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Convex set : characteristic cone

  1. Jun 15, 2011 #1
    Hello :)
    I have been giving a mathematical problem. But I find difficulties solving this. Therefore, I will be very grateful if anybody might wanted to help?
    The problem is

    "Let K be a compact convex set in R^n and C a closed convex cone in R^n. Show that
    ccone (K + C) = C."

    - Julie.
  2. jcsd
  3. Jun 15, 2011 #2
    Hi wjulie and welcome to PF! :smile:

    I'm not familiar with the terminology "characteristic cone", is it perhaps the same thing as a recession cone? ( http://planetmath.org/encyclopedia/DirectionOfAConvexSet.html [Broken] )

    It is obvious that [itex]C\subseteq ccone(K+C)[/itex]. Assume that this inclusion was strict, then there would be a direction d which is not in C. This d has a >0 distance from C. Thus the multiples of d grow further away from C. That is, the distance from d to C becomes arbitrarily big. But we still have that d is in ccone(K+C). Can you find a contradiction with that?
    Last edited by a moderator: May 5, 2017
  4. Jun 15, 2011 #3
    Thanks for your reply :) and yes, a characteristic cone is the same as a recession cone.
    T S.jpg
    Then I must show that T = S

    I can show that if a belongs to S, then a must belong to T as well.
    Let a=y+d belong to C.
    if y belongs to C, and because C [itex]\subseteq[/itex] C+K, then y must belong to C+K
    Therefore a=y+d must belong to C+K.

    But how about the other way? I'm finding it quite difficult.
  5. Jun 15, 2011 #4
    I think you made a mistake in your picture since T and S are exactly the same there. :biggrin:

    But I see what you mean. Let's prove this in steps. Let's begin with this: let d be a direction not in C. Can you prove that the distance between x+rd and C becomes arbitrarily large as r becomes large?

    I.e. can you show that [itex]d(x+rd,C)\rightarrow +\infty[/itex] as [itex]r\rightarrow +\infty[/itex]?
  6. Jun 15, 2011 #5
    hmm i can't quite see the trick. But K has no direction because it is compact?
  7. Jun 15, 2011 #6
    Do you see intuitively why it must be true?
    Consider for example the cone [itex]C=\{(x,0)~\vert~x\in \mathbb{R}\}[/itex] in [itex]\mathbb{R}^2[/itex]. Take something not in C, for example (1,1). Do you see that multiples of (1,1) are getting further away from C? That is, if [itex]r\rightarrow +\infty[/itex], then the distance between (r,r) becomes arbitrarily large.

    The general case is quite the same...
  8. Jun 15, 2011 #7
    i can see the intuitive behind it now. But when i have shown that this distance grow larger, what's next? Where are we heading?
  9. Jun 15, 2011 #8
    Well, x+rd is getting further away from C. But if d is in ccone(K+C), it must hold that x+rd is in K+C. And thus we must be able to express x+rd=k+c. But as the distance between x+rd and c becomes large, then k must become large. Thus K must be unbounded.
  10. Jun 15, 2011 #9
    Aha! I see. I got it now. Thank you, you have saved my day :)
  11. Jun 16, 2011 #10
    "It is obvious that C⊆ccone(K+C)"

    why is this obvious, please explain ?

  12. Jun 16, 2011 #11
    See post #3.
  13. Jun 16, 2011 #12
    Well, that isn't a useable argument, in my opinion though...

    Isn't that just at proof of y+d belonging to the set (K+C), and not the characteristic cone(K+C)?
  14. Jun 16, 2011 #13
    Well, to see that

    [tex]C\subseteq ccone(C+K)[/tex]

    Take d in C, then for all x in C, we have that x+rd is in C. In particular rd is in C.
    Now, take c+k in C+K, then c+k+rd=k+(c+rd) is in C+K (since C is convex). Thus for every x in C+K, we have that c+rd is in C+K
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