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Hilbert Spaces 329
Z
Moreover, the inner product on = Z induces the metric , since
º²%c & ³Á ²%c & ³» ~ lim º%c & Á %c & »
¦B
~ ²lim % Á & ³
¦B
Z
~ ²²% ³Á ²& ³³
Hence, the metric space isometry ¢= ¦ = Z is an isometry of inner product
spaces, since
Z
º %Á &» ~ ² %Á &³ ~ ²%Á &³ ~ º%Á &»
Thus, = Z is a complete inner product space and = is a dense subspace of = Z
that is isometrically isomorphic to . We leave the issue of uniqueness to the
=
reader.
The next result concerns subspaces of inner product spaces.
Theorem 13.7
)
1 Any complete subspace of an inner product space is closed.
)
2 A subspace of a Hilbert space is a Hilbert space if and only if it is closed.
)
3 Any finite-dimensional subspace of an inner product space is closed and
complete.
Proof. Parts 1 and 2 follow from Theorem 12.6. Let us prove that a finite-
)
)
dimensional subspace of an inner product space is closed. Suppose that
:
=
: ²% ³ ¦ % and
²% ³ is a sequence in , % ¤ :. Let 8 ~ ¸ Á Ã Á ¹ be an
orthonormal Hamel basis for . The Fourier expansion
:
~ º%Á »
~
in has the property that c % £ but
:
º% c Á » ~ º%Á » c º Á » ~
Thus, if we write &~ % c and & ~ % c : , the sequence ²& ³ , which is
in , converges to a vector that is orthogonal to . But this is impossible,
:
:
&
because & & implies that
&
&
) & c & ) ~ ) ) b & )) )) ¦ °
This proves that is closed.
:
To see that any finite-dimensional subspace of an inner product space is
:
complete, let us embed as an inner product space in its own right in its
)
(
:
Z
(
completion . Then or rather an isometric copy of ) is a finite-dimensional
:
:
:
