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Multivariable calculus/analysis problem

  1. Jul 6, 2008 #1
    Hi guys... Haven't been in the forum for a couple years now.
    I have an old analysis problem that I never manage to solve. Would be nice if someone can shed some light on this.


    let [tex]f[/tex] be a [tex]C^1[/tex] function from [tex]\mathbb{R}^n \rightarrow \mathbb{R}^n[/tex], [tex]n>1[/tex]. [tex]df[/tex] is invertible except at isolated points (WLOG assume only at 0), prove that [tex]f[/tex] is locally injective, i.e. there is a neighborhood around each point in the domain such that [tex]f[/tex] is injective.

    thoughts about this problem: contraction principle doesn't work at all, since df gets really small around 0. The theorem is false when n=1 (like y=x^2), this makes me think the problem should involve some (if not mostly) topology of R^n

    I asked my topology professor but he said he can't think of a solution right away, he told me to consider the eigenvectors of the derivative... would be nice if anyone can put this problem to "sleep" once and for all.
     
    Last edited: Jul 6, 2008
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  3. Jul 6, 2008 #2

    quasar987

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    Well, the inverse function theorem implies that f is locally bijective everywhere except perhaps at 0. You must prove that f is injective at 0 also?
     
  4. Jul 6, 2008 #3
    yes, precisely what the problem is about... i think it may be best to look at special cases like [tex]\mathbb{R}^2[/tex] first.
     
  5. Jul 6, 2008 #4

    quasar987

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    Is it true for f(x,y)=x²+y² (the 2-dimensinal version of y=x²)??
     
  6. Jul 6, 2008 #5
    i think you forgot that f has to be a function from [tex]\mathbb{R}^n[/tex] to [tex]\mathbb{R}^n[/tex] :wink:
     
  7. Jul 7, 2008 #6
    In case anyone is still interested in this problem...in the 2D case, i can show (i think) that there exists a neighborhood U of 0, such that f is injective on U-{0} using Jordon curve theorem. However, I cannot proceed any further.

    I don't know if my reasoning is 100% correct. Generally, I first pick a [tex]U[/tex] around 0, small enough such that [tex]|df|[/tex] is bounded by some upper bound. [tex]f(U)[/tex] is bounded, by uniform continuity. I suppose [tex]x,y\neq 0[/tex] and [tex]f(x)=f(y)[/tex], take a path from [tex]x[/tex] to [tex]y[/tex], call it [tex]g:I\rightarrow \mathbb{R}^2[/tex], with non-vanishing derivative. it's image under [tex]f[/tex] will be a loop, call it [tex]h[/tex]. if the loop is self-intersecting, we can cut out the self-intersecting part (the pre-image will become line segments, but that is fine). This cutting process must end because we can cover the image of [tex]g[/tex] by finitely many balls under which [tex]f[/tex] is a homeomorphism by inverse function theorem and compactness, so that the length of each sub-loop (the self-intersecting part) must have a non-zero lower bound. the resulting loop is piecewise [tex]C^1[/tex] and has non-vanishing derivative (when it exists) since [tex]dh=df\cdot dg[/tex]. Then [tex]f(U-\{\textrm{finitely many line segments}\})=f(U)-\{\textrm{the modified loop}\}[/tex] has at least two components by Jordan curve theorem, yet [tex]U-\{\textrm{finitely many line segments}\}[/tex] is path connected, a contradiction.
     
    Last edited: Jul 7, 2008
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