我们从以下定义开始本部分:
定义2.1
我们说一个映射\(\贝丝:\ mho^{z}\右箭头\ mho\)是Prešić型有理数η-收缩,收缩(客运专线 η-C类,简称)如果有\(伽马\in(0,1)\)以便
$$\eta\bigl(\varpi\bigle(\beth(\zeta{1},\dots,\zeta_{z}),\beth zeta{j+1})}:1\leqj\leqz\biggr\}\biggr)\biggr\}^{\gamma}$$
(2.1)
对于每个\((泽塔{1},点,泽塔{z+1})具有\(\beth(\zeta{1},\dots,\zeta_{z})\neq\beth.
应该注意,如果\(\t(r)=e^{\sqrt{r}}\),然后客运专线 η-C类减少到
$$\varpi\bigl(\beth(\zeta{1},\dots,\zeta_{z}),\beth 1\leq j\leq z\biggr\}\biggr)$$
(2.2)
对于每个\((泽塔{1},点,泽塔{z+1}),\(\beth(\zeta{1},\dots,\zeta_{z})\neq\beth.
此外,如果\((泽塔{1},点,泽塔{z+1})是这样的\(\beth(\zeta_{1},\dots,\zeta_{z})=\beth(\zeta_{2},\dots,\zeta_{z+1})\),然后是条件(2.2)比(1.3),所以映射ℶ英寸(2.2)扩展并统一了Cirić–Prešić收缩。
备注2.2
每客运专线 η-C类 ℶ是Prešić映射\((\eta_{1})\)和(1.4)也就是说,
$$\begin{aligned}\varpi\bigl(\beth(\zeta{1},\dots,\zeta_{z}),\beth qj\leqz\biggr\}\\&<max\bigl\{varpi(\zeta{j},\zeta_{j+1}):1\leqj\ leqz\ bigr\}。\结束{对齐}$$
对于每个\((泽塔{1},点,泽塔{z+1})具有\(\beth(\zeta{1},\dots,\zeta_{z})\neq\beth因此,每个客运专线 η-C类 ℶ是一个连续函数。
现在,我们的第一个结果如下:
定理2.3
假设 \(\贝丝:\ mho^{z}\右箭头\ mho\) 是PTR η-C类.然后对于任何选择的点 \(\泽塔{1},\圆点,\泽塔}\ in \ mho\),顺序 \({\zeta_{l}\}\) 中描述的(1.2)收敛于 \(\ zeta^{\ ast}\ in \ mho\) 和 \(\泽塔^{\ast}\) 是的FP ℶ.此外,如果 \(\beth(\zeta^{\ast},\dots,\zeta_{\ast{)\neq\beth 具有
$$\eta\bigl(\varpi\bigle(\beth\bigl-(\zeta^{ast},\dots,\zeta_{ast}\bigr),\beth\ biglγ}$$
对于 \(\ zeta ^{\sast},\ zeta ^{\prime}}\ in \ mho) 这样的话 \(\泽塔^{\ast}\neq\zeta^{\prime}}\),然后是重点 \(\泽塔^{\ast}\) 是独一无二的.
