陰関数定理をBanachの不動点定理を用いて証明しよう（３）

$f$の連続微分可能性を仮定していたので，任意の$x_1\in B[x_0,{\tau }_1]$に対して，ある${\delta }_{x_1}>0$が存在して，任意の$x\in B[x_1,{\delta }_{x_1}]\cap B[x_0,{\tau }_1]$に対して，$f_y(x,\hat{\phi }(x))$も可逆になる．
(実際，$f$の連続微分可能性を用いて，$\Vert f_y(x,\hat{\phi }(x))-f_y(x_1,\hat{\phi }(x_1))\Vert \le (2\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert {)}^{-1}$となるように${\delta }_{x_1}>0$を決めてやればよい．）

$f$は微分可能だから，ある$\delta >0$が存在して，任意の$x\in B[x_1,\delta ]$に対して
$\frac{\Vert ϵ(x,\hat{\phi }(x))\Vert }{\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert }\le \frac{1}{2\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert }$つまり$\Vert ϵ(x,\hat{\phi }(x))\Vert \le \frac{\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert }{2\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert }$
が成立するようにできる．

また$x\in B[x_1,{\delta }_{x_1}^{\prime }]$となるどんな$x\in B[x_0,{\tau }_1]$に対しても，
$\frac{\Vert ϵ(x,\hat{\phi }(x))\Vert }{\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert }\le \frac{{ϵ}_{x_1}}{2\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert (\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert f_x(x_1,\hat{\phi }(x_1))\Vert +1)}$が成り立つように${\delta }_{x_1}^{\prime }>0$をとってくる．
$$\begin{array}{rl}f(x,\hat{\phi }(x))& =0\\ & =f(x_1,\hat{\phi }(x_1))+f_x(x_1,\hat{\phi }(x_1))(x-x_1)+f_y(x_1,\hat{\phi }(x_1))(\hat{\phi }(x)-\hat{\phi }(x_1))+ϵ(x,\hat{\phi }(x))\\ & =f_x(x_1,\hat{\phi }(x_1))(x-x_1)+f_y(x_1,\hat{\phi }(x_1))(\hat{\phi }(x)-\hat{\phi }(x_1))+ϵ(x,\hat{\phi }(x))\end{array}$$
$f_y(x_1,\hat{\phi }(x_1){)}^{-1}{f}_x(x_1,\hat{\phi }(x_1))(x-x_1)+(\hat{\phi }(x)-\hat{\phi }(x_1))+f_y(x_1,\hat{\phi }(x_1){)}^{-1}ϵ(x,\hat{\phi }(x))=0$より，${\delta }^{\prime }:=min\{\delta ,{\delta }_{x_1},{\delta }_{x_1}^{\prime }\}$として，任意の$x\in B[x_1,{\delta }^{\prime }]$に対して
$$\begin{array}{rl}-(\hat{\phi }(x)-\hat{\phi }(x_1))& =f_y(x_1,\hat{\phi }(x_1){)}^{-1}{f}_x(x_1,\hat{\phi }(x_1))(x-x_1)+f_y(x_1,\hat{\phi }(x_1){)}^{-1}ϵ(x,\hat{\phi }(x))\\ \Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert & =\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}{f}_x(x_1,\hat{\phi }(x_1))(x-x_1)+f_y(x_1,\hat{\phi }(x_1){)}^{-1}ϵ(x,\hat{\phi }(x))\Vert \\ & \le \Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert f_x(x_1,\hat{\phi }(x_1))\Vert \Vert x-x_1\Vert +\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert ϵ(x,\hat{\phi }(x))\Vert \\ & \le \Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert f_x(x_1,\hat{\phi }(x_1))\Vert \Vert x-x_1\Vert +\frac{1}{2}(\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert )\end{array}$$

$$\begin{array}{rl}\frac{\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert }{2}& \le \Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert f_x(x_1,\hat{\phi }(x_1))\Vert \Vert x-x_1\Vert +\frac{\Vert x-x_1\Vert }{2}\\ \Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert & \le (2\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert f_x(x_1,\hat{\phi }(x_1))\Vert +1)\cdot \Vert x-x_1\Vert \end{array}$$
後で使うので
$\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert \le 2(\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert f_x(x_1,\hat{\phi }(x_1))\Vert +1)\cdot \Vert x-x_1\Vert$
も記述しておく．

$f_y(x_1,\hat{\phi }(x_1){)}^{-1}{f}_x(x_1,\hat{\phi }(x_1))(x-x_1)+(\hat{\phi }(x)-\hat{\phi }(x_1))=-f_y(x_1,\hat{\phi }(x_1){)}^{-1}ϵ(x,\hat{\phi }(x))$を用いて，

$$\begin{array}{rl}& \frac{\Vert \hat{\phi }(x)-\hat{\phi }(x_1)+f_y(x_1,\hat{\phi }(x_1){)}^{-1}{f}_x(x_1,\hat{\phi }(x_1))(x-x_1)\Vert }{\Vert x-x_1\Vert }\\ & \le \frac{\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert ϵ(x,\hat{\phi }(x))\Vert }{\Vert x-x_1\Vert }\\ & =\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \frac{\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert }{\Vert x-x_1\Vert }\frac{\Vert ϵ(x,\hat{\phi }(x))\Vert }{\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert }\\ & \le \Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \cdot 2(\Vert f_y(x_1,\hat{\phi }(x_1){)}^{-1}\Vert \Vert f_x(x_1,\hat{\phi }(x_1))\Vert +1)\frac{\Vert ϵ(x,\hat{\phi }(x))\Vert }{\Vert x-x_1\Vert +\Vert \hat{\phi }(x)-\hat{\phi }(x_1)\Vert }.\end{array}$$

$x\in B[x_1,{\delta }^{\prime }]\subset B[x_1,{\delta }_{x_1}^{\prime }]$なので，${\delta }_{x_1}^{\prime }>0$の定め方より，
$$\frac{\Vert \hat{\phi }(x)-\hat{\phi }(x_1)+f_y(x_1,\hat{\phi }(x_1){)}^{-1}{f}_x(x_1,\hat{\phi }(x_1))(x-x_1)\Vert }{\Vert x-x_1\Vert }\le {ϵ}_{x_1}$$
が求まる．${ϵ}_{x_1}$は任意の正数であったから，これは$x_1$で$\hat{\phi }$が微分可能で$\mathrm{\nabla }\hat{\phi }(x_1)=f_y(x_1,\hat{\phi }(x_1){)}^{-1}{f}_x(x_1,\hat{\phi }(x_1))$であるということを示している．$◻$