sum_{i_1 \cdots i_\mu=1}^{n} d_{i_1 \cdots i_\mu} z_{i_1} \cdots z_{i_\mu},
\end{equation}
\]
[这意味着假设 {\bf (IV)} 满足]，并设\(p\)是一个正整数满足{eq-determine-p}，那么利用定理{theorem-5-2}，我们可以得到
定理(theorem-5-3)
假设函数 \(f\) 满足 {\rm (I)(V)} 并设 \(h(z,w)\) 是一个在 \((z,w) = (0,0)\) 的某个邻域内解析的函数， \(h(0,0) = 0\)。那么对于在展开式{eq-5-34} 中的每一组 \(d_{j_1 \cdots j_q}, q = 1, 2, \cdots, p\)，方程{eq-5-32} 在 \(z=0\) 的某个邻域内有唯一的局部解析解 \(\varphi\) 满足
\[\varphi(0) = 0, ~ \frac{\partial^q \varphi(z)}{\partial z_{j_1} \cdots \partial z_{j_q}} \bigg|_{z=0} = d_{j_1 \cdots j_q}, ~~ j_1, \cdots, j_q = 1, \cdots, n; ~ q = 1, 2, \cdots, p.
\]
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