证明
让\(\泽塔{1},\点,\泽塔})武断z(z)中的元素℧和用于\(l\in\mathbb{N}\)顺序\({\zeta_{l}\}\)在中定义(1.2). 如果是一些\(l_{0}=\{1,2,\dots,z\}\)一个有\(泽塔{l{0}}=\泽塔{1{0}+1}\),然后
$$\zeta_{l_{0}+z}=\beth(\zeta_2{l_0}},\zeta_3{l_0{0}+1},\ dots,\zeta _{l_0}+z-1})=\besh(\zeta _{l_}0}+z},\fzeta__{l_2{0}+z},\sdots,\fzeta _$$
也就是说\(\泽塔{l{0}+z}\)是的FPℶ不需要进一步的证据。因此,我们认为\(泽塔{l+z}\neq\泽塔{1+z+1}\)为所有人\(l\in\mathbb{N}\).放置\(\gimel{l+z}=\varpi(\zeta{l+z},\zeta}l+z+1})和
$$\phi=\max\biggl\{\frac{\varpi(\zeta_{1},\zeta_2})}{1+\varpi(\zeta _{1{,\zeta _{2}){,\ frac{\ varpi{z+1})}{1+\varpi(\zeta{z},\zeta_{z+1{)}\biggr\}$$
那么,对所有人来说\(l\in\mathbb{N}\)和\(φ>0),我们有\(\gimel_{l+z}>0\)因此,对于\(l \leq z),我们获得
$$\begin{aligned}1&<\eta(\gimel_{z+1})\\&=\eta\bigl(\varpi(\zeta_{z+1},\zeta_{z+2})\bigr)\\&=\eta\bigl(\varpi\bigl(\beth(\zeta_{1},\zeta_{2},\dots,\zeta_{z}),\beth(\zeta_{2},\zeta_{3},\dots,\zeta_{z+1})\bigr)\\&&\leq\biggl[\eta\biggl(\max\biggl\{\frac{\varpi(\zeta_{j},\zeta_{j+1})}{1+\varpi(\zeta{j},\zeta_{j+1})}:1\leqj\leqz\biggr\}\biggr)\biggr]^{gamma}\\&=\bigl[\eta(\phi)\bigr]^{gamma}。\结束{对齐}$$
也,
$$\开始{对齐}1&<\eta(\gimel_{z+2})\\&=\eta\bigl(\varpi(\zeta_{z=2},\zeta_{z+3})\bigr)\\&=\eta\ bigl ta{z+2})\bigr)\&\leq\biggl[\eta\biggl(\max\biggl/{frac{\varpi(\zeta{j},\zeta_{j+1})}{1+\varpi(\zeta{j},\zeta_{j+1})}:2\leqj\leqz+1\biggr\}\biggr)\biggr]^{gamma}\\&=\bigl[\eta(\phi)\bigr]^{gamma^{2}}。\结束{对齐}$$
以相同的模式继续\(1),我们得到
$$\开始{对齐}1&<\t(\gimel_{z+l})2},\dots,\zeta_{l+z})\bigr)\&\leq\bigl[\eta(\phi)\bigr]^{\gamma^{l}}。\结束{对齐}$$
(2.3)
拿\(l\rightarrow\infty\)在(2.3)和使用\((\eta_{2})\),我们有
$$\lim_{l\rightarrow\infty}\eta(\gimel_{z+l})=1\quad\Longleftrightarror\quad\lim_}l\right arrow\ infty{\gimel _{z+l}=0$$
基于\((\eta_{3})\),有\(\ ell\ in(0,1)\)和\(u \ in(0,\ infty)\)以便
$$\lim_{l\rightarrow\infty}\biggl(\frac{\eta(\gimel_{z+l})-1}{\gimel-{z+l}^{\ell}}\bigr)=u$$
假设\(u<\infty)和\(v=\压裂{u}{2}>0\)根据限额的定义,有\(l_{1}\在\mathbb{N}\中)这样的话
$$\biggl\vert\frac{\eta(\gimel_{z+l})-1}{\gimel _{z+l}^{\ell}}-u\biggr\vert\leqv,\quad\forall l>l_{1}$$
由此可见
$$\frac{\eta(\gimel_{z+l})-1}{\gimel-{z+l}^{\ell}}\gequ-v=\frac{u}{2}=v,\quad\forall l>l{1}$$
设置\(压裂{1}{v}=q\),然后
$$l\gimel_{z+l}^{\ell}\leq-lq\bigl(\eta(\gimel _{z+1})-1\bigr),\quad\对于所有l>l_{1}$$
