description/proof of that for differentiable map from open interval into Euclidean \(C^\infty\) manifold with Euclidean norm, norm of point image is upper-bounded by differentiable map between non-negative interval if there is Lipschitz map between non-negative interval that satisfies certain conditions
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of map from open subset of Euclidean \(C^\infty\) manifold into subset of Euclidean \(C^\infty\) manifold differentiable at point.
- The reader knows a definition of Lipschitz map between metric spaces.
- The reader knows a definition of metric subspace.
- The reader admits the proposition that for any finite-dimensional normed real vectors space with the canonical topology, the norm map is continuous.
- The reader admits the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
- The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
- The reader admits the mean-value theorem for any differentiable map from any closed interval into the \(1\)-dimensional Euclidean \(C^\infty\) manifold.
Target Context
- The reader will have a description and a proof of the proposition that for any differentiable map from any open interval into any Euclidean \(C^\infty\) manifold with the Euclidean norm, the norm of each point image is upper-bounded by a differentiable map between the non-negative interval with an initial condition if there is a Lipschitz map between the non-negative interval that satisfies certain conditions.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\) with the Euclidean metric with the topology induced by the metric
\(\mathbb{R}^d\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\) with the Euclidean norm with the topology induced by the metric induced by the norm
\(J\): \(= (r_1, r_2) \subseteq \mathbb{R}\) with the subspace topology
\(r_0\): \(\in J\)
\(J'\): \(= [0, Max (\{\vert r_1 - r_0 \vert, \vert r_2 - r_0 \vert\})) \subseteq \mathbb{R}\)
\([0, \infty)\): \(\subseteq \mathbb{R}\), with the subspace metric
\(f\): \(: J \to \mathbb{R}^d\), \(\in \{\text{ the differentiable maps }\}\)
\(g\): \(: J' \to [0, \infty)\), \(\in \{\text{ the differentiable maps }\}\) such that \(g (0) = \Vert f (r_0) \Vert\)
\(J''\): \(= [0, b) \subseteq \mathbb{R}\) with the subspace metric such that \(\forall r \in J (\Vert f (r) \Vert \in J'')\) and \(\forall r \in J' (g (r) \in J'')\), where \(b \in (0, \infty]\)
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Statements:
\(\exists h: J'' \to [0, \infty) \in \{\text{ the Lipschitz maps with } L\} (\forall r \in J (\Vert \partial_1 f (r) \Vert \le h (\Vert f (r) \Vert)) \land \forall r \in J' (\partial_1 g (r) = h (g (r))))\)
\(\implies\)
\(\forall r \in J (\Vert f (r) \Vert \le g (\vert r - r_0 \vert))\)
//
2: Proof
Whole Strategy: Step 1: take the open \(J^+ := \{r \in J \vert 0 \lt \Vert f (r) \Vert\}\); Step 2: see that for each \(r \in J^+\), \(- h (\Vert f (r) \Vert) \le \partial_1 \Vert f (r) \Vert \le h (\Vert f (r) \Vert)\); Step 3: see that for each \(r \in (r_1, r_2) \setminus J^+\), \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\); Step 4: see that for each \(r \in J^+ \cap [r_0, r_2)\), \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\); Step 5: see that for each \(r \in J^+ \cap (r_1, r_0]\), \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\); Step 6: conclude the proposition.
Step 1:
Let \(J^+ := \{r \in J \vert 0 \lt \Vert f (r) \Vert\}\).
Let us see that \(J^+ \subseteq \mathbb{R}\) is open.
\(\Vert f (r) \Vert: J \to \mathbb{R}\) is continuous, because \(f\) is continuous as being differentiable and the norm map is continuous, by the proposition that for any finite-dimensional normed real vectors space with the canonical topology, the norm map is continuous.
So, \(\Vert f (r) \Vert^{-1} ((0, \infty)) \subseteq J\) is open, and is open on \(\mathbb{R}\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Step 2:
Let \(r \in J^+\) be any.
