ユークリディアンノルム付きユークリディアンベクトルたちスペース（空間）たちマップ（写像）に対するインバース（逆）定理の記述/証明
話題
About: ベクトルたちスペース（空間）
この記事の目次
開始コンテキスト
- 読者は、ユークリディアンノルム付きユークリディアンベクトルたちスペース（空間）の定義を知っている。
- 読者は、マップ（写像）の定義を知っている。
- 読者は、ノルム付きベクトルたちスペース（空間）のオープンサブセット（開部分集合）からノルム付きベクトルたちスペース（空間）のサブセット（部分集合）の中へのマップ（写像）のポイントにおけるデリバティブ（微分係数）の定義を知っている。
- 読者は、コントラクション（収斂）マッピングの法則を認めている。
- 読者は、任意のリニアリーオーダードセット（線形順序集合）および任意のサブセット（部分集合）に対して、当該セット（集合）の任意の要素は当該サブセット（部分集合）のサプリマム（上限）である、もしも、当該要素が当該サブセット（部分集合）の各要素に等しいかそれより大きく、当該要素より小さい当該セット（集合）の各要素に対して、より大きい当該サブセット（部分集合）のある要素がある場合、そしてその場合に限って、という命題を認めている。
- 読者は、任意の\(C^1\)、ユークリディアンノルム付きユークリディアンベクトルたちスペース（空間）たちマップ（写像）のデリバティブ（微分係数）はヤコビアンであるという命題を認めている。
- 読者は、任意のノルム付きベクトルたちスペース（空間）たちマップ（写像）のデリバティブ（微分係数）のレシデュー（残余）は第2引数の任意のポイントにおいてディファレンシャブル（微分可能）である、もしも、元のマップ（写像）が対応するポイントにおいてディファレンシャブル（微分可能）である場合、そして当該デリバティブ（微分係数）は第1引数ポイントにおける元のマップ（写像）デリバティブ（微分係数）のマイナスプラス対応するポイントにおける元のマップ（写像）デリバティブ（微分係数）である、という命題を認めている。
- 読者は、任意のユークリディアン\(C^\infty\)マニフォールド（多様体）の任意のオープンサブセット（開部分集合）から任意のユークリディアン\(C^\infty\)マニフォールド（多様体）の任意のサブセット（部分集合）の中への任意の\(C^1\)マップ（写像）に対する平均値定理を認めている。
ターゲットコンテキスト
- 読者は、ユークリディアンノルム付きユークリディアンベクトルたちスペース（空間）たちマップ（写像）に対するインバース（逆）定理の記述および証明を得る。
オリエンテーション
本サイトにてこれまで議論された定義たちの一覧があります。
本サイトにてこれまで議論された命題たちの一覧があります。
本体
1: 構造化された記述
ここに'構造化された記述'のルールたちがある。
エンティティ（実体）たち:
\(\mathbb{R}^d\): \(= \text{ 当該ユークリディアンノルム付きユークリディアンベクトルたちスペース（空間） }\)
\(v_{1, 1}\): \(\in \mathbb{R}^d\)
\(U_{v_{1, 1}}\): \(\in \{v_{1, 1} \text{ の } \mathbb{R}^d \text{ 上における全てのオープンネイバーフッド（開近傍）たち } \}\)
\(f\): \(: U_{v_{1, 1}} \to \mathbb{R}^d\), \(\in \{ \text{ 全ての } C^k \text{ マップ（写像）たち } \}\)、ここで、\(1 \le k\)
//
ステートメント（言明）たち:
\(Df_{v_{1, 1}} \in \{\text{ 全てのインバーティブル（可逆）マップ（写像）たち }\}\)
\(\implies\)
\(\exists U_{f (v_{1, 1})} \in \{f (v_{1, 1}) \text{ の } \mathbb{R}^d \text{ 上における全てのオープンネイバーフッド（開近傍）たち } \} \text{ で、以下を満たすもの、つまり、 } U_{f (v_{1, 1})} \subseteq f (U_{v_{1, 1}}), \exists g: U_{f (v_{1, 1})} \to U_{v_{1, 1}} (\forall f (v_1) \in U_{f (v_{1, 1})} (g (f (v_1)) = v_1) \land g \in \{\text{ 全ての } C^k \text{ マップ（写像）たち }\})\)
//
2: Proof
Whole Strategy: Step 1: let \(f (v_{1, 1} + v_{1, 2}) = f (v_{1, 1}) + Df_{v_{1, 1}} v_{1, 2} + r (v_{1, 1}, v_{1, 2}) := v_{2, 1} + v_{2, 2}\) and \(v_{1, 2} = Df_{v_{1, 1}}^{-1} (v_{2, 2} - r (v_{1, 1}, v_{1, 2})) := T_{v_{2, 2}} (v_{1, 2})\); Step 2: take any small enough \(v_{2, 2}\) and define \(M_{v_{2, 2}} := Sup \{\Vert Df_{v_{1, 1}}^{-1} r (v_{1, 1}, v_1) \Vert \vert v_1 \in \mathbb{R}^d \land \Vert v_1 \Vert \le 2 \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert\}\) and \(S_{v_{2, 2}} := \{v_1 \in \mathbb{R}^d \vert \Vert v_1 - Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert \le M_{v_{2, 2}}\}\); Step 3: see that \(T_{v_{2, 2}}\) with