假设\(u=信息\)和\(v>0)根据限额的定义,有\(l_{1}\in\mathbb{N}\)这样的话
$$v\leq\frac{\eta(\gimel_{z+l})-1}{\gimel _{z+l}^{ell}},\quad\forall l>l_{1}$$
这意味着服用后\(压裂{1}{v}=q\)那个
$$l\gimel_{z+l}^{\ell}\leq-lq\bigl(\eta(\gimel _{z+1})-1\bigr),\quad\对于所有l>l_{1}$$
因此,在这两种情况下,都有\(l_{1}\在\mathbb{N}\中)和\(q>0)以便
$$l\gimel_{z+l}^{\ell}\leq-lq\bigl(\eta(\gimel _{z+1})-1\bigr),\quad\对于所有l>l_{1}$$
正在应用(2.3),我们得到
$$l\gimel_{z+l}^{ell}\leq-lq\bigl(\bigl[\eta(\phi)\bigr]^{gamma^{l}}-1\bigr),\quad\forall l>l_{1}$$
以及,当\(l\rightarrow\infty\),有
$$\lim_{l\rightarrow\infty}l\gimel_{z+l}^{ell}=0$$
因此,有\(l_{2}\在\mathbb{N}\中)和\(q>0)这样的话
$$l\gimel_{z+l}^{ell}\leq1,\quad\对于所有l>l{2}$$
因此我们可以写
$$\gimel_{z+l}\leq\frac{1}{l^{\frac}{\ell}}},\quad\forall l>l{2}$$
现在,我们澄清一下\({\zeta_{l}\}\)是一个柯西序列。对于\(b>l>l{2}\),一个人可以写
$$开始{对齐}\varpi(\zeta_{z+l},\zeta_{z+b})={}&\varpi\bigl),\beth(\zeta{l+1},\dots,\zeta_{z+l})\bigr)\\&{}+\varpi\bigl(\beth,\beth(\zeta{l+2},\dots,\zeta_{z+l+1})\bigr)\\&{}+\cdots+\varpi\bigl(\beth+\varpi(泽塔{z+l+1},泽塔{z+l+2})+\cdots+\varπ(泽塔{z+b-1},泽塔{z+b})\\={}&\gimel{l+z}+\gimel+z+1}+\cdots+\gimel{z+b-1}\\={}&\sum{s=l}^{b-1}\gimel{s+z}<\sum{s=l}^{\infty}\gi梅尔{s+z}\leq\sum{s=l}^{\effty}$$
因此可以得出以下结论\({\zeta_{l}\}\)是中的Cauchy序列\((\mho,\varpi)\).完整性℧收益率\(\ zeta^{\ ast}\ in \ mho\)这样的话
$$\lim_{l,b\rightarrow\infty}\varpi(\zeta_{l},\zeta_{b})=\lim_{l\rightarrow\infty}\varpi\bigl(\zeta_{1},\ zeta^{ast}\bigr)=0$$
因为ℶ是连续的,我们有
$$\开始{aligned}\hbar&=\lim_{l\rightarrow\infty}\zeta_{l+z}\\&=\lim_{l\ rightarror\infty}\beth(\zeta_{l},\zeta_a{l+1},\ dots,\zeta _{z+l-1})\\&=\ beth\Bigl点,\lim_{l\rightarrow\infty}\zeta_{z+l-1}\Bigr)\\&=\beth\Bigl(\zeta^{\ast},\zeta^{\ast},\dots,\zeta ^{\ast}\biger)。\结束{对齐}$$
对于唯一性,假设\(\泽塔^{\ast}\)和\(\泽塔^{{\素数}}\)是映射的两个不同FPℶ即。,\(\zeta^{\ast}=\beth(\zeta ^{\ast},\zeta `{\ast{,\dots,\zeta`{\asp})\)和\(\zeta^{{prime}}=\beth具有\(\泽塔^{\ast}\neq\zeta^{\prime}}\)因此,通过假设(2.1),我们可以写
$$\begin{aligned}\eta\bigl(\varpi\bigle(\zeta^{\ast},\zetaca^{\prime}}\bigr)\biger)&=\eta\ bigl zeta^{{prime}}\biger)\biger)\biger素数}}\bigr)\biger)\bigr]^{\gamma},\end{aligned}$$
矛盾,如\(伽马\in(0,1)\)因此,\(\泽塔^{\ast}=\泽塔{{\prime}}\)证明到此结束。□
以下示例支持定理2.3.
