1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Continuity, vector function, inverse

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data
    f:Rn->Rn is continuous and satisfies
    |f(x)-f(y)|>=k|x-y|
    for all x, y in Rn and some k>0. Show that F has a continuous inverse.

    2. Relevant equations



    3. The attempt at a solution
    It is easy to show that f is injective, but I've no idea how to prove the surjectivity. I was thinking on the R1->R1 case for a while, and guess that I can show that f is unbounded to deduce the surjectivity. But it seems that boundedness is not that useful in Rn->Rn case.
    Any hint? THanks!
     
  2. jcsd
  3. Dec 19, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can still use boundedness, I think. Suppose f(R^n) is not all of R^n. Then pick a point b on the boundary of f(R^n). So you can pick a sequence x_i such that f(x_i) converges to b. The sequence x_i can't be bounded, otherwise it would have a convergent subsequence. So you can pick a subsequence of x_i, y_i such that |y_{i+1}-y_i| is as large as you like, yet f(y_i) must converge to b.
     
  4. Dec 19, 2008 #3
    i'm working on a similar problem. could you possibly explain why x_i can't be bounded?
     
  5. Dec 19, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If x_i were bounded then would be contained in a compact set. So it has a convergent subsequence y_i->y. So f(y_i)->f(y) by continuity. But f(y_i)->b, so b=f(y). But b was assumed to be outside of the range of f.
     
  6. Dec 20, 2008 #5
    Thanks, but why b should be outside the range of f? I get it by first proving that injective and continuous f should map open ball to open ball, but not quite clear about the whole reason behind it. Is there some explanation more direct/brilliant/inspiring? Thanks a lot...
     
  7. Dec 20, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Good point. Mostly because I drew it that way. :) Better give this some more thought...
     
  8. Dec 20, 2008 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Have you studied Brouwer's invariance of domain theorem? I think it's necessary. f:R^1->R^2 defined by f(x)=(x,0) satisfies your premises (with unequal dimensions of domain and range) but it's not surjective.
     
  9. Dec 20, 2008 #8
    sorry for reply late...last night I was unable to open this forum...
    I havn't learned Brouwer's theorem and I've no idea what causes the differences between unequal dimensions and equal dimensions of domain and range (guess it needs a lot of preliminary knowledge that I havn't learned).
    but yes, I made a mistake when I tried to prove that "surjective and continuous f maps open ball to open ball". if added f:R^n->R^n, is this right? okay, even if it is true, I guess I cannot prove it, since I am possibly unable to make use of the properties of equal dimensions of domain and range....
    so...it seems that the original problem is beyond my knowledge?
     
    Last edited: Dec 20, 2008
  10. Dec 20, 2008 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, if you can prove f maps open balls to open balls then f(R^n) is an open set, then you don't need Brouwer's theorem. If f(R^n) is open then a point on the boundary of f(R^n) is not in f(R^n).
     
  11. Dec 21, 2008 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You could also use the generalization of Jordan's curve theorem. An injective continuous image of S^(n-1) -> R^n splits the complement of the image into an 'inside' and an 'outside'. It's hard to prove as well, but it's easy to visualize. That should also let you prove f(R^n) is open.
     
  12. Dec 22, 2008 #11
    Many things beyond my knowledge:( I'll come back to this problem when I am ready for it. Anyway, thanks very much! I'll keep an eye on Jordan's curve theorem and Brouwer's theorem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Continuity, vector function, inverse
Loading...