# Doubt in the proof of Implicit function theorem by Rudin's PMA,(Theorem 9.28,p224-227)

I have some doubts when I goes through the proof of Rudin's Implicit function theorem

(Theorem 9.28,p224-227 in Rudin's Principles of mathematical Analysis, 3ed).

Notation:If $\textbf{x}=(x_1,x_2,..,x_n)\in \mathbb{R}^n$ and $\textbf{y}=(y_1,y_2,...y_m)\in \mathbb{R^m}$.Then $(\textbf{x},\textbf{y})$ repesent the vector $(x_1,x_2,..,x_n,y_1,y_2,...y_m)\in \mathbb{R}^{n+m}$

1. Let $\textbf{f}$ be a $C'-$ mapping of an open set $E\subset \mathbb{R}^{n+m}$ to $\mathbb{R}^n$. Let $\textbf{F}(\textbf{x}, \textbf{y})=(\textbf{f}((\textbf{x}, \textbf{y}) , \textbf{y}) ,\space ;(\textbf{x}, \textbf{y})\in E$ . Then how can we prove that $\textbf{F}$ is a $C'-$ mapping of $E$ into $\mathbb{R}^{n+m}$?

2. If $A(\textbf{h}, \textbf{k})$ is a linear transformation from $\mathbb{R}^{n+m}$ to $\mathbb{R}^n$ , then how can we prove that $(\textbf{h} , \textbf{k} ) \to (A(\textbf{h}, \textbf{k}) ,\textbf{k} )$ is a linear transformation form $\mathbb{R}^{n+m}$ to $\mathbb{R}^{n+m}$?

3. Suppose $V$ is an open set in $\mathbb{R}^{n+m}$ with $(\textbf{0}, \textbf{b})\in V$. Let $W$ be the set of all $\textbf{y}\in \mathbb{R}^m$ such that $(\textbf{0},\textbf{y})\in V$ . Then how can we prove that $W$ is open ?

4. If $\textbf{G} \in C'$ and $(\textbf{g(y),y)}=\textbf{G(0,y)}$ then how can we prove that $\textbf{g}\in C'$

5. If $\Phi(\textbf{y})=(\textbf{g(y),y)}$ then how can we prove that $\Phi'(\textbf{y})\textbf{k}=(\textbf{g}'(\textbf{y)k,k})$

Sometimes it may be trivial for you, but I am not getting this.So Please help me to understand

• You should review some earlier material. 1. follows from the chain rule. 2. follows by showing the map is linear. 3. follows because the map $y \mapsto (0,y)$ is continuous. 4. follows because the map (x,y) \mapsto x$is linear and hence smooth. 5. follows from the definition of derivative. Jul 8 '18 at 19:32 • I have studied Rudin's fuctions of several variable chapter from the beginning . 1.How we can express$\textbf{F}(\textbf{x},\textbf{y})$as the composition of two functions? I didn't understood your answer related to 4 & 5 . Jul 9 '18 at 1:25 ## 1 Answer 1. If$f:X \to Y$,$g:X \to Z$are differentiable mappings, then the map$F: X \to Y\times Z$given by$F(x) = (f(x),g(x))$is differentiable and$DF(x) = (Df(x), Dg(x))$. In the above case,$g(x) = x$, the identity map. 2. Let$H$be the map$H((h,k)) = (A(h,k), k)$, then it is trivial to check that$H(\lambda(h,k)) = H((\lambda h, \lambda k)) = (A(\lambda h, \lambda k), \lambda k) = \lambda (A(h,k),k) = \lambda H((h,k))$and it is similarly trivial to verify that$H ((h_1,k_1)+(h_2,k_2)) = H((h_1,k_1))+H((h_2,k_2))$. 3. Let$L$be the continuous (in fact linear) map$L(y) = (0,y)$. Since$L$is continuous, we have that$W=L^{-1}(V)$is open. 4. Let$T$be the linear map$T((x,y)) = x$and$S$be the linear map$S((x,y)) = (0,y)$. Then$g = T \circ G \circ S$, and since$T,G,S$are differentiable it follows that$G$is. 5. This follows from 1. which shows that$\Phi' = (g',I)$. • May I ask another question? Why do we need$k$in the final step of the proof? In the proof, I think when we get$A \Phi'(b)=0$we can directly get the result. Why bother involving$k$? Dec 31 '19 at 21:13 • I don't have the book, so I'm not exactly sure what you are referring to. If you mean the$k$in Part 5. of the question then it is just the application of the derivative to a point$k$. That is,$\Phi' = (g',I)$and$\Phi'(y)k = (g'(x)k,k)$are saying the same thing (on the appropriate domain). Dec 31 '19 at 21:24 • Yeah, that's what I am confused about. In the book, Rudin argues if$A \Phi'(b) k=0$for every$k$, then we know$g'(b)=-(A_{x})^{-1}(A_{y})$. I think we do not need consider any point$k$,just directly get the equation above from the fact$A \Phi'(b)=0$Dec 31 '19 at 21:31 • @Vector: Again, I'm not sure what exactly you are asking, but if$A \Phi'(b)k = 0$for all$k$then$A \Phi'(b) = 0\$. Dec 31 '19 at 22:42
• @ThomasWinckelman: Thanks future person :-). It is good to know it helped someone a little. Jan 4 '20 at 22:23