例2.4
让\({\zeta_{l}\}\)序列定义如下:
$$\textstyle\begin{cases}\zeta{1}=3,\\zeta{2}=3+7,\\vdots\\\zeta_{l}=3+7+11+\cdots+(4l-1)=l(2l+1)。\结束{cases}$$
假设\(\mho=\{\zeta_{l}:l\in\mathbb{N}\}\)和\(\varpi(\widetilde{\zeta},\widehat{\ze塔})=\vert\widetilde{\zeta}-\wideheat{\ze塔}\vert\)显然,\((\mho,\varpi)\)是一个完整的度量空间。定义映射\(\贝丝:\ mho ^{3}\右箭头\ mho)通过
$$\beth(\zeta_{l},\widetilde{\zeta}_{l{,\wedehat{\ze塔}_{1})=\textstyle\begin{cases}\frac{\zeta _{l-1}+\widetelde{\zerta}_}+\wide hat{\zeta{{l-1{3},当}l>1时,&\text{{1}+\widehat{\zeta}{1}}{3},&\text{otherwise.}\end{cases}$$
对于\(l>5\),我们有
$$开始{对齐}和\varpi\bigl{l-3}+\zeta{l-2}+\zeta{l-1}{3}\biggr)\\&\quad=\frac{1}{3+\bigl\vert\bigl((1-5)(2-1-9)+(1-4)(2-l-7)+(l-3)(2l-5)\biger)\\&\qquad{}-\bigl((l-3)(2l-5)+(l-2)(21-3)+(1-1)(21-1)\biger^{2} -45升+88磅以上)-磅(6磅^{2} -21升+22\biger)\bigr\vert\\&\quad=\frac{1}{3}\vert24l-66\vert=8l-22,\end{aligned}$$
和
$$\begin{aligned}和\max\bigl\{\varpi\bigl((\zeta_{l-4},\zeta_{l-3},\ zeta__{l-2})3)(2l-5)-(l-1)(2l-1)\bigr\vert,\bigl\vert(l-2)(21-3)-l(2l+1)\biger\vert\end{B矩阵}\\&\quad=最大值\bigl\{(8l-22),(8l-14),(6l-6)\bigr\}=(8l-14)。\结束{对齐}$$
现在,
$$\lim_{l\rightarrow\infty}\frac{\varpi(\beth(\zeta_{l-4},\zeta_{l-3},\zeta_{l-2}),\beth(\zeta_{l-2},\zeta_{l-1},\zeta_{l}))}{\max\{\varpi((\zeta_{l-4},\zeta_{l-3},\zeta_{l-2}),(\zeta_{l-2},\zeta_{l-1},\zeta_{l}))}=\lim_{l\rightarrow\infty}\frac{8l-22}{8l-14}=1$$
因此,
$$\varpi\bigl(\beth(\zeta_{l-4},\zeta_2},\ zeta__{l-2}),\beth 1},\zeta_{l})\biger)\bigr\}$$
不适用于\(伽马\in(0,1)\),这意味着该假设(1.1)定理的1.1未实现。现在,定义映射\(\ta:(0,\infty)\rightarrow(1,\infcy)\)通过\(eta(s)=e^{\frac{se^{s}}{1+s}}\)。我们可以很容易地验证\(纳布拉语中的)和ℶ是客运专线 η-C类事实上,不平等
$$\开始{对齐}和e^{\sqrt{\varpi(\beth(\zeta{i},\zeta_{i+1},\ zeta_}i+2}),\beth{i+2},\泽塔{i+3},\zeta{i+4})}{1+\varpi,\zeta _{i+4} ) ) }}} \\ &\quad \leq e^{\gamma \sqrt{\varpi ( ( \zeta _{i},\zeta _{i+1}, \zeta _{i+2} ) , ( \zeta _{i+2},\zeta _{i+3},\zeta _{i+4} ) ) \frac{e^{\varpi ( ( \zeta _{i},\zeta _{i+1},\zeta _{i+2} ) , ( \zeta _{i+2},\zeta _{i+3},\zeta _{i+4} ) ) }}{1+\varpi ( ( \zeta _{i},\zeta _{i+1},\zeta _{i+2} ) , ( \zeta _{i+2},\zeta _{i+3},\泽塔{i+4})}}},\结束{对齐}$$
