description/proof of that derivative of \(C^1\), Euclidean-normed Euclidean vectors spaces map is Jacobian
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean-normed Euclidean vectors space.
- The reader knows a definition of derivative of map from open subset of normed vectors space into subset of normed vectors space at point.
- The reader knows a definition of map from open subset of Euclidean \(C^\infty\) manifold into subset of Euclidean \(C^\infty\) manifold \(C^k\) at point, where \(k\) excludes \(0\) and includes \(\infty\).
- The reader knows a definition of matrix norm induced by vector norms.
- 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 \(C^1\) map from any open subset of any Euclidean \(C^\infty\) manifold into any subset of any Euclidean \(C^\infty\) manifold.
Target Context
- The reader will have a description and a proof of the proposition that the derivative of any \(C^1\), Euclidean-normed Euclidean vectors spaces map is the Jacobian.
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:
\(\mathbb{R}^{d_1}\): \(= \text{ the Euclidean-normed Euclidean vectors space }\) also as the Euclidean \(C^\infty\) manifold
\(\mathbb{R}^{d_2}\): \(= \text{ the Euclidean-normed Euclidean vectors space }\) also as the Euclidean \(C^\infty\) manifold
\(f\): \(: \mathbb{R}^{d_1} \to \mathbb{R}^{d_2}\), \(\in \{\text{ the } C^1 \text{ maps }\}\)
\(v_1\): \(\in \mathbb{R}^{d_1}\)
\(D f_{v_1}\): \(= \text{ the derivative of } f \text{ at } v_1\)
\(M_{v_1}\): \(= \begin{pmatrix} \partial_l f^j (v_1) \end{pmatrix}\), the Jacobian of \(f\) at \(v_1\)
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Statements:
\(D f_{v_1} = M_{v_1}\)
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2: Proof
Whole Strategy: Step 1: see that \(lim_{\Vert v'_1 \Vert \to 0} \Vert f (v_1 + v'_1) - f (v_1) - M_{v_1} v'_1 \Vert / \Vert v'_1 \Vert = 0\) using 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.
Step 1:
Let \(M_p\) be the Jacobian of \(f\) at any \(p \in \mathbb{R}^{d_1}\).
Let \(v_2 \in \mathbb{R}^{d_2}\) and \(v_1, v'_1 \in \mathbb{R}^{d_1}\) be any.
\(\langle v_2, f (v_1 + v'_1) - f (v_1) - M_{v_1} v'_1 \rangle = \langle v_2, f (v_1 + v'_1) - f (v_1) \rangle - \langle v_2, M_{v_1} v'_1 \rangle := V1\).
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, there is a \(\mathbb{R}^{d_1}\) vector, \(v''_1\), such that \(v''_1\) is on the interior of the line segment, \(\overline{v_1 v'_1}\), and \(\langle v_2, f (v_1 + v'_1) - f (v_1) \rangle = \langle v_2, M_{v''_1} v'_1 \rangle\).
So, \(V1 = \langle v_2, M_{v''_1} v'_1 \rangle - \langle v_2, M_{v_1} v'_1 \rangle = \langle v_2, (M_{v''_1} - M_{v_1}) v'_1 \rangle\).
But \(v_2\) can be chosen to be the unit vector to the direction of \(f (v_1 + v'_1) - f (v_1) - M_{v_1} v'_1\), so, \(\vert \langle v_2, f (v_1 + v'_1) - f (v_1) - M_{v_1} v'_1 \rangle \vert = \Vert f (v_1 + v'_1) - f (v_1) - M_{v_1} v'_1 \Vert = \vert \langle v_2, (M_{v''_1} - M_{v_1}) v'_1 \rangle \vert \le \Vert v_2 \Vert \Vert (M_{v''_1} - M_{v_1}) v'_1 \Vert\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, \(\le \Vert v_2 \Vert \Vert M_{v''_1} - M_{v_1} \Vert \Vert v'_1 \Vert\), where \(\Vert M_{v''_1} - M_{v_1} \Vert\) is the matrix norm induced by vector norms, \(= \Vert M_{v''_1} - M_{v_1} \Vert \Vert v'_1 \Vert\).
So, \(\Vert f (v_1 + v'_1) - f (v_1) - M_{v_1} v'_1 \Vert / \Vert v'_1 \Vert \le \Vert M_{v''_1} - M_{v_1} \Vert\).
As \(f\) is \(C^1\), when \(\Vert v'_1 \Vert\) nears \(0\), while \(v''_1\) nears \(v_1\), \(M_{v''_1} - M_{v_1}\) nears \(0\), and \(\Vert M_{v''_1} - M_{v_1} \Vert\) nears \(0\).
So, \(\lim_{\Vert v'_1 \Vert \to 0} \Vert f (v_1 + v'_1) - f (v_1) - M_{v_1} v'_1 \Vert / \Vert v'_1 \Vert = 0\).
That means that \(D f_{v_1} = M_{v_1}\).
