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Convex set : characteristic cone

by wjulie
Tags: characteristic, cone, convex
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wjulie
#1
Jun15-11, 08:44 AM
P: 5
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.
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micromass
#2
Jun15-11, 09:08 AM
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Hi wjulie and welcome to PF!

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

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?
wjulie
#3
Jun15-11, 09:34 AM
P: 5
Thanks for your reply :) and yes, a characteristic cone is the same as a recession cone.
Click image for larger version

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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.

micromass
#4
Jun15-11, 09:40 AM
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Convex set : characteristic cone

I think you made a mistake in your picture since T and S are exactly the same there.

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]?
wjulie
#5
Jun15-11, 10:03 AM
P: 5
hmm i can't quite see the trick. But K has no direction because it is compact?
micromass
#6
Jun15-11, 10:12 AM
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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...
wjulie
#7
Jun15-11, 10:17 AM
P: 5
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?
micromass
#8
Jun15-11, 10:20 AM
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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.
wjulie
#9
Jun15-11, 10:22 AM
P: 5
Aha! I see. I got it now. Thank you, you have saved my day :)
Olga-Dahl
#10
Jun16-11, 06:11 AM
P: 2
"It is obvious that C⊆ccone(K+C)"

why is this obvious, please explain ?

/Olga
micromass
#11
Jun16-11, 07:50 AM
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Quote Quote by Olga-Dahl View Post
"It is obvious that C⊆ccone(K+C)"

why is this obvious, please explain ?

/Olga
See post #3.
Olga-Dahl
#12
Jun16-11, 08:11 AM
P: 2
Quote Quote by micromass View Post
See post #3.
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)?
micromass
#13
Jun16-11, 08:21 AM
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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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