\(\partial_1 \Vert f (r) \Vert = \partial_1 (\sum_{j \in \{1, ..., d\}} (f^j (r))^2)^{1 / 2} = 1 / 2 (\sum_{j \in \{1, ..., d\}} (f^j (r))^2)^{- 1 / 2} (\sum_{j \in \{1, ..., d\}} 2 f^j (r) \partial_1 f^j (r)) = 1 / (2 \Vert f (r) \Vert) 2 \langle f (r), \partial_1 f (r) \rangle = 1 / \Vert f (r) \Vert \langle f (r), \partial_1 f (r) \rangle\).
\(\vert \langle f (r), \partial_1 f (r) \rangle \vert \le \Vert f (r) \Vert \Vert \partial_1 f (r) \Vert\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, so, \(- \Vert f (r) \Vert \Vert \partial_1 f (r) \Vert \le \langle f (r), \partial_1 f (r) \rangle \le \Vert f (r) \Vert \Vert \partial_1 f (r) \Vert\).
So, \(- \Vert \partial_1 f (r) \Vert = - 1 / \Vert f (r) \Vert \Vert f (r) \Vert \Vert \partial_1 f (r) \Vert \le 1 / \Vert f (r) \Vert \langle f (r), \partial_1 f (r) \rangle = \partial_1 \Vert f (r) \Vert = 1 / \Vert f (r) \Vert \langle f (r), \partial_1 f (r) \rangle \le 1 / \Vert f (r) \Vert \Vert f (r) \Vert \Vert \partial_1 f (r) \Vert = \Vert \partial_1 f (r) \Vert\).
So, \(- h (\Vert f (r) \Vert) \le \partial_1 \Vert f (r) \Vert \le h (\Vert f (r) \Vert)\).
Step 3:
For each \(r \in (r_1, r_2) \setminus J^+\), \(\Vert f (r) \Vert = 0\), so, \(\Vert f (r) \Vert = 0 \le g (\vert r - r_0 \vert)\).
Step 4:
Let us see that for each \(r \in J^+ \cap [r_0, r_2)\), \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\).
Let us think of \(\phi: [r_0, r_2) \to \mathbb{R}, r \mapsto e^{- L (r - r_0)} (\Vert f (r) \Vert - g (r - r_0))\).
\(\phi (r_0) = e^{- L (r_0 - r_0)} (\Vert f (r_0) \Vert - g (r_0 - r_0)) = 1 (\Vert f (r_0) \Vert - g (0)) = 0\).
\(\partial_1 \phi (r) = - L e^{- L (r - r_0)} (\Vert f (r) \Vert - g (r - r_0)) + e^{- L (r - r_0)} (\partial_1 \Vert f (r) \Vert - \partial_1 g (r - r_0)) \le - L e^{- L (r - r_0)} (\Vert f (r) \Vert - g (r - r_0)) + e^{- L (r - r_0)} (h (\Vert f (r) \Vert) - h (g (r - r_0))) = e^{- L (r - r_0)} (- L (\Vert f (r) \Vert - g (r - r_0)) + (h (\Vert f (r) \Vert) - h (g (r - r_0))))\).
Let us suppose that there was an \(s \in J^+ \cap [r_0, r_2)\) such that \(0 \lt \phi (s)\).
\(- L (\Vert f (s) \Vert - g (s - r_0)) + (h (\Vert f (s) \Vert) - h (g (s - r_0))) \le 0\), because \(h\) was a Lipschitz map with \(L\): \(h (\Vert f (s) \Vert) - h (g (s - r_0)) \le \vert h (\Vert f (s) \Vert) - h (g (s - r_0)) \vert \le L \vert \Vert f (s) \Vert - g (s - r_0) \vert = L (\Vert f (s) \Vert - g (s - r_0))\).
So, \(\partial_1 \phi (s) \le 0\).
