I Continuity of linear map from subspace of Euclidean space

psie
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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.

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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##?
 
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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.
 
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 <br /> \|S\|_1 = \sup \{ \|Sx\| : x \in Y_1, \|x\| \leq 1\}. Then <br /> \|Sx - Sy\| \leq \|S\|_1 \|x - y\| for all x \in Y_1 and y \in Y_1 so that if \|S\|_1 &gt; 0 then if \|x - y\| &lt; \epsilon/\|S\|_1 then \|Sx - Sy\| &lt; \epsilon and S is uniformly continuous, and if \|S\|_1 = 0 then S = 0 is trivially uniformly continuous.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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