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About convex hull and fixed point

  1. Oct 24, 2007 #1

    a(x)={y in X:||y-x||>=1/4}

    b(x)is the convex hull of a(x).

    Identify the set of fixed points.

    My answer is 3/4>=x>=1/4, 3/4>=y>=1/4, but I am not sure...

  2. jcsd
  3. Oct 24, 2007 #2


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    If you take x= [itex](x_0,y_0)[/itex], y= (x,y), then [itex]||x-y||\ge 1/4[/itex] becomes [itex]\sqrt{(x-x_0)^2+ (y-y_0)^2}\ge 1/4}[/itex] which is the same as [itex](x-x_0)^2+ (y-y_0)^2\ge 1/16[/itex], the set of points outside the circle centered at [itex](x_0,y_0)[/itex] with radius 1/4. The "convex hull" of a set, A, is the smallest convex set containing A. To be convex, for any two points the straight line segment between them must be in the set. Draw a picture and start "connecting points". It should be clear what the convex hull of this set is.

    Now, my question is "what does this have to do with fixed points?" (maybe I missed that part of the course!). A fixed point, for a function f, is a point x such that f(x)= x. What function are you talking about?
  4. Oct 24, 2007 #3
    This case is a bit tricky. We are asked to find the fixed points under this correspondence...As you know, the original correspondence is not convex-valued and has no fixed points. Then the convex hull is convex-valued and by Kakutani's theorem there exists at least one fixed point.

    I am not sure about my answer because the convex hull changes when we move from (0,0) (the convex hull is part of the disk) to (0,1/4) (the convex hull is the whole disk).


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