domain \(S_{v_{2, 2}}\) is into \(S_{v_{2, 2}}\) and with \(v_{2, 2}\) further small enough such that \(v_{2, 1} + v_{2, 2} \in B_{f (v_{1, 1}), \epsilon'}\), \(T_{v_{2, 2}}\) is a contraction; Step 4: apply the contraction mapping principle to get \(v_{1, 2} \in S_{v_{2, 2}}\) such that \(T_{v_{2, 2}} (v_{1, 2}) = v_{1, 2}\) and let \(v_{1, 2} = h (v_{2, 2})\) and \(g (v_{2, 1} + v_{2, 2}) = v_{1, 1} + h (v_{2, 2})\); Step 5: see that \(g\) has continuous \(Dg_{v_{2, 1}} = Df_{v_{1, 1}}^{- 1}\); Step 6: see that \(g\) is \(C^k\).
Step 1:
\(f (v_{1, 1} + v_{1, 2}) = f (v_{1, 1}) + Df_{v_{1, 1}} v_{1, 2} + r (v_{1, 1}, v_{1, 2}) = v_{2, 1} + v_{2, 2}\), where \(v_{2, 1} := f (v_{1, 1})\) and \(v_{2, 2} := Df_{v_{1, 1}} v_{1, 2} + r (v_{1, 1}, v_{1, 2})\).
That has to be solved for \(v_{1, 2}\) by \(v_{2, 2}\) as \(v_{1, 2} = h (v_{2, 2})\).
\(v_{1, 2} = Df_{v_{1, 1}}^{-1} (v_{2, 2} - r (v_{1, 1}, v_{1, 2})) := T_{v_{2, 2}} (v_{1, 2})\).
Step 2:
Let \(\Vert Df_{v_{1, 1}}^{- 1} \Vert\) be the matrix norm induced by vector norms, where \(0 \lt \Vert Df_{v_{1, 1}}^{- 1} \Vert\), because \(Df_{v_{1, 1}}^{- 1} \neq 0\), because the matrix norm is a norm.
Let us take \(\epsilon = 1 / (2 \Vert Df_{v_{1, 1}}^{- 1} \Vert)\).
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and for each \(v_1 \in \mathbb{R}^d\) such that \(v_{1, 1} + v_1 \in U_{v_{1, 1}}\) and \(\Vert v_1 \Vert \lt \delta\), \(\Vert r (v_{1, 1}, v_1) \Vert / \Vert v_1 \Vert \lt \epsilon\), by the definition of derivative of map from open subset of normed vectors space into subset of normed vectors space at point.
Let us take any \(v_{2, 2} \in \mathbb{R}^d\) such that \(B_{v_{1, 1}, 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert} \subseteq U_{v_{1, 1}}\) and \(2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \lt \delta\), which exists, because there is a \(\delta' \in \mathbb{R}\) such that \(0 \lt \delta'\) and \(B_{v_{1, 1}, \delta'} \subseteq U_{v_{1, 1}}\), because \(U_{v_{1, 1}}\) is an open neighborhood of \(v_{1, 1}\), and \(2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} \Vert \Vert v_{2, 2} \Vert\), so, any \(\Vert v_{2, 2} \Vert \lt Min (\delta', \delta) / (2 \Vert Df_{v_{1, 1}}^{- 1} \Vert)\) satisfies \(2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} \Vert \Vert v_{2, 2} \Vert \lt 2 \Vert Df_{v_{1, 1}}^{- 1} \Vert Min (\delta', \delta) / (2 \Vert Df_{v_{1, 1}}^{- 1} \Vert) = Min (\delta', \delta)\), so, \(B_{v_{1, 1}, 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert} \subseteq B_{v_{1, 1}, \delta'} \subseteq U_{v_{1, 1}}\) and \(2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \lt Min (\delta', \delta) \le \delta\).
