# Question on Rudin 10.5

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.

## 2 Answers

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$$G(x_1,x_2,x_3,x_4)=(x_1,x_2,x_3,x_4)+\bigl(0,0,g(x_3)-x_3,0\bigr).$$

• $$G(x_1,x_2,x_3,x_4)=\bigl(x_1,x_2,g(x_3),x_4\bigr).$$ can you explain this line more explicitly? Dec 4 '21 at 10:23
• 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. Dec 4 '21 at 10:27
• What has your second comment to do with your question? Dec 4 '21 at 10:28
• it's part of $10.5$ Dec 4 '21 at 10:28
• 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 $1$. Dec 4 '21 at 10:37

You should understand that i$$\neq m$$ doesn't mean that i $$\lt m$$