2

Rudin states

9.32 Theorem: Suppose ,, are nonnegative integers, $≥$,$≥$,$$ is a $^{1}$ mapping of an open set ⊂$^{}$ into $^{}$, and $′()$ has rank $r$ for every $∈$.

Fix $∈$, put $=′()$, Let $_{1}$ be the range of $A$, and let $$ be the projection in $^{}$ whose range if $_{1}$. Let $_{2}$ be the null space of $$.

Then there are open sets $$ and $$ in $^{}$, with $∈$,$⊂$ and there is a 1-1 mapping $$ of $$ onto $$(whose inverse is also of class $^{1}$) such that

$$(())=+() (∈) $$ where $$ is a $^{1}$ mapping of open set $()⊂_{1}$ into $_{2}$.

Here is his first sentence of proof

If $r=0$, Theorem 9.19 shows that $F(x)$ is constant in a neighborhood $U$ of $a$, and the equation above holds trivially. With $V=U$,$H(x)=x$,$ \phi(0)=F(a)$

Could someone help explain the intuition behind this question to me? I totally don't understand what the theorem is saying. I also want to know in the first sentence of proof, how do we show $\phi$ is a $C^{1}$ mapping from $Y_{1}$ to $Y_{2}$?

Thanks in advance

Community
  • 1
Brown
  • 699
  • 1
    In that beginning of the proof $F$ is constant, $A$ has zero rank, i.e. it is zero, $Y_1={0}$. So, $\phi$ only needs to map $0$ to $F(a)$. If you want you could extend it to all $\mathbb{R}^m$ as the constant function $F(a)$. – MoonLightSyzygy Jan 01 '20 at 14:56
  • Thanks, and I have another question. $Y_{2}$ is the null space of $P$, so I think if $\phi$ maps 0 to $F(a)$, does it mean $P(F(a))=0$? If it is, why? – Brown Jan 01 '20 at 15:01
  • 1
    $P$ is the projection onto $Y_1$ (which in that first case is ${0}$). So, the null space of $P$ is all of $\mathbb{R}^m$. – MoonLightSyzygy Jan 01 '20 at 15:03
  • This answer may help. – Matematleta Jan 01 '20 at 22:40

0 Answers0