# Baby rudin theorem 10.7

analysiscomplex-analysisfunctional-analysislinear algebrareal-analysis

We need this definition for the proof of the theorem:

Here is the theorem:
Suppose $$F$$ is a $$\mathscr C'$$ – mapping ( that means continuously differentiability) of an open set E $$\subset R^n$$ into $$R^n$$, $$0 \in E$$, $$F(0) = 0$$, and $$F'(0)$$ is invertible.

Then there is a neighborhood of $$0$$ in $$R^n$$ in which a representation: $$\mathbf{F}(\mathbf{x})=B_1\cdots B_{n-1}\mathbf{G}_n\circ \cdots \mathbf{G}_1(\mathbf{x})$$.

is valid.
with each $$\mathbf{G}_i$$ being a primitive $$\mathscr{C'}$$ mapping in some neighborhood of $$0$$, $$\mathbf{G}_i(\mathbf{0})=0$$, and $$\mathbf{G'}_i(0)$$ is invertible, and each $$B_i$$ is either a flip or the identity operator.

Here is the proof:

Put $$F = F_1$$. Assume $$1 \leq m \leq n – 1,$$ and make the following induction hypothesis ( which evidently holds for $$m$$= 1):

$$V_m$$ is a neighborhood of $$0$$, $$F_m$$ $$\in$$ $$\mathscr C'(V_m)$$, $$F_m(0)$$ = $$0$$, $$F_m'(0)$$ is invertible, and

$$P_{m-1}F_m(x) = P_{m-1}x ( x \in V_m). \tag{\star}$$

by ($$\star$$), we have:

$$F_m(x) = P_{m-1}x + \sum_{i=m}^n \alpha_i(x)e_i\tag{\ast}$$

where $$\alpha_m,…,\alpha_n$$ are real $$\mathscr C'$$-functions in $$V_m$$.

I don't understand from where does the last equality ($$\ast$$) comes.

I would be grateful for any kind of help.

Since $$F$$ maps $$V_m$$ into $$\mathbb R^n,$$ it has the form
$$\tag 1 F_m(x)=\alpha_1(x)\vec e_1+\cdots \alpha_{m-1}(x)\vec e_{m-1}+\alpha_m(x)\vec e_m\cdots +\alpha_n(x)\vec e_n.$$
$$\tag 2 P_{m-1}F_{m}=\alpha_1(x)\vec e_1+\cdots \alpha_{m-1}(x)\vec e_{m-1}=P_{m-1}x.$$
Now, combine $$(1)$$ and $$(2).$$