description/proof of mean-value theorem for \(C^1\) map from open subset of Euclidean \(C^\infty\) manifold into subset of Euclidean \(C^\infty\) manifold
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\).
- The reader knows a definition of matrix norm induced by vector norms.
- The reader admits the mean-value theorem for any differentiable map from any closed interval into the \(1\)-dimensional Euclidean \(C^\infty\) manifold.
- The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
Target Context
- The reader will have a description and a proof of 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.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}^{d_1}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\) with the Euclidean inner product
\(\mathbb{R}^{d_2}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\) with the Euclidean inner product
\(U\): \(\in \{\text{ the open subsets of } \mathbb{R}^{d_1}\}\)
\(S\): \(\in \{\text{ the subsets of } \mathbb{R}^{d_2}\}\)
\(f\): \(: U \to S\), \(\in \{\text{ the } C^1 \text{ maps }\}\)
\(\{r_1, r_2\}\): \(\subseteq U\), such that \(\overline{r_1 r_2} \subseteq U\), where \(\overline{r_1 r_2}\) is the line segment
\(M\): \(: U \to \{\text{ the } d_2 \times d_1 \mathbb{R} \text{ matrices }\}\), \(= \begin{pmatrix} \partial_1 f^1 & ... & \partial_{d_1} f^1 \\ ... \\ \partial_1 f^{d_2} & ... & \partial_{d_1} f^{d_2} \end{pmatrix}\)
\(v\): \(\in \mathbb{R}^{d_2}\)
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Statements:
\(\exists r \in \text{ the interior of } \overline{r_1 r_2} (\langle v, f (r_2) - f (r_1) \rangle = \langle v, M_r (r_2 - r_1) \rangle)\)
\(\land\)
\(\Vert f (r_2) - f (r_1) \Vert \le \Vert M_r \Vert \Vert r_2 - r_1 \Vert\)
//
2: Note
This proposition requires \(f\) to be \(C^1\) not just differentiable, because it uses the chain rule for derivatives with some multiple intermediate variables.
3: Proof
Whole Strategy: Step 1: take \(f': [0, 1] \to S, t \mapsto \langle v, f (r_1 + t (r_2 - r_1)) \rangle\), and see that \(\partial_1 f' = \langle v, M (r_2 - r_1) \rangle\); Step 2: see that there is an \(t' \in (0, 1)\) such that \({\partial_1 f'}_{t'} = (f' (1) - f' (0)) / 1\); Step 3: take \(r = r_1 + t' (r_2 - r_1)\), and see that \(\langle v, M_r (r_2 - r_1) \rangle = \langle v, f (r_2) - f (r_1) \rangle\); Step 4: take \(v\) as the unit vector in the direction of \(f (r_2) - f (r_1)\).
Step 1:
Let us take \(f': [0, 1] \to S, t \mapsto \langle v, f (r_1 + t (r_2 - r_1)) \rangle\).
\(f'\) is \(C^1\) with\(\partial_1 f' = \partial_1 (\sum_{j \in \{1, ..., d_2\}} v^j f^j (r_1 + t (r_2 - r_1))) = \sum_{j \in \{1, ..., d_2\}} \partial_1 (v^j f^j (r_1 + t (r_2 - r_1))) = \sum_{j \in \{1, ..., d_2\}} v^j \partial_1 f^j (r_1 + t (r_2 - r_1)) = \sum_{j \in \{1, ..., d_2\}} v^j \partial_l f^j (r_1 + t (r_2 - r_1)) \partial_1 (r_1 + t (r_2 - r_1))^l\), by the chain rule, \(= \sum_{j \in \{1, ..., d_2\}} v^j \partial_l f^j (r_1 + t (r_2 - r_1)) (r_2 - r_1)^l = \sum_{j \in \{1, ..., d_2\}} v^j M^j_l (r_2 - r_1)^l = \langle v, M (r_2 - r_1) \rangle\).
Step 2:
There is an \(t' \in (0, 1)\) such that \({\partial_1 f'}_{t'} = (f' (1) - f' (0)) / (1 - 0) = f' (1) - f' (0)\), by the mean-value theorem for any differentiable map from any closed interval into the \(1\)-dimensional Euclidean \(C^\infty\) manifold.
Let \(r = r_1 + t' (r_2 - r_1)\).
\(\langle v, M_r (r_2 - r_1) \rangle = {\partial_1 f'}_{t'} = f' (1) - f' (0) = \langle v, f (r_2) \rangle - \langle v, f (r_1) \rangle = \langle v, f (r_2) - f (r_1) \rangle\).
Step 4:
When \(f (r_2) - f (r_1) = 0\), \(0 = \Vert f (r_2) - f (r_1) \Vert \le \Vert M_r \Vert \Vert r_2 - r_1 \Vert\) holds.
Let \(f (r_2) - f (r_1) \neq 0\) hereafter.
Let \(v\) be the unit vector in the direction of \(f (r_2) - f (r_1)\).
\(\langle v, f (r_2) - f (r_1) \rangle = \Vert f (r_2) - f (r_1) \Vert\).
\(\langle v, M_r (r_2 - r_1) \rangle \le \Vert v \Vert \Vert M_r (r_2 - r_1) \Vert = \Vert M_r (r_2 - r_1) \Vert\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
So, \(\Vert f (r_2) - f (r_1) \Vert \le \Vert M_r (r_2 - r_1) \Vert \le \Vert M_r \Vert \Vert r_2 - r_1 \Vert\).
