description/proof of that derivative of \(C^1\), Euclidean-normed Euclidean vectors spaces map is the Jacobian
Topics
About: normed 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 matrix norm induced by vector norms.
- The reader knows a definition of map.
- The reader knows a definition of derivative of normed vectors spaces map.
- The reader admits the mean value theorem for \(C^1\), Euclidean-normed Euclidean vectors spaces function.
Target Context
- The reader will have a description and a proof of the proposition that derivative of \(C^1\), 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: Description
For any Euclidean-normed Euclidean vectors spaces, \(\mathbb{R}^{d1}\) and \(\mathbb{R}^{d2}\), and any \(C^1\) map, \(f: \mathbb{R}^{d1} \to \mathbb{R}^{d2}\), the derivative of the map, \(Df\), is the Jacobian, that is \(\left[\frac{\partial f_i}{\partial v_{1j}}\right]\) where {\(v_{1j}\)} are the components of vector on \(\mathbb{R}^{d1}\).
2: Proof
For any \(\mathbb{R}^{d_2}\) vector, \(v_{2}\), and any \(\mathbb{R}^{d_1}\) vectors, \(v_{11}\) and \(v_{12}\), $$v_{2} \left(f (v_{11} + v_{12}) - f (v_{11}) - f' (v_{11}) v_{12}\right) = v_{2} \left(f (v_{11} + v_{12}) - f (v_{11})\right) - v_{2} \left(f' (v_{11}) v_{12}\right) := V1$$, but by the mean value theorem for \(C^1\), Euclidean-normed Euclidean vectors spaces function, there is a \(\mathbb{R}^{d_1}\) vector, \(v_{13}\), such that \(v_{13}\) is on the line segment, \(\overline{v_{11}v_{12}}\), and $$v_{2} \left(f (v_{11} + v_{12}) - f (v_{11})\right) = v_{2} \left(f' (v_{13}) v_{12}\right)$$. So, $$V1 = v_{2} \left(f' (v_{13}) v_{12}\right) - v_{2} \left(f' (v_{11}) v_{12}\right) = v_{2} \left(f' (v_{13}) - f' (v_{11})\right) v_{v12}$$. But \(v_{2}\) can be chosen to be the unit vector to the direction of \(f (v_{11} + v_{12}) - f (v_{11}) - f' (v_{11}) v_{12}\), which does not depend on \(v_{13}\), so, $$\Vert v_{2} \left(f (v_{11} + v_{12}) - f (v_{11}) - f' (v_{11}) v_{12}\right)\Vert = \Vert f (v_{11} + v_{12}) - f (v_{11}) - f' (v_{11}) v_{12}\Vert = \Vert v_{2} \left(f' (v_{13}) - f' (v_{11})\right) v_{12}\Vert$$ $$\le \Vert v_{2}\Vert \Vert f' (v_{13}) - f' (v_{11})\Vert \Vert v_{12}\Vert = \Vert f' (v_{13}) - f' (v_{11})\Vert \Vert v_{12}\Vert$$. So, $$\frac{\Vert f (v_{11} + v_{12}) - f (v_{11}) - f' (v_{11}) v_{12}\Vert}{\Vert v_{12}\Vert} \le \Vert f' (v_{13}) - f' (v_{11})\Vert$$. As $$\lim_{v_{12} \Rightarrow 0} \Vert f' (v_{13}) - f' (v_{11})\Vert = 0$$, $$\lim_{\Vert v_{12}\Vert \to 0} \frac{\Vert f (v_{11} + v_{12}) -f (v_{11}) - f' (v_{11}) v_{12}\Vert}{\Vert v_{12}\Vert} = 0$$.