Let us define \(M_{v_{2, 2}} := Sup \{\Vert Df_{v_{1, 1}}^{-1} r (v_{1, 1}, v_1) \Vert \vert v_1 \in \mathbb{R}^d \land \Vert v_1 \Vert \le 2 \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert\}\): \(\Vert v_1 \Vert \le 2 \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert\) implies that \(v_{1, 1} + v_1 \in U_{v_{1, 1}}\), because \(B_{v_{1, 1}, 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert} \subseteq U_{v_{1, 1}}\), so, \(r (v_{1, 1}, v_1)\) is valid.
Then, for each \(v_1 \in \mathbb{R}^d\) such that \(\Vert v_1 \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert\), \(\Vert Df_{v_{1, 1}}^{- 1} r (v_{1, 1}, v_1) \Vert \le \Vert Df_{v_{1, 1}}^{- 1} \Vert \Vert r (v_{1, 1}, v_1) \Vert \lt \Vert Df_{v_{1, 1}}^{- 1} \Vert \epsilon \Vert v_1 \Vert \le \Vert Df_{v_{1, 1}}^{- 1} \Vert \epsilon 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert = \Vert Df_{v_{1, 1}}^{- 1} \Vert 1 / (2 \Vert Df_{v_{1, 1}}^{- 1} \Vert) 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert = \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert\), so, \(M_{v_{2, 2}} \le \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert\).
For each \(v'_1 \in \mathbb{R}^d\) such that \(\Vert v'_1 - Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le M_{v_{2, 2}}\), \(\Vert v'_1 \Vert = \Vert v'_1 - Df_{v_{1, 1}}^{- 1} v_{2, 2} + Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le \Vert v'_1 - Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert + \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le M_{v_{2, 2}} + \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert\), which means that \(v_{1, 1} + v'_1 \in B_{v_{1, 1}, 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert} \subseteq U_{v_{1, 1}}\).
Let us define the set, \(S_{v_{2, 2}} := \{v_1 \in \mathbb{R}^d \vert \Vert v_1 - Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert \le M_{v_{2, 2}}\}\).
Step 3:
The map, \(T_{v_{2, 2}}\), with the domain as \(S_{v_{2, 2}}\), is in fact, \(T_{v_{2, 2}}: S_{v_{2, 2}} \to S_{v_{2, 2}}\), because \(\Vert v_1 \Vert = \Vert v_1 - Df_{v_{1, 1}}^{-1} v_{2, 2} + Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert \le \Vert v_1 - Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert + \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert \le M_{v_{2, 2}} + \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert \le \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert + \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert = 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert\), so, \(v_1\) is in the subset that defines \(M_{v_{2, 2}}\), so, \(\Vert T_{v_{2, 2}} (v_1) - Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert = \Vert Df_{v_{1, 1}}^{-1} r (v_{1, 1}, v_1) \Vert \le M_{v_{2, 2}}\), by the definition of \(M_{v_{2, 2}}\).
For each \(v_{1, 2, 1}, v_{1, 2, 2} \in S_{v_{2, 2}}\), \(\Vert T_{v_{2, 2}} (v_{1, 2, 1}) - T_{v_{2, 2}} (v_{1, 2, 2}) \Vert = \Vert Df_{v_{1, 1}}^{-1} (v_{2, 2} - r (v_{1, 1}, v_{1, 2, 1})) - Df_{v_{1, 1}}^{-1} (v_{2, 2} - r (v_{1, 1}, v_{1, 2, 2})) \Vert = \Vert Df_{v_{1, 1}}^{-1} (r (v_{1, 1}, v_{1, 2, 2}) - r (v_{1, 1}, v_{1, 2, 1})) \Vert \le \Vert Df_{v_{1, 1}}^{-1} \Vert \Vert r (v_{1, 1}, v_{1, 2, 2}) - r (v_{1, 1}, v_{1, 2, 1}) \Vert\).
\(r\) is differentiable with respect to the 2nd argument and the derivative is continuous, by the proposition that the residue of the derivative of any normed vectors spaces map is differentiable at any point of the 2nd argument if the original map is differentiable at the corresponding point with derivative as minus the derivative of the original map at the 1st argument point plus the derivative of the original map at the corresponding point, so, by the mean-value theorem for any \(C^1\) map from any open subset of any Euclidean \(C^\infty\) manifold into any subset of any Euclidean \(C^\infty\) manifold, \(\Vert r (v_{1, 1}, v_{1, 2, 2}) - r (v_{1, 1}, v_{1, 2, 1}) \Vert \le \Vert Dr_{v_{1, 1}, v_{1, 2, 3}} \Vert \Vert v_{1, 2, 2} - v_{1, 2, 1} \Vert\) where \(v_{1, 2, 3}\) is in the interior of \(\overline{v_{1, 2, 1} v_{1, 2, 2}}\), so, \(\Vert T_{v_{2, 2}} (v_{1, 2, 1}) - T_{v_{2, 2}} (v_{1, 2, 2}) \Vert \le \Vert Df_{v_{1, 1}}^{-1} \Vert \Vert Dr_{v_{1, 1}, v_{1, 2, 3}} \Vert \Vert v_{1, 2, 2} - v_{1, 2, 1} \Vert\).
