definition of derivative of map from arbitrary subset of normed vectors space into subset of normed vectors space at point
Topics
About: vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will have a definition of derivative of map from arbitrary subset of normed vectors space into subset of normed vectors space at point.
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:
\( V_1\): \(\in \{\text{ the normed vectors spaces }\}\) with the topology induced by the metric induced by the norm
\( V_2\): \(\in \{\text{ the normed vectors spaces }\}\)
\( S_1\): \(\in \{\text{ the subsets of } V_1\}\)
\( S_2\): \(\in \{\text{ the subsets of } V_2\}\)
\( f\): \(: S_1 \to S_2\)
\( s_1\): \(\in S_1\)
\(*{D f}_{s_1}\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
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Conditions:
\(\exists U'_{s_1} \subseteq V_1 \in \{\text{ the open neighborhoods of } s_1\}, \exists f': U'_{s_1} \to V_2 (f' \vert_{U'_{s_1} \cap S_1} = f \vert_{U'_{s_1} \cap S_1} \land {D f'}_{s_1} \text{ exists independently from the choice of } f' \land {D f}_{s_1} = {D f'}_{s_1})\)
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2: Note
\({D f}_{s_1}\) does not necessarily exist.
When \({D f}_{s_1}\) exists, it is unique, because Conditions requires that \({D f'}_{s_1}\) does not depend on the choice of \(f'\).
A typical case that \({D f'}_{s_1}\) does not depend on the choice of \(f'\) is that \(\{r s'_1 \in V_1 \vert s_1 + s'_1 \in S_1 \land r \in F\} = V_1\), where \(F\) is the field of \(V_1\) and \(V_2\), and for each \(s'_1 \in V_1\) such that \(s_1 + s'_1 \in S_1\), \(s_1 + r s'_1 \in S_1\) for any small enough \(r \in \mathbb{R}\) such that \(0 \lt r\).
Some typical examples of the typical case are half-closed subsets like \(V_1 = \mathbb{R}\), \(S_1 = [a, b)\), and \(s_1 = a\) and \(V_1 = \mathbb{R}^2\), \(S_1 = \{(r^1, r^2) \in \mathbb{R}^2 \vert a \le r^2 \lt b\}\), and \(s_1 = (a, c)\).
An example that \(s_1 + r s'_1 \in S_1\) is not satisfied is \(V_1 = \mathbb{R}\), \(S_1 = \{0\} \cup [1, 2)\), \(s_1 = 0\), and \(f: 0 \mapsto 0, r \in [1, 2) \mapsto 2 r\): \(s_1 + s'_1 = 0 + 1 = 1 \in S_1\), but \(r s'_1 \notin S_1\), then, for \(f' = 3 r\) and \(f'' = 4 r\) (which satisfies \(f' \vert_{U'_{s_1} \cap S_1} = f'' \vert_{U'_{s_1} \cap S_1} = f \vert_{U'_{s_1} \cap S_1}\)), \({D f'}_{s_1} = 3 \neq 4 = {D f''}_{s_1}\).
Let us see that the typical case guarantees that \({D f'}_{s_1}\) does not depend on the choice of \(f'\).
Let \(f'': U''_{s_1} \to V_2\) be any other extension.
Let \(s'_1 \in V_1\) be any such that \(s_1 + s'_1 \in S_1\).
There is a small enough \(r \in \mathbb{R}\) such that \(0 \lt r\) and \(s_1 + r s'_1 \in U'_{s_1} \cap U''_{s_1}\), but the typical case dictates that \(r\) can be chosen smaller enough such that \(s_1 + r s'_1 \in U'_{s_1} \cap U''_{s_1} \cap S_1\).
\(f' (s_1 + r s'_1) = f' (s_1) + {D f'}_{s_1} (r s'_1) + r' (s_1, r s'_1)\) and \(f'' (s_1 + r s'_1) = f'' (s_1) + {D f''}_{s_1} (r s'_1) + r'' (s_1, r s'_1)\).
But as \(f' (s_1 + r s'_1) = f'' (s_1 + r s'_1)\) and \(f' (s_1) = f'' (s_1)\), \({D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) + r' (s_1, r s'_1) - r'' (s_1, r s'_1) = 0\).
\(\Vert {D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) \Vert = \Vert {D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) + r' (s_1, r s'_1) - r'' (s_1, r s'_1) - (r' (s_1, r s'_1) - r'' (s_1, r s'_1)) \Vert = \Vert 0 - (r' (s_1, r s'_1) - r'' (s_1, r s'_1)) \Vert \le \Vert r' (s_1, r s'_1) - r'' (s_1, r s'_1) \Vert \le \Vert r' (s_1, r s'_1) \Vert + \Vert r'' (s_1, r s'_1) \Vert\).
\(\lim_{\Vert r s'_1 \Vert \to 0} \Vert {D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) \Vert / \Vert r s'_1 \Vert \le \lim_{\Vert r s'_1 \Vert \to 0} (\Vert r' (s_1, r s'_1) \Vert + \Vert r'' (s_1, r s'_1) \Vert) / \Vert r s'_1 \Vert = \lim_{\Vert r s'_1 \Vert \to 0} \Vert r' (s_1, r s'_1) \Vert / \Vert r s'_1 \Vert + \lim_{\Vert r s'_1 \Vert \to 0} \Vert r'' (s_1, r s'_1) \Vert) / \Vert r s'_1 \Vert = 0 + 0 = 0\).
