Continuity of linear map from subspace of Euclidean space

[psie]

TL;DR

I'm reading about the (constant) rank theorem in Rudin's Principle of Mathematical Analysis (Theorem 9.32). I am stuck on a small detail in that proof concerning the continuity of linear maps with domain being a subspace of Euclidean space.

The statement of the rank theorem can be found below.

It's a bit messy, but the relevant details are that in the statement of that theorem, we have a function $F:E\subset\mathbb R^n\to\mathbb R^m$, where $F'(x)$ has rank $r$ for every $x\in E$ ($r\leq m,r\leq n$). Now fix $a\in E$ and put $A=F'(a)$. Let $Y_1$ be the range of $A$.

In the proof of the theorem, Rudin first treats the case when $r=0$. That is quite trivial. Then, when $r>0$, he constructs a right inverse $S:Y_1\to\mathbb R^m$ of $A$, that is, the map $S$ that satisfies $ASy=y$ for every $y\in Y_1$. In a later claim in the proof of the theorem, we require the continuity of $S$ (to show $A(V)$ is open in $Y_1$). Rudin has an earlier theorem where he states that linear maps from $\mathbb R^n\to\mathbb R^m$ have finite operator norm (that is, the $\sup$ of $|Ax|$ over all $|x|\leq 1$) and are uniformly continuous (this is Theorem 9.7 in the text). I wonder, does Theorem 9.7 also apply to $S$, whose domain $Y_1$ seems to be a subspace of $\mathbb R^m$?

 

[Office_Shredder]

You can effectively apply 9.7 by construction. Let $S'=SQ$ where $Q$ is a projection onto $Y_1$. Then $|SQx|$ is bounded when $|x|<1$. But if $x\in Y_1$, $Qx=x$ so $|Sx|$ satisfies the same bound.

I also suspect the proof of 9.7 can be transformed to handle a subspace domain without much more effort than this to begin with.

 

[pasmith]

Note that if $S$ is a right inverse of $A : \mathbb{R}^n \to Y_1$ then $S : Y_1 \to \mathbb{R}^n$ not $\mathbb{R}^m$ as you have in your post.

We can define a subspace operator norm $$\|S\|_1 = \sup \{ \|Sx\| : x \in Y_1, \|x\| \leq 1\}.$$ Then $$\|Sx - Sy\| \leq \|S\|_1 \|x - y\|$$ for all $x \in Y_1$ and $y \in Y_1$ so that if $\|S\|_1 > 0$ then if $\|x - y\| < \epsilon/\|S\|_1$ then $\|Sx - Sy\| < \epsilon$ and $S$ is uniformly continuous, and if $\|S\|_1 = 0$ then $S = 0$ is trivially uniformly continuous.