While \(v_{1, 2, 3} = v_{1, 2, 1} + s (v_{1, 2, 2} - v_{1, 2, 1}) = (1 - s) v_{1, 2, 1} + s v_{1, 2, 2}\) where \(0 \lt s \lt 1\), \(\Vert v_{1, 2, 3} \Vert = \Vert (1 - s) v_{1, 2, 1} + s v_{1, 2, 2} \Vert \le \Vert (1 - s) v_{1, 2, 1} \Vert + \Vert s v_{1, 2, 2} \Vert = (1 - s) \Vert v_{1, 2, 1} \Vert + s \Vert v_{1, 2, 2} \Vert \le (1 - s) 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert + s 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert = 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} \Vert \Vert v_{2, 2} \Vert\), because \(\Vert v_{1, 2, 1} \Vert, \Vert v_{1, 2, 2} \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert\), as has been seen above.
So, if \(\Vert v_{2, 2} \Vert\) is small enough, \(\Vert v_{1, 2, 3} \Vert\) is small enough, and as \(Dr_{v_{1, 1}, 0} = - Df_{v_{1, 1}} + Df_{v_{1, 1} + 0} = 0\) and \(Dr_{v_{1, 1}, \bullet}\) is continuous, \(\Vert Df_{v_{1, 1}}^{-1} \Vert \Vert Dr_{v_{1, 1}, v_{1, 2, 3}} \Vert \lt 1\).
Let \(B_{f (v_{1, 1}), \epsilon'}\) be any area on which \(\Vert v_{2, 2} \Vert\) is such small enough and let \(U_{f (v_{1, 1})} := B_{f (v_{1, 1}), \epsilon'}\).
So, \(T_{v_{2, 2}}\) with the domain is a contraction with norm of subtraction as distance.
Step 4:
By the contraction mapping principle, there is the unique fixed point, \(v_{1, 2} \in S_{v_{2, 2}}\), such that \(T_{v_{2, 2}} (v_{1, 2}) = v_{1, 2}\), which is the desired value.
\(U_{f (v_{1, 1})} \subseteq f (U_{v_{1, 1}})\), because for each \(v_{2, 1} + v_{2, 2} \in U_{f (v_{1, 1})}\), \(T_{v_{2, 2}} (v_{1, 2}) = v_{1, 2}\) holds, so, \(v_{1, 2} = Df_{v_{1, 1}}^{-1} (v_{2, 2} - r (v_{1, 1}, v_{1, 2}))\), so, \(Df_{v_{1, 1}} v_{1, 2} = v_{2, 2} - r (v_{1, 1}, v_{1, 2})\), so, \(f (v_{1, 1} + v_{1, 2}) = f (v_{1, 1}) + Df_{v_{1, 1}} v_{1, 2} + r (v_{1, 1}, v_{1, 2}) = v_{2, 1} + v_{2, 2}\).
