Page 430 - Advanced Linear Algebra
P. 430
414 Advanced Linear Algebra
Definition
)
1 Let %Á Ã Á % s . Any linear combination of the form
! % bÄb! %
where ! and ! bÄb! ~ is called a convex combination of
.
the vectors %Á Ã Á %
)
2 A subset ? s is convex if whenever %Á & ? , then the line segment
between and also lies in , in symbols,
&
?
%
¸!% b ² c !³& ! ¹?
)
3 A subset ? s is closed if whenever ²% ³ is a convergent sequence of
elements of , then the limit is also in .
?
?
)
4 A subset ? s is compact if it is both closed and bounded.
)
5 A subset ? s is a cone if %? implies that %? for all .
We will also have need of the following facts from analysis.
)
1 A continuous function that is defined on a compact set in s takes on
?
maximum and minimum values at some points within the set .
?
2 A subset ? ) of s is compact if and only if every sequence in ? has a
subsequence that converges to a point in .
?
Theorem 15.2 Let and be subsets of s . Define
@
?
? b @ ~ ¸ b ?Á @ ¹
)
1 If and are convex, then so is ? b @
@
?
)
2 If is compact and is closed, then ? b @ is closed.
?
@
Proof. For 1), let %b & and %b & be in ? b @ . The line segment between
these two points is
!²% b & ³ b ² c !³²% b & ³
~ ´!% b ² c !³% µ b ´!& b ² c !³& µ ? b @
for ! and so ? b @ is convex.
be a convergent sequence in ? b @ . Suppose that
For part 2), let %b &
%b & ¦ '. We must show that ' ? b @ . Since % is a sequence in the
compact set ? , it has a convergent subsequence % whose limit lies in .
%
?
Since % b & ¦ ' and % ¦ % we can conclude that & ¦ ' c % . Since @
is closed, it follows that 'c % @ and so ' ~ % b ²'c %³ ? b @ .
Convex Hulls
We will also have use for the notion of the smallest convex set containing a
given set.
