268: Inverse Theorem for Euclidean-Normed Spaces Map
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A description/proof of the inverse theorem for Euclidean-normed spaces map
Topics
About:
normed space
About:
map
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the inverse theorem for Euclidean-normed spaces map.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any Euclidean-normed spaces, and , and any neighborhood, , of point, , any map, , such that it is (where ) in and its derivative at , , is invertible, is invertible in a neighborhood of , and the inverse map is .
2: Proof
which has to be solved for by as . Define the set, where . The map, , with the domain as , is in fact if is small enough by the following reason: as , , so, , but for any , there is a such that for , , so, , so, if , , then, take , then, , so, , then, . with the domain is a contraction with norm of subtraction as distance for a small enough , because , but r is differentiable with respect to the 2nd argument and the Jacobian is continuous because it is expressed by some Jacobians of f, which are continuous, so, by the mean value theorem, it is , but as and continuous, if is small enough, is small enough, and . So, by the contraction mapping principle, there is the unique fixed point, , such that , which is the desired value. As , , but (which comes from the definition of and ), which exactly means that is differentiable at 0 with the derivative as , which means that is differentiable with the same derivative because . , is continuous with respect to , because is continuous being differentiable, the Jacobian is continuous with respect to , having the reverse matrix preserves the continuousness, and the compound of such continuous maps is continuous.
Now that the theorem has been proved to be true for , suppose that the theorem is true for and f is . As , the right hand side is with respect to because is as f is and taking the reverse matrix preserves the -ness, and as is by the theorem for , as the compound map, the right hand side is with respect to , also the left hand side is so, which means that is .
3: Note
The area of possible s is a concern, and is determined as this, although this is not very straightforward: 1st, take and take a such that for , and take as ; 2nd, belittle as for .
References
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