24: Derivative of C1, Euclidean-Normed Euclidean Vectors Spaces Map Is Jacobian

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description/proof of that derivative of $C^1$, Euclidean-normed Euclidean vectors spaces map is the Jacobian

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Topics

About: normed vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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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 ${\mathbb{R}}^{d1}$.

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2: Proof

For any ${\mathbb{R}}^{d_2}$ vector, $v_2$, and any ${\mathbb{R}}^{d_1}$ vectors, $v_{11}$ and $v_{12}$, $$v_2(f(v_{11}+v_{12})-f(v_{11})-f^{\prime }(v_{11})v_{12})=v_2(f(v_{11}+v_{12})-f(v_{11}))-v_2(f^{\prime }(v_{11})v_{12}):=V1$$, 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 $$v_2(f(v_{11}+v_{12})-f(v_{11}))=v_2(f^{\prime }(v_{13})v_{12})$$. So, $$V1=v_2(f^{\prime }(v_{13})v_{12})-v_2(f^{\prime }(v_{11})v_{12})=v_2(f^{\prime }(v_{13})-f^{\prime }(v_{11}))v_{v12}$$. But $v_2$ can be chosen to be the unit vector to the direction of $f(v_{11}+v_{12})-f(v_{11})-f^{\prime }(v_{11})v_{12}$, which does not depend on $v_{13}$, so, $$\Vert v_2(f(v_{11}+v_{12})-f(v_{11})-f^{\prime }(v_{11})v_{12})\Vert =\Vert f(v_{11}+v_{12})-f(v_{11})-f^{\prime }(v_{11})v_{12}\Vert =\Vert v_2(f^{\prime }(v_{13})-f^{\prime }(v_{11}))v_{12}\Vert$$ $$\le \Vert v_2\Vert \Vert f^{\prime }(v_{13})-f^{\prime }(v_{11})\Vert \Vert v_{12}\Vert =\Vert f^{\prime }(v_{13})-f^{\prime }(v_{11})\Vert \Vert v_{12}\Vert$$. So, $$\frac{\Vert f(v_{11}+v_{12})-f(v_{11})-f^{\prime }(v_{11})v_{12}\Vert }{\Vert v_{12}\Vert }\le \Vert f^{\prime }(v_{13})-f^{\prime }(v_{11})\Vert$$. As $$\underset{v_{12}\Rightarrow 0}{lim}\Vert f^{\prime }(v_{13})-f^{\prime }(v_{11})\Vert =0$$, $$\underset{\Vert v_{12}\Vert \to 0}{lim}\frac{\Vert f(v_{11}+v_{12})-f(v_{11})-f^{\prime }(v_{11})v_{12}\Vert }{\Vert v_{12}\Vert }=0$$.

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References

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