\({D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1)\) is linear with respect to \(r s'_1\), so, \({D f'}_{s_1} (t r s'_1) - {D f''}_{s_1} (t r s'_1) = t ({D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1))\), and \(\lim_{t \to 0} \Vert {D f'}_{s_1} (t r s'_1) - {D f''}_{s_1} (t r s'_1) \Vert / \Vert t r s'_1 \Vert = \lim_{t \to 0} \Vert t ({D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1)) \Vert / (\vert t \vert \Vert r s'_1 \Vert) = \lim_{t \to 0} \vert t \vert \Vert {D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) \Vert / (\vert t \vert \Vert r s'_1 \Vert) = \lim_{t \to 0} \Vert {D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) \Vert / \Vert r s'_1 \Vert = 0\), so, \(\Vert {D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) \Vert = 0\), which implies that \({D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) = 0\).
But as \({D f'}_{s_1} (r s'_1) - {D f''}_{s_1} (r s'_1) = r ({D f'}_{s_1} (s'_1) - {D f''}_{s_1} (s'_1))\), \({D f'}_{s_1} (s'_1) - {D f''}_{s_1} (s'_1) = 0\).
Then, for each \(s''_1 \in V_1\), the typical case dictates that \(s''_1 = r s'_1\) for an \(s'_1\), so, \({D f'}_{s_1} (s''_1) = {D f'}_{s_1} (r s'_1) = r {D f'}_{s_1} (s'_1) = r {D f''}_{s_1} (s'_1) = {D f''}_{s_1} (r s'_1) = {D f''}_{s_1} (s''_1)\).
So, \({D f'}_{s_1} = {D f''}_{s_1}\).
For the half-closed subset case, \(V_1 = \mathbb{R}\), \(S_1 = [a, b)\) (in fact, that can be \([a, b]\)), and \(s_1 = a\), let us see that \({D f}_{s_1}\) is the one-sided derivative.
Let us suppose that any one-sided derivative, \(d \in V_2\), exists.
That means that \(d = lim_{s'_1 \to 0+} (f (a + s'_1) - f (a)) / s'_1\), which means that \(lim_{s'_1 \to 0+} \Vert d - (f (a + s'_1) - f (a)) / s'_1 \Vert = 0\).
Let \(U'_a := (a - \delta, a + \delta)\) and \(f': U'_a \to V_2, r \in (a - \delta, a) \mapsto f (a) + (r - a) d, r \in [a, a + \delta) \mapsto f (r)\).
Let \(f' (a + s'_1) = f' (a) + s'_1 d + r' (a, s'_1)\).
\(r' (a, s'_1) = f' (a + s'_1) - f' (a) - s'_1 d\), and \(\Vert r' (a, s'_1) \Vert / \Vert s'_1 \Vert = \Vert f' (a + s'_1) - f' (a) - s'_1 d \Vert / \Vert s'_1 \Vert = \Vert f' (a + s'_1) - f' (a) - s'_1 d \Vert / \vert s'_1 \vert = \Vert (f' (a + s'_1) - f' (a) - s'_1 d) / s'_1 \Vert = \Vert (f' (a + s'_1) - f' (a)) / s'_1 - d \Vert\).
\(lim_{s'_1 \to 0+} \Vert (f' (a + s'_1) - f' (a)) / s'_1 - d \Vert = lim_{s'_1 \to 0+} \Vert (f (a + s'_1) - f (a)) / s'_1 - d \Vert = 0\), as is seen above.
\(lim_{s'_1 \to 0-} \Vert (f' (a + s'_1) - f' (a)) / s'_1 - d \Vert = lim_{s'_1 \to 0-} \Vert (f (a) + s'_1 d - (f (a) + 0)) / s'_1 - d \Vert = lim_{s'_1 \to 0-} \Vert (s'_1 d) / s'_1 - d \Vert = lim_{s'_1 \to 0-} \Vert d - d \Vert = lim_{s'_1 \to 0-} \Vert 0 \Vert = lim_{s'_1 \to 0-} 0 = 0\).
That means that \(lim_{s'_1 \to 0} \Vert r' (a, s'_1) \Vert / \Vert s'_1 \Vert = 0\).
That means that \({D f'}_a = d\), the linear map that maps \(r \in \mathbb{R}\) to \(r d\).
Let us suppose that a \(U'_a \subseteq \mathbb{R}\) and an \(f': U'_a \to V_2\) exist with \({D f'}_a\).
That means that \(f' (a + s'_1) = f' (a) + {D f'}_a s'_1 + r' (a, s'_1)\) and \(lim_{s'_1 \to 0} \Vert r' (a, s'_1) \Vert / \Vert s'_1 \Vert = 0\).
So, \(lim_{s'_1 \to 0+} \Vert r' (a, s'_1) \Vert / \Vert s'_1 \Vert = 0\), but \(lim_{s'_1 \to 0+} \Vert r' (a, s'_1) \Vert / \Vert s'_1 \Vert = lim_{s'_1 \to 0+} \Vert f (a + s'_1) - f (a) - {D f'}_a s'_1 \Vert / \vert s'_1 \vert = lim_{s'_1 \to 0+} \Vert (f (a + s'_1) - f (a) - {D f'}_a s'_1) / s'_1 \Vert = lim_{s'_1 \to 0+} \Vert (f (a + s'_1) - f (a)) / s'_1 - {D f'}_a s'_1 / s'_1 \Vert\), where \(d := {D f'}_a s'_1 / s'_1 \in V_2\) is the constant vector as \({D f'}_a\) is a linear map, which means that \(lim_{s'_1 \to 0+} (f (a + s'_1) - f (a)) / s'_1 = d\).
That means that \(d\) is the one-sided derivative.
\({D f'}_a s'_1 = s'_1 d\), so, \({D f'}_a\) is the linear map that maps \(r\) to \(r d\).
It is likewise for the case, \(V_1 = \mathbb{R}\), \(S_1 = (a, b] \text{ or } [a, b]\), and \(s_1 = b\).