(2.4)
等待\(\beth(\zeta{i},\zeta}i+1},\ zeta{i+2})\neq\beth,\(i=1,2,\点\),对一些人来说\(伽马\in(0,1)\).不平等(1.1)相当于
$$开始{对齐}和\varpi\bigl i+2},\泽塔{i+3},\zeta{i+4})}{1+\varpi((泽塔{i},泽塔{i+1},泽塔{i+2})。\结束{对齐}$$
所以,对一些人来说\(伽马\in(0,1)\),我们可以写
$$\frac{\varpi(\beth(\zeta{i},\zeta_{i+1},\ zeta{i+2}),\beth{i+3},\zeta{i+4})}{1+\varpi)}}{max\{varpi((泽塔{i},泽塔{1+1},泽塔{i+2}),泽塔{i+3},\zeta{i+4}))\}}}}\leq\gamma^{2}$$
现在,我们将讨论以下情况:
(i) 如果\(i=l=1\),我们得到
$$开始{对齐}和\frac{\varpi(\beth(\zeta{1},\zeta},\ zeta{3}),\beth{5})}{1+\varpi(\beth(\zeta{1},\zeta},\ zeta{3}),\beth)}}{max\{varpi((泽塔{1},泽塔{2},泽塔{3}),){1+\max\{varpi((\zeta{1},\zeta},\ zeta{3}))\}}}}\\&\quad=\frac{\varpi(\frac}\zeta{1}+\ zeta{2}+\泽塔{3}}{3},\frac\\zeta{3}+\泽塔{4}+\ zeta{5}}{3})e^{\frac{\varpi(\frac{\zeta}+\压裂{泽塔{3}+\泽塔{4}+\泽塔{5}}{3})}{1+\varpi(\frac{\zeta{1}+\ zeta{2}+\ zeta{3}{3{,\frac}\zeta}+\}}{max\{varpi((泽塔{1},泽塔{2},泽塔{3}))}{1+max\{varpi((泽塔{1},泽塔{2},泽塔{3}),(泽塔}3},\泽塔{4},\ze塔{5}))\}}}}\\&\quad=\frac{\varpi}e^{{frac{max\{varpi((3,10,21),(21,36,55))}}}}\\&\quad\leq\frac{26e^{26}}{34e^{34}}=\ frac{13}{17} e(电子)^{-8}<e^{-2}。\结束{对齐}$$
(ii)如果\(i=l>1),我们获得
$$开始{对齐}和\frac{瓦尔比2},\泽塔{l+3},\zeta{l+4})}{1+\varpi{l+2},\zeta{l+3},\ zeta{1+4})\}}{1+\max\{(\zeta_{l},\fzeta{l+2}),(\fzeta_1+2}}}}}}\\&\quad=\frac{\varpi(\frac{\zeta{l-1}+\zeta_{l}+\zeta_{1}}{3},\frac{\ zeta{{l+1}+\泽塔{l+2}+\泽塔{l+3}}{3})e^{\frac}1}}{3},\frac{\zeta{l+1}+\zeta}{l+2}+\zeta{l+3}}{3)}{1+\varpi+\泽塔{l+3}}{3})}}{max\{varpi,\泽塔{l+4})}}}}{3}-\压裂{6l^{2}+27l+34}{3}{3}-\压裂{6l^{2}+27l+34}{3}\vert}{1+vert\frac{6l|{2}+3l+4}{3}-\{6l^{2}+27l+34}{3}\vert}}}{max\{vert8l+10\vert,\vert81+18\vert、\vert9l+26\vert\}e^{max\vert7l+10\ vert,\ vert8l+18\ vert、\ vert 8l+26\ vert}{1+max\vert 8l+10\vert,\Vert8l+26 \vert{}}}}四元=frac{(8l+10)e^{压裂{(81+10)}{1+(8l+1)}}}{(18l+26)e^{压裂{(8l+26)}{1+(8l+66)}}}$$
具有\(\gamma=\frac{1}{e}\)因此,定理的所有要求2.3已实现,并且该点\((1,1,1)\)是的唯一FPℶ.
