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Tangents of a Convex Region

  1. Feb 16, 2012 #1
    I am having trouble proving the following:

    Suppose that [itex]E[/itex] is a convex region in the plane bounded by a curve [itex]C[/itex]. Show that [itex]C[/itex] has a tangent line except at a countable number of points.

    [itex]E[/itex] is convex iff for every [itex]x, y \in E,[/itex] and for every [itex]\lambda \in [0,1], (1-\lambda) x + \lambda y \in E[/itex].

    I am considering an approach where I parametrize [itex]C[/itex] in a fixed orientation and then look at the places where it is not differentiable, showing somehow that corners with some angular measure [itex]a \in [0,\pi)[/itex] are the only flavor of non-differentiable parts on this curve, and then showing that the number of corners is bounded by [itex]\frac{2\pi}{\pi - a}[/itex] for the largest [itex]a[/itex].

    Any thoughts?
     
    Last edited: Feb 16, 2012
  2. jcsd
  3. Feb 16, 2012 #2

    Dick

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    I think you are on the right general track. But there can be a countably infinite number of nondifferentiable points, yes? Can you show that an uncountable sum of positive numbers must be infinite?
     
  4. Feb 16, 2012 #3
    Ah, right, we could have a polygon(?) with angles ##\pi, \frac{3\pi}{2}, \frac{7\pi}{4}...##

    How do you take an uncountable sum? That sounds quite exotic! I am trying to imagine a proof by contradiction and pigeon-hole.
     
  5. Feb 16, 2012 #4

    Dick

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    It's not that exotic. Suppose you are summing c_i over an uncountable index i belonging to a set I. Define I_n to be the set of all i such that c_i>1/n for n a positive integer. Then if the sum is finite, I_n must be finite for all n, right? What's the union of all of the I_n? Forgive me for not TeXing this.
     
  6. Feb 16, 2012 #5
    It would be all the positive values in I, so I has at most a countable subset of nonzero values when the value of the sum is finite, as the union of countably many finite sets is at most countable. Thanks for the help!
     
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