description/proof of that for continuous map with compact parameter space from subspace of Euclidean metric space into subspace of Euclidean metric space that locally satisfies Lipschitz estimates, restriction on compact subspace satisfies Lipschitz estimates
Topics
About: metric space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean metric space.
- The reader knows a definition of continuous, topological spaces map.
- The reader admits the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
- The reader admits the proposition that the product of any finite number of compact topological spaces is compact.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
- The reader admits the proposition that for any topological space induced by any metric and any subset, the subset as the topological subspace equals the subset as the topological space induced by the metric subspace.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
- The reader admits the Heine-Borel theorem: any subset of any Euclidean topological space is compact if and only if it is closed and bounded.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map with any compact topological parameter space from any subspace of any Euclidean metric space with the induced topology into any any subspace of any Euclidean metric space with the induced topology, that (the map) locally satisfies Lipschitz estimates, the restriction of the map on any compact subspace domain satisfies Lipschitz estimates.
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:
\(\mathbb{R}^{d_1}\): \(= \text{ the Euclidean metric space }\)
\(\mathbb{R}^{d_2}\): \(= \text{ the Euclidean metric space }\)
\(M_{1, 1}\): \(\in \{\text{ the metric subspaces of } \mathbb{R}^{d_1}\}\) with the induced topology
\(M_2\): \(\in \{\text{ the metric subspaces of } \mathbb{R}^{d_2}\}\) with the induced topology
\(T\): \(\in \{\text{ the compact topological spaces }\}\)
\(M_{1, 2}\): \(\in \{\text{ the metric subspaces of } M_{1, 1}\}\) with the induced topology, \(\in \{\text{ the compact topological subspaces of } M_{1, 1}\}\)
\(f\): \(: M_{1, 1} \times T \to M_2\), \(\in \{\text{ the continuous maps }\}\)
\(f \vert_{M_{1, 2} \times T}\): \(: M_{1, 2} \times T \to M_2\)
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Statements:
\(\forall m \in M_{1, 1} (\exists U_m \in \{\text{ the open neighborhoods of } m \text{ on } M_{1, 1}\}, \exists L_m \in \mathbb{R} (\forall m_1, m_2 \in U_m, \forall t \in T (dist (f (m_1, t), f (m_2, t)) \le L_m dist (m_1, m_2))))\)
\(\implies\)
\(\exists L \in \mathbb{R} (\forall m_1, m_2 \in M_{1, 2}, \forall t \in T (dist (f (m_1, t), f (m_2, t)) \le L dist (m_1, m_2)))\)
//
2: Note
The topology of \(M_{1, 2}\) induced by the subspace metric is nothing but the subspace topology of \(M_{1, 1}\), by the proposition that for any topological space induced by any metric and any subset, the subset as the topological subspace equals the subset as the topological space induced by the metric subspace, so, \(M_{1, 2}\) is compact with the induced topology.
3: Proof
Whole Strategy: see that \(f (M_{1, 2} \times T)\) is bounded with the diameter, \(D\); Step 2: take a finite cover of \(M_{1, 2}\), \(\{B_{m_j, \delta (m_j)}\}\), such that \(f\) satisfies Lipschitz estimates with \(L_{m_j}\) over \(\{B_{m_j, 2 \delta (m_j)}\}\) and define \(\delta := Min \{\delta (m_j)\}\) and \(L' := Max \{L_{m_j}\}\); Step 3: take \(L := Max \{L', D / \delta\}\) and see that \(L\) satisfies the requirement.
Step 1:
\(M_{1, 2} \times T\) as the topological product of \(M_{1, 2}\) and \(T\) is the topological subspace of \(M_{1, 1} \times T\) as the topological product of \(M_{1, 1}\) and \(T\), by the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
\(M_{1, 2} \times T\) is a compact topological space, by the proposition that the product of any finite number of compact topological spaces is compact.
So, \(M_{1, 2} \times T\) is a compact subspace of \(M_{1, 1} \times T\).
\(M_{1, 2} \times T\) is a compact subset of \(M_{1, 1} \times T\), by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
\(f (M_{1, 2} \times T)\) is compact on \(M_2\), by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
\(M_2\) is the topological subspace of \(\mathbb{R}^{d_2}\), by the proposition that for any topological space induced by any metric and any subset, the subset as the topological subspace equals the subset as the topological space induced by the metric subspace.
\(f (M_{1, 2} \times T)\) is compact on \(\mathbb{R}^{d_2}\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
\(f (M_{1, 2} \times T)\) is bounded, by the Heine-Borel theorem: any subset of any Euclidean topological space is compact if and only if it is closed and bounded.
So, there is a constant, \(D\), such that for any \(m_1, m_2 \in f (M_{1, 2} \times T)\), \(dist (m_1, m_2) \le D\).
Step 2:
For each \(m \in M_{1, 2}\), there are some \(U_m \subseteq M_{1, 1}\) and \(L_m\) by the supposition, but \(U_m\) can be taken to be an open ball, \(B_{m, 2 \delta (m)} \subseteq M_{1, 1}\), because if \(U_m\) is not so, \(B_{m, 2 \delta (m)}\) can be taken to be such that \(B_{m, 2 \delta (m)} \subseteq U_m\), and \(B_{m, 2 \delta (m)}\) can be used instead of \(U_m\) without moving \(L_m\).
\(\{B_{m, \delta (m)} \subseteq M_{1, 1} \vert m \in M_{1, 2}\}\) is an open cover of \(M_{1, 2}\) on \(M_{1, 1}\), and there is a finite subcover, \(\{B_{m_j, \delta (m_j)}\}\).
Let us define \(\delta := Min \{\delta (m_j)\}\) and \(L' := Max \{L_{m_j}\}\).
Step 3:
Let us take \(L := Max \{L', D / \delta\}\).
For each \(p_1, p_2 \in M_{1, 2}\), there are these 2 cases: 1) \(dist (p_1, p_2) \lt \delta\); 2) \(\delta \le dist (p_1, p_2)\).
For the case 1), \(p_1 \in B_{m_j, \delta (m_j)}\) for a \(j\), and \(dist (p_2, m_j) \le dist (p_2, p_1) + dist (p_1, m_j) \lt \delta + \delta (m_j) \le 2 \delta (m_j)\). So, \(p_1, p_2 \in B_{m_j, 2 \delta (m_j)}\). As \(B_{m_j, 2 \delta (m_j)}\) is a subset where the Lipschitz estimates are satisfied, \(dist (f (p_1, t) - f (p_2, t)) \le L_{m_j} dist (p_1, p_2) \le L' dist (p_1, p_2) \le L dist (p_1, p_2)\).
For the case 2), \(dist (f (p_1, t) - f (p_2, t)) \le D \le D / \delta dist (p_1, p_2) \le L dist (p_1, p_2)\).
So, \(dist (f (p_1, t) - f (p_2, t)) \le L dist (p_1, p_2)\) for any case.