例2.5
假设\(mho=[0,1]\),\(\varpi(\widetilde{\zeta},\widehat{\ze塔})=\vert\widetilde{\zeta}-\wideheat{\ze塔}\vert\)、和\(\贝丝:\ mho^{z}\右箭头\ mho\)由描述
$$\beth(\zeta_{1},\dots,\zeta_{l})=\frac{\zeta_1}+\zeta{l}}{8l},\ quad\forall \ zeta_},\todts,\zeta_{l}\in\mho$$
让\(\ta:(0,\infty)\rightarrow(1,\infcy)\)是由定义的映射\(\t(s)=e^{\sqrt{\frac{s}{1+s}}}\)。自\(e^{\sqrt{\frac{s}{1+s}}}\leqe^{\sqrt{s}}\),我们可以从中看到[15]那个\(纳布拉语中的)。现在,为了\(泽塔{1},泽塔{2},圆点,泽塔}l+1}in),一个人可以写
$$\varpi\bigl(\beth(\zeta{1},\dots,\zeta_{l}),\beth$$
和
$$开始{对齐}和\eta\bigl ta{l+1})}{1+\varpi(\beth(\zeta{1},\dots,\zeta}),\beth)}}}\\&\quad=e^{\sqrt{\frac{(\frac{1}{8l})\vert(\zeta_{1}-\泽塔{2})+(\泽塔_{l}-\zeta{l+1})\vert}{1+\vert(\zeta_{1}-\泽塔{2})+(\泽塔_{l}-\zeta{l+1})\vert}}}\\&\quad=e^{(\frac{1}{2\sqrt{2l}})\sqrt{\frac}\vert(\zeta_{1}-\泽塔{2})+(\泽塔_{l}-\ζ_{l+1})\vert}{1+\vert(\ζ_{1}-\泽塔{2})+(\泽塔_{l}-\zeta{l+1})\vert}}}\\&\quad\leqe^{(\frac{1}{\sqrt{2}})\sqrt{\frac}\max\{\varpi(\zeta{1},\zeta_{2}),\ varpi{l},\zeta{l+1})}}(\frac{1}{\sqrt{2})}{1+\varpi(\zeta_{j},\zeta_{j+1})}:1\leq-j\leqz\}}\\&\quad=\biggl[\eta\biggl(\max\biggl/{frac{\varpi(\zeta _{j{,\泽塔_{j+1})}\biggr)\biggr]^{\gamma},\end{aligned}$$
具有\(\gamma=\frac{1}{\sqrt{2}}\)此外,对于所有人\(\泽塔^{\ast},\泽塔{\prime}\ in \mho\)具有\(\泽塔^{\ast}\neq\zeta^{\prime}\),我们获得
$$\varpi\bigl(\beth\bigle(\zeta^{\ast},\zeta_{\ast{,\dots,\zeta ^{\est}\bigr$$
和
$$\begin{aligned}\eta\bigl(\varpi\bigle(\beth\bigl-(\zeta^{\ast},\zeta_{\ast{,\dots,\zeta ^{\last}\bigr),\beth\ bigl素数}\vert}{8l}\biggr)\\&=e^{\sqrt{(\frac{\frac{vert\zeta^{ast}-\泽塔^{\prime}\vert}{8l}}{1+\frac{\vert\zeta^{\ast}-\zeta^}\\prime}\\vert}}{8l}})}\\&\leqe^{(\frac}{1}{2\sqrt{2l}}}\vert})}\&\leqe^{\frac{1}{\sqrt{2}}\sqrt{(\frac}\vert\zeta^{\ast}-\zeta^}\prime}{vert}{1+\vert\zeta^{\ast}-\zeta^{\prime}\vert})}}\&=\bigl[\eta\bigl(\varpi\bigl(\zeta^{\ast},\zeta^{\prime}\bigr)\bigr]^{\gamma},\end{aligned}$$
具有\(\gamma=\frac{1}{\sqrt{2}}\)因此,定理的所有假设2.3实现了。此外,对于一些选定的\(\泽塔{1},\点,\泽塔_{l}\ in \ mho\),序列\({\zeta_{l}\}\)定义于(2.3)收敛到\(\zeta^{\ast}=0\),它是的唯一FPℶ.