\(lim_{\Vert v_{2, 2} \Vert \to 0} M_{v_{2, 2}} / \Vert v_{2, 2} \Vert = 0\), because for any \(0 \lt \epsilon''\), there is a \(v_1 \in \{v_1 \in \mathbb{R}^d \vert \Vert v_1 \Vert \le 2 \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert\}\) such that \(M_{v_{2, 2}} - \epsilon'' \lt \Vert Df_{v_{1, 1}}^{-1} r (v_{1, 1}, v_1) \Vert \le \Vert Df_{v_{1, 1}}^{-1} \Vert \Vert r (v_{1, 1}, v_1) \Vert\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger, and \(M_{v_{2, 2}} / \Vert v_{2, 2} \Vert \lt (\Vert Df_{v_{1, 1}}^{-1} \Vert \Vert r (v_{1, 1}, v_1) \Vert + \epsilon'') / \Vert v_{2, 2} \Vert = \Vert Df_{v_{1, 1}}^{-1} \Vert \Vert r (v_{1, 1}, v_1) \Vert / \Vert v_{2, 2} \Vert + \epsilon'' / \Vert v_{2, 2} \Vert\), then, take \(\epsilon'' = \Vert v_{2, 2} \Vert^2\), so, \(= \Vert Df_{v_{1, 1}}^{-1} \Vert \Vert r (v_{1, 1}, v_1) \Vert / \Vert v_{2, 2} \Vert + \Vert v_{2, 2} \Vert\), but as \(\Vert v_1 \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} v_{2, 2} \Vert \le 2 \Vert Df_{v_{1, 1}}^{- 1} \Vert \Vert v_{2, 2} \Vert\), \(\Vert v_1 \Vert / (2 \Vert Df_{v_{1, 1}}^{- 1} \Vert) \le \Vert v_{2, 2} \Vert \), \(M_{v_{2, 2}} / \Vert v_{2, 2} \Vert \lt \Vert Df_{v_{1, 1}}^{-1} \Vert \Vert r (v_{1, 1}, v_1) \Vert / \Vert v_1 \Vert 2 \Vert Df_{v_{1, 1}}^{- 1} \Vert + \Vert v_{2, 2} \Vert\), and as \(\Vert v_{2, 2} \Vert \to 0\), \(\Vert v_1 \Vert \to 0\), so, \(lim_{\Vert v_{2, 2} \Vert \to 0} M_{v_{2, 2}} / \Vert v_{2, 2} \Vert \lt lim_{\Vert v_1 \Vert \to 0} \Vert Df_{v_{1, 1}}^{-1} \Vert \Vert r (v_{1, 1}, v_1) \Vert / \Vert v_1 \Vert 2 \Vert Df_{v_{1, 1}}^{- 1} \Vert + lim_{\Vert v_{2, 2} \Vert \to 0} \Vert v_{2, 2} \Vert = 0 + 0 = 0\).
That means that \(v_{1, 2} = h (v_{2, 2})\) is differentiable at \(0\) with the derivative as \(Df_{v_{1, 1}}^{- 1}\), because \(h (0 + v_{2, 2}) = v_{1, 2} = Df_{v_{1, 1}}^{- 1} v_{2, 2} - Df_{v_{1, 1}}^{- 1} r (v_{1, 1}, v_{1, 2} (v_{2, 2}))\), and \(- Df_{v_{1, 1}}^{- 1} r (v_{1, 1}, v_{1, 2} (v_{2, 2}))\) is the residue, and as \(v_{1, 2} \in S_{v_{2, 2}}\), \(v_{1, 2} \in \{v_1 \in \mathbb{R}^d \vert \Vert v_1 \Vert \le 2 \Vert Df_{v_{1, 1}}^{-1} v_{2, 2} \Vert\}\), and \(\Vert - Df_{v_{1, 1}}^{- 1} r (v_{1, 1}, v_{1, 2} (v_{2, 2})) \Vert \le M_{v_{2, 2}}\).
Then, \(g\) can be defined such that \(g (v_{2, 1} + v_{2, 2}) = v_{1, 1} + h (v_{2, 2}) = v_{1, 1} + Df_{v_{1, 1}}^{- 1} v_{2, 2} - Df_{v_{1, 1}}^{- 1} r (v_{1, 1}, v_{1, 2} (v_{2, 2}))\) where the derivative at \(v_{2, 1}\) is \(Df_{v_{1, 1}}^{- 1}\).
Step 5:
\(Dg_{v_{2, 1}} = Df_{v_{1, 1}}^{- 1} = Df_{g (v_{2, 1})}^{- 1}\), is continuous with respect to \(v_{2, 1}\), because \(g\) is continuous being differentiable, \(Df_{v_{1, 1}}\) is continuous with respect to \(v_{1, 1}\), having the reverse matrix preserves the continuousness, and the composition of such continuous maps is continuous.
Now that the theorem has been proved to be true for \(C^1\).
Step 6:
Let us suppose that the theorem holds for \(C^{k - 1}\) and f is \(C^k\).
While \(Dg = \begin{pmatrix} \partial f^j / \partial {v_1}^l \end{pmatrix}^{- 1}\), the right hand side is \(C^{k - 1}\) with respect to \(v_1\), because \(\begin{pmatrix} \partial f^j / \partial {v_1}^l \end{pmatrix}\) is \(C^{k - 1}\) as \(f\) is \(C^k\) and taking the reverse matrix preserves the \(C^{k - 1}\)-ness, and as \(v_1 = g (v_2)\) is \(C^{k - 1}\) by the theorem for \(C^{k - 1}\), as the composition map, the right hand side is \(C^{k - 1}\) with respect to \(v_2\), so, also the left hand side is so, which means that \(g\) is \(C^k\).
