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I don't understand what does it mean that "A primitive mapping is thus one that changes at most once coordinate" ( what do we mean in darkened $x$ in the $10.5$ ? is it the set of all points $x$ $\in$ $E$ where $g(x)$ $\neq$ $0$ ? )

Hence, I don't understand from where does the $(10)$ inequality come. $G(x) = x + [g(x) - x_m]e_m$.

Any help would be appreciated.

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The notation $\mathbf x$ stands for a vector in $\Bbb R^n$. Suppose, say, that $n=4$ and that $m=3$. Then$$G(x_1,x_2,x_3,x_4)=\bigl(x_1,x_2,g(x_3),x_4\bigr).$$So, only the third coordinate of $(x_1,x_2,x_3,x_4)$ is changed. And then$$G(x_1,x_2,x_3,x_4)=(x_1,x_2,x_3,x_4)+\bigl(0,0,g(x_3)-x_3,0\bigr).$$

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  • $\begingroup$ $$G(x_1,x_2,x_3,x_4)=\bigl(x_1,x_2,g(x_3),x_4\bigr).$$ can you explain this line more explicitly? $\endgroup$
    – vendetta
    Dec 4 '21 at 10:23
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    $\begingroup$ Rudin is saying that there is a function $g\colon\Bbb R\longrightarrow\Bbb R$ such that$$\bigl(\forall(x_1,x_2,x_3,x_4)\in\Bbb R^4\bigr):G(x_1,x_2,x_3,x_4)=\bigl(x_1,x_2,g(x_3),x_4\bigr).$$So, $G$ leaves the first, second and fourth coordinates unchanged and only changes (eventually) the third coordinate. $\endgroup$ Dec 4 '21 at 10:27
  • $\begingroup$ What has your second comment to do with your question? $\endgroup$ Dec 4 '21 at 10:28
  • $\begingroup$ it's part of $10.5$ $\endgroup$
    – vendetta
    Dec 4 '21 at 10:28
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    $\begingroup$ For instance (taking again my previous assumption that $n=4$ and that $m=4$), the second component of $G(x_1,x_2,x_3,x_4)$ is just $x_2$. The partial derivative of this with respect to $x_1$, $x_3$ and $x_4$ is $0$, whereas its partial derivative of this with respect to $x_2$ is $1$. $\endgroup$ Dec 4 '21 at 10:37
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You should understand that i$\neq m $ doesn't mean that i $\lt m $

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