Page 349 - Advanced Linear Algebra

Hilbert Spaces 333

Proof. Only the uniqueness remains to be established. Suppose that
) V )%c% ~ ~ ) Z ) % c%

Then, by the parallelogram law,
Z
) V Z )% c % ~ ²% c % ³ c ) ²% c %³ V
)
Z Z
)
) ~ )%c% V ) b % c% ) c %c%c% ) V
V
%b% Z
Z
) ~ )%c% V ) b % c% ) c h h %c
2
 b c ~
V
and so %~% Z .
Since any subspace of an inner product space is convex, Theorem 13.9
:
=
applies to complete subspaces. However, in this case, we can say more.
:
Theorem 13.10 Let = be an inner product space and let be a complete
=
:
%
subspace of . Then for any %  = , the best approximation to in is the
Z
Z
unique vector % : for which % c %ž : .
Z
Z
Proof. Suppose that %c% ž : , where %  : . Then for any  : , we have
Z
%c% ž c% and so
Z
) ) %c ~ ) Z )%c% b ) Z )% c ‚ ) Z )%c%
Z
Hence %~ % V is the best approximation to in . Now we need only show that
:
%
%c% ž :, where is the best approximation to % in :. For any  :, a little
% V
V
computation reminiscent of completing the square gives
2
) )%c ~ º%c Á % c »
~ % ) )
)) c º%Á »c º Á %»b
~ % ) ) 8 c º%Á » c º%Á » 9
)) b

)) ))
~% ) ) 8 c º%Á » 9 8 c º%Á 9 c ( » (º%Á »
)) b

)) )) ))

~% b ) ) c º%Á » c ( º%Á »( e
))
e
))
))
Now, this is smallest when
º%Á »
~ •

))