\(\{r \in [r_0, s] \vert \phi (r) \le 0\} \neq \emptyset\), because \(r_0\) belonged there, and let us take \(t := Sup (\{r \in [r_0, s] \vert \phi (r) \le 0\})\).
Let us see that \(\phi (t) = 0\).
When \(t = r_0\), \(\phi (t) = \phi (r_0) = 0\).
Let us suppose otherwise.
\(t \neq s\), because as \(\phi\) was continuous, there would be a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(\phi ((s - \delta, s]) \subseteq (0, \infty)\), so, \(s - \delta / 2 \in Ub (\{r \in [r_0, s] \vert \phi (r) \le 0\})\), so, \(s\) would not be the minimum of \(Ub (\{r \in [r_0, s] \vert \phi (r) \le 0\})\).
\(\phi ((t, s]) \subseteq (0, \infty)\), because otherwise, \(t\) would not be in \(Ub (\{r \in [r_0, s] \vert \phi (r) \le 0\})\).
If \(0 \lt \phi (t)\), there would be a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(\phi ((t - \delta, t + \delta)) \subseteq (0, \infty)\), but \(t\) would not be the minimum of \(Ub (\{r \in [r_0, s] \vert \phi (r) \le 0\})\), because \(t - \delta / 2 \in Ub (\{r \in [r_0, s] \vert \phi (r) \le 0\})\), a contradiction, so, \(\phi (t) \le 0\).
If \(\phi (t) \lt 0\), there would be a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(\phi ((t - \delta, t + \delta)) \subseteq (- \infty, 0)\), but \(t\) would not be in \(Ub (\{r \in [r_0, s] \vert \phi (r) \le 0\})\), because \(\phi (t + \delta / 2) \le 0\), a contradiction, so, \(0 \le \phi (t)\).
So, \(\phi (t) = 0\).
There would be an \(r \in (t, s)\) such that \(\partial_1 \phi (r) = (\phi (s) - \phi (t)) / (s - t)\), which would be positive, by the mean-value theorem for any differentiable map from any closed interval into the \(1\)-dimensional Euclidean \(C^\infty\) manifold, but \(0 \lt \phi (r)\), which would imply that \(0 \lt \Vert f (r) \Vert\), which would imply that \(r \in J^+\), a contradiction against \(\partial_1 \phi (r) \le 0\) whenever \(r \in J^+ \cap [r_0, r_2)\) and \(0 \lt \phi (r)\).
So, there is no such \(s \in J^+ \cap [r_0, r_2)\).
So, \(\phi (r) \le 0\) for each \(r \in J^+ \cap [r_0, r_2)\).
That implies that \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\) for each \(r \in J^+ \cap [r_0, r_2)\).
Step 5:
Let us see that for each \(r \in J^+ \cap (r_1, r_0]\), \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\).
Let us think of \(\phi: (r_1, r_0] \to \mathbb{R}, r \mapsto e^{- L (r_0 - r)} (\Vert f (r) \Vert - g (r_0 - r))\).
\(\phi (r_0) = e^{- L (r_0 - r_0)} (\Vert f (r_0) \Vert - g (r_0 - r_0)) = 1 (\Vert f (r_0) \Vert - g (0)) = 0\).
\(\partial_1 \phi (r) = L e^{- L (r_0 - r)} (\Vert f (r) \Vert - g (r_0 - r)) + e^{- L (r_0 - r)} (\partial_1 \Vert f (r) \Vert + \partial_1 g (r_0 - r)) = e^{- L (r_0 - r)} (L (\Vert f (r) \Vert - g (r_0 - r)) + \partial_1 \Vert f (r) \Vert + \partial_1 g (r_0 - r))\).
\(e^{- L (r_0 - r)} (L (\Vert f (r) \Vert - g (r_0 - r)) - h (\Vert f (r) \Vert) + h (g (r_0 - r))) \le e^{- L (r_0 - r)} (L (\Vert f (r) \Vert - g (r_0 - r)) + \partial_1 \Vert f (r) \Vert + \partial_1 g (r_0 - r))) = \partial_1 \phi (r)\).
