Prove that a hyperplane and its associated halfspaces are convex sets

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Clarification of Question by dime365-ga on 03 Oct 2003 14:10 PDT

halfspaces are H+ and H-
and part of R^n

Hi dime365!
By definition, an hyperplane is the set of all vectors such that the
product of a non-zero scalar vector and any memeber of the set is a
scalar "b". A more formal definition (with symbols I can't write here)
is given in the following page:

Hyperplane
http://mathworld.wolfram.com/Hyperplane.html

In this page, however, a hyperplane is defined as the set of vectors
such that the product specified in the previous paragraph is 0, which
seems to be the case of your question, because you then define the
subspaces to be H+ and H-. In any case, the more general definition of
a hyperplane can be found in this page:

Convex Sets
http://www.stanford.edu/class/ee364/cvx-sets.pdf

In this last link you can also find the definition of convexity and
halfspaces. Now, let's get to the proof that a hyperplane is a convex
set.

Let X and Y be vectors that belong to the hyperplane. Since they
belong to the hyperplane, A.X=0 and A.Y=0 (where A is the scalar
vector). In order to prove the convexity of the set we must show that:

L.X + (1-L).Y

belongs to the hyperplane, where L is a scalar that belongs to (0,1).
In particular, it will belong to the hyperplane if it's true that:

A[L.X + (1-L).Y] = 0

This can be easily proven. We can use the distributive property on the
lefthand side to get:

 A.L.X + A.(1-L).Y
=L.A.X + (1-L).A.Y

And since we started by saying that X and Y belong to the hyperplane,
then A.X=0 and A.Y=0. Therefore, we've shown that:

L.A.X + (1-L).A.Y
   =0          =0   = 0

So an hyperplane is convex.


The proof is exactly the same for the halfspaces. One of the
halfspaces is the set of vectors such that A.X<0

In order to prove convexity of this halfspace, we have to show that if
X and Y belong to this halfspace, then L.X+(1-L).Y belongs to the
halfspace. The proof follows just as before. We must show that:

A(L.X + (1-L).Y) < 0

The lefthand side becomes:

 A.L.X + A.(1-L).Y
=L.A.X + (1-L).A.Y
    <0          <0   <0

Thus, since A.X<0 and A.Y<0, then L.A.X+(1-L).A.Y<0. This shows that
H- is a convex set. The proof is, of course, analogous for the
halfspace H+. Also, the proof is still the same if you use the first
definition of hyperplane (and its associated halfspaces) I gave here,
with the product of the vectors being 'b'.


Google search strategy:
convex hyperplane set
://www.google.com/search?sourceid=navclient&ie=UTF-8&oe=UTF-8&q=convex+hyperplane+set


I hope this helps! If you find anything unclear about my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.

Best wishes!
elmarto

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Request for Answer Clarification by dime365-ga on 06 Oct 2003 18:02 PDT

In particular, it will belong to the hyperplane if it's true that:
A[L.X + (1-L).Y] = 0
do you know why this is true ?

sorry for the lateness of the followup, ill be going though this more
in depth tomorrow aswell !! =)

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Clarification of Answer by elmarto-ga on 07 Oct 2003 06:36 PDT

Hi again dime365!
The statement is true because of the very definition of an hyperplane.
Recall that, by definition, a hyperplane is the set of all vectors
such that the product of a non-zero scalar vector and any memeber of
the set is 0. Therefore, the vector L.X+(1-L).X belongs to the
hyperplane only if the product of this vector and a scalar vector
("A") is zero. Thus, it belongs to the hyperplane if:

A.(L.X+(1-L).X = 0


I hope this clarified the answer. If you still have any doubts, please
don't hesitate to request further clarification.


Best regards!
elmarto