如果我们把\(\t(s)=e^{\sqrt{s}}\)在定理中2.3,我们得到以下结果。
推论2.6
考虑 \(\贝丝:\ mho^{z}\右箭头\ mho\) 是一个给定的映射,假设有 \(伽马\in(0,1)\) 这样的话
$$\varpi\bigl(\beth(\zeta_{1},\dots,\zeta_{z}),\beth(\zeta_{2},\dots,\zeta_{z+1})\bigr)\leq\gamma^{2}\biggl(\max\biggl\{\frac{\varpi(\zeta_{j},\zeta_{j+1})}{1+\varpi(\zeta_{j},\zeta_{j+1})}:1\leq j\leq z\biggr\ biggr)$$
(2.5)
然后对于任何选定的点 \(\ζ_{1},\ dots,\ζ_{z}\ in \ mho \),顺序 \({\zeta_{l}\}\) 中描述的(1.2)收敛到 \(\ zeta^{\ ast}\ in \ mho\) 和 \(\zeta^{\ast}=\beth(\zeta ^{\asp},\dots,\zeta®{\ast}).此外,如果
$$\varpi\bigl(\beth\bigle(\zeta^{ast},\dots,\zeta_{ast}\bigr),\beth\ bigl$$
为所有人保留 \(\泽塔^{\ast},\泽塔{{\prime}}\在\ mho\中) 具有 \(\泽塔^{\ast}\neq\zeta^{\prime}}\),然后是重点 \(上一页) 是映射的唯一FP ℶ.
推论2.7
假设 \(\贝丝:\ mho^{z}\右箭头\ mho\) 是一个给定的映射,并且有非负常量 \(\gamma_{1}、\gamma_2}、\点、\gama_{z}) 具有 \(伽玛{1}+\gamma{2}+\cdots+\ga玛{z}<1) 这样的话
$$\开始{对齐}\varpi\bigl(\beth(\zeta{1},\dots,\zeta_{z}),\beth \frac{\varpi(\zeta{2},\zeta_{3})}{1+\ varpi\frac{\varpi(\zeta{z},\zeta_{z+1})}{1+\varpi(\zeta{z},\zeta _{z+1}){,\end{aligned}$$
(2.6)
对于每个 \((泽塔{1},点,泽塔{z+1}) 具有 \(\beth(\zeta{1},\dots,\zeta_{z})\neq\beth.然后对于任何选择的点 \(\泽塔{1},\圆点,\泽塔}\ in \ mho\),顺序 \({\zeta_{l}\}\),由提供(1.2)收敛到 \(在\ mho中\ zeta ^{\ ast}\),哪里 \(\泽塔^{\ast}\) 是唯一的FP ℶ.
证明
很明显(2.6)暗示(2.5)带有\(\gamma^{2}=\gamma_{1}+\gamma_2}+\cdots+\gama_{z}\).