Let us suppose that there was an \(s \in J^+ \cap (r_1, r_0]\) such that \(0 \lt \phi (s)\).
\(0 \le L (\Vert f (s) \Vert - g (r_0 - s)) - h (\Vert f (s) \Vert) + h (g (r_0 - s))\), because \(h\) is a Lipschitz map: \(h (\Vert f (s) \Vert) - h (g (r_0 - s)) \le \vert h (\Vert f (s) \Vert) - h (g (r_0 - s)) \vert \le L \vert \Vert f (s) \Vert - g (r_0 - s) \vert = L (\Vert f (s) \Vert - g (r_0 - s))\).
So, \(0 \le \partial_1 \phi (s)\).
\(\{r \in [s, r_0] \vert \phi (r) \le 0\} \neq \emptyset\), because \(r_0\) belonged there, and let us take \(t := Inf (\{r \in [s, r_0] \vert \phi (r) \le 0\})\).
Let us see that \(\phi (t) = 0\).
When \(t = r_0\), \(\phi (t) = \phi (r_0) = 0\).
Let us suppose otherwise.
\(t \neq s\), because as \(\phi\) was continuous, there would be a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(\phi ([s, s + \delta)) \subseteq (0, \infty)\), so, \(s + \delta / 2 \in Lb (\{r \in [s, r_0] \vert \phi (r) \le 0\})\), so, \(s\) would not be the maximum of \(Lb (\{r \in [s, r_0] \vert \phi (r) \le 0\})\).
\(\phi ([s, t)) \subseteq (0, \infty)\), because otherwise, \(t\) would not be in \(Lb (\{r \in [s, r_0] \vert \phi (r) \le 0\})\).
If \(0 \lt \phi (t)\), there would be a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(\phi ((t - \delta, t + \delta)) \subseteq (0, \infty)\), but \(t\) would not be the maximum of \(Lb (\{r \in [s, r_0] \vert \phi (r) \le 0\})\), because \(t + \delta / 2 \in Lb (\{r \in [s, r_0] \vert \phi (r) \le 0\})\), a contradiction, so, \(\phi (t) \le 0\).
If \(\phi (t) \lt 0\), there would be a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(\phi ((t - \delta, t + \delta)) \subseteq (- \infty, 0)\), but \(t\) would not be in \(Lb (\{r \in [s, r_0] \vert \phi (r) \le 0\})\), because \(\phi (t - \delta / 2) \le 0\), a contradiction, so, \(0 \le \phi (t)\).
So, \(\phi (t) = 0\).
There would be an \(r \in (s, t)\) such that \(\partial_1 \phi (r) = (\phi (t) - \phi (s)) / (t - s)\), which would be negative, by the mean-value theorem for any differentiable map from any closed interval into the \(1\)-dimensional Euclidean \(C^\infty\) manifold, but \(0 \lt \phi (r)\), which would imply that \(0 \lt \Vert f (r) \Vert\), which would imply that \(r \in J^+\), a contradiction against \(0 \le \partial_1 \phi (r)\) whenever \(r \in J^+ \cap (r_1, r_0]\) and \(0 \lt \phi (r)\).
So, there is no such \(s \in J^+ \cap (r_1, r_0]\).
So, \(\phi (r) \le 0\) for each \(r \in J^+ \cap (r_1, r_0]\).
That implies that \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\) for each \(r \in J^+ \cap (r_1, r_0]\).
Step 6:
\((r_1, r_2) = ((r_1, r_2) \setminus J^+) \cup (J^+ \cap [r_0, r_2)) \cup (J^+ \cap (r_1, r_0])\).
So, \(\Vert f (r) \Vert \le g (\vert r - r_0 \vert)\) for each \(r \in (r_1, r_2)\).
