Page 264 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 24: Orthogonal Complements

Now S is a closed subspace of the Hilbert space H. Notes
So, S is complete and hence a Hilbert space. This completes the proof of the theorem.
Theorem 3: If M is a linear subspace of a Hilbert space H, then M is closed
M = M .

Proof: Let us assume that M = M ,
M being a subspace of H.
by theorem (2), (M ) is closed subspace of H.
Therefore M = M is a closed subspace of H. Conversely, let M be a closed subspace of H. We
shall show that M = M .
We know that M M .
Now suppose that M M .

Now M is a proper closed subspace of Hilbert space M . a non-zero vector z in M such that
o
z M or z M .
o o
Now z M and M gives z M M … (1)
o o
Since M is a subspace of H, we have
M M = {0} … (2)
(by theorem 1 (iii))
From (1) and (2) we conclude that z = 0, a contradiction to the fact that z is a non-zero vector.
o
M M can be a proper inclusion.
Hence M = M .
This completes the proof of the theorem.
Cor. If M is a non-empty subset of a Hilbert space H, then M = M .

Proof: By theorem (2), M is a closed subspace of H. So by theorem (3),
M = (M ) = M .
Theorem 4: If M and N are closed linear subspace of a Hilbert space H such that M N, then the
linear subspace M N is closed.
Proof: To prove: M + N is closed, we have to prove that it contains all its limit point.
Let z be a limit point of M + N,
a sequence (z ) in M + N such that z z in H.
n n
Since M N, M N = {0} and M + N is the direct sum of the subspace M and N, z can be written
n
uniquely as
z = x + y where x M and y N.
n n n n n
Taking two points z = x + y and z = x + y , we have
m m m n n n
z – z = (x – x ) + (y – y ).
m n m n m n
Since x – x M and y – y N, we get
m n m n
(x – x ) (y – y )
m n m n

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