25: Chain Rule for Derivative of Composition of C^1, Euclidean-Normed Euclidean Vectors Spaces Maps

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description/proof of the chain rule for derivative of composition of $C^1$, Euclidean-normed Euclidean vectors spaces maps

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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 norm.
• The reader knows a definition of map.
• The reader knows a definition of derivative of normed vectors spaces map.
• The reader admits some properties of matrix norm induced by vector norm.

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Target Context

• The reader will have a description and a proof of the chain rule for derivative of composition map of $C^1$, Euclidean-normed Euclidean vectors spaces maps.

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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, ${\mathbb{R}}^{d_1}$, ${\mathbb{R}}^{d_2}$, and ${\mathbb{R}}^{d_3}$, any $C^1$ maps, $f_1:{\mathbb{R}}^{d_1}\to {\mathbb{R}}^{d_2}$ and $f_2:{\mathbb{R}}^{d_2}\to {\mathbb{R}}^{d_3}$, and the composition map, $f_3:{\mathbb{R}}^{d_1}\to {\mathbb{R}}^{d_3}$, such that $f_3=f_2(f_1)$, the derivative of $f_3$, $D{f}_3$, is $D{f}_{2}D{f}_1$.

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2: Proof

For any vectors, $v_{1,1},v_{1,2}\in {\mathbb{R}}^{d_1}$, $f_3(v_{1,1}+v_{1,2})=f_2(f_1(v_{1,1}+v_{1,2}))=f_2(f_1(v_{1,1})+f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2}))=f_2(f_1(v_{1,1}))+f_2^{\prime }(f_1(v_{1,1}))(f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2}))+r_2(f_1(v_{1,1}),f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2})):=V_1$. As $f_2^{\prime }$ is linear, $V_1=f_2(f_1(v_{1,1}))+f_2^{\prime }(f_1(v_{1,1}))f_1^{\prime }(v_{1,1})v_{1,2}+f_2^{\prime }(f_1(v_{1,1}))r_1(v_{1,1},v_{1,2})+r_2(f_1(v_{1,1}),f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2}))$.

But as $\Vert f_2^{\prime }(f_1(v_{1,1}))r_1(v_{1,1},v_{1,2})\Vert \le \Vert f_2^{\prime }(f_1(v_{1,1}))\Vert \Vert r_1(v_{1,1},v_{1,2})\Vert$, $li{m}_{v_{1,2}\to 0}\frac{\Vert f_2^{\prime }(f_1(v_{1,1}))r_1(v_{1,1},v_{1,2})\Vert }{\Vert v_{1,2}\Vert }=0$, and $\frac{\Vert r_2(f_1(v_{1,1}),f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2}))\Vert }{\Vert f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2})\Vert }=\frac{\Vert r_2(f_1(v_{1,1}),f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2}))\Vert }{\Vert v_{1,2}\Vert }\frac{\Vert v_{1,2}\Vert }{\Vert f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2})\Vert }:=V_2$.

But $\frac{\Vert f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2})\Vert }{\Vert v_{1,2}\Vert }\le \frac{\Vert f_1^{\prime }(v_{1,1})v_{1,2}\Vert }{\Vert v_{1,2}\Vert }+\frac{\Vert r_1(v_{1,1},v_{1,2})\Vert }{\Vert v_{1,2}\Vert }$ where $\underset{v_{1,2}\to 0}{lim}\frac{\Vert r_1(v_{1,1},v_{1,2})\Vert }{\Vert v_{12}\Vert }=0$ and $\frac{\Vert f_1^{\prime }(v_{1,1})v_{1,2}\Vert }{\Vert v_{1,2}\Vert }\le \frac{\Vert f_1^{\prime }(v_{1,1})\Vert \Vert v_{1,2}\Vert }{\Vert v_{1,2}\Vert }=\Vert f_1^{\prime }(v_{1,1})\Vert$. So, as $v_{1,2}\to 0$, $f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2})\to 0$, and so $V_2\to 0$, but as $\frac{\Vert v_{1,2}\Vert }{\Vert f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2})\Vert }$ does not approach 0, $li{m}_{v_{1,2}\to 0}\frac{\Vert r_2(f_1(v_{1,1}),f_1^{\prime }(v_{1,1})v_{1,2}+r_1(v_{1,1},v_{1,2}))\Vert }{\Vert v_{1,2}\Vert }=0$.

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References

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