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

  • 1
    $\begingroup$ 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. $\endgroup$
    – copper.hat
    Jul 8 '18 at 19:32
  • $\begingroup$ 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 . $\endgroup$
    –  Mathan
    Jul 9 '18 at 1:25
  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 $\Phi' = (g',I)$.

  • $\begingroup$ 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 $k$? $\endgroup$
    – Brown
    Dec 31 '19 at 21:13
  • $\begingroup$ 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). $\endgroup$
    – copper.hat
    Dec 31 '19 at 21:24
  • $\begingroup$ 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 $A \Phi'(b)=0$ $\endgroup$
    – Brown
    Dec 31 '19 at 21:31
  • $\begingroup$ @Vector: Again, I'm not sure what exactly you are asking, but if $A \Phi'(b)k = 0$ for all $k$ then $A \Phi'(b) = 0$. $\endgroup$
    – copper.hat
    Dec 31 '19 at 22:42
  • 1
    $\begingroup$ @ThomasWinckelman: Thanks future person :-). It is good to know it helped someone a little. $\endgroup$
    – copper.hat
    Jan 4 '20 at 22:23

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