Page 344 - Advanced Linear Algebra

328 Advanced Linear Algebra

any norm that comes from an inner product must satisfy the parallelogram law

) )% ) & b )% b ) & c ) ~ ) % & )b
(
)
But the norm in 13.1 does not satisfy this law. To see this, take
% & ~ ² Á Á Á Ã ³ and ~ ² Á c Á Á Ã ³. Then
) & b )% Á ) ~ & c )% ~

and

&
)) ~ ° Á ) ) ~ °
%

Thus, the left side of the parallelogram law is and the right side is h 2° ,

which equals if and only if ~ .

Just as any metric space has a completion, so does any inner product space.
Theorem 13.6 Let be an inner product space. Then there exists a Hilbert
=
/
/
space and an isometry ¢ = ¦ / for which = is dense in . Moreover, /
is unique up to isometric isomorphism.

Proof. We know that the metric space ²= Á ³ , where is induced by the inner
Z
Z
product, has a unique completion ²= Á ³ , which consists of equivalence classes
of Cauchy sequences in . If ²%³  ²%³  = Z and ²& ³  ²& ³  = Z , then we
=

set
²% ³ b ²& ³ ~ ²% b & ³Á ²% ³ ~ ² % ³

and
º²% ³Á ²& ³» ~ lim º% Á & »

¦B
It is easy to see that since ²% ³ and ²& ³ are Cauchy sequences, so are ²% b & ³

and ² % ³ . In addition, these definitions are well-defined, that is, they are

independent of the choice of representative from each equivalence class. For
instance, if ²% ³  ²% ³ , then

V
lim ) )%c % V ~
¦B
and so
( V (º% Á& » c º% Á& ( (» ~ º% c % V ) ) )Á& »  % c % V & ) ¦
(The Cauchy sequence ²& ³ is bounded. Hence,
³

º²% ³Á ²& ³» ~ lim º% Á & » ~ lim º% Á & » ~ º²% ³Á ²& ³»
V

V

¦B ¦B
We leave it to the reader to show that = Z is an inner product space under these
operations.