现在,假设\(\泽塔^{\ast},\泽塔{{\prime}}\在\ mho\中)具有\(\zeta^{\ast}\neq\zeta\),基于(2.6),可以获得
$$\begon{aligned}&\ varpi\bigl(\ beth\bigl(\ zeta^{\ast},\ zeta^{\ast},\ dots,\ zeta^{\ast}\bigr),\ beth\bigl(\ zeta^{\prime}},\ zeta^{\prime}},\ dots,\ zeta^{\prime}\bigr)\ bigr)\&&\ quad=\ varpi\bigl(\ beth\bigl(\ zeta^{\ast},\ dots,\ zeta^{\ast}\bigr),\ beth\ bigl(\ζ^{\ast},\ dots,\ζ^{\ast},\ζ^{\prime}\biger)\bigr)\\&\qquad{}+\varpi\bigl h\bigl(\zeta^{\ast},\dots,\zeta_{\prime},\ zeta__{\prime}\bigr),\beth\bigl(\ zeta^{\prime},\ dots,\ zeta^{\prime},\ zeta^{\prime}\bigr)\\&&quad\leq(\ gamma{z}+\ gamma{z-1}+\ cdots+\ gamma{z})\ frac{\varpi(\ zeta^{\sast},\ zeta^{\prime})}{1+\ varpi(\ zeta^{\sast},\ zeta^{\prime})}\\&&quad\leq \ gamma ^{2}\ varpi\bigh l(ζ^{\ast},ζ^{\prime}\bigr)。\结束{对齐}$$
因此,推论的条件2.6保持。□
如果我们使用一大类函数∇例如,
$$\eta(s)=2-\frac{2}{\pi}\arctan\biggl(\frac{1}{s^{\theta}}\biggr)$$
哪里\(θ在(0,1)中)和\(s>0\),我们从定理中得到以下定理2.3.
定理2.8
假设 \(\贝丝:\ mho^{z}\右箭头\ mho\) 是给定的映射.如果有映射 \(纳布拉语中的) 和常量 \(伽马,θ,单位(0,1)) 这样的话
$$开始{对齐}&2-\frac{2}{\pi}\arctan\biggl(\frac}1}{[\varpi(\beth(\zeta{1},\dots,\zeta_{z}),\beth压裂{1}{[\max\{\frac{\varpi(\zeta{j},\zeta_{j+1}}:1\leqj\leqz\}]^{theta}}\biggr)\biggr]^{gamma},\end{aligned}$$
对于每个 \((泽塔{1},点,泽塔{z+1}) 具有 \(\beth(\zeta{1},\dots,\zeta_{z})\neq\beth,那么对于任何选定的点 \(\泽塔{1},\圆点,\泽塔}\ in \ mho\),顺序 \({\zeta_{l}\}\),由提供(1.2)收敛到 \(\ zeta^{\ ast}\ in \ mho\).然后 \(\泽塔^{\ast}\) 是唯一的FP ℶ.此外,如果
$$\开始{对齐}和2-\frac{2}{\pi}\arctan\biggl(\frac}{[\varpi(\beth(\zeta^{\ast},\dots,\zeta_{\ast{),\beth biggl(\frac{1}{(\varpi(\zeta^{ast},\zeta_{prime}))^{theta}}\biggr)\biggr]^{gamma},\结束{对齐}$$
持有 \(\泽塔^{\ast},\泽塔{{\prime}}\在\ mho\中) 具有 \(\泽塔^{\ast}\neq\zeta^{\prime}}\),然后是重点 \(\泽塔^{\ast}\) 是映射的唯一FP ℶ.
备注2.9
应注意:
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我们的定理2.3统一并推广了中的定理1.3[10]和定理1.2[9]。
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推论1英寸[15]可以直接从定理中得到2.3推杆\(伽马=1)忽略了收缩条件的分母(2.1).
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如果我们采取\(伽马=1)忽略推论收缩性条件的分母2.6和2.7,我们得到BCP[1]。