1
$\begingroup$

I'm seeking for an injective piecewise continuous function $f:\mathbb S^n\rightarrow[0,1]$ where $\mathbb S^N$ is the set of vectors with $L_2$ norm equals $1$.

The piecewise continuity requirement can be replaced by the following two other conditions: that $f$ is integrable, and there exists positive $K$ such that $|f(x)-f(y)|\geq K|x-y|$, i.e. something like the opposite of Lipschitz condition.

Aleph thanks!

  • 0
    You are looking for an injective continuous function from the unit n-sphere to the unit interval? Sounds topologically impossible.2012-12-30
  • 0
    you're right, btw i'm still not sure its impossible, i'll be glad if you can show me that. anyway, this is why I gave an option to give up continuity.2012-12-30
  • 0
    Take any curve on the sphere (so with sufficient dimensions). Consider the image of it. If you require continuity, you must have 2 points that have the same value, since you're looping back.2012-12-30
  • 0
    right. i also found this http://mathoverflow.net/questions/34232/injective-maps-mathbbrn-to-mathbbrm so we're left without the continuity, and with the other two conditions2012-12-30
  • 0
    Yes, the fancy argument looks at the algebraic topology, which I didn't know if you understood it. The 2-D version is easy to understand. The other condition seems weird, since it's like 'anti-continuous', and I don't have a good sense about it.2012-12-30
  • 0
    see comments there, there are simpler arguments. the rationale for the anti-lipschitz requirement came when I build a Gram matrix by some manipulation that uses that $f$, and it turns out that this constant $K$ has a lot to do with the condition number of the matrix.2012-12-30
  • 0
    in addition, note that this is not anti-continuity at all: take $f(x)=x(e^{x^2}+1)$.2012-12-30
  • 0
    I was just indicating that I had no sense of the condition. That's partly why I posted the continuous part as a comment, and not as an answer.2012-12-30
  • 0
    The anti-Lipschitz condition does not seem possible. For example, if we let $ \mathbf{x} = (1,0) $ and $ \mathbf{y} = (-1,0) $ (two points that lie on the unit circle $ \mathbb{S}^{1} $), then $ \| \mathbf{x} - \mathbf{y} \| = \| (2,0) \| = 2 $. However, as we require $ f(\mathbf{x}),f(\mathbf{y}) \in [0,1] $, we necessarily have $ |f(\mathbf{x}) - f(\mathbf{y})| \leq 1 $. Therefore, $ |f(\mathbf{x}) - f(\mathbf{y})| \geq \| \mathbf{x} - \mathbf{y} \| $ cannot hold.2012-12-30
  • 0
    you're right, I omitted K in the last edit, I'll put it back with some explanation2012-12-30
  • 0
    Of course, by my comment, that would force $ K \in \left[ 0,\dfrac{1}{2} \right] $.2012-12-30
  • 0
    generally I don't care if you take the whole space rather than only the sphere, I just thought it'll make things simpler2012-12-30
  • 0
    How do you characterize piecewise continuous on $\mathbb{S}^n$? (I am asking about how you describe the 'boundaries'.)2012-12-30
  • 0
    sorry, i didn't understand your question2012-12-30

0 Answers 0